Instructors Solution Manual Introduction

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Instructor’s Solution Manual

Introduction to Electrodynamics

Fourth Edition

David J. Griﬃths

2014

Contents

1 Vector Analysis 4

2 Electrostatics 26

3 Potential 53

4 Electric Fields in Matter 92

5 Magnetostatics 110

6 Magnetic Fields in Matter 133

7 Electrodynamics 145

8 Conservation Laws 168

9 Electromagnetic Waves 185

10 Potentials and Fields 210

11 Radiation 231

12 Electrodynamics and Relativity 262

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Preface

Although I wrote these solutions, much of the typesetting was done by Jonah Gollub, Christopher Lee, and

James Terwilliger (any mistakes are, of course, entirely their fault). Chris also did many of the ﬁgures, and I

would like to thank him particularly for all his help. If you ﬁnd errors, please let me know (griﬃth@reed.edu).

David Griﬃths

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reproduced, in any form or by any means, without permission in writing from the publisher.

4CHAPTER 1. VECTOR ANALYSIS

Chapter 1

Vector Analysis

Problem 1.1

CHAPTER 1. VECTOR ANALYSIS 3

Chapter 1

Vector Analysis

Problem 1.1

✲

✒

✯

✣

B+C

! "# $

|B|cos θ1!"# $

|C|cos θ2

}

|B|sin θ1

}

|C|sin θ2

θ1

θ2

θ3

(a) From the diagram, |B+C|cos θ3=|B|cos θ1+|C|cos θ2.

|A||B+C|cos θ3=|A||B|cos θ1+|A||C|cos θ2.

So: A·(B+C) = A·B+A·C. (Dot product is distributive)

Similarly: |B+C|sin θ3=|B|sin θ1+|C|sin θ2. Mulitply by |A|ˆn.

|A||B+C|sin θ3ˆn =|A||B|sin θ1ˆn +|A||C|sin θ2ˆn.

If ˆn is the unit vector pointing out of the page, it follows that

A×(B+C) = (A×B) + (A×C). (Cross product is distributive)

(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and

Section 8 (cross product)

Problem 1.2

✲A=B

✻

❂

B×C❄A×(B×C)

The triple cross-product is not in general associative. For example,

suppose A=Band Cis perpendicular to A, as in the diagram.

Then (B×C) points out-of-the-page, and A×(B×C) points down,

and has magnitude ABC. But (A×B) = 0, so (A×B)×C= 0 ̸=

A×(B×C).

Problem 1.3

✲y

✻

✰

✣B

❲

A= +1 ˆx + 1 ˆy −1ˆz;A=√3; B= 1 ˆx + 1 ˆy + 1 ˆz;

A·B= +1 + 1 −1 = 1 = AB cos θ=√3√3 cos θ⇒cos θ.

θ= cos−1%1

3&≈70.5288◦

Problem 1.4

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,

we might pick the base (A) and the left side (B):

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(a) From the diagram, |B+C|cos ✓3=|B|cos ✓1+|C|cos ✓2. Multiply by |A|.

|A||B+C|cos ✓3=|A||B|cos ✓1+|A||C|cos ✓2.

So: A·(B+C)=A·B+A·C. (Dot product is distributive)

Similarly: |B+C|sin ✓3=|B|sin ✓1+|C|sin ✓2. Mulitply by |A|ˆn .

|A||B+C|sin ✓3ˆn =|A||B|sin ✓1ˆn +|A||C|sin ✓2ˆn .

If ˆn is the unit vector pointing out of the page, it follows that

A⇥(B+C)=(A⇥B)+(A⇥C). (Cross product is distributive)

(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and

Section 8 (cross product)

Problem 1.2

CHAPTER 1. VECTOR ANALYSIS 3

Chapter 1

Vector Analysis

Problem 1.1

✲

✒

✯

✣

B+C

! "# $

|B|cos θ1! "# $

|C|cos θ2

}

|B|sin θ1

}

|C|sin θ2

θ1

θ2

θ3

(a) From the diagram, |B+C|cos θ3=|B|cos θ1+|C|cos θ2. Multiply by |A|.

|A||B+C|cos θ3=|A||B|cos θ1+|A||C|cos θ2.

So: A·(B+C) = A·B+A·C. (Dot product is distributive)

Similarly: |B+C|sin θ3=|B|sin θ1+|C|sin θ2. Mulitply by |A|ˆn.

|A||B+C|sin θ3ˆn =|A||B|sin θ1ˆn +|A||C|sin θ2ˆn.

If ˆn is the unit vector pointing out of the page, it follows that

A×(B+C) = (A×B) + (A×C). (Cross product is distributive)

(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and

Section 8 (cross product)

Problem 1.2

✲A=B

✻

❂

B×C❄A×(B×C)

The triple cross-product is not in general associative. For example,

suppose A=Band Cis perpendicular to A, as in the diagram.

Then (B×C) points out-of-the-page, and A×(B×C) points down,

and has magnitude ABC. But (A×B) = 0, so (A×B)×C= 0 ̸=

A×(B×C).

Problem 1.3

✲y

✻

✰

✣B

❲

A= +1 ˆx + 1 ˆy −1ˆz;A=√3; B= 1 ˆx + 1 ˆy + 1 ˆz;B=√3.

A·B= +1 + 1 −1 = 1 = AB cos θ=√3√3 cos θ⇒cos θ=1

θ= cos−1%1

3&≈70.5288◦

Problem 1.4

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,

we might pick the base (A) and the left side (B):

A=−1ˆx + 2 ˆy + 0 ˆz;B=−1ˆx + 0 ˆy + 3 ˆz.

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reproduced, in any form or by any means, without permission in writing from the publisher.

The triple cross-product is not in general associative. For example,

suppose A=Band Cis perpendicular to A, as in the diagram.

Then (B⇥C) points out-of-the-page, and A⇥(B⇥C) points down,

and has magnitude ABC. But (A⇥B)=0, so (A⇥B)⇥C=06=

A⇥(B⇥C).

Problem 1.3

CHAPTER 1. VECTOR ANALYSIS 3

Chapter 1

Vector Analysis

Problem 1.1

✲

✒

✯

✣

B+C

! "# $

|B|cos θ1!"# $

|C|cos θ2

}

|B|sin θ1

}

|C|sin θ2

θ1

θ2

θ3

(a) From the diagram, |B+C|cos θ3=|B|cos θ1+|C|cos θ2.

|A||B+C|cos θ3=|A||B|cos θ1+|A||C|cos θ2.

So: A·(B+C) = A·B+A·C. (Dot product is distributive)

Similarly: |B+C|sin θ3=|B|sin θ1+|C|sin θ2. Mulitply by |A|ˆn.

|A||B+C|sin θ3ˆn =|A||B|sin θ1ˆn +|A||C|sin θ2ˆn.

If ˆn is the unit vector pointing out of the page, it follows that

A×(B+C) = (A×B) + (A×C). (Cross product is distributive)

(b) For the general case, see G. E. Hay’s Vector and Tensor Analysis, Chapter 1, Section 7 (dot product) and

Section 8 (cross product)

Problem 1.2

✲A=B

✻

❂

B×C❄A×(B×C)

The triple cross-product is not in general associative. For example,

suppose A=Band Cis perpendicular to A, as in the diagram.

Then (B×C) points out-of-the-page, and A×(B×C) points down,

and has magnitude ABC. But (A×B) = 0, so (A×B)×C= 0 ̸=

A×(B×C).

Problem 1.3

✲y

✻

✰

✣B

❲

A= +1 ˆx + 1 ˆy −1ˆz;A=√3; B= 1 ˆx + 1 ˆy + 1 ˆz;

A·B= +1 + 1 −1 = 1 = AB cos θ=√3√3 cos θ⇒cos θ.

θ= cos−1%1

3&≈70.5288◦

Problem 1.4

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,

we might pick the base (A) and the left side (B):

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A=+1ˆx +1ˆy 1ˆz ;A=p3; B=1ˆx +1ˆy +1ˆz ;B=p3.

A·B= +1 + 1 1=1=AB cos ✓=p3p3 cos ✓)cos ✓=1

✓= cos11

3⇡70.5288

Problem 1.4

The cross-product of any two vectors in the plane will give a vector perpendicular to the plane. For example,

we might pick the base (A) and the left side (B):

A=1ˆx +2ˆy +0ˆz ;B=1ˆx +0ˆy +3ˆz .

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CHAPTER 1. VECTOR ANALYSIS 5

A⇥B=

ˆx ˆy ˆz

120

103

=6ˆx +3ˆy +2ˆz .

This has the right direction, but the wrong magnitude. To make a unit vector out of it, simply divide by its

length:

|A⇥B|=p36 + 9 + 4 = 7. ˆn =A⇥B

|A⇥B|=6

7ˆx +3

7ˆy +2

7ˆz .

Problem 1.5

A⇥(B⇥C)=

ˆx ˆy ˆz

AxAyAz

(ByCzBzCy)(BzCxBxCz)(BxCyByCx)

=ˆx [Ay(BxCyByCx)Az(BzCxBxCz)] + ˆy () + ˆz ()

(I’ll just check the x-component; the others go the same way)

=ˆx (AyBxCyAyByCxAzBzCx+AzBxCz)+ˆy () + ˆz ().

B(A·C)C(A·B)=[Bx(AxCx+AyCy+AzCz)Cx(AxBx+AyBy+AzBz)] ˆx + () ˆy + () ˆz

=ˆx (AyBxCy+AzBxCzAyByCxAzBzCx)+ˆy () + ˆz (). They agree.

Problem 1.6

A⇥(B⇥C)+B⇥(C⇥A)+C⇥(A⇥B)=B(A·C)C(A·B)+C(A·B)A(C·B)+A(B·C)B(C·A)=0.

So: A⇥(B⇥C)(A⇥B)⇥C=B⇥(C⇥A)=A(B·C)C(A·B).

If this is zero, then either Ais parallel to C(including the case in which they point in opposite directions, or

one is zero), or else B·C=B·A= 0, in which case Bis perpendicular to Aand C(including the case B=0.)

Conclusion: A⇥(B⇥C)=(A⇥B)⇥C() either Ais parallel to C, or Bis perpendicular to Aand C.

Problem 1.7

= (4 ˆx +6ˆy +8ˆz )(2 ˆx +8ˆy +7ˆz )= 2ˆx 2ˆy +ˆz

=p4+4+1= 3

3ˆx 2

3ˆy +1

3ˆz

Problem 1.8

(a) ¯

Ay¯

By+¯

Az¯

Bz= (cos Ay+ sin Az)(cos By+ sin Bz)+(sin Ay+ cos Az)(sin By+ cos Bz)

= cos2AyBy+sincos (AyBz+AzBy) + sin2AzBz+sin

2AyBysin cos (AyBz+AzBy)+

cos2AzBz

= (cos2+ sin2)AyBy+ (sin2+ cos2)AzBz=AyBy+AzBz.X

(b) (Ax)2+(Ay)2+(Az)2=⌃3

i=1AiAi=⌃3

i=1 ⌃3

j=1Rij Aj⌃3

k=1RikAk=⌃j,k (⌃iRij Rik)AjAk.

This equals A2

x+A2

y+A2

zprovided ⌃3

i=1Rij Rik =1if j =k

0if j 6=k

Moreover, if Ris to preserve lengths for all vectors A, then this condition is not only suﬃcient but also

necessary. For suppose A= (1,0,0). Then ⌃j,k (⌃iRij Rik )AjAk=⌃iRi1Ri1, and this must equal 1 (since we

want A2

x+A2

y+A2

z= 1). Likewise, ⌃3

i=1Ri2Ri2=⌃3

i=1Ri3Ri3= 1. To check the case j6=k, choose A=(1,1,0).

Then we want 2 = ⌃j,k (⌃iRij Rik)AjAk=⌃iRi1Ri1+⌃iRi2Ri2+⌃iRi1Ri2+⌃iRi2Ri1. But we already

know that the ﬁrst two sums are both 1; the third and fourth are equal, so ⌃iRi1Ri2=⌃iRi2Ri1= 0, and so

on for other unequal combinations of j,k.XIn matrix notation: ˜

RR = 1, where ˜

Ris the transpose of R.

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6CHAPTER 1. VECTOR ANALYSIS

Problem 1.9

CHAPTER 1. VECTOR ANALYSIS 5

✲x

✻

✠

❃

✿

Looking down the axis:

✻

✰

✻

z′

&y′

✰

x′

❄

■

✒

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Looking down the axis:

CHAPTER 1. VECTOR ANALYSIS 5

✲x

✻

✠

❃

✿

Looking down the axis:

✻

✰

✻

z′

&y′

✰

x′

❄

■

✒

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reproduced, in any form or by any means, without permission in writing from the publisher.

A 120rotation carries the zaxis into the y(= z) axis, yinto x(= y), and xinto z(= x). So Ax=Az,

Ay=Ax,Az=Ay.

R=



001

100

010





Problem 1.10

(a) No change. (Ax=Ax,Ay=Ay,Az=Az)

(b) A! A, in the sense (Ax=Ax,Ay=Ay,Az=Az)

behavior of an “ordinary” vector, as given by (b). If Aand Bare pseudovectors, then (A⇥B)! (A)⇥(B)=

(A⇥B). So the cross-product of two pseudovectors is again a pseudovector. In the cross-product of a vector

and a pseudovector, one changes sign, the other doesn’t, and therefore the cross-product is itself a vector.

Angular momentum (L=r⇥p) and torque (N=r⇥F) are pseudovectors.

(d) A·(B⇥C)! (A)·((B)⇥(C)) = A·(B⇥C). So, if a=A·(B⇥C), then a! a; a pseudoscalar

changes sign under inversion of coordinates.

Problem 1.11

(a)rf=2xˆx +3y2ˆy +4z3ˆz

(b)rf=2xy3z4ˆx +3x2y2z4ˆy +4x2y3z3ˆz

(c)rf=exsin yln zˆx +excos yln zˆy +exsin y(1/z)ˆz

Problem 1.12

(a) rh= 10[(2y6x18) ˆx + (2x8y+ 28) ˆy ]. rh= 0 at summit, so

2y6x18 = 0

2x8y+ 28 = 0 =)6x24y+ 84 = 0 2y18 24y+ 84 = 0.

22y=66=)y=3=)2x24 + 28 = 0 =)x=2.

Top is 3 miles north, 2 miles west, of South Hadley.

(b) Putting in x=2, y= 3:

h= 10(12 12 36 + 36 + 84 + 12) = 720 ft.

|rh|= 220p2⇡311 ft/mile; direction: northwest.

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CHAPTER 1. VECTOR ANALYSIS 7

Problem 1.13

=(xx0)ˆx +(yy0)ˆy +(zz0)ˆz ;

=(xx0)2+(yy0)2+(zz0)2.

(a) r(

2)= @

@x[(xx0)2+(yy0)2+(zz0)2]ˆx +@

@y() ˆy +@

@z() ˆz = 2(xx0)ˆx +2(yy0)ˆy +2(zz0)ˆz =2

(b) r(1

)= @

@x[(xx0)2+(yy0)2+(zz0)2]1

2ˆx +@

@y()1

2ˆy +@

@z()1

2ˆz

=1

2()3

22(xx0)ˆx 1

2()3

22(yy0)ˆy 1

2()3

22(zz0)ˆz

=()3

2[(xx0)ˆx +(yy0)ˆy +(zz0)ˆz ]=(1/

=(1/

2)ˆ

@x(

n)=n

n1@

@x=n

n1(1

x)=n

n1ˆ

x,so r(

n)=n

n1ˆ

Problem 1.14

y=+ycos +zsin ; multiply by sin :ysin =+ysin cos +zsin2.

z=ysin +zcos ; multiply by cos :zcos =ysin cos +zcos2.

Add: ysin +zcos =z(sin2+ cos2)=z. Likewise, ycos zsin =y.

So @y

@y= cos ;@y

@z=sin ;@z

@y= sin ;@z

@z= cos . Therefore

(rf)y=@f

@y=@f

@y+@f

@y= + cos (rf)y+ sin (rf)z

(rf)z=@f

@z=@f

@z+@f

@z=sin (rf)y+ cos (rf)zSo rftransforms as a vector. qed

Problem 1.15

(a)r·va=@

@x(x2)+ @

@y(3xz2)+ @

@z(2xz)=2x+02x=0.

(b)r·vb=@

@x(xy)+ @

@y(2yz)+ @

@z(3xz)=y+2z+3x.

(c)r·vc=@

@x(y2)+ @

@y(2xy +z2)+ @

@z(2yz) = 0 + (2x) + (2y) = 2(x+y)

Problem 1.16

r·v=@

@x(x

r3)+ @

@y(y

r3)+ @

@z(z

r3)= @

@xx(x2+y2+z2)3

2

@yy(x2+y2+z2)3

2+@

@zz(x2+y2+z2)3

2

= ()3

2+x(3/2)()5

22x+ ()3

2+y(3/2)()5

22y+ ()3

+z(3/2)()5

22z=3r33r5(x2+y2+z2)=3r33r3=0.

This conclusion is surprising, because, from the diagram, this vector ﬁeld is obviously diverging away from the

origin. How, then, can r·v= 0? The answer is that r·v= 0 everywhere except at the origin, but at the

origin our calculation is no good, since r= 0, and the expression for vblows up. In fact, r·vis inﬁnite at

that one point, and zero elsewhere, as we shall see in Sect. 1.5.

Problem 1.17

vy= cos vy+ sin vz;vz=sin vy+ cos vz.

@vy

@y=@vy

@ycos +@vz

@ysin =@vy

@y+@vy

@ycos +@vz

@y+@vz

@ysin .Use result in Prob. 1.14:

=@vy

@ycos +@vy

@zsin cos +@vz

@ycos +@vz

@zsin sin .

@vz

@z=@vy

@zsin +@vz

@zcos =@vy

@z+@vy

@zsin +@vz

@z+@vz

@zcos 

=@vy

@ysin +@vy

@zcos sin +@vz

@ysin +@vz

@zcos cos . So

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8CHAPTER 1. VECTOR ANALYSIS

@vy

@y+@vz

@z=@vy

@ycos2+@vy

@zsin cos +@vz

@ysin cos +@vz

@zsin2+@vy

@ysin2@vy

@zsin cos 

@vz

@ysin cos +@vz

@zcos2

=@vy

@ycos2+ sin2+@vz

@zsin2+ cos2=@vy

@y+@vz

@z.X

Problem 1.18

(a) r⇥va=

ˆx ˆy ˆz

x23xz22xz =ˆx (0 6xz)+ˆy (0 + 2z)+ˆz (3z20) = 6xz ˆx +2zˆy +3z2ˆz .

(b) r⇥vb=

ˆx ˆy ˆz

xy 2yz 3xz =ˆx (0 2y)+ˆy (0 3z)+ˆz (0 x)= 2yˆx 3zˆy xˆz .

ˆx ˆy ˆz

y2(2xy +z2)2yz =ˆx (2z2z)+ˆy (0 0) + ˆz (2y2y)= 0.

Problem 1.19

As we go from point Ato point B(9 o’clock to 10 o’clock), x

increases, yincreases, vxincreases, and vydecreases, so @vx/@y>

0, while @vy/@y<0. On the circle, vz= 0, and there is no

dependence on z, so Eq. 1.41 says

r⇥v=ˆz @vy

@x@vx

@y

points in the negative zdirection (into the page), as the right

hand rule would suggest. (Pick any other nearby points on the

circle and you will come to the same conclusion.) [I’m sorry, but I

cannot remember who suggested this cute illustration.]

Problem 1.20

v=yˆx +xˆy ; or v=yz ˆx +xz ˆy +xy ˆz ; or v= (3x2zz3)ˆx +3ˆy +(x33xz2)ˆz ;

or v= (sin x)(cosh y)ˆx (cos x)(sinh y)ˆy ;etc.

Problem 1.21

(i) r(fg)= @(fg)

@xˆx +@(fg)

@yˆy +@(fg)

@zˆz =f@g

@x+g@f

@xˆx +f@g

@y+g@f

@yˆy +f@g

@z+g@f

@zˆz

=f@g

@xˆx +@g

@yˆy +@g

@zˆz +g@f

@xˆx +@f

@yˆy +@f

@zˆz =f(rg)+g(rf). qed

(iv) r·(A⇥B)= @

@x(AyBzAzBy)+ @

@y(AzBxAxBz)+ @

@z(AxByAyBx)

=Ay@Bz

@x+Bz@Ay

@xAz@By

@xBy@Az

@x+Az@Bx

@y+Bx@Az

@yAx@Bz

@yBz@Ax

+Ax@By

@z+By@Ax

@zAy@Bx

@zBx@Ay

=Bx@Az

@y@Ay

@z+By@Ax

@z@Az

@x+Bz@Ay

@x@Ax

@yAx@Bz

@y@By

@z

Ay@Bx

@z@Bz

@xAz@By

@x@Bx

@y=B·(r⇥A)A·(r⇥B). qed

(v) r⇥(fA)=@(fAz)

@y@(fAy)

@zˆx +@(fAx)

@z@(fAz)

@xˆy +@(fAy)

@x@(fAx)

@yˆz

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CHAPTER 1. VECTOR ANALYSIS 9

=f@Az

@y+Az@f

@yf@Ay

@zAy@f

@zˆx +f@Ax

@z+Ax@f

@zf@Az

@xAz@f

@xˆy

+f@Ay

@x+Ay@f

@xf@Ax

@yAx@f

@yˆz

=f@Az

@y@Ay

@zˆx +@Ax

@z@Az

@xˆy +@Ay

@x@Ax

@yˆz 

Ay@f

@zAz@f

@yˆx +Az@f

@xAx@f

@zˆy +Ax@f

@yAy@f

@xˆz 

=f(r⇥A)A⇥(rf). qed

Problem 1.22

(a) (A·r)B=Ax@Bx

@x+Ay@Bx

@y+Az@Bx

@zˆx +Ax@By

@x+Ay@By

@y+Az@By

@zˆy

+Ax@Bz

@x+Ay@Bz

@y+Az@Bz

@zˆz .

(b) ˆr =r

r=xˆx +yˆy +zˆz

px2+y2+z2. Let’s just do the xcomponent.

[(ˆr ·r)ˆr ]x=1

px@

@x+y@

@y+z@

@zx

px2+y2+z2

rx1

p+x(1

2)1

(p)32x+yx 1

(p)32y+zx1

(p)32z

rx

r1

r3x3+xy2+xz2=1

rx

rx

r3x2+y2+z2=1

rx

rx

r= 0.

Same goes for the other components. Hence: (ˆr ·r)ˆr =0.

@x+3xz2@

@y2xz @

@z(xy ˆx +2yz ˆy +3xz ˆz )

=x2(yˆx +0ˆy +3zˆz )+3xz2(xˆx +2zˆy +0ˆz )2xz (0 ˆx +2yˆy +3xˆz )

=x2y+3x2z2ˆx +6xz34xyzˆy +3x2z6x2zˆz

=x2y+3z2ˆx +2xz 3z22yˆy 3x2zˆz

Problem 1.23

(ii) [r(A·B)]x=@

@x(AxBx+AyBy+AzBz)=@Ax

@xBx+Ax@Bx

@x+@Ay

@xBy+Ay@By

@x+@Az

@xBz+Az@Bz

[A⇥(r⇥B)]x=Ay(r⇥B)zAz(r⇥B)y=Ay@By

@x@Bx

@yAz@Bx

@z@Bz

@x

[B⇥(r⇥A)]x=By@Ay

@x@Ax

@yBz@Ax

@z@Az

@x

[(A·r)B]x=Ax@

@x+Ay@

@y+Az@

@zBx=Ax@Bx

@x+Ay@Bx

@y+Az@Bx

[(B·r)A]x=Bx@Ax

@x+By@Ax

@y+Bz@Ax

So [A⇥(r⇥B)+B⇥(r⇥A)+(A·r)B+(B·r)A]x

=Ay@By

@xAy@Bx

@yAz@Bx

@z+Az@Bz

@x+By@Ay

@xBy@Ax

@yBz@Ax

@z+Bz@Az

+Ax@Bx

@x+Ay@Bx

@y+Az@Bx

@z+Bx@Ax

@x+By@Ax

@y+Bz@Ax

=Bx@Ax

@x+Ax@Bx

@x+By@Ay

@x@Ax

/+@Ax

/+Ay@By

@x@Bx

/+@Bx

/

+Bz@Ax

/+@Az

@x+@Ax

/+Az@Bx

/+@Bz

@x+@Bx

/

=[r(A·B)]x(same for yand z)

(vi) [r⇥(A⇥B)]x=@

@y(A⇥B)z@

@z(A⇥B)y=@

@y(AxByAyBx)@

@z(AzBxAxBz)

=@Ax

@yBy+Ax@By

@y@Ay

@yBxAy@Bx

@y@Az

@zBxAz@Bx

@z+@Ax

@zBz+Ax@Bz

[(B·r)A(A·r)B+A(r·B)B(r·A)]x

=Bx@Ax

@x+By@Ax

@y+Bz@Ax

@zAx@Bx

@xAy@Bx

@yAz@Bx

@z+Ax@Bx

@x+@By

@y+@Bz

@zBx@Ax

@x+@Ay

@y+@Az

@z

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10 CHAPTER 1. VECTOR ANALYSIS

=By@Ax

@y+Ax@Bx

/+@Bx

/+@By

@y+@Bz

@z+Bx@Ax

/@Ax

/@Ay

@y@Az

@z

+Ay@Bx

@y+Az@Bx

@z+Bz@Ax

@z

=[r⇥(A⇥B)]x(same for yand z)

Problem 1.24

r(f/g)= @

@x(f/g)ˆx +@

@y(f/g)ˆy +@

@z(f/g)ˆz

=g@f

@xf@g

g2ˆx +g@f

@yf@g

g2ˆy +g@f

@zf@g

g2ˆz

g2g@f

@xˆx +@f

@yˆy +@f

@zˆz f@g

@xˆx +@g

@yˆy +@g

@zˆz =grffrg

g2.qed

r·(A/g)= @

@x(Ax/g)+ @

@y(Ay/g)+ @

@z(Az/g)

=g@Ax

@xAx@g

g2+g@Ay

@yAy@g

g2+g@Az

@zAz@g

g2g@Ax

@x+@Ay

@y+@Az

@zAx@g

@x+Ay@g

@y+Az@g

@z=gr·AA·rg

g2.qed

[r⇥(A/g)]x=@

@y(Az/g)@

@z(Ay/g)

=g@Az

@yAz@g

g2g@Ay

@zAy@g

g2g@Az

@y@Ay

@zAz@g

@yAy@g

@z

=g(r⇥A)x+(A⇥rg)x

g2(same for yand z).qed

Problem 1.25

(a) A⇥B=

ˆx ˆy ˆz

x2y3z

3y2x0=ˆx (6xz)+ˆy (9zy)+ˆz (2x26y2)

r·(A⇥B)= @

@x(6xz)+ @

@y(9zy)+ @

@z(2x26y2)=6z+9z+ 0 = 15z

r⇥A=ˆx @

@y(3z)@

@z(2y)+ˆy @

@z(x)@

@x(3z)+ˆz @

@x(2y)@

@y(x)= 0; B·(r⇥A)=0

r⇥B=ˆx @

@y(0) @

@z(2x)+ˆy @

@z(3y)@

@x(0)+ˆz @

@x(2x)@

@y(3y)=5ˆz ;A·(r⇥B)=15z

r·(A⇥B)?

=B·(r⇥A)A·(r⇥B)=0(15z) = 15z. X

(b) A·B=3xy 4xy =xy ;r(A·B)=r(xy)=ˆx @

@x(xy)+ˆy @

@y(xy)=yˆx xˆy

A⇥(r⇥B)=

ˆx ˆy ˆz

x2y3z

005=ˆx (10y)+ˆy (5x); B⇥(r⇥A)=0

(A·r)B=x@

@x+2y@

@y+3z@

@z(3yˆx 2xˆy )=ˆx (6y)+ˆy (2x)

(B·r)A=3y@

@x2x@

@y(xˆx +2yˆy +3zˆz )=ˆx (3y)+ˆy (4x)

A⇥(r⇥B)+B⇥(r⇥A)+(A·r)B+(B·r)A

=10yˆx +5xˆy +6yˆx 2xˆy +3yˆx 4xˆy =yˆx xˆy =r·(A·B).X

@y(2x26y2)@

@z(9zy)+ˆy @

@z(6xz)@

@x(2x26y2)+ˆz @

@x(9zy)@

@y(6xz)

=ˆx (12y9y)+ˆy (6x+4x)+ˆz (0) = 21yˆx + 10xˆy

r·A=@

@x(x)+ @

@y(2y)+ @

@z(3z) = 1 + 2 + 3 = 6; r·B=@

@x(3y)+ @

@y(2x)=0

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CHAPTER 1. VECTOR ANALYSIS 11

(B·r)A(A·r)B+A(r·B)B(r·A)=3yˆx 4xˆy 6yˆx +2xˆy 18yˆx + 12xˆy =21yˆx + 10xˆy

=r⇥(A⇥B).X

Problem 1.26

(a) @2Ta

@x2= 2; @2Ta

@y2=@2Ta

@z2=0 ) r2Ta=2.

(b) @2Tb

@x2=@2Tb

@y2=@2Tb

@z2=Tb) r2Tb=3Tb=3 sin xsin ysin z.

@x2= 25Tc;@2Tc

@y2=16Tc;@2Tc

@z2=9Tc) r2Tc=0.

(d) @2vx

@x2=2;@2vx

@y2=@2vx

@z2=0 )r

2vx=2

@2vy

@x2=@2vy

@y2=0;@2vy

@z2=6x)r

2vy=6x

@2vz

@x2=@2vz

@y2=@2vz

@z2=0 )r

2vz=0









r2v=2ˆx +6xˆy .

Problem 1.27

r·(r⇥v)= @

@x@vz

@y@vy

@z+@

@y@vx

@z@vz

@x+@

@z@vy

@x@vx

@y

=@2vz

@x@y@2vz

@y@x+@2vx

@y@z@2vx

@z@y+@2vy

@z@x@2vy

@x@z= 0, by equality of cross-derivatives.

From Prob. 1.18: r⇥va=6xz ˆx +2zˆy +3z2ˆz )r·(r⇥va)= @

@x(6xz)+ @

@y(2z)+ @

@z(3z2)=6z+6z=0.

Problem 1.28

r⇥(rt)=

ˆx ˆy ˆz

=ˆx @2t

@y@z@2t

@z@y+ˆy @2t

@z@x@2t

@x@z+ˆz @2t

@x@y@2t

@y@x

= 0, by equality of cross-derivatives.

In Prob. 1.11(b), rf=2xy3z4ˆx +3x2y2z4ˆy +4x2y3z3ˆz ,so

r⇥(rf)=

ˆx ˆy ˆz

2xy3z43x2y2z44x2y3z3

=ˆx (3 ·4x2y2z34·3x2y2z3)+ˆy (4 ·2xy3z32·4xy3z3)+ˆz (2 ·3xy2z43·2xy2z4)=0.X

Problem 1.29

(a) (0,0,0) ! (1,0,0).x:0!1,y =z= 0; dl=dx ˆx ;v·dl=x2dx;v·dl=1

0x2dx =(x3/3)|1

0=1/3.

(1,0,0) ! (1,1,0).x=1,y :0!1,z = 0; dl=dy ˆy ;v·dl=2yz dy = 0; v·dl=0.

(1,1,0) ! (1,1,1).x=y=1,z :0!1; dl=dz ˆz ;v·dl=y2dz =dz;v·dl=1

0dz =z|1

0=1.

Total: v·dl=(1/3) + 0 + 1 = 4/3.

(b) (0,0,0) ! (0,0,1).x=y=0,z :0!1; dl=dz ˆz ;v·dl=y2dz = 0; v·dl=0.

(0,0,1) ! (0,1,1).x=0,y :0!1,z = 1; dl=dy ˆy ;v·dl=2yz dy =2y dy;v·dl=1

02y dy =y2|1

0=1.

(0,1,1) ! (1,1,1).x:0!1,y =z= 1; dl=dx ˆx ;v·dl=x2dx;v·dl=1

0x2dx =(x3/3)|1

0=1/3.

Total: v·dl=0+1+(1/3) = 4/3.

v·dl=1

04x2dx = (4x3/3)|1

0= 4/3.

(d) v·dl=(4/3) (4/3) = 0.

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12 CHAPTER 1. VECTOR ANALYSIS

Problem 1.30

x, y :0!1,z = 0; da=dx dy ˆz ;v·da=y(z23) dx dy =3y dx dy;v·da=32

0dx 2

0y dy =

3(x|2

0)(y2

2|2

0)=3(2)(2) = 12. In Ex. 1.7 we got 20, for the same boundary line (the square in the

xy-plane), so the answer is no: the surface integral does not depend only on the boundary line. The total ﬂux

for the cube is 20 + 12 = 32.

Problem 1.31

Td⌧=z2dx dy dz. You can do the integrals in any order—here it is simplest to save zfor last:

z2dxdydz.

The sloping surface is x+y+z= 1, so the xintegral is (1yz)

0dx =1yz. For a given z,yranges from 0 to

1z, so the yintegral is (1z)

0(1 yz)dy = [(1 z)y(y2/2)]|(1z)

0= (1 z)2[(1 z)2/2] = (1 z)2/2=

(1/2) z+(z2/2).Finally, the zintegral is 1

0z2(1

2z+z2

2)dz =1

0(z2

2z3+z4

2)dz =(

6z4

4+z5

10 )|1

61

4+1

10 = 1/60.

Problem 1.32

T(b) = 1 + 4 + 2 = 7; T(a)=0.)T(b)T(a)=7.

rT=(2x+4y)ˆx + (4x+2z3)ˆy + (6yz2)ˆz ;rT·dl=(2x+4y)dx + (4x+2z3)dy + (6yz2)dz

(a) Segment 1: x:0!1,y=z=dy =dz =0.rT·dl=1

0(2x)dx =x21

0=1.

Segment 2: y:0!1,x=1,z=0, dx =dz =0.rT·dl=1

0(4) dy =4y|1

0=4.

Segment 3: z:0!1,x=y=1, dx =dy =0.rT·dl=1

0(6z2)dz =2z31

0=2.









b

arT·dl=7.X

(b) Segment 1: z:0!1,x=y=dx =dy =0.rT·dl=1

0(0) dz =0.

Segment 2: y:0!1,x=0,z=1, dx =dz =0.rT·dl=1

0(2) dy =2y|1

0=2.

Segment 3: x:0!1,y=z=1, dy =dz =0.rT·dl=1

0(2x+ 4) dx

=(x2+4x)1

0=1+4=5.









b

arT·dl=7.X

rT·dl=(2x+4x)dx + (4x+2x6)dx + (6xx4)2xdx= (10x+ 14x6)dx.

b

arT·dl=1

0(10x+ 14x6)dx =(5x2+2x7)1

0=5+2=7.X

Problem 1.33

r·v=y+2z+3x

(r·v)d⌧=(y+2z+3x)dx dy dz =2

0(y+2z+3x)dxdy dz

,!(y+2z)x+3

2x22

0= 2(y+2z)+6

=2

0(2y+4z+ 6)dydz

,!y2+ (4z+ 6)y2

0= 4 + 2(4z+ 6) = 8z+ 16

=2

0(8z+ 16)dz =(4z2+ 16z)2

0= 16 + 32 = 48.

Numbering the surfaces as in Fig. 1.29:

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CHAPTER 1. VECTOR ANALYSIS 13

(i) da=dy dz ˆx ,x =2.v·da=2y dy dz. v·da=2y dy dz =2y22

0=8.

(ii) da=dy dz ˆx ,x =0.v·da=0.v·da=0.

(iii) da=dx dz ˆy ,y =2.v·da=4z dx dz. v·da=4z dx dz = 16.

(iv) da=dx dz ˆy ,y =0.v·da=0.v·da=0.

(v) da=dx dy ˆz ,z =2.v·da=6x dx dy. v·da=24.

(vi) da=dx dy ˆz ,z =0.v·da=0.v·da=0.

)v·da= 8 + 16 + 24 = 48 X

Problem 1.34

r⇥v=ˆx (0 2y)+ˆy (0 3z)+ˆz (0 x)=2yˆx 3zˆy xˆz .

da=dy dz ˆx , if we agree that the path integral shall run counterclockwise. So

(r⇥v)·da=2y dy dz.

(r⇥v)·da=2z

0(2y)dydz

,!y22z

0=(2 z)2

=2

0(4 4z+z2)dz =4z2z2+z3

3

=88+8

3=8

@@@@@

y=2z

Meanwhile, v·dl=(xy)dx + (2yz)dy + (3zx)dz. There are three segments.

@@@@@

(1)

@I (2)

(3)

(1) x=z=0; dx =dz =0.y:0!2.v·dl=0.

(2) x= 0; z=2y;dx =0, dz =dy, y :2!0.v·dl=2yz dy.

v·dl=0

22y(2 y)dy =2

0(4y2y2)dy =2y22

3y32

0=82

3·8=8

(3) x=y= 0; dx =dy = 0; z:2!0.v·dl=0.v·dl=0.So v·dl=8

3.X

Problem 1.35

By Corollary 1, (r⇥v)·dashould equal 4

3.r⇥v=(4z22x)ˆx +2zˆz .

(i) da=dy dz ˆx ,x= 1; y, z :0!1.(r⇥v)·da=(4z22)dy dz;(r⇥v)·da=1

0(4z22)dz

3z32z)1

0=4

32=2

(ii) da=dx dy ˆz ,z= 0; x, y :0!1.(r⇥v)·da=0; (r⇥v)·da=0.

(iii) da=dx dz ˆy ,y= 1; x, z :0!1.(r⇥v)·da=0; (r⇥v)·da=0.

(iv) da=dx dz ˆy ,y= 0; x, z :0!1.(r⇥v)·da=0; (r⇥v)·da=0.

(v) da=dx dy ˆz ,z= 1; x, y :0!1.(r⇥v)·da=2dx dy;(r⇥v)·da=2.

)(r⇥v)·da=2

3+2= 4

3.X

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14 CHAPTER 1. VECTOR ANALYSIS

Problem 1.36

(a) Use the product rule r⇥(fA)=f(r⇥A)A⇥(rf):

S

f(r⇥A)·da=Sr⇥(fA)·da+S

[A⇥(rf)] ·da=P

fA·dl+S

[A⇥(rf)] ·da.qed

(I used Stokes’ theorem in the last step.)

(b) Use the product rule r·(A⇥B)=B·(r⇥A)A·(r⇥B):

V

B·(r⇥A)d⌧=Vr·(A⇥B)d⌧+V

A·(r⇥B)d⌧=S

(A⇥B)·da+V

A·(r⇥B)d⌧.qed

(I used the divergence theorem in the last step.)

Problem 1.37 r=x2+y2+z2;✓= cos1z

px2+y2+z2;= tan1y

x.

Problem 1.38

There are many ways to do this one—probably the most illuminating way is to work it out by trigonometry

from Fig. 1.36. The most systematic approach is to study the expression:

r=xˆx +yˆy +zˆz =rsin ✓cos ˆx +rsin ✓sin ˆy +rcos ✓ˆz .

If I only vary rslightly, then dr=@

@r(r)dr is a short vector pointing in the direction of increase in r. To make

it a unit vector, I must divide by its length. Thus:

ˆr =

@r

@r;ˆ

✓=

@✓

@r

@✓ ;ˆ

=

@

@r

@.

@r= sin ✓cos ˆx + sin ✓sin ˆy + cos ✓ˆz ;@r

@r2= sin2✓cos2+ sin2✓sin2+ cos2✓=1.

@✓ =rcos ✓cos ˆx +rcos ✓sin ˆy rsin ✓ˆz ;@r

@✓ 2=r2cos2✓cos2+r2cos2✓sin2+r2sin2✓=r2.

@ =rsin ✓sin ˆx +rsin ✓cos ˆy ;@r

@2=r2sin2✓sin2+r2sin2✓cos2=r2sin2✓.

)

ˆr = sin ✓cos ˆx + sin ✓sin ˆy + cos ✓ˆz .

✓= cos ✓cos ˆx + cos ✓sin ˆy sin ✓ˆz .

=sin ˆx + cos ˆy .

Check: ˆr ·ˆr = sin2✓(cos2+ sin2) + cos2✓= sin2✓+ cos2✓=1,X

✓·ˆ

=cos ✓sin cos + cos ✓sin cos =0,Xetc.

sin ✓ˆr = sin2✓cos ˆx + sin2✓sin ˆy + sin ✓cos ✓ˆz .

cos ✓ˆ

✓= cos2✓cos ˆx + cos2✓sin ˆy sin ✓cos ✓ˆz .

Add these:

(1) sin ✓ˆr + cos ✓ˆ

✓= + cos ˆx + sin ˆy ;

(2) ˆ

=sin ˆx + cos ˆy .

Multiply (1) by cos , (2) by sin , and subtract:

ˆx = sin ✓cos ˆr + cos ✓cos ˆ

✓sin ˆ

.

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CHAPTER 1. VECTOR ANALYSIS 15

Multiply (1) by sin , (2) by cos , and add:

ˆy = sin ✓sin ˆr + cos ✓sin ˆ

✓+ cos ˆ

.

cos ✓ˆr = sin ✓cos ✓cos ˆx + sin ✓cos ✓sin ˆy + cos2✓ˆz .

sin ✓ˆ

✓= sin ✓cos ✓cos ˆx + sin ✓cos ✓sin ˆy sin2✓ˆz .

Subtract these:

ˆz = cos ✓ˆr sin ✓ˆ

✓.

Problem 1.39

(a) r·v1=1

@r(r2r2)= 1

r24r3=4r

(r·v1)d⌧=(4r)(r2sin ✓dr d✓d) = (4) R

0r3dr ⇡

0sin ✓d✓2⇡

0d= (4) R4

4(2)(2⇡)=4⇡R4

v1·da=(r2ˆr )·(r2sin ✓d✓dˆr )=r4⇡

0sin ✓d✓2⇡

0d=4⇡R4X(Note: at surface of sphere r=R.)

(b) r·v2=1

@rr21

r2=0 )(r·v2)d⌧=0

v2·da=1

r2ˆr (r2sin ✓d✓dˆr )=sin ✓d✓d= 4⇡.

They don’t agree! The point is that this divergence is zero except at the origin, where it blows up, so our

calculation of (r·v2) is incorrect. The right answer is 4⇡.

Problem 1.40

r·v=1

@r(r2rcos ✓)+ 1

rsin ✓

@✓ (sin ✓rsin ✓)+ 1

rsin ✓

@(rsin ✓cos )

r23r2cos ✓+1

rsin ✓r2 sin ✓cos ✓+1

rsin ✓rsin ✓(sin )

= 3 cos ✓+ 2 cos ✓sin = 5 cos ✓sin 

(r·v)d⌧=(5 cos ✓sin )r2sin ✓dr d✓d=R

0r2dr ✓

02⇡

0(5 cos ✓sin )dd✓sin ✓

,!2⇡(5 cos ✓)

=R3

3(10⇡)⇡

0sin ✓cos ✓d✓

,!sin2✓

2

⇡

0=1

=5⇡

3R3.

Two surfaces—one the hemisphere: da=R2sin ✓d✓dˆr ;r=R;:0!2⇡,✓:0!⇡

v·da=(rcos ✓)R2sin ✓d✓d=R3⇡

0sin ✓cos ✓d✓2⇡

0d=R31

2(2⇡)=⇡R3.

other the ﬂat bottom: da=(dr)(rsin ✓d)(+ˆ

✓)=r dr dˆ

✓(here ✓=⇡

2). r:0!R, :0!2⇡.

v·da=(rsin ✓)(r dr d)=R

0r2dr 2⇡

0d=2⇡R3

Total: v·da=⇡R3+2

3⇡R3=5

3⇡R3.X

Problem 1.41 rt= (cos ✓+ sin ✓cos )ˆr +(sin ✓+ cos ✓cos )ˆ

✓+1

sin ✓

/(sin ✓

/sin )ˆ



r2t=r·(rt)

@rr2(cos ✓+ sin ✓cos )+1

rsin ✓

@✓ (sin ✓(sin ✓+ cos ✓cos )) + 1

rsin ✓

@(sin )

r22r(cos ✓+ sin ✓cos )+ 1

rsin ✓(2 sin ✓cos ✓+ cos2✓cos sin2✓cos )1

rsin ✓cos 

rsin ✓[2 sin ✓cos ✓+ 2 sin2✓cos 2 sin ✓cos ✓+ cos2✓cos sin2✓cos cos ]

rsin ✓(sin2✓+ cos2✓) cos cos =0.

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16 CHAPTER 1. VECTOR ANALYSIS

) r2t=0

Check: rcos ✓=z, r sin ✓cos =x)in Cartesian coordinates t=x+z. Obviously Laplacian is zero.

Gradient Theorem: b

art·dl=t(b)t(a)

Segment 1: ✓=⇡

2,=0,r:0!2.dl=dr ˆr ;rt·dl= (cos ✓+ sin ✓cos )dr = (0 + 1)dr =dr.

rt·dl=2

0dr =2.

Segment 2: ✓=⇡

2,r=2,:0!⇡

2.dl=rsin ✓dˆ

=2dˆ

.

rt·dl=(sin )(2 d)=2 sin d.rt·dl=⇡

02 sin d= 2 cos |⇡

0=2.

Segment 3: r=2,=⇡

2;✓:⇡

2!0.

dl=rd✓ˆ

✓=2d✓ˆ

✓;rt·dl=(sin ✓+ cos ✓cos )(2 d✓)=2 sin ✓d✓.

rt·dl=0

⇡

22 sin ✓d✓= 2 cos ✓|0

⇡

2=2.

Total: b

art·dl=22+2= 2 . Meanwhile, t(b)t(a) = [2(1 + 0)] [0( )] = 2.X

Problem 1.42 From Fig. 1.42, ˆs = cos ˆx + sin ˆy ;ˆ

=sin ˆx + cos ˆy ;ˆz =ˆz

Multiply ﬁrst by cos , second by sin , and subtract:

ˆs cos ˆ

sin = cos2ˆx + cos sin ˆy + sin2ˆx sin cos ˆy =ˆx (sin2+ cos2)=ˆx .

So ˆx = cos ˆs sin ˆ

.

Multiply ﬁrst by sin , second by cos , and add:

ˆs sin +ˆ

cos = sin cos ˆx + sin2ˆy sin cos ˆx + cos2ˆy =ˆy (sin2+ cos2)=ˆy .

So ˆy = sin ˆs + cos ˆ

.ˆz =ˆz .

Problem 1.43

(a) r·v=1

@sss(2 + sin2)+1

@(ssin cos )+ @

@z(3z)

s2s(2 + sin2)+1

ss(cos2sin2)+3

= 4 + 2 sin2+ cos2sin2+3

= 4 + sin2+ cos2+3= 8.

(b) (r·v)d⌧=(8)sdsddz =82

0sds⇡

0d5

0dz = 8(2) ⇡

2(5) = 40⇡.

Meanwhile, the surface integral has ﬁve parts:

top: z=5,da=sdsdˆz ;v·da=3zsdsd=15sdsd.v·da= 15 2

0sds⇡

0d= 15⇡.

bottom: z=0,da=sdsdˆz ;v·da=3zsdsd=0.v·da=0.

back: =⇡

2,da=ds dz ˆ

;v·da=ssin cos ds dz =0.v·da=0.

left: =0,da=ds dz ˆ

;v·da=ssin cos ds dz =0.v·da=0.

front: s=2,da=sddz ˆs ;v·da=s(2 + sin2)sddz = 4(2 + sin2)ddz.

v·da=4⇡

0(2 + sin2)d5

0dz = (4)(⇡+⇡

4)(5) = 25⇡.

So v·da= 15⇡+ 25⇡= 40⇡.X

@(3z)@

@z(ssin cos )ˆs +@

@zs(2 + sin2)@

@s(3z)ˆ



s@

@s(s2sin cos )@

@ s(2 + sin2)ˆz

s(2ssin cos s2 sin cos )ˆz =0.

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CHAPTER 1. VECTOR ANALYSIS 17

Problem 1.44

(a) 3(32)2(3) 1 = 27 61= 20.

(b) cos ⇡= -1.

(d) ln(2 + 3) = ln 1 = zero.

Problem 1.45

(a) 2

2(2x+ 3)1

3(x)dx =1

3(0 + 3) = 1.

(b) By Eq. 1.94, (1 x)=(x1), so 1 + 3 + 2 = 6.

19x21

3(x+1

3)dx =91

321

3=1

(d) 1 (if a>b), 0 (if a<b).

Problem 1.46

(a) 1

1 f(x)xd

dx (x)dx =xf(x)(x)|1

1 1

1

dx (xf(x)) (x)dx.

The ﬁrst term is zero, since (x) = 0 at ±1;d

dx (xf(x)) = xdf

dx +dx

dx f=xdf

dx +f.

So the integral is 1

1 xdf

dx +f(x)dx =0f(0) = f(0) = 1

1 f(x)(x)dx.

So, xd

dx (x)=(x).qed

(b) 1

1 f(x)d✓

dx dx =f(x)✓(x)|1

1 1

1

dx ✓(x)dx =f(1)1

dx dx =f(1)(f(1)f(0))

=f(0) = 1

1 f(x)(x)dx. So d✓

dx =(x).qed

Problem 1.47

(a) ⇢(r)=q3(rr0).Check: ⇢(r)d⌧=q3(rr0)d⌧=q. X

(b) ⇢(r)=q3(ra)q3(r).

Q=⇢d⌧=A(rR)4⇡r2dr =A4⇡R2.So A=Q

4⇡R2.⇢(r)= Q

4⇡R2(rR).

Problem 1.48

(a) a2+a·a+a2= 3a2.

(b) (rb)21

533(r)d⌧=1

125 b2=1

125 (42+3

2)= 1

(d) (e(2 ˆx +2ˆy +2ˆz ))2=(1ˆx +0ˆy +(1) ˆz )2=1+1=2<(1.5)2=2.25, so eis inside V,

and hence the integral is e·(de) = (3,2,1)·(2,0,2) = 6+0+2= -4.

Problem 1.49

First method: use Eq. 1.99 to write J=er4⇡3

(r)d⌧=4⇡e0= 4⇡.

Second method: integrating by parts (use Eq. 1.59).

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18 CHAPTER 1. VECTOR ANALYSIS

J=

ˆr

r2·r(er)d⌧+

erˆr

r2·da.But rer=@

@rerˆr =erˆr .

=1

r2er4⇡r2dr +erˆr

r2·r2sin ✓d✓dˆr =4⇡



erdr +eRsin ✓d✓d

=4⇡erR

0+4⇡eR=4⇡eR+e0+4⇡eR=4⇡.XHere R=1,so eR=0.

Problem 1.50 (a) r·F1=@

@x(0) + @

@y(0) + @

@zx2= 0 ; r·F2=@x

@x+@y

@y+@z

@z=1+1+1= 3

r⇥F1=

ˆx ˆy ˆz

00x2=ˆy @

@xx2=2xˆy ;r⇥F2=

ˆx ˆy ˆz

xyz

=0

F2is a gradient; F1is a curl U2=1

2x3+y2+z2would do (F2=rU2).

For A1, we want @Ay

@z@Az

@y=@Ax

@z@Az

@x= 0; @Ay

@x@Ax

@y=x2.A

y=x3

3,A

x=Az= 0 would do it.

A1=1

3x2ˆy (F1=r⇥A1).(But these are not unique.)

(b) r·F3=@

@x(yz)+ @

@y(xz)+ @

@z(xy) = 0; r⇥F3=

ˆx ˆy ˆz

yz xz xy =ˆx (xx)+ˆy (yy)+ˆz (zz)=0.

So F3can be written as the gradient of a scalar (F3=rU3) and as the curl of a vector (F3=r⇥A3). In

fact, U3=xyz does the job. For the vector potential, we have











@Az

@y@Ay

@z=yz, which suggests Az=1

4y2z+f(x, z); Ay=1

4yz2+g(x, y)

@Ax

@z@Az

@x=xz, suggesting Ax=1

4z2x+h(x, y); Az=1

4zx2+j(y, z)

@Ay

@x@Ax

@y=xy, so Ay=1

4x2y+k(y, z); Ax=1

4xy2+l(x, z)











Putting this all together: A3=1

4xz2y2ˆx +yx2z2ˆy +zy2x2ˆz (again, not unique).

Problem 1.51

(d) )(a): r⇥F=r⇥(rU)=0(Eq. 1.44 – curl of gradient is always zero).

(a) )(c): F·dl=(r⇥F)·da= 0 (Eq. 1.57–Stokes’ theorem).

aIF·dlb

aIIF·dl=b

aIF·dl+a

bIIF·dl=F·dl=0,so

b

F·dl=b

aII

F·dl.

(b) )(c): same as (c) )(b), only in reverse; (c) )(a): same as (a))(c).

Problem 1.52

(d) )(a): r·F=r·(r⇥W) = 0 (Eq 1.46—divergence of curl is always zero).

(a) )(c): F·da=(r·F)d⌧= 0 (Eq. 1.56—divergence theorem).

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CHAPTER 1. VECTOR ANALYSIS 19

I

F·da=II

F·da.

(Note: sign change because for F·da,dais outward, whereas for surface II it is inward.)

(b) )(c): same as (c) )(b), in reverse; (c))(a): same as (a))(c) .

Problem 1.53

In Prob. 1.15 we found that r·va= 0; in Prob. 1.18 we found that r⇥vc=0. So

vccan be written as the gradient of a scalar; vacan be written as the curl of a vector.

(a) To ﬁnd t:

(1) @t

@x=y2)t=y2x+f(y, z)

(2) @t

@y=2xy +z2

(3) @t

@z=2yz

From (1) & (3) we get @f

@z=2yz )f=yz2+g(y))t=y2x+yz2+g(y), so @t

@y=2xy +z2+@g

@y=

2xy +z2(from (2)) )@g

@y= 0. We may as well pick g= 0; then t=xy2+yz2.

(b) To ﬁnd W:@Wz

@y@Wy

@z=x2;@Wx

@z@Wz

@x=3z2x;@Wy

@x@Wx

@y=2xz.

Pick Wx= 0; then

@Wz

@x=3xz2)Wz=3

2x2z2+f(y, z)

@Wy

@x=2xz )Wy=x2z+g(y, z).

@Wz

@y@Wy

@z=@f

@y+x2@g

@z=x2)@f

@y@g

@z= 0. May as well pick f=g= 0.

W=x2zˆy 3

2x2z2ˆz .

Check: r⇥W=

ˆx ˆy ˆz

0x2z3

2x2z2=ˆx x2+ˆy 3xz2+ˆz (2xz).X

You can add any gradient (rt) to Wwithout changing its curl, so this answer is far from unique. Some

other solutions:

W=xz3ˆx x2zˆy ;

W=2xyz +xz3ˆx +x2yˆz ;

W=xyz ˆx 1

2x2zˆy +1

2x2y3z2ˆz .

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20 CHAPTER 1. VECTOR ANALYSIS

Problem 1.54

r·v=1

@rr2r2cos ✓+1

rsin ✓

@✓ sin ✓r2cos +1

rsin ✓

@ r2cos ✓sin 

r24r3cos ✓+1

rsin ✓cos ✓r2cos +1

rsin ✓r2cos ✓cos 

=rcos ✓

sin ✓[4 sin ✓+ cos cos ]=4rcos ✓.

(r·v)d⌧=(4rcos ✓)r2sin ✓dr d✓d=4



r3dr

⇡/2



cos ✓sin ✓d✓

⇡/2



d

=R41

2⇡

2=⇡R4

Surface consists of four parts:

(1) Curved: da=R2sin ✓d✓dˆr ;r=R. v·da=R2cos ✓R2sin ✓d✓d.

v·da=R4

⇡/2



cos ✓sin ✓d✓

⇡/2



d=R41

2⇡

2=⇡R4

(2) Left: da=r dr d✓ˆ

;=0.v·da=r2cos ✓sin (r dr d✓)=0.v·da=0.

(3) Back: da=r dr d✓ˆ

;=⇡/2.v·da=r2cos ✓sin (r dr d✓)=r3cos ✓dr d✓.

v·da=



r3dr

⇡/2



cos ✓d✓=1

4R4(+1) = 1

4R4.

(4) Bottom: da=rsin ✓dr dˆ

✓;✓=⇡/2.v·da=r2cos (r dr d).

v·da=



r3dr

⇡/2



cos d=1

4R4.

Total: v·da=⇡R4/4+01

4R4+1

4R4=⇡R4

4.X

Problem 1.55

r⇥v=

ˆx ˆy ˆz

ay bx 0=ˆz (ba).So (r⇥v)·da=(ba)⇡R2.

v·dl=(ay ˆx +bx ˆy )·(dx ˆx +dy ˆy +dz ˆz )=ay dx +bx dy;x2+y2=R2)2xdx+2y dy = 0,

so dy =(x/y)dx. So v·dl=ay dx +bx(x/y)dx =1

yay2bx2dx.

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CHAPTER 1. VECTOR ANALYSIS 21

For the “upper” semicircle, y=pR2x2,sov·dl=a(R2x2)bx2

pR2x2dx.

v·dl=

R



aR2(a+b)x2

pR2x2dx =aR2sin1x

R(a+b)x

2R2x2+R2

2sin1x

R

R

2R2(ab) sin1(x/R)

R

2R2(ab)sin1(1) sin1(+1)=1

2R2(ab)⇡

2⇡

2

2⇡R2(ba).

And the same for the lower semicircle (ychanges sign, but the limits on the integral are reversed) so

v·dl=⇡R2(ba). X

Problem 1.56

(1) x=z=0; dx =dz = 0; y:0!1.v·dl=(yz2)dy = 0; v·dl=0.

(2) x= 0; z=22y;dz =2dy;y:1!0.v·dl=(yz2)dy +(3y+z)dz =y(22y)2dy (3y+22y)2 dy;

v·dl=2



(2y34y2+y2) dy =2y4

24y3

3+y2

22y

=14

(3) x=y= 0; dx =dy = 0; z:2!0.v·dl=(3y+z)dz =z dz;

v·dl=



z dz =z2

2

2=2.

Total: v·dl=0+14

32= 8

Meanwhile, Stokes’ thereom says v·dl=(r⇥v)·da. Here da=dy dz ˆx , so all we need is

(r⇥v)x=@

@y(3y+z)@

@z(yz2)=32yz. Therefore

(r⇥v)·da=(3 2yz)dy dz =1

022y

(3 2yz)dzdy

=1

03(2 2y)2y1

2(2 2y)2dy =1

(4y3+8y210y+ 6) dy

=y4+8

3y35y2+6y

0=1+8

35+6= 8

3.X

Problem 1.57

Start at the origin.

(1) ✓=⇡

2,=0; r:0!1.v·dl=rcos2✓(dr)=0.v·dl=0.

(2) r=1,✓=⇡

2;:0!⇡/2.v·dl= (3r)(rsin ✓d)=3d.v·dl=3

⇡/2

0

d=3⇡

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22 CHAPTER 1. VECTOR ANALYSIS

(3) =⇡

2;rsin ✓=y=1,so r=1

sin ✓, dr =1

sin2✓cos ✓d✓,✓:⇡

2!✓0⌘tan1(1/2).

v·dl=rcos2✓(dr)(rcos ✓sin ✓)(rd✓)= cos2✓

sin ✓cos ✓

sin2✓d✓cos ✓sin ✓

sin2✓d✓

=cos3✓

sin3✓+cos ✓

sin ✓d✓=cos ✓

sin ✓cos2✓+ sin2✓

sin2✓d✓=cos ✓

sin3✓d✓.

Therefore v·dl=

✓0



⇡/2

cos ✓

sin3✓d✓=1

2 sin2✓

✓0

⇡/2

2·(1/5) 1

2·(1) =5

21

2=2.

(4) ✓=✓0,=⇡

2;r:p5!0.v·dl=rcos2✓(dr)=4

5r dr.

v·dl=4



r dr =4

2

=4

5·5

2=2.

Total:

v·dl=0+3⇡

2+22= 3⇡

Stokes’ theorem says this should equal (r⇥v)·da

r⇥v=1

rsin ✓@

@✓(sin ✓3r)@

@(rsin ✓cos ✓)ˆr +1

r1

sin ✓

@ rcos2✓@

@r(r3r)ˆ

✓

r@

@r(rr cos ✓sin ✓)@

@✓ rcos2✓ˆ



rsin ✓[3rcos ✓]ˆr +1

r[6r]ˆ

✓+1

r[2rcos ✓sin ✓+2rcos ✓sin ✓]ˆ



= 3 cot ✓ˆr 6ˆ

✓.

(1) Back face: da=r dr d✓ˆ

;(r⇥v)·da=0.(r⇥v)·da=0.

(2) Bottom: da=rsin ✓dr dˆ

✓;(r⇥v)·da=6rsin ✓dr d.✓=⇡

2,so(r⇥v)·da=6r dr d

(r⇥v)·da=



6r dr

⇡/2



d=6·1

2·⇡

2=3⇡

2.X

Problem 1.58

v·dl=y dz.

(1) Left side: z=ax;dz =dx;y= 0. Therefore v·dl=0.

(2) Bottom: dz = 0. Therefore v·dl=0.

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CHAPTER 1. VECTOR ANALYSIS 23

(3) Back: z=a1

2y;dz =1/2dy;y:2a!0.v·dl=



y1

2dy=1

2

2a=4a2

4=a2.

Meanwhile, r⇥v=ˆx ,so(r⇥v)·dais the projection of this surface on the xy plane = 1

2·a·2a=a2.X

Problem 1.59

r·v=1

@rr2r2sin ✓+1

rsin ✓

@✓ sin ✓4r2cos ✓+1

rsin ✓

@ r2tan ✓

r24r3sin ✓+1

rsin ✓4r2cos2✓sin2✓=4r

sin ✓sin2✓+ cos2✓sin2✓

=4rcos2✓

sin ✓.

(r·v)d⌧=4rcos2✓

sin ✓r2sin ✓dr d✓d=



4r3dr

⇡/6



cos2✓d✓

2⇡



d=R4(2⇡)✓

2+sin 2✓

4

⇡/6

=2⇡R4⇡

12 +sin 60

4=⇡R4

6⇡+3p3

2=⇡R4

12 2⇡+3

p3.

Surface coinsists of two parts:

(1) The ice cream: r=R;:0!2⇡;✓:0!⇡/6; da=R2sin ✓d✓dˆr ;v·da=R2sin ✓R2sin ✓d✓d=

R4sin2✓d✓d.

v·da=R4

⇡/6



sin2✓d✓

2⇡



d=R4(2⇡)1

2✓1

4sin 2✓⇡/6

=2⇡R4⇡

12 1

4sin 60=⇡R4

6⇡3p3

2

(2) The cone: ✓=⇡

6;:0!2⇡;r:0!R;da=rsin ✓ddr ˆ

✓=p3

2r dr dˆ

✓;v·da=p3r3dr d

v·da=p3



r3dr

2⇡



d=p3·R4

4·2⇡=p3

2⇡R4.

Therefore v·da=⇡R4

2⇡

3p3

2+p3=⇡R4

12 2⇡+3

p3.X.

Problem 1.60

(a) Corollary 2 says (rT)·dl=0.Stokes’ theorem says (rT)·dl=[r⇥(rT)]·da.So [r⇥(rT)]·da=0,

and since this is true for any surface, the integrand must vanish: r⇥(rT)=0,conﬁrming Eq. 1.44.

(b) Corollary 2 says (r⇥v)·da=0.Divergence theorem says (r⇥v)·da=r·(r⇥v)d⌧.So r·(r⇥v)d⌧

=0,and since this is true for any volume, the integrand must vanish: r(r⇥v)=0,conﬁrming Eq. 1.46.

Problem 1.61

(a) Divergence theorem: v·da=(r·v)d⌧.Let v=cT, where cis a constant vector. Using product

rule #5 in front cover: r·v=r·(cT)=T(r·c)+c·(rT).But cis constant so r·c= 0. Therefore we have:

c·(rT)d⌧=Tc·da.Since cis constant, take it outside the integrals: c·rTd⌧=c·Tda.But c

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24 CHAPTER 1. VECTOR ANALYSIS

is any constant vector—in particular, it could be be ˆx , or ˆy , or ˆz —so each component of the integral on left

equals corresponding component on the right, and hence

rTd⌧=Tda.qed

(b) Let v!(v⇥c) in divergence theorem. Then r·(v⇥c)d⌧=(v⇥c)·da.Product rule #6 )

r·(v⇥c)=c·(r⇥v)v·(r⇥c)=c·(r⇥v).(Note: r⇥c=0, since cis constant.) Meanwhile vector

indentity (1) says da·(v⇥c)=c·(da⇥v)=c·(v⇥da).Thus c·(r⇥v)d⌧=c·(v⇥da).Take c

outside, and again let cbe ˆx ,ˆy ,ˆz then:

(r⇥v)d⌧=v⇥da.qed

Tr·(rU)+(rU)·(rT)=Tr2U+(rU)·(rT). Therefore

Tr2U+(rU)·(rT)d⌧=(TrU)·da.qed

(d) Rewrite (c) with T$U:Ur2T+(rT)·(rU)d⌧=(UrT)·da. Subtract this from (c), noting

that the (rU)·(rT) terms cancel:

Tr2UUr2Td⌧=(TrUUrT)·da.qed

(e) Stokes’ theorem: (r⇥v)·da=v·dl. Let v=cT. By Product Rule #(7): r⇥(cT)=T(r⇥c)

c⇥(rT)=c⇥(rT) (since cis constant). Therefore, (c⇥(rT)) ·da=Tc·dl. Use vector indentity

#1 to rewrite the ﬁrst term (c⇥(rT)) ·da=c·(rT⇥da). So c·(rT⇥da)=c·Tdl. Pull coutside,

and let c!ˆx ,ˆy ,and ˆz to prove: rT⇥da=Tdl.qed

Problem 1.62

(a) da=R2sin ✓d✓dˆr .Let the surface be the northern hemisphere. The ˆx and ˆy components clearly integrate

to zero, and the ˆz component of ˆr is cos ✓,so

a=R2sin ✓cos ✓d✓dˆz =2⇡R2ˆz ⇡/2

sin ✓cos ✓d✓=2⇡R2ˆz sin2✓

2

⇡/2

0=⇡R2ˆz .

(b) Let T= 1 in Prob. 1.61(a). Then rT= 0, so da=0.qed

da=a1a26=0.

(d) For one such triangle, da=1

2(r⇥dl) (since r⇥dlis the area of the parallelogram, and the direction is

perpendicular to the surface), so for the entire conical surface, a=1

2r⇥dl.

(e) Let T=c·r, and use product rule #4: rT=r(c·r)=c⇥(r⇥r)+(c·r)r.But r⇥r= 0, and

(c·r)r=(cx@

@x+cy@

@y+cz@

@z)(xˆx +yˆy +zˆz )=cxˆx +cyˆy +czˆz =c.So Prob. 1.61(e) says

Tdl=(c·r)dl=(rT)⇥da=c⇥da=c⇥da=c⇥a=a⇥c.qed

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CHAPTER 1. VECTOR ANALYSIS 25

Problem 1.63

(1)

r·v=1

@rr2·1

r=1

@r(r)= 1

r2.

For a sphere of radius R:

v·da=1

Rˆr ·R2sin ✓d✓dˆr =Rsin ✓d✓d=4⇡R.

(r·v)d⌧=1

r2r2sin ✓dr d✓d=R

0

drsin ✓d✓d=4⇡R. 









So divergence

theorem checks.

Evidently there is no delta function at the origin.

r⇥(rnˆr )= 1

@rr2rn=1

@rrn+2=1

r2(n+ 2)rn+1 = (n+ 2)rn1

(except for n=2, for which we already know (Eq. 1.99) that the divergence is 4⇡3

(r)).

(2) Geometrically, it should be zero. Likewise, the curl in the spherical coordinates obviously gives zero.

To be certain there is no lurking delta function here, we integrate over a sphere of radius R, using

Prob. 1.61(b): If r⇥(rnˆr )=0, then (r⇥v)d⌧=0?

=v⇥da.Butv=rnˆr and da=

R2sin ✓d✓dˆr are both in the ˆr directions, so v⇥da=0.X

Problem 1.64

(a) Since the argument is not a function of angle, Eq. 1.73 says

D=1

4⇡

dr r21

22r

(r2+✏2)3/2=1

4⇡r2

dr r3

(r2+✏2)3/2

4⇡r23r2

(r2+✏2)3/23

r32r

(r2+✏2)5/3=1

4⇡r2

3r2

(r2+✏2)5/2r2+✏2r2=3✏2

4⇡(r2+✏2)5/2.X

(b) Setting r!0:

D(0,✏)= 3✏2

4⇡✏5=3

4⇡✏3,

which goes to inﬁnity as ✏!0. X

(d) D(r, ✏)4⇡r2dr =3✏21

(r2+✏2)5/2dr =3✏21

3✏2=1.X

(I looked up the integral.) Note that (b), (c), and (d) are the deﬁning conditions for 3(r).

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26 CHAPTER 2. ELECTROSTATICS

Chapter 2

Electrostatics

Problem 2.1

(a) Zero.

(b) F=1

4⇡✏0

r2,where ris the distance from center to each numeral. Fpoints toward the missing q.

Explanation: by superposition, this is equivalent to (a), with an extra qat 6 o’clock—since the force of all

twelve is zero, the net force is that of qonly.

(d) 1

4⇡✏0

r2,pointing toward the missing q. Same reason as (b). Note, however, that if you explained (b) as

a cancellation in pairs of opposite charges (1 o’clock against 7 o’clock; 2 against 8, etc.), with one unpaired q

doing the job, then you’ll need a di↵erent explanation for (d).

Problem 2.2

This time the “vertical” components cancel, leaving

E=1

4⇡✏02q

2sin ✓ˆx , or

E=1

4⇡✏0

z2+d

223/2ˆx .-x

q





✓

-E





From far away, (zd), the ﬁeld goes like E⇡1

4⇡✏0

z3ˆz , which, as we shall see, is the ﬁeld of a dipole. (If we

set d!0, we get E=0, as is appropriate; to the extent that this conﬁguration looks like a single point charge

from far away, the net charge is zero, so E!0.)

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CHAPTER 2. ELECTROSTATICS 27

Problem 2.3

✲

✻

!"# $

✴

dq =λdx

❑Ez=1

4πϵ0%L

λdx

η2cos θ; (η2=z2+x2; cos θ=z

η)

4πϵ0λz%L

(z2+x2)3/2dx

4πϵ0λz&1

√z2+x2'(((

0=1

4πϵ0

√z2+L2.

Ex=−1

4πϵ0%L

λdx

η2sin θ=−1

4πϵ0λ%x dx

(x2+z2)3/2

=−1

4πϵ0λ&−1

√x2+z2'(((

0=−1

4πϵ0λ&1

z−1

√z2+L2'.

E=1

4πϵ0

z)*−1 + z

√z2+L2+ˆx +*L

√z2+L2+ˆz,.

For z≫Lyou expect it to look like a point charge q=λL:E→1

4πϵ0

λL

z2ˆz. It checks, for with z≫Lthe ˆx

term →0, and the ˆz term →1

4πϵ0

zˆz.

Problem 2.4

From Ex. 2.1, with L→a

2and z→-z2+.a

2/2(distance from center of edge to P), ﬁeld of one edge is:

E1=1

4πϵ0

λa

-z2+a2

4-z2+a2

4+a2

There are 4 sides, and we want vertical components only, so multiply by 4 cos θ= 4 z

qz2+a2

E=1

4πϵ0

4λaz

.z2+a2

4/-z2+a2

ˆz.

Problem 2.5

❑

“Horizontal” components cancel, leaving: E=1

4πϵ00%λdl

η2cos θ1ˆz.

Here, η2=r2+z2, cos θ=z

η(both constants), while %dl = 2πr. So

E=1

4πϵ0

λ(2πr)z

(r2+z2)3/2ˆz.

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Ez=1

4⇡✏0L

dx

2cos ✓;(

2=z2+x2; cos ✓=z

)

4⇡✏0zL

(z2+x2)3/2dx

4⇡✏0z1

pz2+x2

0=1

4⇡✏0



pz2+L2.

Ex=1

4⇡✏0L

dx

2sin ✓=1

4⇡✏0x dx

(x2+z2)3/2

=1

4⇡✏01

px2+z2

0=1

4⇡✏01

z1

pz2+L2.

E=1

4⇡✏0



z1+ z

pz2+L2ˆx +L

pz2+L2ˆz .

For zLyou expect it to look like a point charge q=L:E!1

4⇡✏0

L

z2ˆz . It checks, for with zLthe ˆx

term !0, and the ˆz term !1

4⇡✏0



zˆz .

Problem 2.4

From Ex. 2.2, with L!a

2and z!z2+a

22(distance from center of edge to P), ﬁeld of one edge is:

E1=1

4⇡✏0

a

z2+a2

4z2+a2

4+a2

There are 4 sides, and we want vertical components only, so multiply by 4 cos ✓=4 z

qz2+a2

E=1

4⇡✏0

4az

z2+a2

4z2+a2

ˆz .

Problem 2.5

Problem 2.3

✲

✻

! "# $

x✴

dq =λdx

ηθ

❑Ez=1

4πϵ0%L

λdx

η2cos θ; (η2=z2+x2; cos θ=z

η)

4πϵ0λz%L

(z2+x2)3/2dx

4πϵ0λz&1

√z2+x2'(((

0=1

4πϵ0

√z2+L2.

Ex=−1

4πϵ0%L

λdx

η2sin θ=−1

4πϵ0λ%x dx

(x2+z2)3/2

=−1

4πϵ0λ&−1

√x2+z2'(((

0=−1

4πϵ0λ&1

z−1

√z2+L2'.

E=1

4πϵ0

z)*−1 + z

√z2+L2+ˆx +*L

√z2+L2+ˆz,.

For z≫Lyou expect it to look like a point charge q=λL:E→1

4πϵ0

λL

z2ˆz. It checks, for with z≫Lthe ˆx

term →0, and the ˆz term →1

4πϵ0

zˆz.

Problem 2.4

From Ex. 2.1, with L→a

2and z→-z2+.a

2/2(distance from center of edge to P), ﬁeld of one edge is:

E1=1

4πϵ0

λa

-z2+a2

4-z2+a2

4+a2

There are 4 sides, and we want vertical components only, so multiply by 4 cos θ= 4 z

qz2+a2

E=1

4πϵ0

4λaz

.z2+a2

4/-z2+a2

ˆz.

Problem 2.5

❑

“Horizontal” components cancel, leaving: E=1

4πϵ00%λdl

η2cos θ1ˆz.

Here, η2=r2+z2, cos θ=z

η(both constants), while %dl = 2πr. So

E=1

4πϵ0

λ(2πr)z

(r2+z2)3/2ˆz.

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“Horizontal” components cancel, leaving: E=1

4⇡✏0dl

2cos ✓ˆz .

Here,

2=r2+z2, cos ✓=z

(both constants), while dl =2⇡r. So

E=1

4⇡✏0

(2⇡r)z

(r2+z2)3/2ˆz .

Problem 2.6

Break it into rings of radius r, and thickness dr, and use Prob. 2.5 to express the ﬁeld of each ring. Total

charge of a ring is ·2⇡r·dr =·2⇡r,so=dr is the “line charge” of each ring.

Ering =1

4⇡✏0

(dr)2⇡rz

(r2+z2)3/2;Edisk =1

4⇡✏0

2⇡zR

(r2+z2)3/2dr.

Edisk =1

4⇡✏0

2⇡z1

z1

pR2+z2ˆz .

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28 CHAPTER 2. ELECTROSTATICS

For Rzthe second term !0, so Eplane =1

4⇡✏02⇡ˆz =

2✏0

ˆz .

For zR,1

pR2+z2=1

z1+R2

z21/2⇡1

z11

z2, so [ ] ⇡1

z1

z+1

z3=R2

2z3,

and E=1

4⇡✏0

2⇡R2

2z2=1

4⇡✏0

z2, where Q=⇡R2.X

Problem 2.7

Eis clearly in the zdirection. From the diagram,

dq =da =R2sin ✓d✓d,

2=R2+z22Rz cos ✓,

cos =zRcos ✓

Contents

❂

✲y

✻

Ez=1

4πϵ0!σR2sin θdθdφ(z−Rcos θ)

(R2+z2−2Rz cos θ)3/2."dφ= 2π.

4πϵ0

(2πR2σ)!π

(z−Rcos θ) sin θ

(R2+z2−2Rz cos θ)3/2dθ.Let u= cos θ;du =−sin θdθ;#θ= 0 ⇒u= +1

θ=π⇒u=−1$.

4πϵ0

(2πR2σ)!1

−1

z−Ru

(R2+z2−2Rzu)3/2du. Integral can be done by partial fractions—or look it up.

4πϵ0

(2πR2σ)%1

zu −R

√R2+z2−2Rzu &1

−1

4πϵ0

2πR2σ

z2#(z−R)

|z−R|−(−z−R)

|z+R|$.

For z > R (outside the sphere), Ez=1

4πϵ0

4πR2σ

z2=1

4πϵ0

z2, so E=1

4πϵ0

z2ˆz.

For z < R (inside), Ez= 0, so E= 0.

Problem 2.8

According to Prob. 2.7, all shells interior to the point (i.e. at smaller r) contribute as though their charge

were concentrated at the center, while all exterior shells contribute nothing. Therefore:

E(r) = 1

4πϵ0

Qint

r2ˆr,

where Qint is the total charge interior to the point. Outside the sphere, all the charge is interior, so

E=1

4πϵ0

r2ˆr.

Inside the sphere, only that fraction of the total which is interior to the point counts:

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Ez=1

4⇡✏0R2sin ✓d✓d(zRcos ✓)

(R2+z22Rz cos ✓)3/2.d=2⇡.

4⇡✏0

(2⇡R2)⇡

(zRcos ✓) sin ✓

(R2+z22Rz cos ✓)3/2d✓.Let u= cos ✓;du =sin ✓d✓;✓=0)u=+1

✓=⇡)u=1.

4⇡✏0

(2⇡R2)1

1

zRu

(R2+z22Rzu)3/2du. Integral can be done by partial fractions—or look it up.

4⇡✏0

(2⇡R2)1

zu R

pR2+z22Rzu1

1

4⇡✏0

2⇡R2

z2(zR)

|zR|(zR)

|z+R|.

For z>R(outside the sphere), Ez=1

4⇡✏0

4⇡R2

z2=1

4⇡✏0

z2,so E=1

4⇡✏0

z2ˆz .

For z<R(inside), Ez= 0, so E=0.

Problem 2.8

According to Prob. 2.7, all shells interior to the point (i.e. at smaller r) contribute as though their charge

were concentrated at the center, while all exterior shells contribute nothing. Therefore:

E(r)= 1

4⇡✏0

Qint

r2ˆr ,

where Qint is the total charge interior to the point. Outside the sphere, all the charge is interior, so

E=1

4⇡✏0

r2ˆr .

Inside the sphere, only that fraction of the total which is interior to the point counts:

Qint =

3⇡r3

3⇡R3Q=r3

R3Q, so E=1

4⇡✏0

R3Q1

r2ˆr =1

4⇡✏0

R3r.

Problem 2.9

(a) ⇢=✏0r·E=✏01

@rr2·kr3=✏01

r2k(5r4)= 5✏0kr2.

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CHAPTER 2. ELECTROSTATICS 29

(b) By Gauss’s law: Qenc =✏0E·da=✏0(kR3)(4⇡R2)= 4⇡✏0kR5.

By direct integration: Qenc =⇢d⌧=R

0(5✏0kr2)(4⇡r2dr) = 20⇡✏0kR

0r4dr =4⇡✏0kR5.X

Problem 2.10

Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface

of this larger cube gets the same ﬂux as every other one, so:

Qint =

3πr3

3πR3Q=r3

R3Q, so E=1

4πϵ0

R3Q1

r2ˆr =1

4πϵ0

R3r.

Problem 2.9

(a) ρ=ϵ0∇·E=ϵ01

∂

∂r!r2·kr3"=ϵ01

r2k(5r4) = 5ϵ0kr2.

(b) By Gauss’s law: Qenc =ϵ0#E·da=ϵ0(kR3)(4πR2) = 4πϵ0kR5.

By direct integration: Qenc =$ρdτ=$R

0(5ϵ0kr2)(4πr2dr) = 20πϵ0k$R

0r4dr = 4πϵ0kR5.!

Problem 2.10

Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface

of this larger cube gets the same ﬂux as every other one, so:

one

face

E·da=1

24 %

whole

large

cube

E·da.

The latter is 1

ϵ0q, by Gauss’s law. Therefore %

one

face

E·da=q

24ϵ0

Problem 2.11

✲

✙Gaussian surface: Inside:#E·da=E(4πr2) = 1

ϵ0Qenc = 0 ⇒E= 0.

✲Gaussian surface: Outside:E(4πr2) = 1

ϵ0(σ4πR2)⇒E=σR2

ϵ0r2ˆr.}(As in Prob. 2.7.)

Problem 2.12

✒

❄

✙

Gaussian surface #E·da=E·4πr2=1

ϵ0Qenc =1

ϵ0

3πr3ρ.So

E=1

3ϵ0

ρrˆr.

Since Qtot =4

3πR2ρ,E=1

4πϵ0

R3ˆr (as in Prob. 2.8).

Problem 2.13

✻

&'( )

✠

Gaussian surface #E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0λl. So

E=λ

2πϵ0sˆs (same as Ex. 2.1).

Problem 2.14

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

one

face

E·da=1

24 

whole

large

cube

E·da.

The latter is 1

✏0q, by Gauss’s law. Therefore 

one

face

E·da=q

24✏0

Problem 2.11

Qint =

3πr3

3πR3Q=r3

R3Q, so E=1

4πϵ0

R3Q1

r2ˆr =1

4πϵ0

R3r.

Problem 2.9

(a) ρ=ϵ0∇·E=ϵ01

∂

∂r!r2·kr3"=ϵ01

r2k(5r4) = 5ϵ0kr2.

(b) By Gauss’s law: Qenc =ϵ0#E·da=ϵ0(kR3)(4πR2) = 4πϵ0kR5.

By direct integration: Qenc =$ρdτ=$R

0(5ϵ0kr2)(4πr2dr) = 20πϵ0k$R

0r4dr = 4πϵ0kR5.!

Problem 2.10

Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface

of this larger cube gets the same ﬂux as every other one, so:

one

face

E·da=1

24 %

whole

large

cube

E·da.

The latter is 1

ϵ0q, by Gauss’s law. Therefore %

one

face

E·da=q

24ϵ0

Problem 2.11

✲

✙Gaussian surface: Inside:#E·da=E(4πr2) = 1

ϵ0Qenc = 0 ⇒E= 0.

✲Gaussian surface: Outside:E(4πr2) = 1

ϵ0(σ4πR2)⇒E=σR2

ϵ0r2ˆr.}(As in Prob. 2.7.)

Problem 2.12

✒

❄

✙

Gaussian surface #E·da=E·4πr2=1

ϵ0Qenc =1

ϵ0

3πr3ρ.So

E=1

3ϵ0

ρrˆr.

Since Qtot =4

3πR2ρ,E=1

4πϵ0

R3ˆr (as in Prob. 2.8).

Problem 2.13

✻

&'( )

✠

Gaussian surface #E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0λl. So

E=λ

2πϵ0sˆs (same as Ex. 2.1).

Problem 2.14

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Problem 2.12

Qint =

3πr3

3πR3Q=r3

R3Q, so E=1

4πϵ0

R3Q1

r2ˆr =1

4πϵ0

R3r.

Problem 2.9

(a) ρ=ϵ0∇·E=ϵ01

∂

∂r!r2·kr3"=ϵ01

r2k(5r4) = 5ϵ0kr2.

(b) By Gauss’s law: Qenc =ϵ0#E·da=ϵ0(kR3)(4πR2) = 4πϵ0kR5.

By direct integration: Qenc =$ρdτ=$R

0(5ϵ0kr2)(4πr2dr) = 20πϵ0k$R

0r4dr = 4πϵ0kR5.!

Problem 2.10

Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface

of this larger cube gets the same ﬂux as every other one, so:

one

face

E·da=1

24 %

whole

large

cube

E·da.

The latter is 1

ϵ0q, by Gauss’s law. Therefore %

one

face

E·da=q

24ϵ0

Problem 2.11

✲

✙Gaussian surface: Inside:#E·da=E(4πr2) = 1

ϵ0Qenc = 0 ⇒E= 0.

✲Gaussian surface: Outside:E(4πr2) = 1

ϵ0(σ4πR2)⇒E=σR2

ϵ0r2ˆr.}(As in Prob. 2.7.)

Problem 2.12

✒

❄

✙

Gaussian surface #E·da=E·4πr2=1

ϵ0Qenc =1

ϵ0

3πr3ρ.So

E=1

3ϵ0

ρrˆr.

Since Qtot =4

3πR2ρ,E=1

4πϵ0

R3ˆr (as in Prob. 2.8).

Problem 2.13

✻

&'( )

✠

Gaussian surface #E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0λl. So

E=λ

2πϵ0sˆs (same as Ex. 2.1).

Problem 2.14

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E·da=E·4⇡r2=1

✏0Qenc =1

✏0

3⇡r3⇢.So

E=1

3✏0

⇢rˆr .

Since Qtot =4

3⇡R3⇢,E=1

4⇡✏0

R3ˆr (as in Prob. 2.8).

Problem 2.13

Contents

Problem 2.13

✻

!"# $

✠

Gaussian surface %E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0λl. So

E=λ

2πϵ0sˆs (same as Ex. 2.1).

Problem 2.14

✒

r✙

Gaussian surface

Problem 2.15

(i) Qenc = 0, so E= 0.

(ii) %E·da=E(4πr2) = 1

ϵ0Qenc =1

ϵ0&ρdτ=1

ϵ0&k

¯r2¯r2sin θd¯r dθdφ

=4πk

ϵ0&r

ad¯r=4πk

ϵ0(r−a)∴E=k

ϵ0'r−a

r2(ˆr.

(iii) E(4πr2) = 4πk

ϵ0&b

ad¯r=4πk

ϵ0(b−a),so

E=k

ϵ0'b−a

r2(ˆr.

✲

✻

|E|

a b

Problem 2.16

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E·da=E·2⇡s·l=1

✏0Qenc =1

✏0l. So

E=

2⇡✏0sˆs (same as Eq. 2.9).

Problem 2.14

Contents

Problem 2.13

✻

!"# $

✠

Gaussian surface %E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0λl. So

E=λ

2πϵ0sˆs (same as Ex. 2.1).

Problem 2.14

✒

r✙

Gaussian surface

Problem 2.15

(i) Qenc = 0, so E= 0.

(ii) %E·da=E(4πr2) = 1

ϵ0Qenc =1

ϵ0&ρdτ=1

ϵ0&k

¯r2¯r2sin θd¯r dθdφ

=4πk

ϵ0&r

ad¯r=4πk

ϵ0(r−a)∴E=k

ϵ0'r−a

r2(ˆr.

(iii) E(4πr2) = 4πk

ϵ0&b

ad¯r=4πk

ϵ0(b−a),so

E=k

ϵ0'b−a

r2(ˆr.

✲

✻

|E|

a b

Problem 2.16

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E·da=E·4⇡r2=1

✏0Qenc =1

✏0⇢d⌧=1

✏0(k¯r)(¯r2sin ✓d¯rd✓d)

✏0k4⇡r

0¯r3d¯r=4⇡k

✏0

4=⇡k

✏0r4.

)E=1

4⇡✏0

⇡kr2ˆr .

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30 CHAPTER 2. ELECTROSTATICS

Problem 2.15

(i) Qenc = 0, so E=0.

(ii) E·da=E(4⇡r2)= 1

✏0Qenc =1

✏0⇢d⌧=1

✏0k

¯r2¯r2sin ✓d¯rd✓d

=4⇡k

✏0r

ad¯r=4⇡k

✏0(ra))E=k

✏0ra

r2ˆr .

(iii) E(4⇡r2)=4⇡k

✏0b

ad¯r=4⇡k

✏0(ba),so

E=k

✏0ba

r2ˆr .

Contents

Problem 2.13

✻

!"# $

✠

Gaussian surface %E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0λl. So

E=λ

2πϵ0sˆs (same as Ex. 2.1).

Problem 2.14

✒

r✙

Gaussian surface

Problem 2.15

(i) Qenc = 0, so E= 0.

(ii) %E·da=E(4πr2) = 1

ϵ0Qenc =1

ϵ0&ρdτ=1

ϵ0&k

¯r2¯r2sin θd¯r dθdφ

=4πk

ϵ0&r

ad¯r=4πk

ϵ0(r−a)∴E=k

ϵ0'r−a

r2(ˆr.

(iii) E(4πr2) = 4πk

ϵ0&b

ad¯r=4πk

ϵ0(b−a),so

E=k

ϵ0'b−a

r2(ˆr.

✲

✻

|E|

a b

Problem 2.16

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Problem 2.16

(i)

Contents

Problem 2.13

Problem 2.16

(i)

✛Gaussian surface !E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπs2l;

E=ρs

2ϵ0

ˆs.

(ii)

✛Gaussian surface

✻

s!E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπa2l;

E=ρa2

2ϵ0sˆs.

(iii)

✛Gaussian surface

✻

!E·da=E·2πs·l=1

ϵ0Qenc = 0;

E= 0.

✲

✻

|E|

a b

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E·da=E·2⇡s·l=1

✏0Qenc =1

✏0⇢⇡s2l;

E=⇢s

2✏0

ˆs .

(ii)

Contents

Problem 2.13

Problem 2.16

(i)

✛Gaussian surface !E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπs2l;

E=ρs

2ϵ0

ˆs.

(ii)

✛Gaussian surface

✻

s!E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπa2l;

E=ρa2

2ϵ0sˆs.

(iii)

✛Gaussian surface

✻

!E·da=E·2πs·l=1

ϵ0Qenc = 0;

E= 0.

✲

✻

|E|

a b

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E·da=E·2⇡s·l=1

✏0Qenc =1

✏0⇢⇡a2l;

E=⇢a2

2✏0sˆs .

(iii)

Contents

Problem 2.13

Problem 2.16

(i)

✛Gaussian surface !E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπs2l;

E=ρs

2ϵ0

ˆs.

(ii)

✛Gaussian surface

✻

s!E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπa2l;

E=ρa2

2ϵ0sˆs.

(iii)

✛Gaussian surface

✻

!E·da=E·2πs·l=1

ϵ0Qenc = 0;

E= 0.

✲

✻

|E|

a b

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E·da=E·2⇡s·l=1

✏0Qenc = 0;

E=0.

Contents

Problem 2.13

Problem 2.16

(i)

✛Gaussian surface !E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπs2l;

E=ρs

2ϵ0

ˆs.

(ii)

✛Gaussian surface

✻

s!E·da=E·2πs·l=1

ϵ0Qenc =1

ϵ0ρπa2l;

E=ρa2

2ϵ0sˆs.

(iii)

✛Gaussian surface

✻

!E·da=E·2πs·l=1

ϵ0Qenc = 0;

E= 0.

✲

✻

|E|

a b

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Problem 2.17 On the xz plane E= 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane

and the other at y.

Contents

Problem 2.17

On the x z plane E= 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane and the

other at y.

✲ ✛

❂

Gaussian pillbox

Qenc =1

ϵ0Adρ⇒E=ρ

ϵ0

dˆy (for y > d).

✲

✻

−d

ρd

ϵ0

Problem 2.18

From Prob. 2.12, the ﬁeld inside the positive sphere is E+=ρ

3ϵ0r+, where r+is the vector from the positive

center to the point in question. Likewise, the ﬁeld of the negative sphere is −ρ

3ϵ0r−. So the total ﬁeld is

E=ρ

3ϵ0

(r+−r−)

But (see diagram) r+−r−=d. So E=ρ

3ϵ0

✕✯

✛

−

r−

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E·da=E·A=1

✏0Qenc =1

✏0Ay⇢;

E=⇢

✏0

yˆy (for |y|<d).

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CHAPTER 2. ELECTROSTATICS 31

Qenc =1

✏0Ad⇢)E=⇢

✏0

dˆy (for y>d).

Problem 2.17

On the x z plane E= 0 by symmetry. Set up a Gaussian “pillbox” with one face in this plane and the

other at y.

✲ ✛

❂

Gaussian pillbox !E·da=E·A=1

ϵ0Qenc =1

ϵ0Ayρ;

E=ρ

ϵ0

yˆy (for |y|< d).

Qenc =1

ϵ0Adρ⇒E=ρ

ϵ0

dˆy (for y > d).

✲

✻

−d

ρd

ϵ0

Problem 2.18

From Prob. 2.12, the ﬁeld inside the positive sphere is E+=ρ

3ϵ0r+, where r+is the vector from the positive

center to the point in question. Likewise, the ﬁeld of the negative sphere is −ρ

3ϵ0r−. So the total ﬁeld is

E=ρ

3ϵ0

(r+−r−)

But (see diagram) r+−r−=d. So E=ρ

3ϵ0

✕✯

✛

−

r−

Problem 2.19

∇×E=1

4πϵ0

∇×"ˆηˆη

ˆη

η2ρdτ=1

4πϵ0"#∇×$ˆηˆη

ˆη

η2%&ρdτ(since ρdepends on r′, not r)

= 0 (since ∇×$ˆηˆη

ˆη

η2%= 0, from Prob. 1.62).

Problem 2.20

(1) ∇×E1=k'

ˆx ˆy ˆz

∂

∂x

∂

∂y

∂

∂z

xy 2yz 3zx

=k[ˆx(0 −2y) + ˆy(0 −3z) + ˆz(0 −x)] ̸= 0,

so E1is an impossible electrostatic ﬁeld.

(2) ∇×E2=k'

ˆx ˆy ˆz

∂

∂x

∂

∂y

∂

∂z

y22xy +z22yz

=k[ˆx(2z−2z) + ˆy(0 −0) + ˆz(2y−2y)] = 0,

so E2is a possible electrostatic ﬁeld.

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Problem 2.18

From Prob. 2.12, the ﬁeld inside the positive sphere is E+=⇢

3✏0r+,wherer+is the vector from the positive

center to the point in question. Likewise, the ﬁeld of the negative sphere is ⇢

3✏0r. So the total ﬁeld is

E=⇢

3✏0

(r+r)

But (see diagram) r+r=d. So E=⇢

3✏0

d.r





*





r

Problem 2.19

r⇥E=1

4⇡✏0r⇥ˆ

2⇢d⌧=1

4⇡✏0r⇥ˆ

2⇢d⌧(since ⇢depends on r0, not r)

=0(since r⇥ˆ

2=0, from Prob. 1.63).

Problem 2.20

(1) r⇥E1=k

ˆx ˆy ˆz

xy 2yz 3zx=k[ˆx (0 2y)+ˆy (0 3z)+ˆz (0 x)] 6=0,

so E1is an impossible electrostatic ﬁeld.

(2) r⇥E2=k

ˆx ˆy ˆz

y22xy +z22yz=k[ˆx (2z2z)+ˆy (0 0) + ˆz (2y2y)] = 0,

so E2is a possible electrostatic ﬁeld.

Let’s go by the indicated path:

⇢

⇢=

-y

⇢

⇢=

III

r(x0,y

0,z

E·dl=(y2dx + (2xy +z2)dy +2yz dz)k

Step I: y=z=0;dy =dz =0.E·dl=ky2dx =0.

Step II: x=x0,y:0!y0,z=0.dx=dz =0.

E·dl=k(2xy +z2)dy =2kx0y dy.

II E·dl=2kx0y0

0y dy =kx0y2

Step III :x=x0,y=y0,z:0!z0;dx =dy =0.

E·dl=2kyz dz =2ky0z dz.

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32 CHAPTER 2. ELECTROSTATICS

III E·dl=2y0kz0

0z dz =ky0z2

V(x0,y

0,z

0)= 

(x0,y0,z0)

0

E·dl=k(x0y2

0+y0z2

0), or V(x, y, z)=k(xy2+yz2).

Check:rV=k[@

@x(xy2+yz2)ˆx +@

@y(xy2+yz2)ˆy +@

@z(xy2+yz2)ˆz ]=k[y2ˆx +(2xy+z2)ˆy +2yz ˆz ]=E.X

Problem 2.21

V(r)=r

1E·dl.





Outside the sphere (r>R): E=1

4⇡✏0

r2ˆr .

Inside the sphere (r<R): E=1

4⇡✏0

R3rˆr .

So for r>R:V(r)=r

11

4⇡✏0

¯r2d¯r=1

4⇡✏0q1

¯r

1=q

4⇡✏0

and for r<R:V(r)=R

11

4⇡✏0

¯r2d¯rr

R1

4⇡✏0

R3¯rd¯r=q

4⇡✏01

R1

R3r2R2

2

4⇡✏0

2R3r2

R2.

When r>R,rV=q

4⇡✏0

@r1

rˆr =q

4⇡✏0

r2ˆr ,soE=rV=q

4⇡✏0

r2ˆr .X

When r<R,rV=q

4⇡✏0

@r3r2

R2ˆr =q

4⇡✏0

2R2r

R2ˆr =q

4⇡✏0

R3ˆr ;soE=rV=1

4⇡✏0

R3rˆr .X

0.5 1 1.5 2 2.5 3

0.2

0.4

0.6

0.8

1.2

1.4

1.6

V(r)

(In the ﬁgure, ris in units of R, and V(r) is in units of q/4⇡✏0R.)

Problem 2.22

E=1

4⇡✏0

2

sˆs (Prob. 2.13). In this case we cannot set the reference point at 1, since the charge itself

extends to 1. Let’s set it at s=a. Then

V(s)=s

a1

4⇡✏0

2

¯sd¯s=1

4⇡✏0

2ln s

a.

(In this form it is clear why a=1would be no good—likewise the other “natural” point, a= 0.)

rV=1

4⇡✏02@

@sln s

aˆs =1

4⇡✏021

sˆs =E.X

Problem 2.23

V(0) = 0

1E·dl=b

1k

✏0

(ba)

r2dr a

bk

✏0

(ra)

r2dr 0

a(0)dr =k

✏0

(ba)

bk

✏0ln a

b+a1

a1

b

✏01a

bln a

b1+a

b=k

✏0

ln b

a.

Problem 2.24

Using Eq. 2.22 and the ﬁelds from Prob. 2.16:

V(b)V(0) = b

0E·dl=a

0E·dlb

aE·dl=⇢

2✏0a

0sds⇢a2

2✏0b

sds

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CHAPTER 2. ELECTROSTATICS 33

=⇢

2✏0s2

2

0+⇢a2

2✏0ln s|b

a=⇢a2

4✏01 + 2 ln b

a.

Problem 2.25

(a) V=1

4⇡✏0

z2+d

22.

(b) V=1

4⇡✏0L

L

dx

pz2+x2=

4⇡✏0ln(x+pz2+x2)L

L

=

4⇡✏0

ln L+pz2+L2

L+pz2+L2=

2⇡✏0

ln L+pz2+L2

z.

Contents

Problem 2.25

(a) V=1

4πϵ0

!z2+"d

2#2.

(b) V=1

4πϵ0$L

−L

λdx

√z2+x2=λ

4πϵ0ln(x+√z2+x2)%%

−L

=λ

4πϵ0

ln &L+√z2+L2

−L+√z2+L2'=λ

2πϵ0ln (L+√z2+L2

z).

4πϵ0$R

σ2πr dr

√r2+z2=1

4πϵ02πσ (√r2+z2)%%

0=σ

2ϵ0(*R2+z2−z).

In each case, by symmetry ∂V

∂y=∂V

∂x= 0.∴E=−

∂V

∂zˆz.

(a) E=−1

4πϵ02q"−1

2#2z

“z2+(d

2)2”3/2ˆz =1

4πϵ0

2qz

"z2+"d

2#2#3/2ˆz (agrees with Prob. 2.2a).

(b) E=−

4πϵ0+1

(L+√z2+L2)

√z2+L22z−1

(−L+√z2+L2)

√z2+L22z,ˆz

=−

4πϵ0

√z2+L2+−L+√z2+L2−L−√z2+L2

(z2+L2)−L2,ˆz =2Lλ

4πϵ0

z√z2+L2ˆz (agrees with Ex. 2.1).

2ϵ0+1

√R2+z22z−1,ˆz =σ

2ϵ0-1−

√R2+z2.ˆz (agrees with Prob. 2.6).

If the right-hand charge in (a) is −q, then V= 0 , which, naively, suggests E=−∇V= 0, in contradiction

with the answer to Prob. 2.2b. The point is that we only know Von the zaxis, and from this we cannot

hope to compute Ex=−

∂V

∂xor Ey=−

∂V

∂y. That was OK in part (a), because we knew from symmetry that

Ex=Ey= 0. But now Epoints in the xdirection, so knowing Von the zaxis is insuﬃcient to determine E.

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4⇡✏0R

2⇡rdr

pr2+z2=1

4⇡✏0

2⇡ (r2+z2)

0=

2✏0R2+z2z.

In each case, by symmetry @V

@y=@V

@x=0.)E=@V

@zˆz .

(a) E=1

4⇡✏0

2q1

22z

z2+d

223/2ˆz =1

4⇡✏0

2qz

z2+d

223/2ˆz (agrees with Ex. 2.1).

(b) E=

4⇡✏01

(L+pz2+L2)

pz2+L22z1

(L+pz2+L2)

pz2+L22zˆz

=

4⇡✏0

pz2+L2L+pz2+L2Lpz2+L2

(z2+L2)L2ˆz =2L

4⇡✏0

zpz2+L2ˆz (agrees with Ex. 2.2).

2✏01

pR2+z22z1ˆz =

2✏01z

pR2+z2ˆz (agrees with Prob. 2.6).

If the right-hand charge in (a) is q, then V=0 , which, naively, suggests E=rV=0, in contradiction

with the answer to Prob. 2.2. The point is that we only know Von the zaxis, and from this we cannot hope

to compute Ex=@V

@xor Ey=@V

@y. That was OK in part (a), because we knew from symmetry that

Ex=Ey= 0. But now Epoints in the xdirection, so knowing Von the zaxis is insuﬃcient to determine E.

Problem 2.26

✲

✻

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V(a)= 1

4⇡✏0p2h

02⇡r

d

=2⇡

4⇡✏0

p2(p2h)= h

2✏0

(where r=

/p2)

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34 CHAPTER 2. ELECTROSTATICS

V(b)= 1

4⇡✏0p2h

02⇡r

d

(where ¯

=h2+

2p2h

)

=2⇡

4⇡✏0

p2p2h

h2+

2p2h

=

2p2✏0h2+

2p2h

p2ln(2h2+

2p2h

p2h)

p2h

=

2p2✏0h+h

p2ln(2h+2

p2hp2h)hh

p2ln(2hp2h)

=

2p2✏0

p2ln(2h+p2h)ln(2hp2h)=h

4✏0

ln 2+p2

2p2=h

4✏0

ln (2 + p2)2

2

=h

2✏0

ln(1 + p2).)V(a)V(b)= h

2✏01ln(1 + p2).

Problem 2.27

! "# $

z−L

#$! "

!"# $

✲ ✛

Cut the cylinder into slabs, as shown in the ﬁgure, and

use result of Prob. 2.25c, with z→xand σ→ρdx:

V=ρ

2ϵ0

z+L/2

z−L/2&√R2+x2−x'dx

=ρ

2ϵ0

2(x√R2+x2+R2ln(x+√R2+x2)−x2)**z+L/2

z−L/2

=ρ

4ϵ08

(z+L

2)qR2+(z+L

2)2−(z−L

2)qR2+(z−L

2)2+R2ln2

z+L

2+rR2+(z+L

2)2

z−

2+rR2+(z−

2)23

5−2zL9

;

(Note: −&z+L

2'2+&z−L

2'2=−z2−zL −L2

4+z2−zL +L2

4=−2zL.)

E=−∇V=−ˆz ∂V

∂z=−

ˆzρ

4ϵ0+,R2+-z+L

2.2

+&z+L

2'2

/R2+&z+L

2'2−,R2+-z−L

2.2

−&z−L

2'2

/R2+&z−L

2'2

+R201 + z+L

qR2+(z+L

2)2

z+L

2+/R2+&z+L

2'2−

1 + z−L

qR2+(z−L

2)2

z−L

2+/R2+&z−L

2'2

#$! "

/R2+&z+L

2'2−1

/R2+&z−L

2'2

1−2L2

E=−

ˆzρ

4ϵ0⎧

⎨

⎩

2,R2+-z+L

2.2

−2,R2+-z−L

2.2

−2L⎫

⎬

⎭

=ρ

2ϵ0⎡

⎣L−,R2+-z+L

2.2

+,R2+-z−L

2.2⎤

⎦ˆz.

Problem 2.28

Orient axes so Pis on zaxis.

❂

✲y

✻

zθ

V=1

4πϵ0%ρ

ηdτ.=Here ρis constant, dτ=r2sin θdr dθdφ,

η=√z2+r2−2rz cos θ.

V=ρ

4πϵ0%r2sin θdr dθdφ

√z2+r2−2rz cos θ;%2π

0dφ= 2π.

%π

sin θ

√z2+r2−2rz cos θdθ=1

rz &√r2+z2−2rz cos θ'**π

0=1

rz &√r2+z2+ 2rz −√r2+z2−2rz'

rz (r+z−|r−z|) = =2/z , if r < z,

2/r , if r > z. >

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Cut the cylinder into slabs, as shown in the ﬁgure, and

use result of Prob. 2.25c, with z!xand !⇢dx:

V=⇢

2✏0

z+L/2



zL/2pR2+x2xdx

=⇢

2✏0

2xpR2+x2+R2ln(x+pR2+x2)x2z+L/2

zL/2

=⇢

4✏08

(z+L

2)qR2+(z+L

2)2(zL

2)qR2+(zL

2)2+R2ln2

z+L

2+rR2+(z+L

2)2

zL

2+rR2+(zL

2)23

52zL9

;

(Note: z+L

22+zL

22=z2zL L2

4+z2zL +L2

4=2zL.)

E=rV=ˆz @V

@z=ˆz ⇢

4✏0R2+z+L

22

+z+L

22

R2+z+L

22R2+zL

22

zL

22

R2+zL

22

+R21+ z+L

qR2+(z+L

2)2

z+L

2+R2+z+L

22

1+ zL

qR2+(zL

2)2

zL

2+R2+zL

22

  

R2+z+L

221

R2+zL

22

2L

E=ˆz ⇢

4✏0



2R2+z+L

22

2R2+zL

22

2L





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CHAPTER 2. ELECTROSTATICS 35

=⇢

2✏0

LR2+z+L

22

+R2+zL

22

ˆz .

Problem 2.28

Orient axes so Pis on zaxis.

Hello

❂

✲y

✻

zθ

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V=1

4⇡✏0⇢

d⌧.Here ⇢is constant, d⌧=r2sin ✓dr d✓d,

=pz2+r22rz cos ✓.

V=⇢

4⇡✏0r2sin ✓dr d✓d

pz2+r22rz cos ✓;2⇡

0d=2⇡.

⇡

sin ✓

pz2+r22rz cos ✓d✓=1

rz pr2+z22rz cos ✓⇡

0=1

rz pr2+z2+2rz pr2+z22rz

rz (r+z|rz|)=2/z , if r<z,

2/r , if r>z.

)V=⇢

4⇡✏0·2⇡·2z

0

zr2dr +

z

rr2dr=⇢

✏01

3+R2z2

2=⇢

2✏0R2z2

3.

But ⇢=q

3⇡R3,soV(z)= 1

2✏0

4⇡R3R2z2

3=q

8⇡✏0R3z2

R2;V(r)= q

8⇡✏0R3r2

R2.X

Problem 2.29

r2V=1

4⇡✏0r2⇢

d⌧=1

4⇡✏0⇢(r0)r21

d⌧(since ⇢is a function of r0, not r)

4⇡✏0⇢(r0)[4⇡3(rr0)] d⌧=1

✏0⇢(r).X

Problem 2.30.

(a) Ex. 2.5: Eabove =

2✏0ˆn ;Ebelow =

2✏0ˆn (ˆn always pointing up); Eabove Ebelow =

✏0ˆn .X

Ex. 2.6: At each surface, E= 0 one side and E=

✏0other side, so E=

✏0.X

Prob. 2.11: Eout =R2

✏0r2ˆr =

✏0ˆr ;Ein =0;soE=

✏0ˆr .X

(b) Outside:E·da=E(2⇡s)l=1

✏0Qenc =

✏0(2⇡R)l)E=

✏0

sˆs =

✏0ˆs (at surface).

Inside:Qenc = 0, so E= 0. )E=

✏0ˆs .X

Contents

Problem 2.30.

(a) Ex. 2.4: Eabove =σ

2ϵ0

ˆn;Ebelow =−σ

2ϵ0

ˆn (ˆn always pointing up); Eabove −Ebelow =σ

ϵ0

ˆn.!

Ex. 2.5: At each surface, E= 0 one side and E=σ

ϵ0other side, so ∆E=σ

ϵ0.!

Prob. 2.11: Eout =σR2

ϵ0r2ˆr =σ

ϵ0

ˆr ;Ein = 0 ; so ∆E=σ

ϵ0

ˆr.!

(b) Outside:!E·da=E(2πs)l=1

ϵ0Qenc =1

ϵ0(2πR)l⇒E=σ

ϵ0

sˆs =σ

ϵ0

ˆs (at surface).

Inside:Qenc = 0, so E= 0. ∴∆E=σ

ϵ0

ˆs.!

✻

s✻

"#$%

ϵ0r=Rσ

ϵ0(at surface); Vin =Rσ

ϵ0; so Vout =Vin.!

∂Vout

∂r=−R2σ

ϵ0r2=−σ

ϵ0(at surface); ∂Vin

∂r= 0 ; so ∂Vout

∂r−∂Vin

∂r=−σ

ϵ0.!

Problem 2.31

(a) V=1

4πϵ0&qi

rij =1

4πϵ0'−q

a+q

√2a+−q

a(=q

4πϵ0a)−2 + 1

√2*.

∴W4=qV =q2

4πϵ0a+−2 + 1

√2,.

(2)

(3)

−

(4)

(1)

−

(b) W1= 0, W2=1

4πϵ0)−q2

a*;W3=1

4πϵ0)q2

√2a−q2

a*;W4= (see (a)).

Wtot =1

4πϵ0

a'−1 + 1

√2−1−2 + 1

√2(=1

4πϵ0

2q2

a+−2 + 1

√2,.

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✏0r=R

✏0(at surface); Vin =R

✏0; so Vout =Vin.X

@Vout

@r=R2

✏0r2=

✏0(at surface); @Vin

@r= 0 ; so @Vout

@r@Vin

@r=

✏0.X

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36 CHAPTER 2. ELECTROSTATICS

Problem 2.31

(a) V=1

4⇡✏0qi

rij =1

4⇡✏0q

a+q

p2a+q

a=q

4⇡✏0a2+ 1

p2.

)W4=qV =q2

4⇡✏0a2+ 1

p2.

(2)

+r(3)



r(4)

(1)



(b) W1= 0, W2=1

4⇡✏0q2

a;W3=1

4⇡✏0q2

p2aq2

a;W4= (see (a)).

Wtot =1

4⇡✏0

a1+ 1

p212+ 1

p2=1

4⇡✏0

2q2

a2+ 1

p2.

Problem 2.32

Conservation of energy (kinetic plus potential):

2mAv2

A+1

2mBv2

B+1

4⇡✏0

qAqB

r=E.

At release vA=vB=0,r=a,so

E=1

4⇡✏0

qAqB

When they are very far apart (r!1) the potential energy is zero, so

2mAv2

A+1

2mBv2

B=1

4⇡✏0

qAqB

Meanwhile, conservation of momentum says mAvA=mBvB, or vB=(mA/mB)vA. So

2mAv2

A+1

2mBmA

mB2

A=1

2mA

mB(mA+mB)v2

A=1

4⇡✏0

qAqB

vA=1

2⇡✏0

qAqB

(mA+mB)amA

mB;vB=1

2⇡✏0

qAqB

(mA+mB)amB

mA.

Problem 2.33

From Eq. 2.42, the energy of one charge is

W=1

2qV =1

2(2) 1



n=1

4⇡✏0

(1)nq2

na =q2

4⇡✏0a



(1)n

(The factor of 2 out front counts the charges to the left as well as to the right of q.) The sum is ln 2 (you

can get it from the Taylor expansion of ln(1 + x):

ln(1 + x)=x1

2x2+1

3x31

4x4+···

with x= 1. Evidently ↵=ln2.

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CHAPTER 2. ELECTROSTATICS 37

Problem 2.34

(a) W=1

2⇢Vd⌧. From Prob. 2.21 (or Prob. 2.28): V=⇢

2✏0R2r2

3=1

4⇡✏0

2R3r2

R2

W=1

2⇢1

4⇡✏0

2RR

03r2

R24⇡r2dr =q⇢

4✏0R3r3

31

5

=q⇢

4✏0RR3R3

5

=q⇢

5✏0

R2=qR2

5✏0

3⇡R3=1

4⇡✏03

R.

(b) W=✏0

2E2d⌧.Outside (r>R)E=1

4⇡✏0

r2ˆr ;Inside (r<R)E=1

4⇡✏0

R3rˆr .

)W=✏0

(4⇡✏0)2q21

r4(r24⇡dr)+R

0r

R32(4⇡r2dr)

4⇡✏0

21

r

R6r5

5

0=1

4⇡✏0

21

R+1

5R=1

4⇡✏0

R.X

2SVE·da+VE2d⌧, where Vis large enough to enclose all the charge, but otherwise

arbitrary. Let’s use a sphere of radius a>R. Here V=1

4⇡✏0

W=✏0

2

r=a1

4⇡✏0

r 1

4⇡✏0

r2r2sin ✓d✓d+R

E2d⌧+a

R1

4⇡✏0

r22

(4⇡r2dr)

=✏0

2q2

(4⇡✏0)2

a4⇡+q2

(4⇡✏0)2

4⇡

5R+1

(4⇡✏0)24⇡q21

r

R

4⇡✏0

21

a+1

5R1

a+1

R=1

4⇡✏0

R.X

As a!1, the contribution from the surface integral 1

4⇡✏0

2agoes to zero, while the volume integral

1

4⇡✏0

2a(6a

5R1)picks up the slack.

Problem 2.35

Problem 2.32

(a) W=1

2!ρV dτ. From Prob. 2.21 (or Prob. 2.28): V=ρ

2ϵ0"R2−r2

3#=1

4πϵ0

2R"3−r2

R2#

W=1

2ρ1

4πϵ0

2R$R

0%3−

R2&4πr2dr =qρ

4ϵ0R'3r3

3−1

5())))

=qρ

4ϵ0R%R3−

=qρ

5ϵ0

R2=qR2

5ϵ0

3πR3=1

4πϵ0%3

R&.

(b) W=ϵ0

2!E2dτ.Outside (r > R)E=1

4πϵ0

r2ˆr ;Inside (r < R)E=1

4πϵ0

R3rˆr.

∴W=ϵ0

(4πϵ0)2q2*$∞

r4(r24πdr) + $R

0"r

R3#2

(4πr2dr)+

4πϵ0

2*%−1

r&))))

∞

R6%r5

5&))))

0+=1

4πϵ0

2%1

R+1

5R&=1

4πϵ0

R."

2,-SVE·da+!VE2dτ., where Vis large enough to enclose all the charge, but otherwise

arbitrary. Let’s use a sphere of radius a > R. Here V=1

4πϵ0

W=ϵ0

2*$

r=a%1

4πϵ0

r&% 1

4πϵ0

r2&r2sin θdθdφ+$R

E2dτ+$a

R%1

4πϵ0

r2&2

(4πr2dr)+

=ϵ0

2/q2

(4πϵ0)2

a4π+q2

(4πϵ0)2

4π

5R+1

(4πϵ0)24πq2%−1

r&))))

4πϵ0

2/1

a+1

5R−1

a+1

R0=1

4πϵ0

R."

As a→ ∞, the contribution from the surface integral "1

4πϵ0

2a#goes to zero, while the volume integral

4πϵ0

2a(6a

5R−1)#picks up the slack.

Problem 2.33

¯q

✯

✮

d¯q

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dW =d¯qV =d¯q1

4⇡✏0¯q

r,(¯q= charge on sphere of radius r).

¯q=4

3⇡r3⇢=qr3

R3(q= total charge on sphere).

d¯q=4⇡r2dr ⇢=4⇡r2

3⇡R3q dr =3q

R3r2dr.

dW =1

4⇡✏0qr3

R31

r3q

R3r2dr=1

4⇡✏0

3q2

R6r4dr

W=1

4⇡✏0

3q2

R6R

r4dr =1

4⇡✏0

3q2

5=1

4⇡✏03

R.X

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38 CHAPTER 2. ELECTROSTATICS

Problem 2.36

(a) W=✏0

2E2d⌧.E=1

4⇡✏0

r2(a<r<b),zero elsewhere.

W=✏0

2q

4⇡✏02b

a1

r224⇡r2dr =q2

8⇡✏0b

r2=q2

8⇡✏01

a1

b.

(b) W1=1

8⇡✏0

a,W

2=1

8⇡✏0

b,E1=1

4⇡✏0

r2ˆr (r>a),E2=1

4⇡✏0q

r2ˆr (r>b).So

E1·E2=1

4⇡✏02q2

r4,(r>b),and hence E1·E2d⌧=1

4⇡✏02q21

r44⇡r2dr =q2

4⇡✏0b.

Wtot =W1+W2+✏0E1·E2d⌧=1

8⇡✏0q21

a+1

b2

b=q2

8⇡✏01

a1

b.X

Problem 2.37

E1=1

4⇡✏0

r2ˆr ;E2=1

4⇡✏0

2ˆ

;Wi=✏0

q1q2

(4⇡✏0)21

2cos r2sin ✓dr d✓d,

where (from the ﬁgure)

=r2+a22ra cos ✓,cos =(racos ✓)

Therefore

Wi=q1q2

(4⇡)2✏0

2⇡(racos ✓)

3sin ✓dr d✓.

It’s simplest to do the rintegral ﬁrst, changing variables to

=(2r2acos ✓)dr )(racos ✓)dr =

As r:0!1,

:a!1,so

Wi=q1q2

8⇡✏0⇡

01

sin ✓d✓.

The

integral is 1/a,so

Wi=q1q2

8⇡✏0a⇡

sin ✓d✓=q1q2

4⇡✏0a.

Of course, this is precisely the interaction energy of two point charges.

Problem 2.38

(a) R=q

4⇡R2;a=q

4⇡a2;b=q

4⇡b2.

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CHAPTER 2. ELECTROSTATICS 39

(b) V(0) = 0

1E·dl=b

11

4⇡✏0

r2dr a

b(0)dr R

a1

4⇡✏0

r2dr 0

R(0)dr =1

4⇡✏0q

b+q

Rq

a.

1(0)dr R

a1

4⇡✏0

r2dr 0

R(0)dr =1

4⇡✏0q

Rq

a.

Problem 2.39

(a) a=qa

4⇡a2;b=qb

4⇡b2;R=qa+qb

4⇡R2.

(b) Eout =1

4⇡✏0

qa+qb

r2ˆr ,where r= vector from center of large sphere.

4⇡✏0

ˆr a,Eb=1

4⇡✏0

ˆr b,where ra(rb) is the vector from center of cavity a(b).

(d) Zero.

(e) Rchanges (but not aor b); Eoutside changes (but not Eaor Eb); force on qaand qbstill zero.

Problem 2.40

(a) No. For example, if it is very close to the wall, it will induce charge of the opposite sign on the wall,

and it will be attracted.

(b) No. Typically it will be attractive, but see footnote 12 for an extraordinary counterexample.

Problem 2.41

Between the plates, E= 0; outside the plates E=/✏0=Q/✏0A. So

P=✏0

2E2=✏0

✏2

0A2=Q2

2✏0A2.

Problem 2.42

Inside, E=0; outside, E=1

4⇡✏0

r2ˆr ; so

Problem 2.34

(a) W=ϵ0

2!E2dτ.E=1

4πϵ0

r2(a < r < b),zero elsewhere.

W=ϵ0

2"q

4πϵ0#2!b

a$1

r2%24πr2dr =q2

8πϵ0!b

r2=q2

8πϵ0&1

−1

b'.

(b) W1=1

8πϵ0

a, W2=1

8πϵ0

b,E1=1

4πϵ0

r2ˆr (r > a),E2=1

4πϵ0

−q

r2ˆr (r > b).So

E1·E2="1

4πϵ0#2−q2

r4,(r > b),and hence !E1·E2dτ=−"1

4πϵ0#2

q2!∞

r44πr2dr =−q2

4πϵ0b.

Wtot =W1+W2+ϵ0!E1·E2dτ=1

8πϵ0q2$1

a+1

b−2

b%=q2

8πϵ0$1

a−1

b%.!

Problem 2.35

(a) σR=q

4πR2;σa=−q

4πa2;σb=q

4πb2.

(b) V(0) = −!0

∞E·dl=−!b

∞$1

4πϵ0

r2%dr −!a

b(0)dr −!R

a$1

4πϵ0

r2%dr −!0

R(0)dr =1

4πϵ0"q

b+q

−q

a#.

∞(0)dr −!R

a$1

4πϵ0

r2%dr −!0

R(0)dr =1

4πϵ0"q

−q

a#.

Problem 2.36

(a) σa=−qa

4πa2;σb=−qb

4πb2;σR=qa+qb

4πR2.

(b) Eout =1

4πϵ0

qa+qb

r2ˆr,where r= vector from center of large sphere.

4πϵ0

ˆra,Eb=1

4πϵ0

ˆrb,where ra(rb) is the vector from center of cavity a(b).

(d) Zero.

(e) σRchanges (but not σaor σb); Eoutside changes (but not Eaor Eb); force on qaand qbstill zero.

Problem 2.37

Between the plates, E= 0; outside the plates E=σ/ϵ0=Q/ϵ0A. So

P=ϵ0

2E2=ϵ0

ϵ2

0A2=Q2

2ϵ0A2.

Problem 2.38

Inside, E= 0; outside, E=1

4πϵ0

r2ˆr; so

✻

✒

Eave =1

4πϵ0

R2ˆr;fz=σ(Eave)z;σ=Q

4πR2.

Fz=!fzda =!$ Q

4πR2%1

2$1

4πϵ0

R2%cos θR2sin θdθdφ

2ϵ0$Q

4πR%22π!π/2

0sin θcos θdθ=1

πϵ0$Q

4R%2$1

2sin2θ%(

(

π/2

0=1

2πϵ0$Q

4R%2=Q2

32πR2ϵ0

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Eave =1

4⇡✏0

R2ˆr ;fz=(Eave)z;=Q

4⇡R2.

Fz=fzda =Q

4⇡R21

21

4⇡✏0

R2cos ✓R2sin ✓d✓d

2✏0Q

4⇡R22⇡⇡/2

0sin ✓cos ✓d✓=1

⇡✏0Q

4R21

2sin2✓⇡/2

0=1

2⇡✏0Q

4R2=Q2

32⇡R2✏0

Problem 2.43

Say the charge on the inner cylinder is Q, for a length L. The ﬁeld is given by Gauss’s law:

E·da=E·2⇡s·L=1

✏0Qenc =1

✏0Q)E=Q

2⇡✏0L

sˆs . Potential di↵erence between the cylinders is

V(b)V(a)=b

E·dl=Q

2⇡✏0Lb

sds =Q

2⇡✏0Lln b

a.

As set up here, ais at the higher potential, so V=V(a)V(b)= Q

2⇡✏0Lln b

a.

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40 CHAPTER 2. ELECTROSTATICS

C=Q

V=2⇡✏0L

ln(b

a), so capacitance per unit length is 2⇡✏0

ln b

a.

Problem 2.44

(a) W= (force)⇥(distance) = (pressure)⇥(area)⇥(distance) = ✏0

2E2A✏.

(b) W= (energy per unit volume)⇥(decrease in volume) = ✏0E2

2(A✏). Same as (a), conﬁrming that the

energy lost is equal to the work done.

Problem 2.45

Problem 2.39

Say the charge on the inner cylinder is Q, for a length L. The ﬁeld is given by Gauss’s law:

!E·da=E·2πs·L=1

ϵ0Qenc =1

ϵ0Q⇒E=Q

2πϵ0L

sˆs. Potential diﬀerence between the cylinders is

V(b)−V(a) = −"b

E·dl=−Q

2πϵ0L"b

sds =−Q

2πϵ0Lln #b

a$.

As set up here, ais at the higher potential, so V=V(a)−V(b) = Q

2πϵ0Lln %b

a&.

C=Q

V=2πϵ0L

ln(b

a), so capacitance per unit length is 2πϵ0

ln %b

a&.

Problem 2.40

(a) W= (force)×(distance) = (pressure)×(area)×(distance) = ϵ0

2E2Aϵ.

(b) W= (energy per unit volume)×(decrease in volume) = 'ϵ0E2

2((Aϵ). Same as (a), conﬁrming that the

energy lost is equal to the work done.

Problem 2.41

✲ ✛ da

✛ ✲

a+da

✲ ✛ da

From Prob. 2.4, the ﬁeld at height zabove the center of a square loop (side a) is

E=1

4πϵ0

4λaz

%z2+a2

4&)z2+a2

ˆz.

Here λ→σda

2(see ﬁgure), and we integrate over afrom 0 to ¯a:

E=1

4πϵ0

2σz"¯a

a da

%z2+a2

4&)z2+a2

. Let u=a2

4, so a da = 2 du.

4πϵ0

4σz"¯a2/4

(u+z2)√2u+z2=σz

πϵ0*2

ztan−1+√2u+z2

z,-¯a2/4

=2σ

πϵ0.tan−1+)¯a2

2+z2

z,−tan−1(1)/;

E=2σ

πϵ0*tan−101 + a2

2z2−

4-ˆz.

a→ ∞ (inﬁnite plane): E=2σ

πϵ01tan−1(∞)−π

42=2σ

πϵ0%π

2−π

4&=σ

2ϵ0.!

z≫a(point charge): Let f(x) = tan−1√1 + x−π

4, and expand as a Taylor series:

f(x) = f(0) + xf ′(0) + 1

2x2f′′(0) + · · ·

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From Prob. 2.4, the ﬁeld at height zabove the center of a square loop (side a) is

E=1

4⇡✏0

4az

z2+a2

4z2+a2

ˆz .

Here !da

2(see ﬁgure), and we integrate over afrom 0 to ¯a:

E=1

4⇡✏0

2z¯a

ada

z2+a2

4z2+a2

. Let u=a2

4, so ada=2du.

4⇡✏0

4z¯a2/4

(u+z2)p2u+z2=z

⇡✏02

ztan1p2u+z2

z¯a2/4

=2

⇡✏0tan1¯a2

2+z2

ztan1(1);

E=2

⇡✏0tan11+ a2

2z2⇡

4ˆz =

⇡✏0

tan1a2

4zz2+(a2/2) ˆz .

a!1(inﬁnite plane): E=2

⇡✏0tan1(1)⇡

4=2

⇡✏0⇡

2⇡

4=

2✏0.X

za(point charge): Let f(x) = tan1p1+x⇡

4, and expand as a Taylor series:

f(x)=f(0) + xf0(0) + 1

2x2f00(0) + ···

Here f(0) = tan1(1) ⇡

4=⇡

4⇡

4= 0; f0(x)= 1

1+(1+x)

p1+x=1

2(2+x)p1+x,sof0(0) = 1

4,so

f(x)=1

4x+()x2+()x3+···

Thus (since a2

2z2=x⌧1), E⇡2

⇡✏01

2z2=1

4⇡✏0

a2

z2=1

4⇡✏0

z2.X

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CHAPTER 2. ELECTROSTATICS 41

Problem 2.46

⇢=✏0r·E=✏01

@rr23k

r+1

rsin ✓

@✓ sin ✓2ksin ✓cos ✓sin 

r+1

rsin ✓

@ ksin ✓cos 

r

=✏01

r23k+1

rsin ✓

2ksin (2 sin ✓cos2✓sin3✓)

r+1

rsin ✓

(ksin ✓sin )

r

=k✏0

r23 + 2 sin (2 cos2✓sin2✓)sin =k✏0

r23+sin(4 cos2✓2 + 2 cos2✓1)

=3k✏0

r21+sin(2 cos2✓1)=3k✏0

r2(1 + sin cos 2✓).

Problem 2.47

From Prob. 2.12, the ﬁeld inside a uniformly charged sphere is: E=1

4⇡✏0

R3r. So the force per unit volume

is f=⇢E=Q

3⇡R3 Q

4⇡✏0R3r=3

✏0Q

4⇡R32r, and the force in the zdirection on d⌧is:

dFz=fzd⌧=3

✏0Q

4⇡R32

rcos ✓(r2sin ✓dr d✓d).

The total force on the “northern” hemisphere is:

Fz=fzd⌧=3

✏0Q

4⇡R32R

r3dr ⇡/2

cos ✓sin ✓d✓2⇡

d

✏0Q

4⇡R32R4

4sin2✓

2

⇡/2

0(2⇡)= 3Q2

64⇡✏0R2.

Problem 2.48

Here f(0) = tan−1(1) −π

4=π

4−π

4= 0; f′(x) = 1

1+(1+x)

√1+x=1

2(2+x)√1+x, so f′(0) = 1

4, so

f(x) = 1

4x+ ( )x2+ ( )x3+···

Thus (since a2

2z2=x≪1), E≈2σ

πϵ0!1

2z2"=1

4πϵ0

σa2

z2=1

4πϵ0

z2.!

Problem 2.42

ρ=ϵ0∇·E=ϵ0#1

∂

∂r$r2A

r%+1

rsin θ

∂

∂φ $Bsin θcos φ

r%&

=ϵ0'1

r2A+1

rsin θ

Bsin θ

r(−sin φ)(=ϵ0

r2(A−Bsin φ).

Problem 2.43

From Prob. 2.12, the ﬁeld inside a uniformly charged sphere is: E=1

4πϵ0

R3r. So the force per unit volume

is f=ρE=)Q

3πR3*) Q

4πϵ0R3*r=3

ϵ0)Q

4πR3*2r, and the force in the zdirection on dτis:

dFz=fzdτ=3

ϵ0$Q

4πR3%2

rcos θ(r2sin θdr dθdφ).

The total force on the “northern” hemisphere is:

Fz=+fzdτ=3

ϵ0$Q

4πR3%2+R

r3dr +π/2

cos θsin θdθ+2π

dφ

ϵ0$Q

4πR3%2$R4

4%,sin2θ

2----

π/2

0.(2π) = 3Q2

64πϵ0R2.

Problem 2.44

Rθ

Vcenter =1

4πϵ0+σ

ηda =1

4πϵ0

R+da =1

4πϵ0

R(2πR2) = σR

2ϵ0

Vpole =1

4πϵ0+σ

ηda , with /da = 2πR2sin θdθ,

η2=R2+R2−2R2cos θ= 2R2(1 −cos θ).

4πϵ0

σ(2πR2)

R√2+π/2

sin θdθ

√1−cos θ=σR

2√2ϵ0

(2√1−cos θ)---

π/2

=σR

√2ϵ0

(1 −0) = σR

√2ϵ0

.∴Vpole −Vcenter =σR

2ϵ0

(√2−1).

Problem 2.45

First let’s determine the electric ﬁeld inside and outside the sphere, using Gauss’s law:

ϵ00E·da=ϵ04πr2E=Qenc =+ρdτ=+(k¯r)¯r2sin θd¯r dθdφ= 4πk+r

¯r3d¯r=/πkr4(r < R),

πkR4(r > R).

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Vcenter =1

4⇡✏0

da =1

4⇡✏0



Rda =1

4⇡✏0



R(2⇡R2)=R

2✏0

Vpole =1

4⇡✏0

da , with da =2⇡R2sin ✓d✓,

2=R2+R22R2cos ✓=2R2(1 cos ✓).

4⇡✏0

(2⇡R2)

Rp2⇡/2

sin ✓d✓

p1cos ✓=R

2p2✏0

(2p1cos ✓)

⇡/2

=R

p2✏0

(1 0) = R

p2✏0

.)Vpole Vcenter =R

2✏0

(p21).

Problem 2.49

First let’s determine the electric ﬁeld inside and outside the sphere, using Gauss’s law:

✏0E·da=✏04⇡r2E=Qenc =⇢d⌧=(k¯r)¯r2sin ✓d¯rd✓d=4⇡kr

¯r3d¯r=⇡kr4(r<R),

⇡kR4(r>R).

So E=k

4✏0r2ˆr (r<R); E=kR4

4✏0r2ˆr (r>R).

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42 CHAPTER 2. ELECTROSTATICS

Method I :

W=✏0

2E2d⌧(Eq.2.45) = ✏0

2R

0kr2

4✏02

4⇡r2dr +✏0

21

RkR4

4✏0r22

4⇡r2dr

=4⇡✏0

2k

4✏02R

r6dr +R81

r2dr=⇡k2

8✏0R7

7+R81

r

R=⇡k2

8✏0R7

7+R7

=⇡k2R7

7✏0

Method II :

W=1

2⇢Vd⌧(Eq.2.43).

For r<R,V(r)=r

E·dl=R

1kR4

4✏0r2dr r

Rkr2

4✏0dr =k

4✏0R41

r

+r3

3

R

=k

4✏0R3+r3

3R3

3=k

3✏0R3r3

4.

)W=1

2R

(kr)k

3✏0R3r3

44⇡r2dr =2⇡k2

3✏0R

0R3r31

4r6dr

=2⇡k2

3✏0R3R4

41

7=⇡k2R7

2·3✏06

7=⇡k2R7

7✏0

Problem 2.50

E=rV=A@

@rer

rˆr =Ar()erer

r2ˆr =Aer(1 + r)ˆr

r2.

⇢=✏0r·E=✏0Aer(1 + r)r·ˆr

r2+ˆr

r2·rer(1 + r).Butr·ˆr

r2=4⇡3(r) (Eq. 1.99), and

er(1 + r)3(r)=3(r) (Eq. 1.88). Meanwhile,

rer(1 + r)=ˆr @

@rer(1 + r)=ˆr er(1 + r)+er=ˆr (2rer).

So ˆr

r2·rer(1 + r)=2

rer, and ⇢=✏0A4⇡3(r)2

rer.

Q=⇢d⌧=✏0A4⇡3(r)d⌧2er

r4⇡r2dr=✏0A4⇡24⇡1

rerdr.

But 1

0rerdr =1

2, so Q=4⇡✏0A12

2= zero.

Problem 2.51

V=1

4⇡✏0

da =

4⇡✏0R

02⇡

R2+s22Rs cos sdsd.

Let u⌘s/R. Then

V=2R

4⇡✏01

0⇡

1+u22ucos ddu.

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CHAPTER 2. ELECTROSTATICS 43

The (double) integral is a pure number; Mathematica says it is 2. So

V=R

⇡✏0

Problem 2.52

(a) Potential of +is V+=

2⇡✏0ln s+

a, where s+is distance from +(Prob. 2.22).

Potential of is V=+ 

2⇡✏0ln s

a, where sis distance from .

)Total V=

2⇡✏0

ln s

s+.

Now s+=(ya)2+z2, and s=(y+a)2+z2,so

V(x, y, z)= 

2⇡✏0ln p(y+a)2+z2

p(ya)2+z2=

4⇡✏0

ln (y+a)2+z2

(ya)2+z2.

∴Total V=λ

2πϵ0

ln !s−

s+".

Now s+=#(y−a)2+z2, and s−=#(y+a)2+z2, so

V(x, y, z) = λ

2πϵ0ln !√(y+a)2+z2

√(y−a)2+z2"=λ

4πϵ0

ln $(y+a)2+z2

(y−a)2+z2%.

✲y

✻

−λ λ

(x, y, z)

a a

s−

(b) Equipotentials are given by (y+a)2+z2

(y−a)2+z2=e(4πϵ0V0/λ)=k= constant. That is:

y2+ 2ay +a2+z2=k(y2−2ay +a2+z2)⇒y2(k−1) + z2(k−1) + a2(k−1) −2ay(k+ 1) = 0, or

y2+z2+a2−2ay &k+1

k−1'= 0. The equation for a circle, with center at (y0,0) and radius R, is

(y−y0)2+z2=R2, or y2+z2+ (y2

0−R2)−2yy0= 0.

Evidently the equipotentials are circles, with y0=a&k+1

k−1'and

a2=y2

0−R2⇒R2=y2

0−a2=a2&k+1

k−1'2

−a2=a2(k2+2k+1−k2+2k−1)

(k−1)2=a24k

(k−1)2, or

R=2a√k

|k−1|; or, in terms of V0:

y0=ae4πϵ0V0/λ+ 1

e4πϵ0V0/λ−1=ae2πϵ0V0/λ+e−2πϵ0V0/λ

e2πϵ0V0/λ−e−2πϵ0V0/λ=acoth !2πϵ0V0

λ".

R= 2ae2πϵ0V0/λ

e4πϵ0V0/λ−1=a2

(e2πϵ0V0/λ−e−2πϵ0V0/λ)=a

sinh (2πϵ0V0

λ)=acsch !2πϵ0V0

λ".

✲y

✻

λ−λy0

✒

Problem 2.48

(a) ∇2V=−ρ

ϵ0(Eq. 2.24), so d2V

dx2=−1

ϵ0

ρ.

(b) qV =1

2mv2→v=*2qV

dt =aρdx

dt =Aρv=I(constant). (Note:ρ, hence also I, is negative.)

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(b) Equipotentials are given by (y+a)2+z2

(ya)2+z2=e(4⇡✏0V0/)=k= constant. That is:

y2+2ay +a2+z2=k(y22ay +a2+z2))y2(k1) + z2(k1) + a2(k1) 2ay(k+ 1) = 0, or

y2+z2+a22ay k+1

k1= 0. The equation for a circle, with center at (y0,0) and radius R, is

(yy0)2+z2=R2, or y2+z2+(y2

0R2)2yy0= 0.

Evidently the equipotentials are circles, with y0=ak+1

k1and

a2=y2

0R2)R2=y2

0a2=a2k+1

k12a2=a2(k2+2k+1k2+2k1)

(k1)2=a24k

(k1)2, or

R=2apk

|k1|; or, in terms of V0:

y0=ae4⇡✏0V0/+1

e4⇡✏0V0/1=ae2⇡✏0V0/+e2⇡✏0V0/

e2⇡✏0V0/e2⇡✏0V0/=acoth 2⇡✏0V0

.

R=2ae2⇡✏0V0/

e4⇡✏0V0/1=a2

(e2⇡✏0V0/e2⇡✏0V0/)=a

sinh 2⇡✏0V0

=acsch 2⇡✏0V0

.

∴Total V=λ

2πϵ0

ln !s−

s+".

Now s+=#(y−a)2+z2, and s−=#(y+a)2+z2, so

V(x, y, z) = λ

2πϵ0ln !√(y+a)2+z2

√(y−a)2+z2"=λ

4πϵ0

ln $(y+a)2+z2

(y−a)2+z2%.

✲y

✻

−λ λ

(x, y, z)

a a

s−

(b) Equipotentials are given by (y+a)2+z2

(y−a)2+z2=e(4πϵ0V0/λ)=k= constant. That is:

y2+ 2ay +a2+z2=k(y2−2ay +a2+z2)⇒y2(k−1) + z2(k−1) + a2(k−1) −2ay(k+ 1) = 0, or

y2+z2+a2−2ay &k+1

k−1'= 0. The equation for a circle, with center at (y0,0) and radius R, is

(y−y0)2+z2=R2, or y2+z2+ (y2

0−R2)−2yy0= 0.

Evidently the equipotentials are circles, with y0=a&k+1

k−1'and

a2=y2

0−R2⇒R2=y2

0−a2=a2&k+1

k−1'2

−a2=a2(k2+2k+1−k2+2k−1)

(k−1)2=a24k

(k−1)2, or

R=2a√k

|k−1|; or, in terms of V0:

y0=ae4πϵ0V0/λ+ 1

e4πϵ0V0/λ−1=ae2πϵ0V0/λ+e−2πϵ0V0/λ

e2πϵ0V0/λ−e−2πϵ0V0/λ=acoth !2πϵ0V0

λ".

R= 2ae2πϵ0V0/λ

e4πϵ0V0/λ−1=a2

(e2πϵ0V0/λ−e−2πϵ0V0/λ)=a

sinh (2πϵ0V0

λ)=acsch !2πϵ0V0

λ".

✲y

✻

λ−λy0

✒

Problem 2.48

(a) ∇2V=−ρ

ϵ0(Eq. 2.24), so d2V

dx2=−1

ϵ0

ρ.

(b) qV =1

2mv2→v=*2qV

dt =aρdx

dt =Aρv=I(constant). (Note:ρ, hence also I, is negative.)

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44 CHAPTER 2. ELECTROSTATICS

Problem 2.53

(a) r2V=⇢

✏0(Eq. 2.24), so d2V

dx2=1

✏0

⇢.

(b) qV =1

2mv2!v=2qV

dt =a⇢dx

dt =A⇢v=I(constant). (Note:⇢, hence also I, is negative.)

(d) d2V

dx2=1

✏0⇢=1

✏0

Av =I

✏0Am

2qV )d2V

dx2=V1/2, where =I

✏0Am

2q.

(Note: Iis negative,sois positive; qis positive.)

(e) Multiply by V0=dV

dx :

V0dV 0

dx =V1/2dV

dx )V0dV 0=V1/2dV )1

2V02=2V1/2+ constant.

But V(0) = V0(0) = 0 (cathode is at potential zero, and ﬁeld at cathode is zero), so the constant is zero, and

V02=4V1/2)dV

dx =2

V1/4)V1/4dV =2

dx;

V1/4dV =2

dx )4

3V3/4=2

x+ constant.

But V(0) = 0, so this constant is also zero.

V3/4=3

2x, so V(x)=3

24/3

x4/3,or V(x)=9

42/3

x4/3=81I2m

32✏2

0A2q1/3

x4/3.

Interms of V0(instead of I): V(x)=V0x

d4/3(see graph).

Without space-charge, Vwould increase linearly: V(x)=V0x

d.

⇢=✏0

d2V

dx2=✏0V0

d4/3

3·1

3x2/3=4✏0V0

9(d2x)2/3.

v=2q

mpV=2qV0/m x

d2/3.

(d) d2V

dx2=−1

ϵ0ρ=−1

ϵ0

Av =−I

ϵ0A!m

2qV ⇒d2V

dx2=βV−1/2, where β=−I

ϵ0A!m

2q.

(Note: Iis negative, so βis positive; qis positive.)

(e) Multiply by V′=dV

dx :

V′dV ′

dx =βV−1/2dV

dx ⇒"V′dV ′=β"V−1/2dV ⇒1

2V′2= 2βV1/2+ constant.

But V(0) = V′(0) = 0 (cathode is at potential zero, and ﬁeld at cathode is zero), so the constant is zero, and

V′2= 4βV1/2⇒dV

dx = 2#βV1/4⇒V−1/4dV = 2#βdx;

"V−1/4dV = 2#β"dx ⇒4

3V3/4= 2#βx+ constant.

But V(0) = 0, so this constant is also zero.

V3/4=3

2#βx, so V(x) = $3

2#β%4/3

x4/3,or V(x) = $9

4β%2/3

x4/3=$81I2m

32ϵ2

0A2q%1/3

x4/3.

Interms of V0(instead of I): V(x) = V0&x

d'4/3

(see graph).

Without space-charge, Vwould increase linearly: V(x) = V0(x

d).

ρ=−ϵ0

d2V

dx2=−ϵ0V0

d4/3

3x−2/3=−4ϵ0V0

9(d2x)2/3.

v=*2q

m√V=#2qV0/m &x

d'2/3

.✲x

✻

without

with

(f) V(d) = V0=&81I2m

32ϵ2

0A2q'1/3

d4/3⇒V3

0=81md4

32ϵ2

0A2qI2;I2=32ϵ2

0A2q

81md4V3

I=4√2ϵ0A√q

9√m d2V3/2

0=KV 3/2

0, where K=4ϵ0A

9d2*2q

Problem 2.49

(a) E=1

4πϵ0"ρˆηˆη

ˆη

η2&1 + η

λ'e−η/λdτ.

(b) Yes. The ﬁeld of a point charge at the origin is radial and symmetric, so ∇×E= 0, and hence this is also

true (by superposition) for any collection of charges.

∞

E·dl=−1

4πϵ0

q"r

∞

r2&1 + r

λ'e−r/λdr

4πϵ0

q"∞

r2&1 + r

λ'e−r/λdr =q

4πϵ0+"∞

r2e−r/λdr +1

λ"∞

re−r/λdr,.

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(f) V(d)=V0=81I2m

32✏2

0A2q1/3d4/3)V3

0=81md4

32✏2

0A2qI2;I2=32✏2

0A2q

81md4V3

I=4p2✏0Apq

9pmd

2V3/2

0=KV 3/2

0, where K=4✏0A

9d22q

Problem 2.54

(a) E=1

4⇡✏0⇢ˆ

21+

e

/d⌧.

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CHAPTER 2. ELECTROSTATICS 45

(b) Yes. The ﬁeld of a point charge at the origin is radial and symmetric, so r⇥E=0, and hence this is

also true (by superposition) for any collection of charges.

E·dl=1

4⇡✏0

qr

r21+ r

er/dr

4⇡✏0

q1

r21+ r

er/dr =q

4⇡✏01

r2er/dr +1

1

rer/dr.

Now 1

r2er/dr =er/

r1

er/

rdr  exactly right to kill the last term. Therefore

V(r)= q

4⇡✏0er/

r

r=q

4⇡✏0

er/

(d) S

E·da=1

4⇡✏0

q1

1+R

eR/4⇡R2

=q

✏01+R

eR/.

V

Vd⌧=q

4⇡✏0

R

er/

rr24⇡dr =q

✏0R

rer/dr =q

✏0er/

(1/)2r

1R

=2q

✏0eR/1+R

+1

.

)S

E·da+1

2V

Vd⌧=q

✏01+R

eR/1+R

eR/+1

=q

✏0

.qed

(e) Does the result in (d) hold for a nonspherical surface? Suppose we

make a “dent” in the sphere—pushing a patch (area R2sin ✓d✓d)

from radius Rout to radius S(area S2sin ✓d✓d).

Now !1

r2e−r/λdr =−e−r/λ

r−1

λ!e−r/λ

rdr ←− exactly right to kill the last term. Therefore

V(r) = q

4πϵ0"−e−r/λ

r####

∞

r$=q

4πϵ0

e−r/λ

(d) %S

E·da=1

4πϵ0

!q1

!&1 + R

λ'e−R/λ4π!R2

!=q

ϵ0&1 + R

λ'e−R/λ.

V dτ=q

4πϵ0

!(R

e−r/λ

rr24π!dr =q

ϵ0(R

re−r/λdr =q

ϵ0)e−r/λ

(1/λ)2*−r

λ−1+,R

=λ2q

ϵ0-−e−R/λ&1 + R

λ'+ 1..

∴%S

E·da+1

λ2(V

V dτ=q

ϵ0-&1 + R

λ'e−R/λ−&1 + R

λ'e−R/λ+ 1.=q

ϵ0

.qed

(e) Does the result in (d) hold for a nonspherical surface? Suppose we

make a “dent” in the sphere—pushing a patch (area R2sin θdθdφ)

from radius Rout to radius S(area S2sin θdθdφ). q

∆%E·da=q

4πϵ0-1

S2&1 + S

λ'e−S/λ(S2sin θdθdφ)−1

R2&1 + R

λ'e−R/λ(R2sin θdθdφ).

4πϵ0)&1 + S

λ'e−S/λ−&1 + R

λ'e−R/λ,sin θdθdφ.

∆1

λ2(V dτ=1

λ2

4πϵ0(e−r/λ

rr2sin θdr dθdφ=1

λ2

4πϵ0

sin θdθdφ(S

re−r/λdr

=−q

4πϵ0

sin θdθdφ*e−r/λ*1 + r

λ++###

=−q

4πϵ0)&1 + S

λ'e−S/λ−&1 + R

λ'e−R/λ,sin θdθdφ.

So the change in 1

λ2!V dτexactly compensates for the change in /E·da, and we get 1

ϵ0qfor the total using

the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive

distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside,

the total is 1

ϵ0Qenc. Charges outside do not contribute (in the argument above we found that for this

volume /E·da+1

λ2!V dτ= 0—and, again, the sum is not changed by distortions of the surface, as long as q

remains outside. So the new “Gauss’s Law” holds for any charge conﬁguration.

(f) In diﬀerential form, “Gauss’s law” reads: ∇·E+1

λ2V=1

ϵ0

ρ,or, putting it all in terms of E:

∇·E−1

λ2(E·dl=1

ϵ0

ρ.Since E=−∇V, this also yields “Poisson’s equation”: −∇2V+1

λ2V=1

ϵ0

ρ.

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E·da=q

4⇡✏01

S21+S

eS/(S2sin ✓d✓d)1

R21+R

eR/(R2sin ✓d✓d)

4⇡✏01+ S

eS/1+R

eR/sin ✓d✓d.

1

2Vd⌧=1

2

4⇡✏0er/

rr2sin ✓dr d✓d=1

2

4⇡✏0

sin ✓d✓dS

rer/dr

=q

4⇡✏0

sin ✓d✓der/1+ r



=q

4⇡✏01+ S

eS/1+R

eR/sin ✓d✓d.

So the change in 1

2Vd⌧exactly compensates for the change in E·da, and we get 1

✏0qfor the total using

the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive

distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside,

the total is 1

✏0Qenc. Charges outside do not contribute (in the argument above we found that

Now !1

r2e−r/λdr =−e−r/λ

r−1

λ!e−r/λ

rdr ←− exactly right to kill the last term. Therefore

V(r) = q

4πϵ0"−e−r/λ

r####

∞

r$=q

4πϵ0

e−r/λ

(d) %S

E·da=1

4πϵ0

!q1

!&1 + R

λ'e−R/λ4π!R2

!=q

ϵ0&1 + R

λ'e−R/λ.

V dτ=q

4πϵ0

!(R

e−r/λ

rr24π!dr =q

ϵ0(R

re−r/λdr =q

ϵ0)e−r/λ

(1/λ)2*−r

λ−1+,R

=λ2q

ϵ0-−e−R/λ&1 + R

λ'+ 1..

∴%S

E·da+1

λ2(V

V dτ=q

ϵ0-&1 + R

λ'e−R/λ−&1 + R

λ'e−R/λ+ 1.=q

ϵ0

.qed

(e) Does the result in (d) hold for a nonspherical surface? Suppose we

make a “dent” in the sphere—pushing a patch (area R2sin θdθdφ)

from radius Rout to radius S(area S2sin θdθdφ). q

∆%E·da=q

4πϵ0-1

S2&1 + S

λ'e−S/λ(S2sin θdθdφ)−1

R2&1 + R

λ'e−R/λ(R2sin θdθdφ).

4πϵ0)&1 + S

λ'e−S/λ−&1 + R

λ'e−R/λ,sin θdθdφ.

∆1

λ2(V dτ=1

λ2

4πϵ0(e−r/λ

rr2sin θdr dθdφ=1

λ2

4πϵ0

sin θdθdφ(S

re−r/λdr

=−q

4πϵ0

sin θdθdφ*e−r/λ*1 + r

λ++###

=−q

4πϵ0)&1 + S

λ'e−S/λ−&1 + R

λ'e−R/λ,sin θdθdφ.

So the change in 1

λ2!V dτexactly compensates for the change in /E·da, and we get 1

ϵ0qfor the total using

the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive

distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside,

the total is 1

ϵ0Qenc. Charges outside do not contribute (in the argument above we found that for this

volume /E·da+1

λ2!V dτ= 0—and, again, the sum is not changed by distortions of the surface, as long as q

remains outside. So the new “Gauss’s Law” holds for any charge conﬁguration.

(f) In diﬀerential form, “Gauss’s law” reads: ∇·E+1

λ2V=1

ϵ0

ρ,or, putting it all in terms of E:

∇·E−1

λ2(E·dl=1

ϵ0

ρ.Since E=−∇V, this also yields “Poisson’s equation”: −∇2V+1

λ2V=1

ϵ0

ρ.

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for this volume E·da+1

2Vd⌧= 0—and, again, the sum is not changed by distortions of the surface, as

long as qremains outside). So the new “Gauss’s Law” holds for any charge conﬁguration.

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46 CHAPTER 2. ELECTROSTATICS

(f) In di↵erential form, “Gauss’s law” reads: r·E+1

2V=1

✏0

⇢,or, putting it all in terms of E:

r·E1

2E·dl=1

✏0

⇢.Since E=rV, this also yields “Poisson’s equation”: r2V+1

2V=1

✏0

⇢.

∆1

λ2!V dτ=1

λ2

4πϵ0!e−r/λ

rr2sin θdr dθdφ=1

λ2

4πϵ0

sin θdθdφ!S

re−r/λdr

=−q

4πϵ0

sin θdθdφ"e−r/λ"1 + r

λ##$

=−q

4πϵ0%&1 + S

λ'e−S/λ−&1 + R

λ'e−R/λ(sin θdθdφ.

So the change in 1

λ2)V dτexactly compensates for the change in *E·da, and we get 1

ϵ0qfor the total using

the dented sphere, just as we did with the perfect sphere. Any closed surface can be built up by successive

distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside,

the total is 1

ϵ0Qenc. Charges outside do not contribute (in the argument above we found that for this

volume *E·da+1

λ2)V dτ= 0—and, again, the sum is not changed by distortions of the surface, as long as q

remains outside. So the new “Gauss’s Law” holds for any charge conﬁguration.

(f) In diﬀerential form, “Gauss’s law” reads: ∇·E+1

λ2V=1

ϵ0

ρ,or, putting it all in terms of E:

∇·E−1

λ2!E·dl=1

ϵ0

ρ.Since E=−∇V, this also yields “Poisson’s equation”: −∇2V+1

λ2V=1

ϵ0

ρ.

✣

✢✲

✛

❪

❫

E=−∇V

V=−RE·dl

(∇2−1

λ2)V=−ρ

ϵ0

V=1

4πϵ0Rρe−/λdτ

∇·E−1

λ2RE·dl=ρ

ϵ0;∇×E=0

E=1

4πϵ0Rρ

2(1+λ)e−/λdτ

Problem 2.50

ρ=ϵ0∇·E=ϵ0

∂

∂x(ax) = ϵ0a(constant everywhere).

The same charge density would be compatible (as far as Gauss’s law is concerned) with E=ayˆy, for

instance, or E= (a

3)r, etc. The point is that Gauss’s law (and ∇×E= 0) by themselves do not determine

the ﬁeld—like any diﬀerential equations, they must be supplemented by appropriate boundary conditions.

Ordinarily, these are so “obvious” that we impose them almost subconsciously (“Emust go to zero far from

the source charges”)—or we appeal to symmetry to resolve the ambiguity (“the ﬁeld must be the same—in

magnitude—on both sides of an inﬁnite plane of surface charge”). But in this case there are no natural

boundary conditions, and no persuasive symmetry conditions, to ﬁx the answer. The question “What is the

electric ﬁeld produced by a uniform charge density ﬁlling all of space?” is simply ill-posed: it does not give

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(g) Refer to ”Gauss’s law” in di↵erential form (f). Since Eis zero, inside a conductor (otherwise charge would

move, and in such a direction as to cancel the ﬁeld), Vis constant (inside), and hence ⇢is uniform, throughout

the volume. Any “extra” charge must reside on the surface. (The fraction at the surface depends on , and

on the shape of the conductor.)

Problem 2.55

⇢=✏0r·E=✏0@

@x(ax)= ✏0a(constant everywhere).

The same charge density would be compatible (as far as Gauss’s law is concerned) with E=ayˆy , for

instance, or E=(

3)r, etc. The point is that Gauss’s law (and r⇥E=0) by themselves do not determine

the ﬁeld—like any di↵erential equations, they must be supplemented by appropriate boundary conditions.

Ordinarily, these are so “obvious” that we impose them almost subconsciously (“Emust go to zero far from

the source charges”)—or we appeal to symmetry to resolve the ambiguity (“the ﬁeld must be the same—in

magnitude—on both sides of an inﬁnite plane of surface charge”). But in this case there are no natural

boundary conditions, and no persuasive symmetry conditions, to ﬁx the answer. The question “What is the

electric ﬁeld produced by a uniform charge density ﬁlling all of space?” is simply ill-posed: it does not give

us suﬃcient information to determine the answer. (Incidentally, it won’t help to appeal to Coulomb’s law

E=1

4⇡✏0⇢ˆ

2d⌧—the integral is hopelessly indeﬁnite, in this case.)

Problem 2.56

Compare Newton’s law of universal gravitation to Coulomb’s law:

F=Gm1m2

r2ˆr ;F=1

4⇡✏0

q1q2

r2ˆr .

Evidently 1

4⇡✏0!Gand q!m. The gravitational energy of a sphere (translating Prob. 2.34) is therefore

Wgrav =3

5GM2

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CHAPTER 2. ELECTROSTATICS 47

Now, G=6.67 ⇥1011 Nm

2/kg2, and for the sun M=1.99 ⇥1030 kg, R=6.96 ⇥108m, so the sun’s

gravitational energy is W=2.28 ⇥1041 J. At the current rate this energy would be dissipated in a time

t=W

P=2.28 ⇥1041

3.86 ⇥1026 =5.90 ⇥1014 s= 1.87 ⇥107years.

Problem 2.57

First eliminate z, using the formula for the ellipsoid:

(x, y)= Q

4⇡ab

c2(x2/a4)+c2(y2/b4)+1(x2/a2)(y2/b2).

Now (for parts (a) and (b)) set c!0, “squashing” the ellipsoid down to an ellipse in the xy plane:

(x, y)= Q

2⇡ab

1(x/a)2(y/b)2.

(I multiplied by 2 to count both surfaces.)

(a) For the circular disk, set a=b=Rand let r⌘x2+y2.(r)= Q

2⇡R

pR2r2.

(b) For the ribbon, let Q/2b⌘⇤, and then take the limit b!1:(x)= ⇤

2⇡

pa2x2.

a2+r2

c2=1,with =Q

4⇡ac2

x2/a4+r2/c4.

The charge on a ring of width dx is

dq =2⇡r ds, where ds =dx2+dr2=dx1+(dr/dx)2.

Now 2xdx

a2+2r dr

c2=0)dr

dx =c2x

a2r,sods =dx1+c4x2

a4r2=dxc2

rx2/a4+r2/c4. Thus

(x)= dq

dx =2⇡rQ

4⇡ac2

x2/a4+r2/c4

rx2/a4+r2/c4=Q

2a.(Constant!)

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48 CHAPTER 2. ELECTROSTATICS

Problem 2.58

(0,-a)

( , )

( ,- )

÷3

(a) One such point is on the xaxis (see diagram) at x=r. Here the ﬁeld is

Ex=q

4⇡✏01

(a+r)22cos ✓

b2=0,or 2 cos ✓

b2=1

(a+r)2.

Now,

cos ✓=(a/2) r

b;b2=a

2r2+p3

2a2

=(a2ar +r2).

Therefore

2[(a/2) r]

(a2ar +r2)3/2=1

(a+r)2.To simplify,let r

a⌘u:

(1 2u)

(1 u+u2)3/2=1

(1 + u)2,or (1 2u)2(1 + u)4=(1u+u2)3.

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CHAPTER 2. ELECTROSTATICS 49

Multiplying out each side:

16u24u3+9u4+ 12u5+4u6=13u+6u27u3+6u43u5+u6,

3u12u2+3u3+3u4+ 15u5+3u6=0.

u= 0 is a solution (of course—the center of the triangle); factoring out 3uwe are left with a quintic equation:

14u+u2+u3+5u4+u5=0.

According to Mathematica, this has two complex roots, and one negative root. The two remaining solutions are

u=0.284718 and u=0.626691. The latter is outside the triangle, and clearly spurious. So r=0.284718 a.

(The other two places where E=0are at the symmetrically located points, of course.)

( )

÷2

(b) For the square:

Ex=q

4⇡✏02cos ✓+

+2cos ✓

=0 )cos ✓+

=cos ✓



where

cos ✓±=(a/p2) ±r

b±

;b2

±=a

p22

+a

p2±r2

=a2±p2ar +r2.

Thus

(a/p2) + r

(a2+p2ar +r2)3/2=(a/p2) r

(a2p2ar +r2)3/2.

To simplify, let w⌘p2r/a; then

1+w

(2 + 2w+w2)3/2=1w

(2 2w+w2)3/2,or (1 + w)2(2 2w+w2)3= (1 w)2(2 + 2w+w2)3.

Multiplying out the left side:

88w4w2+ 16w310w42w5+7w64w7+w8= (same thing with w!w).

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50 CHAPTER 2. ELECTROSTATICS

The even powers cancel, leaving

8w16w3+2w5+4w7=0,or 4 8v+v2+2v3=0,

where v⌘w2. According to Mathematica, this cubic equation has one negative root, one root that is spurious

(the point lies outside the square), and v=0.598279, which yields

r=v

2a= 0.546936 a .

(a cos(p/5), a sin(p/5))

(-a cos(2p/5), a sin(2p/5))

For the pentagon:

Ex=q

4⇡✏01

(a+r)2+2cos ✓

b22cos 

c2=0,

where

cos ✓=acos(2⇡/5) + r

b,cos =acos(⇡/5) r

b2=[acos(2⇡/5) + r]2+[asin(2⇡/5)]2=a2+r2+2ar cos(2⇡/5),

c2=[acos(⇡/5) r]2+[asin(⇡/5)]2=a2+r22ar cos(⇡/5).

(a+r)2+2 r+acos(2⇡/5)

[a2+r2+2ar cos(2⇡/5)]3/2+2 racos(⇡/5)

[a2+r22ar cos(⇡/5)]3/2=0.

Mathematica gives the solution r=0.688917 a.

For an n-sided regular polygon there are evidently nsuch points, lying on the radial spokes that bisect

the sides; their distance from the center appears to grow monotonically with n:r(3) = 0.285, r(4) = 0.547,

r(5) = 0.689, . . . . As n!1they ﬁll out a circle that (in the limit) coincides with the ring of charge itself.

Problem 2.59 The theorem is false. For example, suppose the conductor is a neutral sphere and the external

ﬁeld is due to a nearby positive point charge q. A negative charge will be induced on the near side of the sphere

(and a positive charge on the far side), so the force will be attractive (toward q). If we now reverse the sign of

q, the induced charges will also reverse, but the force will still be attractive.

If the external ﬁeld is uniform, then the net force on the induced charges is zero, and the total force on the

conductor is QEe, which does switch signs if Eeis reversed. So the “theorem” is valid in this very special case.

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CHAPTER 2. ELECTROSTATICS 51

Problem 2.60 The initial conﬁguration consists of a point charge qat the center, qinduced on the inner

surface, and +qon the outer surface. What is the energy of this conﬁguration? Imagine assembling it piece-by-

piece. First bring in qand place it at the origin—this takes no work. Now bring in qand spread it over the

surface at a—using the method in Prob. 2.35, this takes work q2/(8⇡✏0a). Finally, bring in +qand spread it

over the surface at b—this costs q2/(8⇡✏0b). Thus the energy of the initial conﬁguration is

Wi=q2

8⇡✏01

a1

b.

The ﬁnal conﬁguration is a neutral shell and a distant point charge—the energy is zero. Thus the work

necessary to go from the initial to the ﬁnal state is

W=WfWi=q2

8⇡✏01

a1

b.

Problem 2.61

(2pj/n)

Suppose the npoint charges are evenly spaced around the circle, with the jth particle at angle j(2⇡/n).

According to Eq. 2.42, the energy of the conﬁguration is

Wn=n1

2qV,

where Vis the potential due to the (n1) other charges, at charge # n(on the xaxis).

V=1

4⇡✏0

n1



j=1

j=2Rsin j⇡

n

(see the ﬁgure). So

Wn=q2

4⇡✏0R

n1



j=1

sin(j⇡/n)=q2

4⇡✏0R⌦n.

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52 CHAPTER 2. ELECTROSTATICS

Mathematica says

⌦10 =10



j=1

sin(j⇡/10) = 38.6245

⌦11 =11



j=1

sin(j⇡/11) = 48.5757

⌦12 =12



j=1

sin(j⇡/12) = 59.8074

If (n1) charges are on the circle (energy ⌦n1q2/4⇡✏0R), and the nth is at the center, the total energy is

Wn=[⌦n1+(n1)] q2

4⇡✏0R.

For

n=11: ⌦10 + 10 = 38.6245 + 10 = 48.6245 >⌦11

n=12: ⌦11 + 11 = 48.5757 + 11 = 59.5757 <⌦12

Thus a lower energy is achieved for 11 charges if they are all at the rim, but for 12 it is better to put one at

the center.

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CHAPTER 3. POTENTIAL 53

Chapter 3

Potential

Problem 3.1

The argument is exactly the same as in Sect. 3.1.4, except that since z<R,pz2+R22zR =(Rz),

instead of (zR). Hence Vave =q

4⇡✏0

2zR [(z+R)(Rz)] = 1

4⇡✏0

R. If there is more than one charge

inside the sphere, the average potential due to interior charges is 1

4⇡✏0

Qenc

R,and the average due to exterior

charges is Vcenter,soVave =Vcenter +Qenc

4⇡✏0R.X

Problem 3.2

A stable equilibrium is a point of local minimum in the potential energy. Here the potential energy is qV .

But we know that Laplace’s equation allows no local minima for V. What looks like a minimum, in the ﬁgure,

must in fact be a saddle point, and the box “leaks” through the center of each face.

Problem 3.3

Laplace’s equation in spherical coordinates, for Vdependent only on r, reads:

r2V=1

dr r2dV

dr =0)r2dV

dr =c(constant) )dV

dr =c

r2)V=c

r+k.

Example: potential of a uniformly charged sphere.

In cylindrical coordinates: r2V=1

ds sdV

ds =0)sdV

ds =c)dV

ds =c

s)V=cln s+k.

Example: potential of a long wire.

Problem 3.4

Refer to Fig. 3.3, letting ↵be the angle between

and the zaxis. Obviously, Eave points in the ˆz

direction, so

Eave =1

4⇡R2Eda =ˆz 1

4⇡R2

4⇡✏01

2cos ↵da.

By the law of cosines,

R2=z2+

22

zcos ↵)cos ↵=z2+

2R2

2=R2+z22Rz cos ✓)cos ↵

2=z2+

2R2

3=zRcos ✓

(R2+z22Rz cos ✓)3/2.

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54 CHAPTER 3. POTENTIAL

Eave =ˆz q

16⇡2R2✏0zRcos ✓

(R2+z22Rz cos ✓)3/2R2sin ✓d✓d

=qˆz

8⇡✏0⇡

zRcos ✓

(R2+z22Rz cos ✓)3/2sin ✓d✓=qˆz

8⇡✏01

1

zRu

(R2+z22Rzu)3/2du

(where u⌘cos ✓). The integral is

I=1

RpR2+z22Rzu

11

2Rz2R2+z22Rzu +R2+z2

pR2+z22Rzu

1

R1

|zR|1

z+R1

2Rz2|zR|(z+R)+(R2+z2)1

|zR|1

z+R.

(a) If z>R,

I=1

R1

zR1

z+R1

2Rz2(zR)(z+R)+(R2+z2)1

zR1

z+R

R2R

z2R21

2Rz22R+(R2+z2)2R

z2R2=2

z2.

Eave =1

4⇡✏0

z2ˆz ,

the same as the ﬁeld at the center. By superposition the same holds for any collection of charges outside the

sphere.

(b) If z<R,

I=1

R1

Rz1

z+R1

2Rz2(Rz)(z+R)+(R2+z2)1

Rz1

z+R

R2z

R2z21

2Rz22z+(R2+z2)2z

R2z2=0.

Eave =0.

By superposition the same holds for any collection of charges inside the sphere.

Problem 3.5

Same as proof of second uniqueness theorem, up to the equation SV3E3·da=V(E3)2d⌧. But on

each surface, either V3= 0 (if Vis speciﬁed on the surface), or else E3?= 0 (if @V

@n=E?is speciﬁed). So

V(E3)2= 0, and hence E2=E1. qed

Problem 3.6

Putting U=T=V3into Green’s identity:

VV3r2V3+rV3·rV3d⌧=S

V3rV3·da.But r2V3=r2V1r2V2=⇢

✏0

+⇢

✏0

= 0, and rV3=E3.

So V

3d⌧=S

V3E3·da, and the rest is the same as before.

Problem 3.7

Place image charges +2qat z=dand qat z=3d. Total force on +qis

F=q

4⇡✏02q

(2d)2+2q

(4d)2+q

(6d)2ˆz =q2

4⇡✏0d21

2+1

81

36ˆz =1

4⇡✏029q2

72d2ˆz .

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CHAPTER 3. POTENTIAL 55

Problem 3.8

(a) From Fig. 3.13:

=pr2+a22ra cos ✓;

0=pr2+b22rb cos ✓.Therefore:

0=R

pr2+b22rb cos ✓(Eq.3.15),while b=R2

a(Eq.3.16).

=q

a

Rr2+R4

a22rR2

acos ✓

=q

ar

R2+R22ra cos ✓

Therefore:

V(r, ✓)= 1

4⇡✏0q

+q0

0=q

4⇡✏01

pr2+a22ra cos ✓1

R2+(ra/R)22ra cos ✓.

Clearly, when r=R,V!0.

(b) =✏0@V

@n(Eq. 2.49). In this case, @V

@n=@V

@rat the point r=R. Therefore,

(✓)=✏0q

4⇡✏01

2(r2+a22ra cos ✓)3/2(2r2acos ✓)

2R2+(ra/R)22ra cos ✓3/2a2

R22r2acos ✓r=R

=q

4⇡(R2+a22Ra cos ✓)3/2(Racos ✓)+R2+a22Ra cos ✓3/2a2

Racos ✓

4⇡(R2+a22Ra cos ✓)3/2Racos ✓a2

R+acos ✓

4⇡R(R2a2)(R2+a22Ra cos ✓)3/2.

qinduced =da =q

4⇡R(R2a2)(R2+a22Ra cos ✓)3/2R2sin ✓d✓d

4⇡R(R2a2)2⇡R21

Ra(R2+a22Ra cos ✓)1/2

⇡

2a(a2R2)1

pR2+a2+2Ra 1

pR2+a22Ra.

But a>R(else qwould be inside),so R2+a22Ra =aR.

2a(a2R2)1

(a+R)1

(aR)=q

2a[(aR)(a+R)] = q

2a(2R)

=qR

a=q0.

F=1

4⇡✏0

qq0

(ab)2=1

4⇡✏0R

aq21

(aR2/a)2=1

4⇡✏0

q2Ra

(a2R2)2.

protected under all copyright laws as they currently exist. No portion of this material may be

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56 CHAPTER 3. POTENTIAL

To bring qin from inﬁnity to a, then, we do work

W=q2R

4⇡✏0



(a2R2)2da =q2R

4⇡✏01

(a2R2)

=1

4⇡✏0

q2R

2(a2R2).

Problem 3.9

4CHAPTER 3. SPECIAL TECHNIQUES

F=1

4πϵ0

qq′

(a−b)2=1

4πϵ0!−

aq2"1

(a−R2/a)2=−

4πϵ0

q2Ra

(a2−R2)2.

To bring qin from inﬁnity to a, then, we do work

W=q2R

4πϵ0

∞

(a2−R2)2da =q2R

4πϵ0$−

(a2−R2)%&&&&

∞

=−

4πϵ0

q2R

2(a2−R2).

Problem 3.8

q′′ q′q

'() *

a−b

)*' (

Place a second image charge, q′′, at the center of the sphere;

this will not alter the fact that the sphere is an equipotential,

but merely increase that potential from zero to V0=1

4πϵ0

q′′

q′′ = 4πϵ0V0Rat center of sphere.

For a neutral sphere, q′+q′′ = 0.

F=1

4πϵ0

q!q′′

a2+q′

(a−b)2"=qq′

4πϵ0!−

a2+1

(a−b)2"

=qq′

4πϵ0

b(2a−b)

a2(a−b)2=q(−Rq/a)

4πϵ0

(R2/a)(2a−R2/a)

a2(a−R2/a)2

=−

4πϵ0!R

a"3(2a2−R2)

(a2−R2)2.

(Drop the minus sign, because the problem asks for the force of attraction.)

Problem 3.9

✲x

✻

✣y

✲y

✻

−

(y, z)s+

s−

(a) Image problem: λabove, −λbelow. Potential was found in Prob. 2.47:

V(y, z) = 2λ

4πϵ0

ln(s−/s+) = λ

4πϵ0

ln(s2

−/s2

=λ

4πϵ0

ln +y2+ (z+d)2

y2+ (z−d)2,

(b) σ=−ϵ0

∂V

∂n. Here ∂V

∂n=∂V

∂z, evaluated at z= 0.

σ(y) = −ϵ0

4πϵ0+1

y2+ (z+d)22(z+d)−

y2+ (z−d)22(z−d),&&&&z=0

=−

2λ

4π+d

y2+d2

−

−d

y2+d2,=−

λd

π(y2+d2).

Check: Total charge induced on a strip of width lparallel to the yaxis:

qind =−

lλd

∞

−∞

y2+d2dy =−

lλd

π$1

dtan−1-y

d.%&&&&

∞

−∞

=−

lλd

π/π

−-−

2.0

=−λl. Therefore λind =−λ,as it should be.

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

Place a second image charge, q00, at the center of the sphere;

this will not alter the fact that the sphere is an equipotential,

but merely increase that potential from zero to V0=1

4⇡✏0

q00

q00 =4⇡✏0V0Rat center of sphere.

For a neutral sphere, q0+q00 = 0.

F=1

4⇡✏0

qq00

a2+q0

(ab)2=qq0

4⇡✏01

a2+1

(ab)2

=qq0

4⇡✏0

b(2ab)

a2(ab)2=q(Rq/a)

4⇡✏0

(R2/a)(2aR2/a)

a2(aR2/a)2

=q2

4⇡✏0R

a3(2a2R2)

(a2R2)2.

(Drop the minus sign, because the problem asks for the force of attraction.)

Problem 3.10

(a) Image problem: above, below. Potential was found in Prob. 2.52:

2CHAPTER 3. SPECIAL TECHNIQUES

Chapter 1

Special Techniques

Problem 3.9

✲x

✻

✣y

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reproduced, in any form or by any means, without permission in writing from the publisher.

4CHAPTER 3. SPECIAL TECHNIQUES

F=1

4πϵ0

qq′

(a−b)2=1

4πϵ0!−

aq2"1

(a−R2/a)2=−

4πϵ0

q2Ra

(a2−R2)2.

To bring qin from inﬁnity to a, then, we do work

W=q2R

4πϵ0

∞

(a2−R2)2da =q2R

4πϵ0$−

(a2−R2)%&&&&

∞

=−

4πϵ0

q2R

2(a2−R2).

Problem 3.8

q′′ q′q

'() *

a−b

)*' (

Place a second image charge, q′′, at the center of the sphere;

this will not alter the fact that the sphere is an equipotential,

but merely increase that potential from zero to V0=1

4πϵ0

q′′

q′′ = 4πϵ0V0Rat center of sphere.

For a neutral sphere, q′+q′′ = 0.

F=1

4πϵ0

q!q′′

a2+q′

(a−b)2"=qq′

4πϵ0!−

a2+1

(a−b)2"

=qq′

4πϵ0

b(2a−b)

a2(a−b)2=q(−Rq/a)

4πϵ0

(R2/a)(2a−R2/a)

a2(a−R2/a)2

=−

4πϵ0!R

a"3(2a2−R2)

(a2−R2)2.

(Drop the minus sign, because the problem asks for the force of attraction.)

Problem 3.9

✲x

✻

✣y

✲y

✻

−

(y, z)s+

s−

(a) Image problem: λabove, −λbelow. Potential was found in Prob. 2.47:

V(y, z) = 2λ

4πϵ0

ln(s−/s+) = λ

4πϵ0

ln(s2

−/s2

=λ

4πϵ0

ln +y2+ (z+d)2

y2+ (z−d)2,

(b) σ=−ϵ0

∂V

∂n. Here ∂V

∂n=∂V

∂z, evaluated at z= 0.

σ(y) = −ϵ0

4πϵ0+1

y2+ (z+d)22(z+d)−

y2+ (z−d)22(z−d),&&&&z=0

=−

2λ

4π+d

y2+d2

−

−d

y2+d2,=−

λd

π(y2+d2).

Check: Total charge induced on a strip of width lparallel to the yaxis:

qind =−

lλd

∞

−∞

y2+d2dy =−

lλd

π$1

dtan−1-y

d.%&&&&

∞

−∞

=−

lλd

π/π

−-−

2.0

=−λl. Therefore λind =−λ,as it should be.

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reproduced, in any form or by any means, without permission in writing from the publisher.

V(y, z)= 2

4⇡✏0

ln(s/s+)= 

4⇡✏0

ln(s2

/s2

=

4⇡✏0

ln y2+(z+d)2

y2+(zd)2

(b) =✏0

@n. Here @V

@n=@V

@z, evaluated at z= 0.

(y)=✏0



4⇡✏01

y2+(z+d)22(z+d)1

y2+(zd)22(zd)z=0

=2

4⇡d

y2+d2d

y2+d2=d

⇡(y2+d2).

Check: Total charge induced on a strip of width lparallel to the yaxis:

qind =ld

⇡



1

y2+d2dy =ld

⇡1

dtan1y

d

1

=ld

⇡⇡

2⇡

2

=l. Therefore ind =,as it should be.

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 3. POTENTIAL 57

Problem 3.11

The image conﬁguration is as shown.

V(x, y)= q

4⇡✏01

(xa)2+(yb)2+z2+1

(x+a)2+(y+b)2+z2

1

(x+a)2+(yb)2+z21

(xa)2+(y+b)2+z2.

CHAPTER 3. SPECIAL TECHNIQUES 5

Problem 3.10

The image conﬁguration is as shown.

V(x, y) = q

4πϵ0!1

"(x−a)2+ (y−b)2+z2+1

"(x+a)2+ (y+b)2+z2

−1

"(x+a)2+ (y−b)2+z2−1

"(x−a)2+ (y+b)2+z2#.

✲x

✻

q−q

For this to work, θmust be and integer divisor of 180◦. Thus 180◦, 90◦, 60◦, 45◦, etc., are OK, but no

others. It works for 45◦, say, with the charges as shown.

(Note the strategy: to make the xaxis an equipotential (V= 0),

you place the image charge (1) in the reﬂection point. To make the

45◦line an equipotential, you place charge (2) at the image point.

But that screws up the xaxis, so you must now insert image (3) to

balance (2). Moreover, to make the 45◦line V= 0 you also need (4),

to balance (1). But now, to restore the xaxis to V= 0 you need (5)

to balance (4), and so on.

✲x

45◦line

−(1)

−

(2)

(4)

−

(5)

(3)

why it works for θ= 45◦

✲x

■

135◦line

(0)

−

(1)

−(2)

+(3)

⇐

No good

why it doesn’t work for θ= 135◦

The reason this doesn’t work for arbitrary angles is that you are even-

tually forced to place an image charge within the original region of

interest, and that’s not allowed—all images must go outside the re-

gion, or you’re no longer dealing with the same problem at all.)

Problem 3.11

From Prob. 2.47 (with y0→d): V=λ

4πϵ0

ln $(x+a)2+y2

(x−a)2+y2%,where a2=y02−R2⇒a=√d2−R2,

and

&acoth(2πϵ0V0/λ) = d

acsch(2πϵ0V0/λ) = R'⇒(dividing) d

R= cosh (2πϵ0V0

λ),or λ=2πϵ0V0

cosh−1(d/R).

Problem 3.12

V(x, y) =

∞

n=1

Cne−nπx/a sin(nπy/a) (Eq.3.30),where Cn=2

V0(y) sin(nπy/a)dy (Eq.3.34).

In this case V0(y) = &+V0,for 0 < y < a/2

−V0,for a/2< y < a '.Therefore,

Cn=2

aV0⎧

⎪

⎨

⎪

⎩

a/2

sin(nπy/a)dy −

a/2

sin(nπy/a)dy⎫

⎪

⎬

⎪

⎭

=2V0

a!−cos(nπy/a)

(nπ/a)3333

a/2

+cos(nπy/a)

(nπ/a)3333

a/2#

=2V0

nπ4−cos 5nπ

26+ cos(0) + cos(nπ)−cos 5nπ

267=2V0

nπ41 + (−1)n−2 cos 5nπ

267.

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

F=q2

4⇡✏01

(2a)2ˆx 1

(2b)2ˆy +1

(2pa2+b2)2[cos ✓ˆx + sin ✓ˆy ],

where cos ✓=a/pa2+b2,sin ✓=b/pa2+b2.

F=q2

16⇡✏0 a

(a2+b2)3/21

a2ˆx +b

(a2+b2)3/21

b2ˆy .

W=1

4⇡✏0q2

(2a)+q2

(2b)+q2

(2pa2+b2)=q2

16⇡✏01

pa2+b21

a1

b.

For this to work, ✓must be an integer divisor of 180. Thus 180, 90, 60, 45, etc., are OK, but no

others. It works for 45, say, with the charges as shown.

(Note the strategy: to make the xaxis an equipotential (V= 0),

you place the image charge (1) in the reﬂection point. To make the

45line an equipotential, you place charge (2) at the image point.

But that screws up the xaxis, so you must now insert image (3) to

balance (2). Moreover, to make the 45line V= 0 you also need (4),

to balance (1). But now, to restore the xaxis to V= 0 you need (5)

to balance (4), and so on.

CHAPTER 3. SPECIAL TECHNIQUES 5

Problem 3.10

The image conﬁguration is as shown.

V(x, y) = q

4πϵ0!1

"(x−a)2+ (y−b)2+z2+1

"(x+a)2+ (y+b)2+z2

−1

"(x+a)2+ (y−b)2+z2−1

"(x−a)2+ (y+b)2+z2#.

✲x

✻

q−q

For this to work, θmust be and integer divisor of 180◦. Thus 180◦, 90◦, 60◦, 45◦, etc., are OK, but no

others. It works for 45◦, say, with the charges as shown.

(Note the strategy: to make the xaxis an equipotential (V= 0),

you place the image charge (1) in the reﬂection point. To make the

45◦line an equipotential, you place charge (2) at the image point.

But that screws up the xaxis, so you must now insert image (3) to

balance (2). Moreover, to make the 45◦line V= 0 you also need (4),

to balance (1). But now, to restore the xaxis to V= 0 you need (5)

to balance (4), and so on.

✲x

45◦line

−(1)

−

(2)

(4)

−

(5)

(3)

why it works for θ= 45◦

✲x

■

135◦line

(0)

−

(1)

−(2)

+(3)

⇐

No good

why it doesn’t work for θ= 135◦

The reason this doesn’t work for arbitrary angles is that you are even-

tually forced to place an image charge within the original region of

interest, and that’s not allowed—all images must go outside the re-

gion, or you’re no longer dealing with the same problem at all.)

Problem 3.11

From Prob. 2.47 (with y0→d): V=λ

4πϵ0

ln $(x+a)2+y2

(x−a)2+y2%,where a2=y02−R2⇒a=√d2−R2,

and

&acoth(2πϵ0V0/λ) = d

acsch(2πϵ0V0/λ) = R'⇒(dividing) d

R= cosh (2πϵ0V0

λ),or λ=2πϵ0V0

cosh−1(d/R).

Problem 3.12

V(x, y) =

∞

n=1

Cne−nπx/a sin(nπy/a) (Eq.3.30),where Cn=2

V0(y) sin(nπy/a)dy (Eq.3.34).

In this case V0(y) = &+V0,for 0 < y < a/2

−V0,for a/2< y < a '.Therefore,

Cn=2

aV0⎧

⎪

⎨

⎪

⎩

a/2

sin(nπy/a)dy −

a/2

sin(nπy/a)dy⎫

⎪

⎬

⎪

⎭

=2V0

a!−cos(nπy/a)

(nπ/a)3333

a/2

+cos(nπy/a)

(nπ/a)3333

a/2#

=2V0

nπ4−cos 5nπ

26+ cos(0) + cos(nπ)−cos 5nπ

267=2V0

nπ41 + (−1)n−2 cos 5nπ

267.

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 3. SPECIAL TECHNIQUES 5

Problem 3.10

The image conﬁguration is as shown.

V(x, y) = q

4πϵ0!1

"(x−a)2+ (y−b)2+z2+1

"(x+a)2+ (y+b)2+z2

−1

"(x+a)2+ (y−b)2+z2−1

"(x−a)2+ (y+b)2+z2#.

✲x

✻

q−q

For this to work, θmust be and integer divisor of 180◦. Thus 180◦, 90◦, 60◦, 45◦, etc., are OK, but no

others. It works for 45◦, say, with the charges as shown.

(Note the strategy: to make the xaxis an equipotential (V= 0),

you place the image charge (1) in the reﬂection point. To make the

45◦line an equipotential, you place charge (2) at the image point.

But that screws up the xaxis, so you must now insert image (3) to

balance (2). Moreover, to make the 45◦line V= 0 you also need (4),

to balance (1). But now, to restore the xaxis to V= 0 you need (5)

to balance (4), and so on.

✲x

45◦line

−(1)

−

(2)

(4)

−

(5)

(3)

why it works for θ= 45◦

✲x

■

135◦line

(0)

−

(1)

−(2)

+(3)

⇐

No good

why it doesn’t work for θ= 135◦

The reason this doesn’t work for arbitrary angles is that you are even-

tually forced to place an image charge within the original region of

interest, and that’s not allowed—all images must go outside the re-

gion, or you’re no longer dealing with the same problem at all.)

Problem 3.11

From Prob. 2.47 (with y0→d): V=λ

4πϵ0

ln $(x+a)2+y2

(x−a)2+y2%,where a2=y02−R2⇒a=√d2−R2,

and

&acoth(2πϵ0V0/λ) = d

acsch(2πϵ0V0/λ) = R'⇒(dividing) d

R= cosh (2πϵ0V0

λ),or λ=2πϵ0V0

cosh−1(d/R).

Problem 3.12

V(x, y) =

∞

n=1

Cne−nπx/a sin(nπy/a) (Eq.3.30),where Cn=2

V0(y) sin(nπy/a)dy (Eq.3.34).

In this case V0(y) = &+V0,for 0 < y < a/2

−V0,for a/2< y < a '.Therefore,

Cn=2

aV0⎧

⎪

⎨

⎪

⎩

a/2

sin(nπy/a)dy −

a/2

sin(nπy/a)dy⎫

⎪

⎬

⎪

⎭

=2V0

a!−cos(nπy/a)

(nπ/a)3333

a/2

+cos(nπy/a)

(nπ/a)3333

a/2#

=2V0

nπ4−cos 5nπ

26+ cos(0) + cos(nπ)−cos 5nπ

267=2V0

nπ41 + (−1)n−2 cos 5nπ

267.

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

The reason this doesn’t work for arbitrary angles is that you are even-

tually forced to place an image charge within the original region of

interest, and that’s not allowed—all images must go outside the re-

gion, or you’re no longer dealing with the same problem at all.)

Problem 3.12

From Prob. 2.52 (with y0!d): V=

4⇡✏0

ln (x+a)2+y2

(xa)2+y2, where a2=y02R2)a=pd2R2,

and acoth(2⇡✏0V0/)=d

acsch(2⇡✏0V0/)=R)(dividing) d

R= cosh 2⇡✏0V0

,or =2⇡✏0V0

cosh1(d/R).

Problem 3.13

V(x, y)= 1



n=1

Cnen⇡x/a sin(n⇡y/a) (Eq.3.30),where Cn=2



V0(y) sin(n⇡y/a)dy (Eq.3.34).

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58 CHAPTER 3. POTENTIAL

In this case V0(y)=+V0,for 0 <y<a/2

V0,for a/2<y<a.Therefore,

Cn=2

aV0









a/2



sin(n⇡y/a)dy 



a/2

sin(n⇡y/a)dy







=2V0

acos(n⇡y/a)

(n⇡/a)

a/2

+cos(n⇡y/a)

(n⇡/a)

a/2

=2V0

n⇡cos n⇡

2+ cos(0) + cos(n⇡)cos n⇡

2=2V0

n⇡1+(1)n2 cos n⇡

2.

The term in curly brackets is:











n=1 :112 cos(⇡/2) = 0,

n=2 :1+12 cos(⇡)=4,

n=3 :112 cos(3⇡/2) = 0,

n=4 :1+12 cos(2⇡)=0,











etc.(Zero if nis odd or divisible by 4,otherwise 4.)

Therefore

Cn=8V0/n⇡,n=2,6,10,14,etc.(in general,4j+2,for j=0,1,2, ...),

0,otherwise.

V(x, y)=8V0

⇡

n=2,6,10,...

en⇡x/a sin(n⇡y/a)

n=8V0

⇡



j=0

e(4j+2)⇡x/a sin[(4j+ 2)⇡y/a]

(4j+ 2) .

Problem 3.14

V(x, y)=4V0

⇡

n=1,3,5,...

nen⇡x/a sin(n⇡y/a) (Eq.3.36); =✏0

@n(Eq.2.49).

(y)=✏0

@x4V0

⇡1

nen⇡x/a sin(n⇡y/a)x=0

=✏0

4V0

⇡1

n(n⇡

a)en⇡x/a sin(n⇡y/a)x=0

=4✏0V0

a

n=1,3,5,...

sin(n⇡y/a).

Or, using the closed form 3.37:

V(x, y)=2V0

⇡tan1sin(⇡y/a)

sinh(⇡x/a))=✏0

2V0

⇡

1+ sin2(⇡y/a)

sinh2(⇡x/a)sin(⇡y/a)

sinh2(⇡x/a)⇡

acosh(⇡x/a)x=0

=2✏0V0

sin(⇡y/a) cosh(⇡x/a)

sin2(⇡y/a) + sinh2(⇡x/a)x=0

=2✏0V0

sin(⇡y/a).

[Comment: Technically, the series solution for is defective, since term-by-term di↵erentiation has produced

a (naively) non-convergent sum. More sophisticated deﬁnitions of convergence permit one to work with series

of this form, but it is better to sum the series ﬁrst and then di↵erentiate (the second method.)]

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CHAPTER 3. POTENTIAL 59

Summation of series Eq. 3.36

V(x, y)=4V0

⇡I, where I⌘

n=1,3,5,...

nen⇡x/a sin(n⇡y/a).

Now sin w=Imeiw,so

I=Im1

nen⇡x/aein⇡y/a =Im1

nZn,

where Z⌘e⇡(xiy)/a.Now



1,3,5,...

nZn=1



j=0

(2j+ 1)Z(2j+1) =Z

0







j=0

u2j



du



1u2du =1

2ln 1+Z

1Z=1

2ln Rei✓=1

2(ln R+i✓),

where Rei✓=1+Z

1Z. Therefore

I=Im1

2(ln R+i✓)=1

2✓.But 1+Z

1Z=1+e⇡(xiy)/a

1e⇡(xiy)/a =1+e⇡(xiy)/a1e⇡(x+iy)/a

1e⇡(xiy)/a1e⇡(x+iy)/a

=1+e⇡x/a ei⇡y/a ei⇡y/ae2⇡x/a

1e⇡(xiy)/a2=1+2ie⇡x/a sin(⇡y/a)e2⇡x/a

1e⇡(xiy)/a2,

tan ✓=2e⇡x/a sin(⇡y/a)

1e2⇡x/a =2 sin(⇡y/a)

e⇡x/a e⇡x/a =sin(⇡y/a)

sinh(⇡x/a).

Therefore

I=1

2tan1sin(⇡y/a)

sinh(⇡x/a),and V(x, y)=2V0

⇡tan1sin(⇡y/a)

sinh(⇡x/a).

Problem 3.15

CHAPTER 3. SPECIAL TECHNIQUES 7

where Reiθ=1+Z

1−Z. Therefore

I=Im!1

2(ln R+iθ)"=1

2θ.But 1 + Z

1−Z=1 + e−π(x−iy)/a

1−e−π(x−iy)/a =#1 + e−π(x−iy)/a$#1−e−π(x+iy)/a$

#1−e−π(x−iy)/a$#1−e−π(x+iy)/a$

=1 + e−πx/a #eiπy/a −e−iπy/a$−e−2πx/a

%1−e−π(x−iy)/a%

2=1 + 2ie−πx/a sin(πy/a)−e−2πx/a

%1−e−π(x−iy)/a%

tan θ=2e−πx/a sin(πy/a)

1−e−2πx/a =2 sin(πy/a)

eπx/a −e−πx/a =sin(πy/a)

sinh(πx/a).

Therefore

I=1

2tan−1&sin(πy/a)

sinh(πx/a)',and V(x, y) = 2V0

πtan−1&sin(πy/a)

sinh(πx/a)'.

Problem 3.14

✲x

✻

✙

V= 0

V0(y)

(a) ∂2V

∂x2+∂2V

∂y2= 0, with boundary conditions

⎧

⎪

⎨

⎪

⎩

(i) V(x, 0) = 0,

(ii) V(x, a) = 0,

(iii) V(0, y) = 0,

(iv) V(b, y) = V0(y).

⎫

⎪

⎬

⎪

⎭

As in Ex. 3.4, separation of variables yields

V(x, y) = #Aekx +Be−kx$(Csin ky +Dcos ky).

Here (i)⇒D= 0,(iii)⇒B=−A, (ii)⇒ka is an integer multiple of π:

V(x, y) = AC /enπx/a −e−nπx/a0sin(nπy/a) = (2AC) sinh(nπx/a) sin(nπy/a).

But (2AC) is a constant, and the most general linear combination of separable solutions consistent with (i),

(ii), (iii) is

V(x, y) =

∞

n=1

Cnsinh(nπx/a) sin(nπy/a).

It remains to determine the coeﬃcients Cnso as to ﬁt boundary condition (iv):

1Cnsinh(nπb/a) sin(nπy/a) = V0(y).Fourier’s trick ⇒Cnsinh(nπb/a) = 2

V0(y) sin(nπy/a)dy.

Therefore

Cn=2

asinh(nπb/a)

V0(y) sin(nπy/a)dy.

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(a) @2V

@x2+@2V

@y2= 0, with boundary conditions











(i) V(x, 0) = 0,

(ii) V(x, a)=0,

(iii) V(0,y)=0,

(iv) V(b, y)=V0(y).











As in Ex. 3.4, separation of variables yields

V(x, y)=Aekx +Bekx(Csin ky +Dcos ky).

Here (i))D=0,(iii))B=A, (ii))ka is an integer multiple of ⇡:

V(x, y)=AC en⇡x/a en⇡x/asin(n⇡y/a) = (2AC) sinh(n⇡x/a) sin(n⇡y/a).

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60 CHAPTER 3. POTENTIAL

But (2AC) is a constant, and the most general linear combination of separable solutions consistent with (i),

(ii), (iii) is

V(x, y)= 1



n=1

Cnsinh(n⇡x/a) sin(n⇡y/a).

It remains to determine the coeﬃcients Cnso as to ﬁt boundary condition (iv):

Cnsinh(n⇡b/a) sin(n⇡y/a)=V0(y).Fourier’s trick )Cnsinh(n⇡b/a)= 2



V0(y) sin(n⇡y/a)dy.

Therefore

Cn=2

asinh(n⇡b/a)



V0(y) sin(n⇡y/a)dy.

(b) Cn=2

asinh(n⇡b/a)V0



sin(n⇡y/a)dy =2V0

asinh(n⇡b/a)⇥0,if nis even,

n⇡,if nis odd.

V(x, y)=4V0

⇡

n=1,3,5,...

sinh(n⇡x/a) sin(n⇡y/a)

nsinh(n⇡b/a).

Problem 3.16

Same format as Ex. 3.5, only the boundary conditions are:











(i) V=0 when x=0,

(ii) V=0 when x=a,

(iii) V=0 when y=0,

(iv) V=0 when y=a,

(v) V=0 when z=0,

(vi) V=V0when z=a.











This time we want sinusoidal fuctions in xand y, exponential in z:

X(x)=Asin(kx)+Bcos(kx),Y(y)=Csin(ly)+Dcos(ly),Z(z)=Eepk2+l2z+Gepk2+l2z.

(i))B= 0; (ii))k=n⇡/a; (iii))D= 0; (iv))l=m⇡/a; (v))E+G=0.Therefore

Z(z)=2Esinh(⇡n2+m2z/a).

Putting this all together, and combining the constants, we have:

V(x, y, z)= 1



n=1



m=1

Cn,m sin(n⇡x/a) sin(m⇡y/a) sinh(⇡n2+m2z/a).

It remains to evaluate the constants Cn,m, by imposing boundary condition (vi):

V0=Cn,m sinh(⇡n2+m2)sin(n⇡x/a) sin(m⇡y/a).

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CHAPTER 3. POTENTIAL 61

According to Eqs. 3.50 and 3.51:

Cn,m sinh ⇡n2+m2=2

a2



sin(n⇡x/a) sin(m⇡y/a)dx dy =0,if nor mis even,

16V0

⇡2nm,if both are odd.

Therefore

V(x, y, z)=16V0

⇡2

n=1,3,5,... 

m=1,3,5,...

nm sin(n⇡x/a) sin(m⇡y/a)sinh ⇡pn2+m2z/a

sinh ⇡pn2+m2.

Consider the superposition of six such cubes, one with V0on each of the six faces. The result is a cube

with V0on its entire surface, so the potential at the center is V0. Evidently the potential at the center of the

original cube (with V0on just one face) is one sixth of this: V0/6.To check it, put in x=y=z=a/2:

V(a/2, a/2, a/2) = 16V0

⇡2

n=1,3,5,... 

m=1,3,5,...

nm sin(n⇡/2) sin(m⇡/2)sinh ⇡pn2+m2/2

sinh ⇡pn2+m2.

Let n⌘2i+ 1, m⌘2j+ 1, and note that sinh(2u) = 2 sinh(u) cosh(u). The double sum is then

S=1



i=0



j=0

(1)i+j

(2i+ 1)(2j+ 1)sech ⇡(2i+ 1)2+ (2j+ 1)2/2.

Setting the upper limits at i=3,j= 3 (or above) Mathematica returns S=0.102808, which (to 6 digits) is

equal to ⇡2/96, conﬁrming (at least, numerically) that V(a/2, a/2, a/2) = V0/6.

Problem 3.17

P3(x)= 1

8·6

dx3x213=1

dx23x2122x=1

dx2xx212

dx x212+2xx212x=1

dx x21x21+4x2

dx x215x21=1

82x5x21+x2110x

45x3x+5x35x=1

410x36x=5

2x33

2x.

We need to show that P3(cos ✓) satisﬁes

sin ✓

d✓sin ✓dP

d✓=l(l+ 1)P, with l=3,

where P3(cos ✓)=1

2cos ✓5 cos2✓3.

dP3

d✓=1

2sin ✓5 cos2✓3+ cos ✓(10 cos ✓(sin ✓)=1

2sin ✓5 cos2✓3 + 10 cos2✓

=3

2sin ✓5 cos2✓1.

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62 CHAPTER 3. POTENTIAL

@✓ sin ✓dP3

d✓=3

d✓sin2✓5 cos2✓1=3

22 sin ✓cos ✓5 cos2✓1+ sin2✓(10 cos ✓sin ✓)

=3 sin ✓cos ✓5 cos2✓15 sin2✓.

sin ✓

d✓sin ✓dP

d✓=3 cos ✓5 cos2151cos2✓=3 cos ✓10 cos2✓6

=3·4·1

2cos ✓5 cos2✓3=l(l+ 1)P3.qed



1

P1(x)P3(x)dx =



1

(x)1

25x33xdx =1

2x5x31

1=1

2(1 1+11) = 0.X

Problem 3.18

(a) Inside: V(r, ✓)= 1



l=0

AlrlPl(cos ✓) (Eq.3.66) where

Al=(2l+ 1)

2Rl

⇡



V0(✓)Pl(cos ✓) sin ✓d✓(Eq.3.69).

In this case V0(✓)=V0comes outside the integral, so

Al=(2l+ 1)V0

2Rl

⇡



Pl(cos ✓) sin ✓d✓.

But P0(cos ✓) = 1, so the integral can be written

⇡



P0(cos ✓)Pl(cos ✓) sin ✓d✓=0,if l6=0

2,if l=0(Eq.3.68).

Therefore

Al=0,if l6=0

V0,if l=0.

Plugging this into the general form:

V(r, ✓)=A0r0P0(cos ✓)= V0.

The potential is constant throughout the sphere.

Outside:V(r, ✓)= 1



l=0

rl+1 Pl(cos ✓) (Eq. 3.72), where

Bl=(2l+ 1)

2Rl+1

⇡



V0(✓)Pl(cos ✓) sin ✓d✓(Eq.3.73).

=(2l+ 1)

2Rl+1V0

⇡



Pl(cos ✓) sin ✓d✓=0,if l6=0

RV0,if l=0.

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CHAPTER 3. POTENTIAL 63

Therefore V(r, ✓)=V0

r(i.e. equals V0at r=R, then falls o↵like 1

r).

(b)

V(r, ✓)=











l=0

AlrlPl(cos ✓),for rR(Eq.3.78)



l=0

rl+1 Pl(cos ✓),for rR(Eq.3.79)











where

Bl=R2l+1Al(Eq.3.81)

and

Al=1

2✏0Rl1

⇡



0(✓)Pl(cos ✓) sin ✓d✓(Eq.3.84)

2✏0Rl10

⇡



Pl(cos ✓) sin ✓d✓=0,if l6=0

R0/✏0,if l=0.

Therefore

V(r, ✓)=









R0

✏0

,for rR

R20

✏0

r,for rR











Note: in terms of the total charge Q=4⇡R20,

V(r, ✓)=









4⇡✏0

R,for rR

4⇡✏0

r,for rR











Problem 3.19

V0(✓)=kcos(3✓)=k4 cos3✓3 cos ✓=k[↵P3(cos ✓)+P1(cos ✓)] .

(I know that any 3rd order polynomial can be expressed as a linear combination of the ﬁrst four Legendre

polynomials; in this case, since the polynomial is odd, I only need P1and P3.)

4 cos3✓3 cos ✓=↵1

25 cos3✓3 cos ✓+cos ✓=5↵

2cos3✓+3

2↵cos ✓,

4=5↵

2)↵=8

5;3=3

2↵=3

2·8

5=12

5)=12

53=3

Therefore

V0(✓)=k

5[8P3(cos ✓)3P1(cos ✓)] .

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64 CHAPTER 3. POTENTIAL

Now

V(r, ✓)=











l=0

AlrlPl(cos ✓),for rR(Eq.3.66)



l=0

rl+1 Pl(cos ✓),for rR(Eq.3.72)











where

Al=(2l+ 1)

2Rl

⇡



V0(✓)Pl(cos ✓) sin ✓d✓(Eq.3.69)

=(2l+ 1)

2Rl

5



8

⇡



P3(cos ✓)Pl(cos ✓) sin ✓d✓3

⇡



P1(cos ✓)Pl(cos ✓) sin ✓d✓





(2l+ 1)

2Rl82

(2l+ 1) l332

(2l+ 1) l1=k

Rl[8 l33l1]

=8k/5R3,if l=3

3k/5R, if l=1(zero otherwise).

Therefore

V(r, ✓)=3k

5RrP1(cos ✓)+ 8k

5R3r3P3(cos ✓)= k

58r

R3P3(cos ✓)3r

RP1(cos ✓),

58r

R31

25 cos3✓3 cos ✓3r

Rcos ✓)V(r, ✓)=k

Rcos ✓4r

R25 cos2✓33

(for rR). Meanwhile, Bl=AlR2l+1 (Eq. 3.81—this follows from the continuity of Vat R). Therefore

Bl=8kR4/5,if l=3

3kR2/5,if l=1(zero otherwise).

V(r, ✓)=3kR2

r2P1(cos ✓)+8kR4

r4P3(cos ✓)= k

58R

r4

P3(cos ✓)3R

r2

P1(cos ✓),

V(r, ✓)=k

5R

r2

cos ✓4R

r25 cos2✓33

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CHAPTER 3. POTENTIAL 65

(for rR). Finally, using Eq. 3.83:

(✓)=✏0



l=0

(2l+ 1)AlRl1Pl(cos ✓)=✏03A1P1+7A3R2P3

=✏033k

5RP1+78k

5R3R2P3=✏0k

5R[9P1(cos ✓) + 56P3(cos ✓)]

=✏0k

5R9 cos ✓+56

25 cos3✓3 cos ✓=✏0k

5Rcos ✓[9 + 28 ·5 cos2✓28 ·3]

=✏0k

5Rcos ✓140 cos2✓93.

Problem 3.20

Use Eq. 3.83: (✓)=✏0



l=0

(2l+1)AlRl1Pl(cos ✓).But Eq. 3.69 says: Al=2l+1

2Rl

⇡



V0(✓)Pl(cos ✓) sin ✓d✓.

Putting them together:

(✓)= ✏0



l=0

(2l+ 1)2ClPl(cos ✓),with Cl=

⇡



V0(✓)Pl(cos ✓) sin ✓d✓.qed

Problem 3.21

Set V= 0 on the equatorial plane, far from the sphere. Then the potential is the same as Ex. 3.8 plus the

potential of a uniformly charged spherical shell:

V(r, ✓)=E0rR3

r2cos ✓+1

4⇡✏0

Problem 3.22

(a) V(r, ✓)= 1



l=0

rl+1 Pl(cos ✓)(r>R),so V(r, 0) = 1



l=0

rl+1 Pl(1) = 1



l=0

rl+1 =

2✏0r2+R2r.

Since r>Rin this region, r2+R2=r1+(R/r)2=r1+ 1

2(R/r)21

8(R/r)4+...

, so



l=0

rl+1 =

2✏0

r1+1

r21

r4+...1=

2✏0R2

2rR4

8r3+...

.

Comparing like powers of r, I see that B0=R2

4✏0

1=0,B

2=R4

16✏0

,.... Therefore

V(r, ✓)=R2

4✏01

rR2

4r3P2(cos ✓)+...

,

=R2

4✏0r11

8R

r23 cos2✓1+...

,

(for r>R).

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66 CHAPTER 3. POTENTIAL

(b) V(r, ✓)= 1



l=0

AlrlPl(cos ✓)(r<R). In the northern hemispere, 0 ✓⇡/2,

V(r, 0) = 1



l=0

Alrl=

2✏0r2+R2r.

Since r<Rin this region, r2+R2=R1+(r/R)2=R1+ 1

2(r/R)21

8(r/R)4+...

.Therefore



l=0

Alrl=

2✏0R+1

R1

R3+...r.

Comparing like powers: A0=

2✏0

R, A1=

2✏0

2=

4✏0R,..., so

V(r, ✓)= 

2✏0RrP1(cos ✓)+ 1

2Rr2P2(cos ✓)+...

,

=R

2✏01r

Rcos ✓+1

4r

R23 cos2✓1+...

,

(for r<R, northern hemisphere).

In the southern hemisphere we’ll have to go for ✓=⇡, using Pl(1) = (1)l.

V(r, ⇡)= 1



l=0

(1)lAlrl=

2✏0r2+R2r.

(I put an overbar on Alto distinguish it from the northern Al). The only di↵erence is the sign of A1:

A1= +(/2✏0), A0=A0, A2=A2. So:

V(r, ✓)= 

2✏0R+rP1(cos ✓)+ 1

2Rr2P2(cos ✓)+...

,

=R

2✏01+r

Rcos ✓+1

4r

R23 cos2✓1+...

,

(for r<R, southern hemisphere).

Problem 3.23

V(r, ✓)=











l=0

AlrlPl(cos ✓),(rR) (Eq.3.78),



l=0

rl+1 Pl(cos ✓),(rR) (Eq.3.79),











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CHAPTER 3. POTENTIAL 67

where Bl=AlR2l+1 (Eq. 3.81) and

Al=1

2✏0Rl1

⇡



0(✓)Pl(cos ✓) sin ✓d✓(Eq.3.84)

2✏0Rl10









⇡/2



Pl(cos ✓) sin ✓d✓

⇡



⇡/2

Pl(cos ✓) sin ✓d✓







(let x= cos ✓)

=0

2✏0Rl1







Pl(x)dx 



1

Pl(x)dx



.

Now Pl(x)=(1)lPl(x), since Pl(x) is even, for even l, and odd, for odd l. Therefore



1

Pl(x)dx =



Pl(x)d(x)=(1)l



Pl(x)dx,

and hence

Al=0

2✏0Rl11(1)l1



Pl(x)dx =









0,if lis even

0

✏0Rl1



Pl(x)dx, if lis odd 









So A0=A2=A4=A6= 0, and all we need are A1,A3, and A5.



P1(x)dx =



xdx=x2

2



P3(x)dx =1



05x33xdx =1

25x4

43x2

2

25

43

2=1



P5(x)dx =1



063x570x3+ 15xdx =1

863x6

670x4

4+ 15x2

2

821

235

2+15

2=1

16(36 35) = 1

16.

Therefore

A1=0

✏01

2;A3=0

✏0R21

8;A5=0

✏0R41

16; etc.

and

B1=0

✏0

R31

2;B3=0

✏0

R51

8;B5=0

✏0

R71

16; etc.

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68 CHAPTER 3. POTENTIAL

Thus

V(r, ✓)=









0r

2✏0P1(cos ✓)1

4r

R2P3(cos ✓)+1

8r

R4P5(cos ✓)+...,(rR),

0R3

2✏0r2P1(cos ✓)1

4R

r2

P3(cos ✓)+1

8R

r4

P5(cos ✓)+...,(rR).











Problem 3.24

@ss@V

@s+1

@2V

@2=0.

Look for solutions of the form V(s, )=S(s)():

sd

ds sdS

ds +1

s2Sd2

d2=0.

Multiply by s2and divide by V=S:

ds sdS

ds +1



d2

d2=0.

Since the ﬁrst term involves sonly, and the second only, each is a constant:

ds sdS

ds =C1,1



d2

d2=C2,with C1+C2=0.

Now C2must be negative (else we get exponentials for , which do not return to their original value—as

geometrically they must— when is increased by 2⇡).

C2=k2.Then d2

d2=k2)=Acos k+Bsin k.

Moreover, since (+2⇡)=(), kmust be an integer: k=0,1,2,3,... (negative integers are just repeats,

but k= 0 must be included, since =A(a constant) is OK).

ds sdS

ds =k2Scan be solved by S=sn, provided nis chosen right:

ds snsn1=ns d

ds (sn)=n2ssn1=n2sn=k2S)n=±k.

Evidently the general solution is S(s)=Csk+Dsk,unless k= 0, in which case we have only one solution

to a second-order equation—namely, S= constant. So we must treat k= 0 separately. One solution is a

constant—but what’s the other? Go back to the diferential equation for S, and put in k= 0:

ds sdS

ds =0)sdS

ds = constant = C)dS

ds =C

s)dS =Cds

s)S=Cln s+D(another constant).

So the second solution in this case is ln s. [How about ? That too reduces to a single solution, =A, in the

case k= 0. What’s the second solution here? Well, putting k= 0 into the equation:

d2

d2=0)d

d= constant = B)=B+A.

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CHAPTER 3. POTENTIAL 69

But a term of the form Bis unacceptable, since it does not return to its initial value when is augmented

by 2⇡.] Conclusion: The general solution with cylindrical symmetry is

V(s, )=a0+b0ln s+1



k=1 sk(akcos k+bksin k)+sk(ckcos k+dksin k).

Yes: the potential of a line charge goes like ln s, which is included.

Problem 3.25

Picking V= 0 on the yz plane, with E0in the xdirection, we have (Eq. 3.74):

(i) V=0,when s=R,

(ii) V!E0x=E0scos ,for sR. 

Contents

1 Special Techniques 2

Problem 3.24

Picking V= 0 on the yz plane, with E0in the xdirection,

!(i) V= 0,when s=R,

(ii) V→ −E0x=−E0scos φ,for s≫R. "✲x

✻

✢

−

=⇒E0

Evidently a0=b0=bk=dk= 0, and ak=ck= 0 except for k= 1:

V(s, φ) = #a1s+c1

s$cos φ.

(i)⇒c1=−a1R2; (ii)→a1=−E0. Therefore

V(s, φ) = %−E0s+E0R2

s&cos φ,or V(s, φ) = −E0s'%R

s&2

−1(cos φ.

σ=−ϵ0

∂V

∂s))))s=R

=−ϵ0E0%−R2

s2−1&cos φ))))s=R

=2ϵ0E0cos φ.

Problem 3.25

Inside: V(s, φ) = a0+

∞

k=1

sk(akcos kφ+bksin kφ).(In this region ln sand s−kare no good—they blow

up at s= 0.)

Outside: V(s, φ) = a0+

∞

k=1

sk(ckcos kφ+dksin kφ). (Here ln sand skare no good at s→ ∞).

σ=−ϵ0%∂Vout

∂s−∂Vin

∂s&))))s=R

(Eq.2.36).

Thus

asin 5φ=−ϵ0

∞

k=1 !−k

Rk+1 (ckcos kφ+dksin kφ)−kRk−1(akcos kφ+bksin kφ)".

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Evidently a0=b0=bk=dk= 0, and ak=ck= 0 except for k= 1:

V(s, )=a1s+c1

scos .

(i))c1=a1R2; (ii)!a1=E0. Therefore

V(s, )=E0s+E0R2

scos ,or V(s, )=E0sR

s2

1cos .

=✏0

@ss=R

=✏0E0R2

s21cos s=R

= 2✏0E0cos .

Problem 3.26

Inside: V(s, )=a0+1



k=1

sk(akcos k+bksin k).(In this region ln sand skare no good—they blow

up at s= 0.)

Outside: V(s, )=a0+1



k=1

sk(ckcos k+dksin k). (Here ln sand skare no good at s!1).

=✏0@Vout

@s@Vin

@ss=R

(Eq.2.36).

Thus

asin 5=✏0



k=1 k

Rk+1 (ckcos k+dksin k)kRk1(akcos k+bksin k).

Evidently ak=ck= 0; bk=dk= 0 except k= 5; a=5✏01

R6d5+R4b5. Also, Vis continuous at s=R:

a0+R5b5sin 5=a0+1

R5d5sin 5.Soa0=a0(might as well choose both zero); R5b5=R5d5, or d5=R10b5.

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70 CHAPTER 3. POTENTIAL

Combining these results: a=5✏0R4b5+R4b5= 10✏0R4b5;b5=a

10✏0R4;d5=aR6

10✏0

. Therefore

V(s, )=asin 5

10✏0s5/R4,for s<R,

R6/s5,for s>R.

Problem 3.27 Since ris on the zaxis, the angle ↵is just the polar angle ✓(I’ll drop the primes, for simplicity).

Monopole term: ⇢d⌧=kR 1

r2(R2r) sin ✓r2sin ✓dr d✓d.

But the rintegral is



(R2r)dr =Rr r2R

0=R2R2=0.

So the monopole term is zero.

Dipole term: rcos ✓⇢ d⌧=kR (rcos ✓)1

r2(R2r) sin ✓r2sin ✓dr d✓d.

But the ✓integral is ⇡



sin2✓cos ✓d✓=sin3✓

3

⇡

3(0 0) = 0.

So the dipole contribution is likewise zero.

Quadrupole term:

r23

2cos2✓1

2⇢d⌧=1

2kR r23 cos2✓11

r2(R2r) sin ✓r2sin ✓dr d✓d.

rintegral: R

r2(R2r)dr =r3

3Rr4

2

=R4

3R4

2=R4

✓integral: ⇡



03 cos2✓1

  

3(1sin2✓)1=23 sin2✓

sin2✓d✓=2

⇡



sin2✓d✓3

⇡



sin4✓d✓

⇡

233⇡

8=⇡19

8=⇡

integral:

2⇡



d=2⇡.

The whole integral is:

2kR R4

6⇡

8(2⇡)=k⇡2R5

48 .

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CHAPTER 3. POTENTIAL 71

For point Pon the zaxis (r!zin Eq. 3.95) the approximate potential is

V(z)⇠

4⇡✏0

k⇡2R5

48z3.(Quadrupole.)

Problem 3.28

For a line charge, ⇢(r0)d⌧0!(r0)dl0, which in this case becomes Rd0.

r=rsin ✓cos ˆx +rsin ✓sin ˆy +rcos ✓ˆz ,

r0=Rcos 0+Rsin 0,so

r·r0=rR sin ✓cos cos 0+rR sin ✓sin sin 0=rR cos ↵,

cos ↵= sin ✓(cos cos 0+ sin sin 0).

n=0: ⇢(r0)d⌧0!R2⇡

d0=2⇡R;V0=1

4⇡✏0

2⇡R

r=

2✏0

n=1:

r0cos ↵⇢(r0)d⌧0!Rcos ↵Rd0=R2sin ✓2⇡

(cos cos 0+ sin sin 0)d0=0;V1= 0.

n=2:

(r0)2P2(cos ↵)⇢(r0)d⌧0!R23

2cos2↵1

2Rd0=R3

23 sin2✓(cos cos 0+ sin sin 0)21d0

=R3

23 sin2✓cos22⇡

cos20d0+ sin22⇡

sin20d0+ 2 sin cos 2⇡

sin 0cos 0d02⇡

d0

=R3

23 sin2✓⇡cos2+⇡sin2+0

2⇡=⇡R3

23 sin2✓2=⇡R33

2cos2✓1

2.

V2=

8✏0

r33 cos2✓1=

4✏0

r3P2(cos ✓).

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72 CHAPTER 3. POTENTIAL

Problem 3.29

p=(3qa qa)ˆz +(2qa 2q(a)) ˆy =2qaˆz . Therefore

V⇠

4⇡✏0

p·ˆr

r2,

and p·ˆr =2qaˆz ·ˆr =2qa cos ✓,so

V⇠

4⇡✏0

2qa cos ✓

r2.(Dipole.)

Problem 3.30

(a) By symmetry, pis clearly in the zdirection: p=pˆz ;p=z⇢d⌧)zda.

p=(Rcos ✓)(kcos ✓)R2sin ✓d✓d=2⇡R3k

⇡



cos2✓sin ✓d✓=2⇡R3kcos3✓

3

⇡

3⇡R3k[1 (1)] = 4⇡R3k

3;p=4⇡R3k

3ˆz .

(b)

V⇠

4⇡✏0

4⇡R3k

cos ✓

r2=kR3

3✏0

cos ✓

r2.(Dipole.)

This is also the exact potential. Conclusion: all multiple moments of this distribution (except the dipole) are

exactly zero.

Problem 3.31

Using Eq. 3.94 with r0=d/2 and ↵=✓(Fig. 3.26):



n=0 d

2rn

Pn(cos ✓);

for

, we let ✓!180+✓, so cos ✓!cos ✓:





n=0 d

2rn

Pn(cos ✓).

But Pn(x)=(1)nPn(x), so

V=1

4⇡✏0

q1

+1

=1

4⇡✏0



n=0 d

2rn

[Pn(cos ✓)Pn(cos ✓)] = 2q

4⇡✏0r

n=1,3,5,... d

2rn

Pn(cos ✓).

Therefore

Vdip =2q

4⇡✏0

2rP1(cos ✓)=qd cos ✓

4⇡✏0r2,while Vquad =0.

Voct =2q

4⇡✏0rd

2r3

P3(cos ✓)= 2q

4⇡✏0

8r4

25 cos3✓3 cos ✓=qd3

4⇡✏0

8r45 cos3✓3 cos ✓.

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CHAPTER 3. POTENTIAL 73

Problem 3.32

(a) (i) Q= 2q, (ii) p= 3qaˆz ,(iii) V⇠

4⇡✏0Q

r+p·ˆr

r2=1

4⇡✏02q

r+3qa cos ✓

r2.

(b) (i) Q= 2q, (ii) p=qaˆz ,(iii) V⇠

4⇡✏02q

r+qa cos ✓

r2.

4⇡✏02q

r+3qa sin ✓sin 

r2(from Eq. 1.64, ˆy ·ˆr = sin ✓sin ).

Problem 3.33

(a) This point is at r=a,✓=⇡

2,= 0, so E=p

4⇡✏0a3ˆ

✓=p

4⇡✏0a3(ˆz ); F=qE=pq

4⇡✏0a3ˆz .

(b) Here r=a, ✓= 0, so E=p

4⇡✏0a3(2ˆr )= 2p

4⇡✏0a3ˆz .F=2pq

4⇡✏0a3ˆz .

4⇡✏0a2cos(0) cos ⇡

2=pq

4⇡✏0a2.

Problem 3.34

Q=q, so Vmono =1

4⇡✏0

q

r;p=qaˆz , so Vdip =1

4⇡✏0

qa cos ✓

r2. Therefore

V(r, ✓)⇠

4⇡✏01

r+acos ✓

r2.E(r, ✓)⇠

4⇡✏01

r2ˆr +a

r32 cos ✓ˆr + sin ✓ˆ

✓.

Problem 3.35 The total charge is zero, so the dominant term is the dipole. We need the dipole moment

of this conﬁguration. It obviously points in the zdirection, and for the southern hemisphere (✓:⇡

2!⇡)⇢

switches sign but so does z, so

p=z⇢d⌧=2⇢0⇡/2

✓=0

rcos ✓r2sin ✓dr d✓d=2⇢0(2⇡)R

r3dr ⇡/2

cos ✓sin ✓d✓

=4⇡⇢0

sin2✓

2

⇡/2

=⇡⇢0R4

Therefore (Eq. 3.103)

E⇡⇡⇢0R4

8⇡✏0r32 cos ✓ˆr + sin ✓ˆ

✓.

Problem 3.36

p=(p·ˆr )ˆr +(p·ˆ

✓)ˆ

✓=pcos ✓ˆr psin ✓ˆ

✓(Fig. 3.36). So 3(p·ˆr )ˆr p=3pcos ✓ˆr pcos ✓ˆr +psin ✓ˆ

✓=

2pcos ✓ˆr +psin ✓ˆ

✓.So Eq. 3.104 ⌘Eq. 3.103. X

Problem 3.37

Vave(R)= 1

4⇡R2V(r)da, where the integral is over the surface of a sphere of radius R. Now da =

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74 CHAPTER 3. POTENTIAL

R2sin ✓d✓d, so Vave(R)= 1

4⇡V(R, ✓,) sin ✓d✓d.

dVave

dR =1

4⇡@V

@Rsin ✓d✓d=1

4⇡(rV·ˆr ) sin ✓d✓d=1

4⇡R2(rV)·(R2sin ✓d✓dˆr )

4⇡R2(rV)·da=1

4⇡R2(r2V)d⌧=0.

(The ﬁnal integral, from the divergence theorem, is over the volume of the sphere, where by assumption the

Laplacian of Vis zero.) So Vave is independent of R—the same for all spheres, regardless of their radius—and

hence (taking the limit as R!0), Vave(R)=V(0). qed

Problem 3.38 At a point (x, y) on the plane the ﬁeld of qis

Eq=1

4⇡✏0

3ˆ

,and

=xˆx +yˆy dˆz ,

so its zcomponent is q

4⇡✏0

(x2+y2+d2)3/2. Meanwhile, the ﬁeld of (just below the surface) is 

2✏0

(Eq. 2.17). (Of course, this is for a uniform surface charge, but as long as we are inﬁnitesimally far away is

e↵ectively uniform.) The total ﬁeld inside the conductor is zero, so

q

4⇡✏0

(x2+y2+d2)3/2

2✏0

=0 )(x, y)=qd

2⇡(x2+y2+d2)3/2.X

Problem 3.39

This integral can also be integrated directly. Let x=u2;dx = 2u du.

√x

√d−xdx = 2

√d

√d−u2du = 2 "−u

2#d−u2+d

2sin−1$u

√d%&''''

√d

=−dsin−1(1) = −dπ

Therefore

t=(d

πd

2=)π2d2

2q216πϵ0m=)2π3d3ϵ0m

q2.

Problem 3.35

✛

x+ + + + + +− − − − − −

F=1

4πϵ0

q2*1

[2(a−x)]2+1

[2a+ 2(a−x)]2+1

[4a+ 2(a−x)]2+...

−1

(2x)2−1

(2a+ 2x)2−1

(4a+ 2x)2−...&

4πϵ0

4"+ 1

(a−x)2+1

(2a−x)2+1

(3a−x)2+. . .,−+1

x2+1

(a+x)2+1

(2a+x)2+...,&.

When a→ ∞ (i.e. a≫x) only the 1

x2term survives: F=−1

4πϵ0

(2x)2!(same as for only one plane—

Eq. 3.12). When x=a/2,

F=1

4πϵ0

4"+ 1

(a/2)2+1

(3a/2)2+1

(5a/2)2+. . .,−+1

(a/2)2+1

(3a/2)2+1

(5a/2)2+. . .,&= 0.!

Problem 3.36

Following Prob. 2.47, we place image line charges −λat y=band +λat y=−b(here yis the horizontal

axis, zvertical).

✲y

✻

☛

−λ λ

+λ

−b

−λ

- ./ 0

a−b

2- ./ 0

a−b

- ./ 0

-./ 0

s2s3s4

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The image conﬁguration is shown in the ﬁgure; the positive image charge forces cancel in pairs. The net

force of the negative image charges is:

F=1

4⇡✏0

q21

[2(ax)]2+1

[2a+ 2(ax)]2+1

[4a+ 2(ax)]2+...

1

(2x)21

(2a+2x)21

(4a+2x)2...



4⇡✏0

4 1

(ax)2+1

(2ax)2+1

(3ax)2+...

1

x2+1

(a+x)2+1

(2a+x)2+...

.

When a!1(i.e. ax) only the 1

x2term survives: F=1

4⇡✏0

(2x)2X(same as for only one plane—

Eq. 3.12). When x=a/2,

F=1

4⇡✏0

4 1

(a/2)2+1

(3a/2)2+1

(5a/2)2+...

1

(a/2)2+1

(3a/2)2+1

(5a/2)2+...

=0.X

Problem 3.40

Following Prob. 2.52, we place image line charges at y=band +at y=b(here yis the horizontal

axis, zvertical).

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CHAPTER 3. POTENTIAL 75

This integral can also be integrated directly. Let x=u2;dx = 2u du.

√x

√d−xdx = 2

√d

√d−u2du = 2 "−u

2#d−u2+d

2sin−1$u

√d%&''''

√d

=−dsin−1(1) = −dπ

Therefore

t=(d

πd

2=)π2d2

2q216πϵ0m=)2π3d3ϵ0m

q2.

Problem 3.35

✛

x+ + + + + +− − − − − −

F=1

4πϵ0

q2*1

[2(a−x)]2+1

[2a+ 2(a−x)]2+1

[4a+ 2(a−x)]2+...

−1

(2x)2−1

(2a+ 2x)2−1

(4a+ 2x)2−...&

4πϵ0

4"+ 1

(a−x)2+1

(2a−x)2+1

(3a−x)2+. . .,−+1

x2+1

(a+x)2+1

(2a+x)2+...,&.

When a→ ∞ (i.e. a≫x) only the 1

x2term survives: F=−1

4πϵ0

(2x)2!(same as for only one plane—

Eq. 3.12). When x=a/2,

F=1

4πϵ0

4"+ 1

(a/2)2+1

(3a/2)2+1

(5a/2)2+. . .,−+1

(a/2)2+1

(3a/2)2+1

(5a/2)2+. . .,&= 0.!

Problem 3.36

Following Prob. 2.47, we place image line charges −λat y=band +λat y=−b(here yis the horizontal

axis, zvertical).

✲y

✻

☛

−λ λ

+λ

−b

−λ

- ./ 0

a−b

2- ./ 0

a−b

- ./ 0

-./ 0

s2s3s4

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In the solution to Prob. 2.52 substitute:

a!ab

2,y

0!a+b

2so ab

22

=a+b

22

R2)b=R2

V=

4⇡✏0ln s2

4+ ln s2

2=

4⇡✏0

ln s2

1s2

4s2

2

=

4⇡✏0

ln [(y+a)2+z2][(yb)2+z2]

[(ya)2+z2][(y+b)2+z2],or,using y=scos ,z=ssin ,

=

4⇡✏0

ln (s2+a2+2as cos )[(as/R)2+R22as cos ]

(a2+a22as cos )[(as/R)2+R2+2as cos ].

Problem 3.41 Same as Problem 3.9, only this time we want q0+q00 =q,soq00 =qq0:

F=q

4⇡✏0q00

a2+q0

(ab)2=q2

4⇡✏0a2+qq0

4⇡✏01

a2+1

(ab)2.

The second term is identical to Problem 3.9, and I’ll just quote the answer from there:

F=q2

4⇡✏0a3aR3(2a2R2)

(a2R2)2.

(a) F=0)a(a2R2)2=R3(2a2R2),or (letting x⌘a/R), x(x21)22x2+ 1 = 0. We want a real

root greater than 1; Mathematica delivers x=(1+p5)/2=1.61803, so a=1.61803R= 5.66311 ˚

(b) Let a0=x0Rbe the minimum value of a. The work necessary is

W=a0

F da =q2

4⇡✏01

a3aR3(2a2R2)

(a2R2)2da =q2

4⇡✏0R1

x01

x2(2x21)

x3(x21)2dx

=q2

4⇡✏0R1+2x02x3

2x2

0(1 x2

0).

Putting in x0=(1+p5)/2, Mathematica says the term in square brackets is 1/2 (this is not an accident; see

footnote 6 on page 127), so W=q2

8⇡✏0R.Numerically,

W=(1.60 ⇥1019)2

8⇡(8.85 ⇥1012)(5.66 ⇥1010)J=2.03 ⇥1019 J= 1.27 eV.

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76 CHAPTER 3. POTENTIAL

Problem 3.42

-b b0

= +

The ﬁrst conﬁguration on the right is precisely Example 3.4, but unfortunately the second conﬁguration is not

the same as Problem 3.15:

We could reconstruct Problem 3.15 with the modiﬁed boundaries, but let’s see if we can’t twist it around by

an astute change of variables. Suppose we let x!y,y!u,a!c,b!a, and V0!V1:

This is closer; making the changes in the solution to Problem 3.15 we have (for this conﬁguration)

V(u, y)=4V1

⇡

n=1,3,5...

sinh(n⇡y/c) sin(n⇡u/c)

nsinh(n⇡a/c).

Now let c!2band u!x+b, and the conﬁguration is just what we want:

-b b0

The potential for this conﬁguration is

V(x, y)=4V1

⇡

n=1,3,5...

sinh(n⇡y/2b) sin(n⇡(x+b)/2b)

nsinh(n⇡a/2b).

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CHAPTER 3. POTENTIAL 77

(If you like, write sin(n⇡(x+b)/2b) as (1)(n1)/2cos(n⇡x/2b).) Combining this with Eq. 3.42,

V(x, y)= 4

⇡

n=1,3,5...

nV0

cosh(n⇡x/a) sin(n⇡y/a)

cosh(n⇡b/a)+V1

sinh(n⇡y/2b) sin(n⇡(x+b)/2b)

sinh(n⇡a/2b).

Here’s a plot of this function, for the case a=b=1,V

0=1/2,V

1=1:

-1.0

-0.5

0.0

0.5

1.0

0.0

0.5

1.0

0.0

0.5

1.0

Problem 3.43

Since the conﬁguration is azimuthally symmetric, V(r, ✓)=Alrl+Bl

rl+1 Pl(cos ✓).

(a) r>b:Al= 0 for all l, since V!0 at 1. Therefore V(r, ✓)=Bl

rl+1 Pl(cos ✓).

a<r<b:V(r, ✓)=Clrl+Dl

rl+1 Pl(cos ✓).r<a:V(r, ✓)=V0.

We need to determine Bl,C

l,D

l,and V0. To do this, invoke boundary conditions as follows: (i) Vis

continuous at a, (ii) Vis continuous at b, (iii) 4@V

@r=1

✏0

(✓) at b.

(ii) )Bl

bl+1 Pl(cos ✓)=Clbl+Dl

bl+1 Pl(cos ✓); Bl

bl+1 =Clbl+Dl

bl+1 )Bl=b2l+1Cl+Dl.(1)

(i) )Clal+Dl

al+1 Pl(cos ✓)=V0;









Clal+Dl

al+1 =0,if l6=0,

C0a0+D0

a1=V0,if l=0;









Dl=a2l+1Cl,l6=0,

D0=aV0aC0.(2)

Putting (2) into (1) gives Bl=b2l+1Cla2l+1Cl,l6=0,B

0=bC0+aV0aC0.Therefore

Bl=b2l+1 a2l+1Cl,l6=0,

B0=(ba)C0+aV0.(10)

(iii) )Bl[(l+ 1)] 1

bl+2 Pl(cos ✓)Cllbl1+Dl(l+ 1)

bl+2 Pl(cos ✓)=k

✏0

P1(cos ✓). So

(l+ 1)

bl+2 BlCllbl1+Dl(l+ 1)

bl+2 =0,if l6= 1;

(l+ 1)BllClb2l+1 +(l+ 1)Dl= 0; (l+ 1)(BlDl)=lb2l+1Cl.

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78 CHAPTER 3. POTENTIAL

B1(+2) 1

b2+C1+D12

b2=k

✏0

,for l= 1; C1+2

b3(B1D1)=k.

Therefore

(l+ 1)(BlDl)+lb2l+1Cl=0,for l6=1,

C1+2

b3(B1D1)= k

✏0

.(3)

Plug (2) and (10) into (3):

For l6= 0 or 1:

(l+1) b2l+1 a2l+1Cl+a2l+1Cl+lb2l+1Cl= 0; (l+1)b2l+1Cl+lb2l+1Cl= 0; (2l+1)Cl=0)Cl=0.

Therefore (10) and (2) )Bl=Cl=Dl= 0 for l>1.

For l= 1: C1+2

b3b3a3C1+a3C1=k;C1+2C1=k)C1=k/3✏0;D1=a3C1)

D1=a3k/3✏0;B1=b3a3C1)B1=b3a3k/3✏0.

For l= 0: B0D0=0)B0=D0)(ba)C0+aV0=aV0aC0,so bC0=0)C0= 0; D0=aV0=B0.

Conclusion: V(r, ✓)=aV0

r+b3a3k

3r2✏0

cos ✓, r b. V (r, ✓)=aV0

r+k

3✏0ra3

r2cos ✓, a rb.

(b)i(✓)=✏0

@ra

=✏0aV0

a2+k

3✏01+2a3

a3cos ✓=✏0V0

a+k

✏0

cos ✓=kcos ✓+V0

✏0

(c)qi=ida =V0✏0

a4⇡a2= 4⇡a✏0V0=Qtot.At large r:V⇡aV0

4⇡✏0

r=1

4⇡✏0

4⇡a✏0V0

r=aV0

r.X

Problem 3.44

dr'

q'=q

q'=p-q

Use multipole expansion (Eq. 3.95): ⇢d⌧0!dr0;

=Q

2a; the r0integral breaks into two pieces:

V(r)= 1

4⇡✏0



n=0

rn+1 





(r0)nPn(cos ✓0)dr0+



(r0)nPn(cos ✓0)dr0

.

In the ﬁrst integral ✓0=✓(see diagram); in the second integral ✓0=⇡✓, so cos ✓0=cos ✓. But Pn(z)=

(1)nPn(z), so the integrals cancel when nis odd, and add when nis even.

V(r)=2 1

4⇡✏0



n=0,2,4,...

rn+1 Pn(cos ✓)



xndx.

The integral is an+1

n+1,so

V=Q

4⇡✏0

r

n=0,2,4,... 1

n+1a

rnPn(cos ✓).

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CHAPTER 3. POTENTIAL 79

Problem 3.45

Use separation of variables in cylindrical coordinates (Prob. 3.24):

V(s, )=a0+b0ln s+1



k=1 sk(akcos k+bksin k)+sk(ckcos k+dksin k).

s<R:V(s, )=1

k=1 sk(akcos k+bksin k) (ln sand skblow up at s= 0);

s>R:V(s, )=1

k=1 sk(ckcos k+dksin k) (ln sand skblow up as s!1).

(We may as well pick constants so V!0 as s!1, and hence a0= 0.) Continuity at s=R)

Rk(akcos k+bksin k)=Rk(ckcos k+dksin k), so ck=R2kak,d

k=R2kbk. Eq. 2.36 says:

@sR+@V

@sR

=1

✏0

. Therefore

k

Rk+1 (ckcos k+dksin k)kRk1(akcos k+bksin k)=1

✏0

,

or: 2kRk1(akcos k+bksin k)=0/✏0(0 <<⇡)

0/✏0(⇡<<2⇡).

Fourier’s trick: multiply by (cos l)dand integrate from 0 to 2⇡, using

2⇡



sin kcos ld= 0;

2⇡



cos kcos ld=0,k6=l

⇡,k=l.

Then

2lRl1⇡al=0

✏0



⇡



cos ld

2⇡



⇡

cos ld

=0

✏0sin l

l

⇡

0sin l

l

2⇡

⇡= 0; al=0.

Multiply by (sin l)dand integrate, using

2⇡

0

sin ksin ld=0,k6=l

⇡,k=l:

2lRl1⇡bl=0

✏0



⇡



sin ld

2⇡



⇡

sin ld

=0

✏0cos l

l

⇡

+cos l

l

2⇡

⇡=0

l✏0

(2 2 cos l⇡)

=0,if lis even

40/l✏0,if lis odd )bl=0,if lis even

20/⇡✏0l2Rl1,if lis odd .

Conclusion:

V(s, )=20R

⇡✏0

k=1,3,5,...

k2sin k(s/R)k(s<R)

(R/s)k(s>R).

Problem 3.46

Use Eq. 3.95, in the form V(r)= 1

4⇡✏0



n=0

Pn(cos ✓)

rn+1 In;In=



a

zn(z)dz.

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80 CHAPTER 3. POTENTIAL

(a) I0=k



a

cos ⇡z

2adz =k2a

⇡sin ⇡z

2a

a

=2ak

⇡sin ⇡

2sin ⇡

2=4ak

⇡. Therefore:

Contents

Problem 3.40

✲z

✻

λ(z)

−a a

✲z

✻

λ(z)

−a a

✲z

✻

λ(z)

−a a

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V(r, ✓)⇠

4⇡✏04ak

⇡1

r.(Monopole.)

(b) I0=0.

I1=k



a

zsin(⇡z/a)dz =ka

⇡2sin ⇡z

aaz

⇡cos ⇡z

a

a

=ka

⇡2[sin(⇡)sin(⇡)] a2

⇡cos(⇡)a2

⇡cos(⇡)=k2a2

⇡;

Contents

Problem 3.40

✲z

✻

λ(z)

−a a

✲z

✻

λ(z)

−a a

✲z

✻

λ(z)

−a a

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V(r, ✓)⇠

4⇡✏02a2k

⇡1

r2cos ✓.(Dipole.)

I2=k



a

z2cos ⇡z

adz =k2zcos(⇡z/a)

(⇡/a)2+(⇡z/a)22

(⇡/a)3sin ⇡z

a

a

=2ka

⇡2[acos(⇡)+acos(⇡)] = 4a3k

⇡2.

Contents

Problem 3.40

✲z

✻

λ(z)

−a a

✲z

✻

λ(z)

−a a

✲z

✻

λ(z)

−a a

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V(r, ✓)⇠

4⇡✏04a3k

⇡21

2r33 cos2✓1.(Quadrupole.)

Problem 3.47

(a) The average ﬁeld due to a point charge qat ris

Eave =1

4

3⇡R3Ed⌧,where E=1

4⇡✏0

2ˆ

so Eave =1

4

3⇡R31

4⇡✏0qˆ

2d⌧.

Problem 3.41

✮

dτ

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(Here ris the source point, d⌧is the ﬁeld point, so

goes from rto d⌧.) The ﬁeld at rdue to uniform

charge ⇢over the sphere is E⇢=1

4⇡✏0⇢ˆ

2d⌧.This time d⌧is the source point and ris the ﬁeld

point, so

goes from d⌧to r, and hence carries the opposite sign. So with ⇢=q/ 4

3⇡R3, the two

expressions agree: Eave =E⇢.

(b) From Prob. 2.12:

E⇢=1

3✏0

⇢r=q

4⇡✏0

R3=p

4⇡✏0R3.

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CHAPTER 3. POTENTIAL 81

sum of the individual dipole moments. So Eave =p

4⇡✏0R3.qed

(d) The same argument, only with qplaced at routside the sphere, gives

Eave =E⇢=1

4⇡✏04

3⇡R3⇢

r2ˆr (ﬁeld at rdue to uniformly charged sphere) = 1

4⇡✏0

q

r2ˆr .

But this is precisely the ﬁeld produced by q(at r) at the center of the sphere. So the average ﬁeld (over

the sphere) due to a point charge outside the sphere is the same as the ﬁeld that same charge produces

at the center. And by superposition, this holds for any collection of exterior charges.

Problem 3.48

(a)

Edip =p

4⇡✏0r3(2 cos ✓ˆr + sin ✓ˆ

✓)

4⇡✏0r3[2 cos ✓(sin ✓cos ˆx + sin ✓sin ˆy + cos ✓ˆz )

+ sin ✓(cos ✓cos ˆx + cos ✓sin ˆy sin ✓ˆz )]

4⇡✏0r3



3 sin ✓cos ✓cos ˆx + 3 sin ✓cos ✓sin ˆy +2 cos2✓sin2✓

  

=3 cos2✓1

ˆz 



.

Eave =1

4

3⇡R3Edip d⌧

4

3⇡R3p

4⇡✏01

r33 sin ✓cos ✓(cos ˆx + sin ˆy )+3 cos2✓1ˆz r2sin ✓dr d✓d.

But

2⇡



cos d=

2⇡



sin d= 0, so the ˆx and ˆy terms drop out, and

2⇡

0

d=2⇡,so

Eave =1

4

3⇡R3p

4⇡✏02⇡



rdr

⇡



03 cos2✓1sin ✓d✓

  

(cos3✓+cos ✓)|⇡

0=11+11=0

Evidently Eave =0,which contradicts the result of Prob. 3.47. [Note, however, that the rintegral, R

rdr,

blows up, since ln r!1as r!0. If, as suggested, we truncate the rintegral at r=✏, then it is ﬁnite, and

the ✓integral gives Eave =0.]

(b) We want Ewithin the ✏-sphere to be a delta function: E=A3(r), with Aselected so that the average

ﬁeld is consistent with the general theorem in Prob. 3.47:

Eave =1

4

3⇡R3A3(r)d⌧=A

4

3⇡R3=p

4⇡✏0R3)A=p

3✏0

,and hence E=p

3✏0

3(r).

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82 CHAPTER 3. POTENTIAL

Problem 3.49 We need to show that the ﬁeld inside the sphere approaches a delta-function with the right

coeﬃcient (Eq. 3.106) in the limit as R!0. From Eq. 3.86, the potential inside is

V=k

3✏0

rcos ✓=k

3✏0

z, so E=rV=k

3✏0

ˆz .

From Prob. 3.30, the dipole moment of this conﬁguration is p= (4⇡R3k/3) ˆz , so kˆz =3p/(4⇡R3), and hence

the ﬁeld inside is

E=1

4⇡✏0R3p.

Clearly E!1as R!0 (if pis held constant); its volume integral is

Ed⌧=1

4⇡✏0R3p4

3⇡R3=1

3✏0

which matches the delta-function term in Eq. 3.106. X

Problem 3.50

(a) I=(rV1)·(rV2)d⌧.Butr·(V1rV2)=(rV1)·(rV2)+V1(r2V2), so

I=r·(V1rV2)d⌧V1(r2V2)=S

V1(rV2)·da+1

✏0V1⇢2d⌧.

But the surface integral is over a huge sphere “at inﬁnity”, where V1and V2!0. So I=1

✏0V1⇢2d⌧. By

the same argument, with 1 and 2 reversed, I=1

✏0V2⇢1d⌧. So V1⇢2d⌧=V2⇢1d⌧. qed

(b) 





Situation (1):Qa=a⇢1d⌧=Q;Qb=b⇢1d⌧= 0; V1b⌘Vab.

Situation (2):Qa=a⇢2d⌧= 0; Qb=b⇢2d⌧=Q;V2a⌘Vba.









V1⇢2d⌧=V1aa⇢2d⌧+V1bb⇢2d⌧=VabQ.

V2⇢1d⌧=V2aa⇢1d⌧+V2bb⇢1d⌧=VbaQ. 





Green’s reciprocity theorem says QVab =QVba,so Vab =Vba.qed

Problem 3.51

Problem 3.41

✮

dτ

✲

V=0 V=0

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(a) Situation (1): actual. Situation (2): right plate at V0, left plate at V= 0, no charge at x.

V1⇢2d⌧=Vl1Ql2+Vx1Qx2+Vr1Qr2.

But Vl1=Vr1= 0 and Qx2= 0, so V1⇢2d⌧= 0.

V2⇢1d⌧=Vl2Ql1+Vx2Qx1+Vr2Qr1.

But Vl2=0Qx1=q, Vr2=V0,Q

r1=Q2, and Vx2=V0(x/d). So 0 = V0(x/d)q+V0Q2, and hence

Q2=qx/d.

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CHAPTER 3. POTENTIAL 83

Situation (1): actual. Situation (2): left plate at V0, right plate at V= 0, no charge at x.

V1⇢2d⌧=0=V2⇢1d⌧=Vl2Ql1+Vx2Qx1+Vr2Qr1=V0Q1+qVx2+0.

But Vx2=V01x

d,so

Q1=q(1 x/d).

(b) Situation (1): actual. Situation (2): inner sphere at V0, outer sphere at zero, no charge at r.

V1⇢2d⌧=Va1Qa2+Vr1Qr2+Vb1Qb2.

But Va1=Vb1=0,Q

r2= 0. So V1⇢2d⌧= 0.

V2⇢1d⌧=Va2Qa1+Vr2Qr1+Vb2Qb1=QaV0+qVr2+0.

But Vr2is the potential at rin conﬁguration 2: V(r)=A+B/r, with V(a)=V0)A+B/a =V0, or

aA +B=aV0, and V(b)=0)A+B/b = 0, or bA +B= 0. Subtract: (ba)A=aV0)A=

aV0/(ba); B1

a1

b=V0=B(ba)

ab )B=abV0/(ba). So V(r)= aV0

(ba)b

r1. Therefore

QaV0+qaV0

(ba)b

r1= 0; Qa=qa

(ba)b

r1.

Now let Situation (2) be: inner sphere at zero, outer at V0, no charge at r.

V1⇢2d⌧=0=V2⇢1d⌧=Va2Qa1+Vr2Qr1+Vb2Qb1=0+qVr2+QbV0.

This time V(r)=A+B

rwith V(a)=0)A+B/a =0; V(b)=V0)A+B/b =V0, so

V(r)= bV0

(ba)1a

r. Therefore qbV0

(ba)1a

r+QbV0= 0; Qb=qb

(ba)1a

r.

Problem 3.52

(a)



i,j=1

ˆr iˆr jQij =1

2



3



i=1

ˆr ir0



j=1

ˆr jr0

j(r0)2

i,j

ˆr iˆr jij 



⇢d⌧0

But



i=1

ˆr ir0

i=ˆr ·r0=r0cos ↵=



j=1

ˆr jr0

j;

i,j

ˆr iˆr jij =ˆr jˆr j=ˆr ·ˆr = 1. So

Vquad =1

4⇡✏0

r31

23r02cos2↵r02⇢d⌧0= the third term in Eq. 3.96. X

(b) Because x2=y2=(a/2)2for all four charges, Qxx =Qyy =1

23(a/2)2(p2a/2)2(qqq+q)=0.

Because z= 0 for all four charges, Qzz =1

2[(p2a/2)2](qqq+q) = 0 and Qxz =Qyz =Qzx =Qzy =0.

This leaves only

Qxy =Qyx =3

2a

2a

2q+a

2a

2(q)+a

2a

2(q)+a

2a

2q=3

2a2q.

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84 CHAPTER 3. POTENTIAL

(c)

2Qij =3(ridi)(rjdj)(rd)2ij ⇢d⌧(I0ll drop the primes,for simplicity.)

=3rirjr2ij ⇢d⌧3dirj⇢d⌧3djri⇢d⌧+3didj⇢d⌧+2d·r⇢d⌧

d2ij ⇢d⌧=Qij 3(dipj+djpi)+3didjQ+2ij d·pd2ij Q.

So if p= 0 and Q= 0 then Qij =Qij .qed

(d) Eq. 3.95 with n= 3:

Voct =1

4⇡✏0

r4(r0)3P3(cos ↵)⇢d⌧0;P3(cos ✓)=1

25 cos3✓3 cos ✓.

Voct =1

4⇡✏0

r4

i,j,k

ˆr iˆr jˆr kQijk.

Deﬁne the “octopole moment” as

Qijk ⌘1

25r0

ir0

jr0

k(r0)2(r0

ijk +r0

jik +r0

kij )⇢(r0)d⌧0.

Problem 3.53

V=1

4⇡✏0q1

11

2+q01

31

4

1=r2+a22ra cos ✓,

2=r2+a2+2ra cos ✓,

3=r2+b22rb cos ✓,

1=r2+b2+2rb cos ✓.

Problem 3.46

−q+q

−q′

+q′

✯

❃

✒✼▼

!"# $

! "# $

!"#$

!"# $

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Expanding as in Ex. 3.10: 1

11

2⇠

=2r

a2cos ✓(we want ar, not ra, this time).

1

31

4⇠

=2b

r2cos ✓(here we want b⌧r, because b=R2/a, Eq.3.16)

r2cos ✓.

But q0=R

aq(Eq. 3.15), so

V(r, ✓)⇠

4⇡✏0q2r

a2cos ✓R

aq2

r2cos ✓=1

4⇡✏02q

a2rR3

r2cos ✓.

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CHAPTER 3. POTENTIAL 85

Set E0=1

4⇡✏0

a2(ﬁeld in the vicinity of the sphere produced by ±q):

V(r, ✓)=E0rR3

r2cos ✓(agrees with Eq.3.76).

Problem 3.54

The boundary conditions are

(i) V= 0 when y=0,

(ii) V=V0when y=a,

(iii) V= 0 when x=b,

(iv) V= 0 when x=b.











Go back to Eq. 3.26 and examine the case k= 0: d2X/dx2=d2Y/dy2= 0, so X(x)=Ax+B, Y (y)=Cy+D.

But this conﬁguration is symmetric in x, so A= 0, and hence the k= 0 solution is V(x, y)=Cy +D. Pick

D= 0, C=V0/a, and subtract o↵this part:

V(x, y)=V0

a+¯

V(x, y).

The remainder (¯

V(x, y)) satisﬁes boundary conditions similar to Ex. 3.4:

(i) ¯

V= 0 when y=0,

(ii) ¯

V= 0 when y=a,

(iii) ¯

V=V0(y/a) when x=b,

(iv) ¯

V=V0(y/a) when x=b.











(The point of peeling o↵V0(y/a) was to recover (ii), on which the constraint k=n⇡/a depends.)

The solution (following Ex. 3.4) is

V(x, y)= 1



n=1

Cncosh(n⇡x/a) sin(n⇡y/a),

and it remains to ﬁt condition (iii):

V(b, y)=Cncosh(n⇡b/a) sin(n⇡y/a)=V0(y/a).

Invoke Fourier’s trick:

Cncosh(n⇡b/a)a

sin(n⇡y/a) sin(n0⇡y/a)dy =V0

aa

ysin(n0⇡y/a)dy,

2Cncosh(n⇡b/a)=V0

aa

ysin(n⇡y/a)dy.

Cn=2V0

a2cosh(n⇡b/a)a

n⇡2sin(n⇡y/a)ay

n⇡cos(n⇡y/a)

=2V0

a2cosh(n⇡b/a)a2

n⇡cos(n⇡)=2V0

n⇡

(1)n

cosh(n⇡b/a).

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86 CHAPTER 3. POTENTIAL

V(x, y)= V0y

a+2

⇡



n=1

(1)n

cosh(n⇡x/a)

cosh(n⇡b/a)sin(n⇡y/a).

Alternatively, start with the separable solution

V(x, y)=(Csin kx +Dcos kx)Aeky +Beky.

Note that the conﬁguration is symmetric in x,soC= 0, and V(x, 0) = 0 )B=A, so (combining the

constants)

V(x, y)=Acos kx sinh ky.

But V(b, y) = 0, so cos kb = 0, which means that kb =±⇡/2,±3⇡/2,..., or k=(2n1)⇡/2b⌘↵n, with

n=1,2,3,... (negative kdoes not yield a di↵erent solution—the sign can be absorbed into A). The general

linear combination is

V(x, y)= 1



n=1

Ancos ↵nxsinh ↵ny,

and it remains to ﬁt the ﬁnal boundary condition:

V(x, a)=V0=1



n=1

Ancos ↵nxsinh ↵na.

Use Fourier’s trick, multiplying by cos ↵n0xand integrating:

V0b

b

cos ↵n0xdx=1



n=1

Ansinh ↵nab

b

cos ↵n0xcos ↵nxdx,

2 sin ↵n0b

↵n0



n=1

Ansinh ↵na(bn0n)=bAn0sinh ↵n0a.

So An=2V0

sin ↵nb

↵nsinh ↵na.But sin ↵nb= sin 2n1

2⇡=(1)n,so

V(x, y)= 2V0



n=1

(1)nsinh ↵ny

↵nsinh ↵nacos ↵nx.

Problem 3.55

(a) Using Prob. 3.15b (with b=a):

V(x, y)=4V0

⇡

nodd

sinh(n⇡x/a) sin(n⇡y/a)

nsinh(n⇡).

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CHAPTER 3. POTENTIAL 87

(y)=✏0

@xx=0 =✏0

4V0

⇡

nodd n⇡

acosh(n⇡x/a) sin(n⇡y/a)

nsinh(n⇡)x=0

=4✏0V0

a

nodd

sin(n⇡y/a)

sinh(n⇡).

=a

(y)dy =4✏0V0

a

nodd

sinh(n⇡)a

sin(n⇡y/a)dy.

But a

sin(n⇡y/a)dy =a

n⇡cos(n⇡y/a)a

0=a

n⇡[1 cos(n⇡)] = 2a

n⇡(since nis odd).

=8✏0V0

⇡

nodd

nsinh(n⇡)=✏0V0

⇡ln 2.

[Summing the series numerically (using Mathematica) gives 0.0866434, which agrees precisely with ln 2/8.

The series can be summed analytically, by manipulation of elliptic integrals—see “Integrals and Series, Vol. I:

Elementary Functions,” by A. P. Prudnikov, Yu. A. Brychkov, and O. I. Marichev (Gordon and Breach, New

York, 1986), p. 721. I thank Ram Valluri for calling this to my attention.]

Using Prob. 3.54 (with b=a/2):

V(x, y)=V0y

a+2

⇡

(1)ncosh(n⇡x/a) sin(n⇡y/a)

ncosh(n⇡/2) .

(x)=✏0

@yy=0 =✏0V01

a+2

⇡

nn⇡

a(1)ncosh(n⇡x/a) cos(n⇡y/a)

ncosh(n⇡/2) y=0

=✏0V01

a+2

a

(1)ncosh(n⇡x/a)

cosh(n⇡/2) =✏0V0

a1+2

(1)ncosh(n⇡x/a)

cosh(n⇡/2) .

=a/2

a/2

(x)dx =✏0V0

aa+2

(1)n

cosh(n⇡/2) a/2

a/2

cosh(n⇡x/a)dx.

But a/2

a/2

cosh(n⇡x/a)dx =a

n⇡sinh(n⇡x/a)

a/2

a/2

=2a

n⇡sinh(n⇡/2).

=✏0V0

aa+4a

⇡

(1)ntanh(n⇡/2)

n=✏0V01+ 4

⇡

(1)ntanh(n⇡/2)

n

=✏0V0

⇡ln 2.

[The numerical value is -0.612111, which agrees with the expected value (ln 2 ⇡)/4.]

(b) From Prob. 3.24:

V(s, )=a0+b0ln s+1



k=1 aksk+bk

sk[ckcos(k)+dksin(k)].

In the interior (s<R)b0and bkmust be zero (ln sand 1/s blow up at the origin). Symmetry )dk=0.So

V(s, )=a0+1



k=1

akskcos(k).

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88 CHAPTER 3. POTENTIAL

Problem 3.48

✲x

✻

Problem 3.49

✻

❄

❦

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At the surface:

V(R, )=

k=0

akRkcos(k)=V0,if ⇡/4<<⇡/4,

0,otherwise.

Fourier’s trick: multiply by cos(k0) and integrate from ⇡to ⇡:



k=0

akRk⇡

⇡

cos(k) cos(k0)d=V0⇡/4

⇡/4

cos(k0)d=





V0sin(k0)/k0

⇡/4

⇡/4=(V0/k0) sin(k0⇡/4),if k06=0,

V0⇡/2,if k0=0.

But

⇡

⇡

cos(k) cos(k0)d=





0,if k6=k0

2⇡,if k=k0=0,

⇡,if k=k06=0.

So 2⇡a0=V0⇡/2)a0=V0/4; ⇡akRk=(2V0/k) sin(k⇡/4) )ak=(2V0/⇡kRk) sin(k⇡/4) (k6= 0); hence

V(s, )=V01

4+2

⇡



k=1

sin(k⇡/4)

ks

Rkcos(k).

Using Eq. 2.49, and noting that in this case ˆn =ˆs :

()=✏0

@ss=R=✏0V0

⇡



k=1

sin(k⇡/4)

kRkksk1cos(k)s=R=2✏0V0

⇡R



k=1

sin(k⇡/4) cos(k).

We want the net (line) charge on the segment opposite to V0(⇡<<3⇡/4 and 3⇡/4<<⇡):

=()Rd=2R⇡

3⇡/4

()d=4✏0V0

⇡



k=1

sin(k⇡/4) ⇡

3⇡/4

cos(k)d

=4✏0V0

⇡



k=1

sin(k⇡/4) sin(k)

k

⇡

3⇡/4=4✏0V0

⇡



k=1

sin(k⇡/4) sin(3k⇡/4)

ksin(k⇡/4) sin(3k⇡/4) product

11/p21/p2 1/2

2 1 -1 -1

31/p21/p2 1/2

40 0 0

5-1/p2 -1/p2 1/2

6 -1 1 -1

7-1/p2 -1/p2 1/2

80 0 0

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CHAPTER 3. POTENTIAL 89

=4✏0V0

⇡1

2

1,3,5...

k

2,6,10,...

k=4✏0V0

⇡1

2

1,3,5...

k1

2

1,3,5,...

k=0.

Ouch! What went wrong? The problem is that the series (1/k) is divergent, so the “subtraction” 11

is suspect. One way to avoid this is to go back to V(s, ), calculate ✏0(@V/@s) at s6=R, and save the limit

s!Runtil the end:

(,s)⌘✏0

@s=2✏0V0

⇡



k=1

sin(k⇡/4)

ksk1

Rkcos(k)

=2✏0V0

⇡R



k=1

xk1sin(k⇡/4) cos(k) (where x⌘s/R !1 at the end).

(x)⌘(,s)Rd=4✏0V0

⇡



k=1

kxk1sin(k⇡/4) sin(3k⇡/4)

=4✏0V0

⇡1

2xx+x3

3+x5

5+···1

xx2

2+x6

6+x10

10 +···

=2✏0V0

⇡xx+x3

3+x5

5+···x2+x6

3+x10

5+···.

But (see math tables) : ln 1+x

1x=2x+x3

3+x5

5+···.

=2✏0V0

⇡x1

2ln 1+x

1x1

2ln 1+x2

1x2=✏0V0

⇡xln 1+x

1x1+x2

1x2

=✏0V0

⇡xln (1 + x)2

1+x2;= lim

x!1(x)= ✏0V0

⇡ln 2.

Problem 3.56

F=qE=qp

4⇡✏0r3(2 cos ✓ˆr + sin ✓ˆ

✓).

Problem 3.48

✲x

✻

Problem 3.49

✻

❄

❦

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Now consider the pendulum: F=mg ˆz Tˆr , where Tmg cos =mv2/l and (by conservation of

energy) mgl cos = (1/2)mv2)v2=2gl cos (assuming it started from rest at = 90, as stipulated). But

cos =cos ✓,soT=mg(cos ✓)+(m/l)(2gl cos ✓)=3mg cos ✓,and hence

F=mg(cos ✓ˆr sin ✓ˆ

✓)+3mg cos ✓ˆr =mg(2 cos ✓ˆr + sin ✓ˆ

✓).

This total force is such as to keep the pendulum on a circular arc, and it is identical to the force on qin the

ﬁeld of a dipole, with mg $qp/4⇡✏0l3. Evidently qalso executes semicircular motion, as though it were on a

tether of ﬁxed length l.

Problem 3.57 Symmetry suggests that the plane of the orbit is perpendicular to the zaxis, and since we need

a centripetal force, pointing in toward the axis, the orbit must lie at the bottom of the ﬁeld loops (Fig. 3.37a),

where the zcomponent of the ﬁeld is zero. Referring to Eq. 3.104,

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90 CHAPTER 3. POTENTIAL

E·ˆz =0)3(p·ˆr )(ˆr ·ˆz )p·ˆz =0,or 3 cos2✓1=0.So cos2✓=1/3,cos ✓=1/p3,sin ✓=2/3, z/s =

tan ✓)z=p2s. The ﬁeld at the orbit is (Eq. 3.103)

E=p

4⇡✏0r321

p3ˆr +2

3ˆ

✓

4⇡✏0r32

3p2(sin ✓cos ˆx + sin ✓sin ˆy + cos ✓ˆz ) + (cos ✓cos ˆx + cos ✓sin ˆy sin ✓ˆz )

4⇡✏0r32

3p2 sin ✓+ cos ✓cos ˆx +p2 sin ✓+ cos ✓sin ˆy +p2 cos ✓sin ✓ˆz 

4⇡✏0r32

3p22

31

p3(cos ˆx + sin ˆy )+p21

p32

3ˆz 

4⇡✏0r32

3p3 (cos ˆx + sin ˆy )=p

4⇡✏0r3p2ˆs =p

3p3⇡✏0s3ˆs .

(I used s=rsin ✓=r2/3, in the last step.)

The centripetal force is

F=qE =qp

3p3⇡✏0s3=mv2

s)v2=qp

3p3⇡✏0ms2)v=1

sqp

3p3⇡✏0m.

The angular momentum is

L=smv =qpm

3p3⇡✏0

the same for all orbits, regardless of their radius (!), and the energy is

W=1

2mv2+qV =1

3p3⇡✏0s2+q

4⇡✏0

pcos ✓

r2=qp

6p3⇡✏0s2qp

4⇡✏0p3(3/2)s2= 0.

Problem 3.58

Potential of q:Vq(r)= 1

4⇡✏0

, where

2=a2+r22ar cos ✓.

Equation 3.94, with r0!aand ↵!✓:1



n=0 a

rnPn(cos ✓).So

Vq(r, ✓)= q

4⇡✏0



n=0 a

rnPn(cos ✓).

Meanwhile, the potential of is (Eq. 3.79) V(r, ✓)= 1



l=0

rl+1 Pl(cos ✓).

Comparing the two (Vq=V) we see that Bl=(q/4⇡✏0)al, and hence (Eq. 3.81) Al=(q/4⇡✏0)al/R2l+1.Then

(Eq. 3.83)

(✓)= q

4⇡R2



l=0

(2l+ 1) a

RlPl(cos ✓)= q

4⇡R221



l=0

la

RlPl(cos ✓)+ 1



l=0 a

RlPl(cos ✓).

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CHAPTER 3. POTENTIAL 91

Now (second line above, with r!R)

pa2+R22aR cos ✓=1



l=0 a

RlPl(cos ✓).

Di↵erentiating with respect to a:

da 1

pa2+R22aR cos ✓=(aRcos ✓)

(a2+R22aR cos ✓)3/2=1



l=0

la

RlPl(cos ✓).

Thus

(✓)= q

4⇡R22aR (aRcos ✓)

(a2+R22aR cos ✓)3/2+R

(a2+R22aR cos ✓)1/2

4⇡R2a(aRcos ✓)+(a2+R22aR cos ✓)

(a2+R22aR cos ✓)3/2=q

4⇡R

(R2a2)

(a2+R22aR cos ✓)3/2.

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92 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Chapter 4

Electric Fields in Matter

Problem 4.1

E=V/x = 500/103=5⇥105. Table 4.1: ↵/4⇡✏0=0.66⇥1030, so ↵=4⇡(8.85⇥1012)(0.66⇥1030)=

7.34 ⇥1041.p=↵E=ed )d=↵E/e =(7.34 ⇥1041)(5 ⇥105)/(1.6⇥1019)=2.29 ⇥1016 m.

d/R =(2.29 ⇥1016)/(0.5⇥1010)= 4.6⇥106.To ionize, say d=R.ThenR=↵E/e =↵V /ex )V=

Rex/↵=(0.5⇥1010)(1.6⇥1019)(103)/(7.34 ⇥1041)= 108V.

Problem 4.2

First ﬁnd the ﬁeld, at radius r, using Gauss’ law: E·da=1

✏0Qenc, or E=1

4⇡✏0

r2Qenc.

Qenc =r

⇢d⌧=4⇡q

⇡a3r

e2r/a r2dr =4q

a3a

2e2r/a r2+ar +a2

2

=2q

a2e2r/a r2+ar +a2

2a2

2=q1e2r/a 1+2r

a+2r2

a2.

[Note: Qenc(r!1)=q.] So the ﬁeld of the electron cloud is Ee=1

4⇡✏0

r21e2r/a 1+2r

a+2r2

a2. The

proton will be shifted from r= 0 to the point dwhere Ee=E(the external ﬁeld):

E=1

4⇡✏0

d21e2d/a 1+2d

a+2d2

a2.

Expanding in powers of (d/a):

e2d/a =12d

a+1

22d

a2

1

3! 2d

a3

+···=12d

a+2d

a2

4

3d

a3

+···

1e2d/a 1+2d

a+2d2

a2=112d

a+2d

a2

4

3d

a3

+···1+2d

a+2d2

a2

=1/1/2d

/2d2

/+2d

/+4d2

/+4d3

/2d2

/4d3

/+4

a3+···

3d

a3

+ higher order terms.

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 93

E=1

4⇡✏0

d24

a3=1

4⇡✏0

3a3(qd)= 1

3⇡✏0a3p. ↵=3⇡✏0a3.

[Not so di↵erent from the uniform sphere model of Ex. 4.1 (see Eq. 4.2). Note that this result predicts

4⇡✏0↵=3

4a3=3

40.5⇥10103=0.09 ⇥1030 m3, compared with an experimental value (Table 4.1) of

0.66 ⇥1030 m3. Ironically the “classical” formula (Eq. 4.2) is slightly closer to the empirical value.]

Problem 4.3

⇢(r)=Ar. Electric ﬁeld (by Gauss’s Law): E·da=E4⇡r2=1

✏0Qenc =1

✏0r

0Ar 4⇡r2dr, or E=

4⇡r2

4⇡A

✏0

4=Ar2

4✏0

. This “internal” ﬁeld balances the external ﬁeld Ewhen nucleus is “o↵-center” an amount

d:ad2/4✏0=E)d=4✏0E/A. So the induced dipole moment is p=ed =2e✏0/ApE. Evidently

pis proportional to E1/2.

For Eq. 4.1 to hold in the weak-ﬁeld limit, E must be proportional to r, for small r, which means that ⇢

must go to a constant (not zero) at the origin: ⇢(0) 6=0 (nor inﬁnite).

Problem 4.4

Contents

Problem 4.4

Problem 4.6

✸

−

❦

−

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Field of q:1

4⇡✏0

r2ˆr .Induced dipole moment of atom: p=↵E=

↵q

4⇡✏0r2ˆr .

Field of this dipole, at location of q(✓=⇡, in Eq. 3.103): E=1

4⇡✏0

r32↵q

4⇡✏0r2(to the right).

Force on qdue to this ﬁeld: F=2↵q

4⇡✏021

r5(attractive).

Problem 4.5

Field of p1at p2(✓=⇡/2 in Eq. 3.103): E1=p1

4⇡✏0r3ˆ

✓(points down).

Torque on p2:N2=p2⇥E1=p2E1sin 90=p2E1=p1p2

4⇡✏0r3(points into the page).

Field of p2at p1(✓=⇡in Eq. 3.103): E2=p2

4⇡✏0r3(2ˆr ) (points to the right).

Torque on p1:N1=p1⇥E2=2p1p2

4⇡✏0r3(points into the page).

Problem 4.6

(a) (b)

Contents

Problem 4.4

Problem 4.6

✸

−

❦

−

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Hello

θ2z

✻

✲

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Use image dipole as shown in Fig. (a).

Redraw, placing piat the origin, Fig. (b).

Ei=p

4⇡✏0(2z)3(2 cos ✓ˆr + sin ✓ˆ

✓); p=pcos ✓ˆr +psin ✓ˆ

✓.

N=p⇥Ei=p2

4⇡✏0(2z)3(cos ✓ˆr + sin ✓ˆ

✓)⇥(2 cos ✓ˆr + sin ✓ˆ

✓)

=p2

4⇡✏0(2z)3cos ✓sin ✓ˆ

+ 2 sin ✓cos ✓(ˆ

)

=p2sin ✓cos ✓

4⇡✏0(2z)3(ˆ

) (out of the page).

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94 CHAPTER 4. ELECTRIC FIELDS IN MATTER

But sin ✓cos ✓= (1/2) sin 2✓,so N=p2sin 2✓

4⇡✏0(16z3)(out of the page).

For 0 <✓<⇡/2, Ntends to rotate pcounterclockwise; for ⇡/2<✓<⇡,Nrotates pclockwise. Thus the

stable orientation is perpendicular to the surface—either "or #.

Problem 4.7

If the potential is zero at inﬁnity, the energy of a point charge Qis (Eq. 2.39) W=QV (r). For a physical

dipole, with qat rand +qat r+d,

U=qV (r+d)qV (r)=q[V(r+d)V(r)] = qr+d

E·dl.

For an ideal dipole the integral reduces to E·d, and

U=qE·d=p·E,

since p=qd.If you do not (or cannot) use inﬁnity as the reference point, the result still holds, as long as

you bring the two charges in from the same point,r0(or two points at the same potential). In that case

W=Q[V(r)V(r0)], and

U=q[V(r+d)V(r0)] q[V(r)V(r0)] = q[V(r+d)V(r)],

as before.

Problem 4.8

U=p1·E2, but E2=1

4⇡✏0

r3[3 (p2·ˆr )ˆr p2]. So U=1

4⇡✏0

r3[p1·p23(p1·ˆr )(p2·ˆr )]. qed

Problem 4.9

(a) F=(p·r)E(Eq. 4.5); E=1

4⇡✏0

r2ˆr =q

4⇡✏0

xˆx +yˆy +zˆz

(x2+y2+z2)3/2.

Fx=px

@x+py

@y+pz

@zq

4⇡✏0

(x2+y2+z2)3/2

4⇡✏0px1

(x2+y2+z2)3/23

2x2x

(x2+y2+z2)5/2+py3

2x2y

(x2+y2+z2)5/2

+pz3

2x2z

(x2+y2+z2)5/2=q

4⇡✏0px

r33x

r5(pxx+pyy+pzz)=q

4⇡✏0p

r33r(p·r)

r5x

F=1

4⇡✏0

r3[p3(p·ˆr )ˆr ].

(b) E=1

4⇡✏0

r3{3[p·(ˆr )] (ˆr )p}=1

4⇡✏0

r3[3(p·ˆr )ˆr p].(This is from Eq. 3.104; the minus signs

are because rpoints toward p, in this problem.)

F=qE=1

4⇡✏0

r3[3(p·ˆr )ˆr p].

[Note that the forces are equal and opposite, as you would expect from Newton’s third law.]

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 95

Problem 4.10

(a) b=P·ˆn =kR;⇢b=r·P=1

@r(r2kr)=1

r23kr2=3k.

(b) For r<R,E=1

3✏0⇢rˆr (Prob. 2.12), so E=(k/✏0)r.

For r>R, same as if all charge at center; but Qtot =(kR)(4⇡R2)+(3k)( 4

3⇡R3) = 0, so E=0.

Problem 4.11

⇢b= 0; b=P·ˆn =±P(plus sign at one end—the one Ppoints toward; minus sign at the other—the one

Ppoints away from).

(i) La. Then the ends look like point charges, and the whole thing is like a physical dipole, of length Land

charge P⇡a2. See Fig. (a).

(ii) L⌧a. Then it’s like a circular parallel-plate capacitor. Field is nearly uniform inside; nonuniform “fringing

ﬁeld” at the edges. See Fig. (b).

(iii) L⇡a. See Fig. (c).

θz

✻

✲

Problem 4.7

✲

✻

✲✒

✻

Problem 4.11

✛

✛✲

(a) Like a dipole

−+

✛

✛✻

❄

✻

❄

✲

(b) Like a parallel-plate capacitor

−+

✛

✲

(c)

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Problem 4.12

V=1

4⇡✏0P·ˆ

2d⌧=P·1

4⇡✏0ˆ

2d⌧. But the term in curly brackets is precisely the ﬁeld of a uniformly

charged sphere, divided by ⇢. The integral was done explicitly in Probs. 2.7 and 2.8:

4⇡✏0ˆ

2d⌧=1

⇢









4⇡✏0

(4/3)⇡R3⇢

r2ˆr ,(r>R),

4⇡✏0

(4/3)⇡R3⇢

R3r,(r<R).











So V(r, ✓)=









3✏0r2P·ˆr =R3Pcos ✓

3✏0r2,(r>R),

3✏0

P·r=Prcos ✓

3✏0

,(r<R).











Problem 4.13

Think of it as two cylinders of opposite uniform charge density ±⇢.Inside, the ﬁeld at a distance sfrom

the axis of a uniformly charge cylinder is given by Gauss’s law: E2⇡s`=1

✏0⇢⇡s2`)E=(⇢/2✏0)s. For

two such cylinders, one plus and one minus, the net ﬁeld (inside) is E=E++E=(⇢/2✏0)(s+s). But

s+s=d, so E=⇢d/(2✏0), where dis the vector from the negative axis to positive axis. In this case

the total dipole moment of a chunk of length `is P⇡a2`=⇢⇡a2`d. So ⇢d=P, and E=P/(2✏0),for

s<a.

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96 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Outside, Gauss’s law gives E2⇡s`=1

✏0⇢⇡a2`)E=⇢a2

2✏0

ˆs

s, for one cylinder. For the combination, E=

E++E=⇢a2

2✏0ˆs +

s+ˆs 

s, where

s±=s⌥d

s±

=s⌥d

2s2+d2

4⌥s·d1

⇠

s2s⌥d

21⌥s·d

s21⇠

s2s⌥d

21±s·d

s2

s2s±s(s·d)

s2⌥d

2(keeping only 1st order terms in d).

ˆs +

s+ˆs 

s=1

s2s+s(s·d)

s2d

2ss(s·d)

s2+s

2=1

s22s(s·d)

s2d.

E(s)= a2

2✏0

s2[2(P·ˆs )ˆs P],for s > a.

Problem 4.14

Total charge on the dielectric is Qtot =Sbda +V⇢bd⌧=SP·daVr·Pd⌧. But the divergence

theorem says SP·da=Vr·Pd⌧,soQenc = 0. qed

Problem 4.15

(a) ⇢b=r·P=1

@rr2k

r=k

r2;b=P·ˆn =+P·ˆr =k/b (at r=b),

P·ˆr =k/a (at r=a).

Gauss’s law )E=1

4⇡✏0

Qenc

r2ˆr . For r<a,Qenc = 0, so E=0. For r>b,Qenc = 0 (Prob. 4.14), so E=0.

For a<r<b,Qenc =k

a4⇡a2+r

ak

r24⇡r2dr =4⇡ka 4⇡k(ra)=4⇡kr;so E=(k/✏0r)ˆr .

(b) D·da=Qfenc =0)D=0everywhere. D=✏0E+P=0)E=(1/✏0)P,so

E=0(for r<aand r>b); E=(k/✏0r)ˆr (for a<r<b).

Problem 4.16

(a) Same as E0minus the ﬁeld at the center of a sphere with uniform polarization P. The latter (Eq. 4.14)

is P/3✏0. So E=E0+1

3✏0

P.D=✏0E=✏0E0+1

3P=D0P+1

3P,so D=D02

3P.

(b) Same as E0minus the ﬁeld of ±charges at the two ends of the “needle”—but these are small, and far

away, so E=E0.D=✏0E=✏0E0=D0P,so D=D0P.

(1/✏0)P,so E=E0+1

✏0P.D=✏0E=✏0E0+P,so D=D0.

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 97

Problem 4.17

Hello

✻✻✻✻✻✻✻

(uniform)

❄ ❄ ❄❄❄ ❄❄ ❄❄

(ﬁeld of two circular plates)

✻

❄✻❄❄

✻

❄✻❄

✻

❄✻

(same as Eoutside, but lines

continuous, since ∇·D=0)

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For more detailed ﬁgures see the solution to Problem 6.14, reading Pfor M,Efor H, and Dfor B.

Problem 4.18

(a) Apply D·da=Qfenc to the gaussian surface shown. DA =A)D=.(Note: D=0inside the

metal plate.) This is true in both slabs; Dpoints down.

Problem 4.18

❖+σ

−σ

+σ/3

−σ/3

+σ/2

−σ/2

+σ

2⃝

1⃝

Problem 4.22

✛

✻

✙

✻

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(b) D=✏E)E=/✏1in slab 1, E=/✏2in slab 2. But ✏=✏0✏r, so ✏1=2✏0;✏2=3

2✏0.E1=/2✏0,

E2=2/3✏0.

r).P1=/2, P2=/3.

(d) V=E1a+E2a=(a/6✏0)(3 + 4) = 7a/6✏0.

(e) ⇢b= 0; b=+P1at bottom of slab (1) = /2,

b=P1at top of slab (1) = /2;

b=+P2at bottom of slab (2) = /3,

b=P2at top of slab (2) = /3.

(f) In slab 1: total surface charge above:(/2) = /2,

total surface charge below:(/2) (/3) + (/3) =/2,=)E1=

2✏0

In slab 2: total surface charge above:(/2) + (/2) (/3) = 2/3,

total surface charge below:(/3) =2/3,=)E2=2

3✏0

Problem 4.18

❖+σ

−σ

+σ/3

−σ/3

+σ/2

−σ/2

+σ

2⃝

1⃝

Problem 4.22

✛

✻

✙

✻

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Problem 4.19

With no dielectric, C0=A✏0/d (Eq. 2.54).

In conﬁguration (a), with +on upper plate, on lower, D=between the plates.

E=/✏0(in air) and E=/✏(in dielectric). So V=

✏0

2+

✏

2=Qd

2✏0A1+✏0

✏.

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98 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Ca=Q

V=✏0A

d2

1+1/✏r=)Ca

=2✏r

1+✏r

In conﬁguration (b), with potential di↵erence V:E=V/d,so=✏0E=✏0V/d (in air).

P=✏0eE=✏0eV/d (in dielectric), so b=✏0eV/d (at top surface of dielectric).

tot =✏0V/d =f+b=f✏0eV/d,sof=✏0V(1 + e)/d =✏0✏rV/d (on top plate above dielectric).

=)Cb=Q

V=1

VA

2+f

2=A

2V✏0

d+✏0

d✏r=A✏0

d1+✏r

2.Cb

=1+✏r

[Which is greater? Cb

C0Ca

C0=1+✏r

22✏r

1+✏r=(1+✏r)24✏r

2(1+✏r)=1+2✏r+4✏2

r4✏r

2(1+✏r)=(1✏r)2

2(1+✏r)>0. So Cb>C

a.]

If the xaxis points down:

E D P

(a) air 2✏r

(✏r+1)

dˆx 2✏r

(✏r+1)

✏0V

dˆx 0

(a) dielectric 2

(✏r+1)

dˆx 2✏r

(✏r+1)

✏0V

dˆx 2(✏r1)

(✏r+1)

✏0V

dˆx

(b) air V

dˆx ✏0V

dˆx 0

(b) dielectric V

dˆx ✏r✏0V

dˆx (✏r1)✏0V

dˆx

b(top surface) f(top plate)

(a) 2(✏r1)

(✏r+1)

✏0V

2✏r

(✏r+1)

✏0V

(b) (✏r1)✏0V

d✏r✏0V

d(left); ✏0V

d(right)

Problem 4.20

D·da=Qfenc )D4⇡r2=⇢4

3⇡r3)D=1

3⇢r)E=(⇢r/3✏)ˆr , for r<R;D4⇡r2=⇢4

3⇡R3)D=

⇢R3/3r2)E=(⇢R3/3✏0r2)ˆr , for r>R.

V=0

E·dl=⇢R3

3✏0

r

1⇢

3✏0

rdr =⇢R2

3✏0

+⇢

3✏

2=⇢R2

3✏01+ 1

2✏r.

Problem 4.21

Let Qbe the charge on a length `of the inner conductor.

D·da=D2⇡s`=Q)D=Q

2⇡s`;E=Q

2⇡✏0s`(a<s<b),E=Q

2⇡✏s`(b<r<c).

V=a

E·dl=b

aQ

2⇡✏0`ds

s+c

bQ

2⇡✏`ds

s=Q

2⇡✏0`ln b

a+✏0

✏ln c

b.

`=Q

V`=2⇡✏0

ln(b/a) + (1/✏r) ln(c/b).

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 99

Problem 4.22

Same method as Ex. 4.7: solve Laplace’s equation for Vin(s, )(s<a) and Vout(s, )(s>a), subject to

the boundary conditions







(i) Vin =Vout at s=a,

(ii) ✏@Vin

@s=✏0@Vout

@sat s=a,

(iii) Vout !E0scos for sa.

Problem 4.18

❖+σ

−σ

+σ/3

−σ/3

+σ/2

−σ/2

+σ

2⃝

1⃝

Problem 4.22

✛

✻

✙

✻

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From Prob. 3.24 (invoking boundary condition (iii)):

Vin(s, )= 1



k=1

sk(akcos k+bksin k),V

out(s, )=E0scos +1



k=1

sk(ckcos k+dksin k).

(I eliminated the constant terms by setting V= 0 on the yz plane.) Condition (i) says

ak(akcos k+bksin k)=E0acos +ak(ckcos k+dksin k),

while (ii) says

✏rkak1(akcos k+bksin k)=E0cos kak1(ckcos k+dksin k).

Evidently bk=dk= 0 for all k,ak=ck= 0 unless k= 1, whereas for k= 1,

aa1=E0a+a1c1,✏ra1=E0a2c1.

Solving for a1,

a1=E0

(1 + e/2),so Vin(s, )=E0

(1 + e/2)scos =E0

(1 + e/2)x,

and hence Ein(s, )=@Vin

@xˆx =E0

(1 + e/2).As in the spherical case (Ex. 4.7), the ﬁeld inside is uniform.

Problem 4.23

P0=✏0eE0;E1=1

3✏0

P0=e

3E0;P1=✏0eE1=✏02

3E0;E2=1

3✏0

P1=2

9E0;.... Evidently

En=e

3nE0,so

E=E0+E1+E2+···=1



n=0 e

3nE0.

The geometric series can be summed explicitly:



n=0

xn=1

1x,so E=1

(1 + e/3) E0,

which agrees with Eq. 4.49. [Curiously, this method formally requires that e<3 (else the inﬁnite series

diverges), yet the result is subject to no such restriction, since we can also get it by the method of Ex. 4.7.]

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100 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Problem 4.24

Potentials:











Vout(r, ✓)=E0rcos ✓+Bl

rl+1 Pl(cos ✓),(r>b);

Vmed(r, ✓)=Alrl+¯

rl+1 Pl(cos ✓),(a<r<b);

Vin(r, ✓)=0,(r<a).

Boundary Conditions:







(i) Vout =Vmed,(r=b);

(ii) ✏@Vmed

@r=✏0@Vout

@r,(r=b);

(iii) Vmed =0,(r=a).

(i) )E0bcos ✓+Bl

bl+1 Pl(cos ✓)=Albl+¯

bl+1 Pl(cos ✓);

(ii) )✏rlAlbl1(l+ 1) ¯

bl+2 Pl(cos ✓)=E0cos ✓(l+ 1) Bl

bl+2 Pl(cos ✓);

(iii) )Alal+¯

al+1 =0 )¯

Bl=a2l+1Al.

For l6=1:

(i) Bl

bl+1 =Albla2l+1Al

bl+1 )Bl=Alb2l+1 a2l+1;

(ii) ✏rlAlbl1+(l+ 1)a2l+1Al

bl+2 =(l+ 1) Bl

bl+2 )Bl=✏rAl l

l+1b2l+1 +a2l+1)Al=Bl=0.

For l=1:

(i) E0b+B1

b2=A1ba3A1

b2)B1E0b3=A1b3a3;

(ii) ✏rA1+2a3A1

b3=E02B1

b3)2B1E0b3=✏rA1b3+2a3.

So 3E0b3=A12b3a3+✏rb3+2a3;A1=3E0

2[1 (a/b)3]+✏r[1 + 2(a/b)3].

Vmed(r, ✓)= 3E0

2[1 (a/b)3]+✏r[1 + 2(a/b)3]ra3

r2cos ✓,

E(r, ✓)=rVmed =3E0

2[1 (a/b)3]+✏r[1 + 2(a/b)3]1+2a3

r3cos ✓ˆr 1a3

r3sin ✓ˆ

✓.

Problem 4.25

There are four charges involved: (i) q, (ii) polarization charge surrounding q, (iii) surface charge (b) on

the top surface of the lower dielectric, (iv) surface charge (0

b) on the lower surface of the upper dielectric.

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 101

In view of Eq. 4.39, the bound charge (ii) is qp=q(0

e/(1 + 0

e), so the total (point) charge at (0,0,d) is

qt=q+qp=q/(1 + 0

e)=q/✏0

r. As in Ex. 4.8,

(a) b=✏0e1

4⇡✏0

qd/✏0

(r2+d2)3

2b

2✏00

2✏0(here b=P·ˆn =+Pz=✏0eEz);

(b) 0

b=✏00

e1

4⇡✏0

qd/✏0

(r2+d2)3

2b

2✏00

2✏0(here b=Pz=✏00

eEz).

Solve for b,0

b: ﬁrst divide by eand 0

e(respectively) and subtract:

0

0

eb

e

2⇡

qd/✏0

(r2+d2)3

2)0

b=0

eb

e

2⇡

qd/✏0

(r2+d2)3

2.

Plug this into (a) and solve for b, using ✏0

r=1+0

b=1

4⇡

qd/✏0

(r2+d2)3

e(1 + 0

e)b

2(e+0

e),so b=1

4⇡

(r2+d2)3

e

[1 + (e+0

e)/2];

0

b=0

e1

4⇡

(r2+d2)3

[1 + (e+0

e)/2] +1

2⇡

qd/✏0

(r2+d2)3

2,so 0

b=1

4⇡

(r2+d2)3

✏r0

e/✏0

[1 + (e+0

e)/2].

The total bound surface charge is t=b+0

b=1

4⇡

(r2+d2)3

(0

ee)

✏0

r[1+(e+0

e)/2] (which vanishes, as it should, when

0

e=e). The total bound charge is (compare Eq. 4.51):

qt=(0

ee)q

2✏0

r[1 + (e+0

e)/2] =✏0

r✏r

✏0

r+✏rq

✏0

,and hence

V(r)= 1

4⇡✏0q/✏0

x2+y2+(zd)2+qt

x2+y2+(z+d)2(for z>0).

Meanwhile, since q

✏0

+qt=q

✏0

r1+✏0

r✏r

✏0

r+✏r=2q

✏0

r+✏r

,V(r)= 1

4⇡✏0

[2q/(✏0

r+✏r)]

x2+y2+(zd)2(for z<0).

Problem 4.26

From Ex. 4.5:

D=0,(r<a)

4⇡r2ˆr ,(r>a),E=









0,(r<a)

4⇡✏r2ˆr ,(a<r<b)

4⇡✏0r2ˆr ,(r>b)











W=1

2D·Ed⌧=1

(4⇡)24⇡1

✏b

r2r2dr +1

✏01

r2dr=Q2

8⇡1

✏1

r

✏01

r

b

=Q2

8⇡✏01

(1 + e)1

a1

b+1

b=Q2

8⇡✏0(1 + e)1

a+e

b.

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102 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Problem 4.27

Using Eq. 4.55: W=✏0

2E2d⌧. From Ex. 4.2 and Eq. 3.103,

E=









1

3✏0

Pˆz ,(r<R)

R3P

3✏0r3(2 cos ✓ˆr + sin ✓ˆ

✓),(r>R)







,so

Wr<R =✏0

2P

3✏024

3⇡R3=2⇡

P2R3

✏0

Wr>R =✏0

2R3P

3✏021

r64 cos2✓+ sin2✓r2sin ✓dr d✓d

=(R3P)2

18✏0

2⇡⇡

(1 + 3 cos2✓) sin ✓d✓1

r4dr =⇡(R3P)2

9✏0

(cos ✓cos3✓)⇡

01

3r3

=⇡(R3P)2

9✏04

3R3=4⇡R3P2

27✏0

Wtot =2⇡R3P2

9✏0

This is the correct electrostatic energy of the conﬁguration, but it is not the “total work necessary to assemble

the system,” because it leaves out the mechanical energy involved in polarizing the molecules.

Using Eq. 4.58: W=1

2D·Ed⌧. For r>R,D=✏0E, so this contribution is the same as before.

For r<R,D=✏0E+P=1

3P+P=2

3P=2✏0E,so1

2D·E=2✏0

2E2, and this contribution is

now (2) 2⇡

P2R3

✏0=4⇡

R3P2

✏0, exactly cancelling the exterior term. Conclusion: Wtot =0.This is not

surprising, since the derivation in Sect. 4.4.3 calculates the work done on the free charge, and in this problem

there is no free charge in sight. Since this is a nonlinear dielectric, however, the result cannot be interpreted as

the “work necessary to assemble the conﬁguration”—the latter would depend entirely on how you assemble it.

Problem 4.28

First ﬁnd the capacitance, as a function of h:

Air part: E=2

4⇡✏0s=)V=2

4⇡✏0ln(b/a),

Oil part: D=20

4⇡s=)E=20

4⇡✏s=)V=20

4⇡✏ ln(b/a),



=)

✏0

=0

✏;0=✏

✏0

=✏r.

Q=0h+(`h)=✏rhh+` =[(✏r1)h+`]=(eh+`), where `is the total height.

C=Q

V=(eh+`)

2ln(b/a)4⇡✏0=2⇡✏0

(eh+`)

ln(b/a).

The net upward force is given by Eq. 4.64: F=1

2V2dC

dh =1

2V22⇡✏0e

ln(b/a).

The gravitational force down is F=mg =⇢⇡(b2a2)gh. h=✏0eV2

⇢(b2a2)gln(b/a).

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 103

Problem 4.29

(a) Eq. 4.5 )F2=(p2·r)E1=p2

@y(E1);

Eq. 3.103 )E1=p1

4⇡✏0r3ˆ

✓=p1

4⇡✏0y3ˆz . Therefore

F2=p1p2

4⇡✏0d

dy 1

y3ˆz =3p1p2

4⇡✏0y4ˆz ,or F2=3p1p2

4⇡✏0r4ˆz (upward).

Problem 4.29 ✻

✲

✰y

p1p2

✻❄

✲

✻

✲

✰

xp1

✛

✻

Problem 4.30

✌✌✌✌✌

−

◆

✍

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Problem 4.29 ✻

✲

✰y

p1p2

✻❄

✲

✻

✲

✰

xp1

✛

✻

Problem 4.30

✌✌✌✌✌

−

◆

✍

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To calculate F1, put p2at the origin, pointing in the zdirection; then p1

is at rˆz , and it points in the ˆy direction. So F1=(p1·r)E2=

p1

@E2

@yx=y=0,z=r

; we need E2as a function of x,y, and z.

From Eq. 3.104: E2=1

4⇡✏0

r33(p2·r)r

r2p2,wherer=xˆx +yˆy +zˆz ,p2=p2ˆz , and hence p2·r=p2z.

E2=p2

4⇡✏03z(xˆx +yˆy +zˆz )(x2+y2+z2)ˆz

(x2+y2+z2)5/2=p2

4⇡✏03xz ˆx +3yz ˆy (x2+y22z2)ˆz

(x2+y2+z2)5/2

@E2

@y=p2

4⇡✏05

r7[3xz ˆx +3yz ˆy (x2+y22z2)ˆz ]+ 1

r5(3zˆy 2yˆz );

@E2

@y(0,0)

=p2

4⇡✏0

r5ˆy ;F1=p1p2

4⇡✏0

3r

r5ˆy =3p1p2

4⇡✏0r4ˆy .

But ˆy in these coordinates corresponds to ˆz in the original system, so these results are consistent with

Newton’s third law: F1=F2.

(b) From the remark following Eq. 4.5, N2=(p2⇥E1)+(r⇥F2). The ﬁrst term was calculated in

Prob. 4.5; the second we get from (a), using r=rˆy :

p2⇥E1=p1p2

4⇡✏0r3(ˆx ); r⇥F2=(rˆy )⇥3p1p2

4⇡✏0r4ˆz =3p1p2

4⇡✏0r3ˆx ; so N2=2p1p2

4⇡✏0r3ˆx .

This is equal and opposite to the torque on p1due to p2, with respect to the center of p1(see Prob. 4.5).

Problem 4.30

Net force is to the right (see diagram). Note that the ﬁeld lines must bulge to the right, as shown, because

Eis perpendicular to the surface of each conductor.

Problem 4.29 ✻

✲

✰y

p1p2

✻❄

✲

✻

✲

✰

xp1

✛

✻

Problem 4.30

✌✌✌✌✌

−

◆

✍

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104 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Problem 4.31 In cylindrical coordinates (in the z= 0 plane), p=pˆ

,p·r=p1

@, and E=1

4⇡✏0

s2ˆs ,so

F=(p·r)E=p

@1

4⇡✏0

s2ˆs =pQ

4⇡✏0s3

@ˆs

@ =pQ

4⇡✏0s3ˆ

=Q

4⇡✏0R3p.X

Qualitatively, the forces on the negative and positive ends, though equal in magnitude, point in slightly di↵erent

directions, and they combine to make a net force in the “forward” direction:

To keep the dipole going in a circle, there must be a centripetal force exerted by the track (we may as well

take it to act at the center of the dipole, and it is irrelevant to the problem), and to keep it aiming in the

tangential direction there must be a torque (which we could model by radial forces of equal magnitude acting at

the two ends). Indeed, if the dipole has the orientation indicated in the ﬁgure, and is moving in the ˆ

direction,

the torque exerted by Qis clockwise, whereas the rotation is counterclockwise, so these constraint forces must

actually be larger than the forces exerted by Q, and the net force will be in the “backward” direction—tending

to slow the dipole down. [If the motion is in the ˆ

direction, then the electrical forces will dominate, and the

net force will be in the direction of p, but this again will tend to slow it down.]

Problem 4.32

(a) According to Eqs. 4.1 and 4.5, F=↵(E·r)E.From the product rule,

rE2=r(E·E)=2E⇥(r⇥E) + 2(E·r)E.

But in electrostatics r⇥E=0,so (E·r)E=1

2r(E2), and hence

F=1

2↵r(E2).X

[It is tempting to start with Eq. 4.6, and write F=rU=r(p·E)=↵r(E·E)=↵r(E2).The

error occurs in the third step: pshould not have been di↵erentiated, but after it is replaced by ↵Ewe are

di↵erentiating both E’s.]

(b) Suppose E2has a local maximum at point P. Then there is a sphere (of radius R) about Psuch that

E2(P0)<E

2(P), and hence |E(P0)|<|E(P)|, for all points on the surface. But if there is no charge inside the

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 105

sphere, then Problem 3.4a says the average ﬁeld over the spherical surface is equal to the value at the center:

4⇡R2Eda =E(P),

or, choosing the zaxis to lie along E(P),

4⇡R2Ezda =E(P).

But if E2has a maximum at P, then

Ezda |E|da < |E(P)|da =4⇡R2E(P),

and it follows that E(P)<E(P), a contradiction. Therefore, E2cannot have a maximum in a charge-free

region. [It can have a minimum, however; at the midpoint between two equal charges the ﬁeld is zero, and this

is obviously a minimum.]

Problem 4.33

P=kr=k(xˆx +yˆy +zˆz )=)⇢b=r·P=k(1 + 1 + 1) = 3k.

Total volume bound charge: Qvol =3ka3.

b=P·ˆn . At top surface, ˆn =ˆz ,z=a/2; so b=ka/2. Clearly, b=ka/2 on all six surfaces.

Total surface bound charge: Qsurf = 6(ka/2)a2=3ka3.Total bound charge is zero. X

Problem 4.34

Say the high voltage is connected to the bottom plate, so the electric ﬁeld points in the xdirection, while

the free charge density (f) is positive on the lower plate and negative on the upper plate. (If you connect

the battery the other way, all the signs will switch.) The susceptibility is e=x

d, and the permittivity is

✏=✏01+x

d.Between the plates

D=fˆx ,E=1

✏D=f

✏0(1 + x/d)ˆx ;V=0

E·dl=f

✏0d

(1 + x/d)dx =f

✏0

dln 1+x

d

0=fd

✏0

ln 2.

f=✏0V

dln 2,E=V

dln 2

(1 + x/d)ˆx ,P=✏0eE=✏0V

d2ln 2

(1 + x/d)ˆx .

The bound charges are therefore

⇢b=r ·P=✏0V

d2ln 2 1

(1 + x/d)x/d

(1 + x/d)2=✏0V

d2ln 2

(1 + x/d)2;b=P·ˆn =









0 (x= 0),

✏0V

2dln 2 (x=d).

The total bound charge is

Qb=⇢bd⌧+bda =✏0V

d2ln 2 d

(1 + x/d)2A dx +✏0V

2dln 2A=✏0VA

dln 2 1

dd

(1 + x/d)

2

=✏0VA

dln 2 1

21+1

2=0,X

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106 CHAPTER 4. ELECTRIC FIELDS IN MATTER

where Ais the area of the plates.

Problem 4.35

D·da=Qfenc )D=q

4⇡r2ˆr ;E=1

✏D=q

4⇡✏0(1 + e)

ˆr

r2;P=✏0eE=qe

4⇡(1 + e)

ˆr

r2.

⇢b=r·P=qe

4⇡(1 + e)r·ˆr

r2=qe

1+e

3(r) (Eq. 1.99); b=P·ˆr =qe

4⇡(1 + e)R2;

Qsurf =b(4⇡R2)= qe

1+e

.The compensating negative charge is at the center:

⇢bd⌧=qe

1+e3(r)d⌧=qe

1+e

Problem 4.36

Ekis continuous (Eq. 4.29); D?is continuous (Eq. 4.26, with f= 0). So Ex1=Ex2,Dy1=Dy2)

✏1Ey1=✏2Ey2, and hence

tan ✓2

tan ✓1

=Ex2/Ey2

Ex1/Ey1

=Ey1

Ey2

=✏2

✏1

.qed

If 1 is air and 2 is dielectric, tan ✓2/tan ✓1=✏2/✏0>1, and the ﬁeld lines bend away from the normal. This is

the opposite of light rays, so a convex “lens” would defocus the ﬁeld lines.

Problem 4.37

In view of Eq. 4.39, the net dipole moment at the center is p0=pe

1+ep=1

✏rp. We want the

potential produced by p0(at the center) and b(at R). Use separation of variables:











Outside: V(r, ✓)= 1



l=0

rl+1 Pl(cos ✓) (Eq.3.72)

Inside: V(r, ✓)= 1

4⇡✏0

pcos ✓

✏rr2+1



l=0

AlrlPl(cos ✓) (Eqs.3.66,3.102)











Vcontinuous at R)









Rl+1 =AlRl,or Bl=R2l+1Al(l6= 1)

R2=1

4⇡✏0

✏rR2+A1R, or B1=p

4⇡✏0✏r+A1R3











@rR+@V

@rR

=(l+ 1) Bl

Rl+2 Pl(cos ✓)+ 1

4⇡✏0

2pcos ✓

✏rR3lAlRl1Pl(cos ✓)=1

✏0

b

=1

✏0

P·ˆr =1

✏0

(✏0eE·ˆr )=e

@rR

=e1

4⇡✏0

2pcos ✓

✏rR3+lAlRl1Pl(cos ✓).

(l+ 1) Bl

Rl+2 lAlRl1=elAlRl1(l6= 1); or (2l+ 1)AlRl1=elAlRl1)Al=0(`6= 1).

For l=1: 2B1

R3+1

4⇡✏0

✏rR3A1=e1

4⇡✏0

✏rR3+A1B1+p

4⇡✏0✏rA1R3

2=1

4⇡✏0

ep

✏r

+e

A1R3

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 107

p

4⇡✏0✏rA1R3+p

4⇡✏0✏rA1R3

2=1

4⇡✏0

ep

✏r

+e

A1R3

2)A1R3

2(3 + e)= 1

4⇡✏0

ep

✏r

)A1=1

4⇡✏0

2ep

R3✏r(3 + e)=1

4⇡✏0

2(✏r1)p

R3✏r(✏r+ 2);B1=p

4⇡✏0✏r1+ 2(✏r1)

(✏r+ 2) =p

4⇡✏0✏r

3✏r

✏r+2.

V(r, ✓)=pcos ✓

4⇡✏0r2 3

✏r+2(rR).

Meanwhile, for rR,V(r, ✓)= 1

4⇡✏0

pcos ✓

✏rr2+1

4⇡✏0

pr cos ✓

2(✏r1)

✏r(✏r+ 2)

=pcos ✓

4⇡✏0r2✏r1+2✏r1

✏r+2r3

R3(rR).

Problem 4.38

Given two solutions, V1(and E1=rV1,D1=✏E1) and V2(E2=rV2,D2=✏E2), deﬁne V3⌘V2V1

(E3=E2E1,D3=D2D1).

Vr·(V3D3)d⌧=SV3D3·da= 0, (V3= 0 on S), so (rV3)·D3d⌧+V3(r·D3)d⌧= 0.

But r·D3=r·D2r·D1=⇢f⇢f= 0, and rV3=rV2rV1=E2+E1=E3,soE3·D3d⌧= 0.

But D3=D2D1=✏E2✏E1=✏E3,so✏(E3)2d⌧= 0. But ✏>0, so E3= 0, so V2V1= constant. But

at surface, V2=V1,soV2=V1everywhere. qed

Problem 4.39

(a) Proposed potential: V(r)=V0

r.If so, then E=rV=V0

r2ˆr ,in which case P=✏0eV0

r2ˆr ,

in the region z<0. (P= 0 for z>0, of course.) Then b=✏0eV0

R2(ˆr ·ˆn )= ✏0eV0

R.(Note: ˆn points out

of dielectric )ˆn =ˆr .) This bis on the surface at r=R. The ﬂat surface z= 0 carries no bound charge,

since ˆn =ˆz ?ˆr . Nor is there any volume bound charge (Eq. 4.39). If Vis to have the required spherical

symmetry, the net charge must be uniform:

tot4⇡R2=Qtot =4⇡✏0RV0(since V0=Qtot/4⇡✏0R), so tot =✏0V0/R. Therefore

f=(✏0V0/R),on northern hemisphere

(✏0V0/R)(1 + e),on southern hemisphere .

(b) By construction, tot =b+f=✏0V0/R is uniform (on the northern hemisphere b=0,f=✏0V0/R;

on the southern hemisphere b=✏0eV0/R,sof=✏V0/R). The potential of a uniformly charged sphere is

V0=Qtot

4⇡✏0r=tot(4⇡R2)

4⇡✏0r=✏0V0

✏0r=V0

r.X

Prob. 4.38 guarantees that this is the solution.

(d) Figure (b) works the same way, but Fig. (a) does not: on the ﬂat surface, Pis not perpendicular to ˆn ,

so we’d get bound charge on this surface, spoiling the symmetry.

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108 CHAPTER 4. ELECTRIC FIELDS IN MATTER

Problem 4.40

Eext =

2⇡✏0sˆs . Since the sphere is tiny, this is essentially constant, and hence P=✏0e

1+e/3Eext (Ex. 4.7).

F=✏0e

1+e/3 

2⇡✏0sd

ds 

2⇡✏0sˆs d⌧=✏0e

1+e/3 

2⇡✏021

s1

s2ˆs d⌧

=e

1+e/32

4⇡2✏01

3⇡R3ˆs =e

3+e2R3

⇡✏0s3ˆs .

Problem 4.41

The density of atoms is N=1

(4/3)⇡R3. The macroscopic ﬁeld Eis Eself +Eelse, where Eself is the average

ﬁeld over the sphere due to the atom itself.

p=↵Eelse )P=N↵Eelse.

[Actually, it is the ﬁeld at the center, not the average over the sphere, that belongs here, but the two are in

fact equal, as we found in Prob. 3.47d.] Now

Eself =1

4⇡✏0

(Eq. 3.105), so

E=1

4⇡✏0

↵

R3Eelse +Eelse =1↵

4⇡✏0R3Eelse =1N↵

3✏0Eelse.

P=N↵

(1 N↵/3✏0)E=✏0eE,

and hence

e=N↵/✏0

(1 N↵/3✏0).

Solving for ↵:

eN↵

3✏0

e=N↵

✏0)N↵

✏01+e

3=e,

↵=✏0

e

(1 + e/3) =3✏0

e

(3 + e

.But e=✏r1,so ↵=3✏0

N✏r1

✏r+2.qed

Problem 4.42

For an ideal gas, N= Avagadro’s number/22.4 liters = (6.02 ⇥1023)/(22.4⇥103)=2.7⇥1025.N↵/✏0=

(2.7⇥1025)(4⇡✏0⇥1030)/✏0=3.4⇥104, where is the number listed in Table 4.1.

H: =0.667,N↵/✏0= (3.4⇥104)(0.67) = 2.3⇥104,e=2.5⇥104

He: =0.205,N↵/✏0= (3.4⇥104)(0.21) = 7.1⇥105,e=6.5⇥105

Ne: =0.396,N↵/✏0= (3.4⇥104)(0.40) = 1.4⇥104,e=1.3⇥104

Ar: =1.64,N↵/✏0= (3.4⇥104)(1.64) = 5.6⇥104,e=5.2⇥104











agreement is quite good.

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CHAPTER 4. ELECTRIC FIELDS IN MATTER 109

Problem 4.43

(a) Doing the (trivial) integral, and changing the remaining integration variable from ✓to u(du =pE sin ✓d✓),

hui=pE

pE ueu/kT du

pE

pE eu/kT du =(kT )2eu/kT [(u/kT )1]|pE

pE

kTeu/kT pE

pE

=kT epE/kT epE/kT +(pE/kT )epE/kT +(pE/kT )epE/kT 

epE/kT epE/kT 

=kT pE epE/kT +epE/kT

epE/kT epE/kT =kT pE coth pE

kT .

P=Nhpi;p=hpcos ✓iˆ

E=hp·Ei(ˆ

E/E)=hui(ˆ

E/E); P=Nphui

pE =Npcoth pE

kT kT

pE .

Let y⌘P/Np, x ⌘pE/kT .Theny= coth x1/x. As x!0, y=1

x+x

3x3

45 +···1

x=x

3x3

45 +···!

0, so the graph starts at the origin, with an initial slope of 1/3. As x!1,y!coth(1) = 1, so the graph

goes asymptotically to y= 1 (see Figure).

Problem 4.40

✻

✲

N p

pE/kT

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(b) For small x,y⇡1

3x, so P

Np ⇡pE

3kT , or P⇡Np2

3kT E=✏0eE)Pis proportional to E, and e=Np2

3✏0kT .

For water at 20= 293 K, p=6.1⇥1030 C m; N=molecules

volume =molecules

mole ⇥moles

gram ⇥grams

volume .

N=6.0⇥1023⇥1

18 ⇥106=0.33 ⇥1029;e=(0.33⇥1029 )(6.1⇥1030)2

(3)(8.85⇥1012)(1.38⇥1023 )(293) = 12.Table 4.2 gives an

experimental value of 79, so it’s pretty far o↵.

For water vapor at 100= 373 K, treated as an ideal gas, volume

mole = (22.4⇥103)⇥373

293 =2.85 ⇥102m3.

N=6.0⇥1023

2.85 ⇥102=2.11 ⇥1025;e=(2.11 ⇥1025)(6.1⇥1030)2

(3)(8.85 ⇥1012)(1.38 ⇥1023)(373) = 5.7⇥103.

Table 4.2 gives 5.9⇥103, so this time the agreement is quite good.

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110 CHAPTER 5. MAGNETOSTATICS

Chapter 5

Magnetostatics

Problem 5.1

Since v⇥Bpoints upward, and that is also the direction of the force, qmust be positive. To ﬁnd R, in

terms of aand d, use the pythagorean theorem:

(Rd)2+a2=R2)R22Rd +d2+a2=R2)R=a2+d2

2d.

2CHAPTER 5. MAGNETOSTATICS

Chapter 1

Magnetostatics

Problem 5.1

! "# $

}

Problem 5.2

✲y

✻

β◆

(b)

✲y

✻

−β

✸

(c)

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The cyclotron formula then gives

p=qBR =qB (a2+d2)

2d.

Problem 5.2

The general solution is (Eq. 5.6):

y(t)=C1cos(!t)+C2sin(!t)+ E

Bt+C3;z(t)=C2cos(!t)C1sin(!t)+C4.

(a) y(0) = z(0) = 0; ˙y(0) = E/B;˙z(0) = 0. Use these to determine C1,C2,C3, and C4.

y(0) = 0 )C1+C3= 0; ˙y(0) = !C2+E/B =E/B )C2= 0; z(0) = 0 )C2+C4=0)C4= 0;

˙z(0) = 0 )C1= 0, and hence also C3=0.So y(t)=Et/B;z(t)=0.Does this make sense? The magnetic

force is q(v⇥B)=q(E/B)Bˆz =qE, which exactly cancels the electric force; since there is no net force,

the particle moves in a straight line at constant speed. X

(b) Assuming it starts from the origin, so C3=C1,C4=C2, we have ˙z(0) = 0 )C1=0)C3= 0;

˙y(0) = E

2B)C2!+E

B=E

2B)C2=E

2!B=C4;y(t)=E

2!Bsin(!t)+ E

Bt;

z(t)=E

2!Bcos(!t)+ E

2!B,or y(t)= E

2!B[2!tsin(!t)] ; z(t)= E

2!B[1 cos(!t)] .Let ⌘E/2!B.

Then y(t)=[2!tsin(!t)] ; z(t)=[1 cos(!t)] ; (y2!t)=sin(!t),(z)=cos(!t))

(y2!t)2+(z)2=2. This is a circle of radius whose center moves to the right at constant speed:

y0=2!t;z0=.

B)C1!=E

B)C1=C3=E

!B;C2!+E

B=E

B)C2=C4=0.

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CHAPTER 5. MAGNETOSTATICS 111

y(t)=E

!Bcos(!t)+ E

Bt+E

!B;z(t)= E

!Bsin(!t). y(t)= E

!B[1 + !tcos(!t)] ; z(t)= E

!Bsin(!t).

Let ⌘E/!B; then [y(1 + !t)] = cos(!t),z=sin(!t); [y(1 + !t)]2+z2=2. This is a circle

of radius whose center is at y0=(1 + !t),z

0=0.

2CHAPTER 5. MAGNETOSTATICS

Chapter 1

Magnetostatics

Problem 5.1

! "# $

}

Problem 5.2

✲y

✻

β◆

(b)

✲y

✻

−β

✸

(c)

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Problem 5.3

(a) From Eq. 5.2, F=q[E+(v⇥B)] = 0 )E=vB )v=E

(b) From Eq. 5.3, mv =qBR )q

m=v

BR =E

B2R.

Problem 5.4

Suppose Iﬂows counterclockwise (if not, change the sign of the answer). The force on the left side (toward

the left) cancels the force on the right side (toward the right); the force on the top is IaB =Iak(a/2) =

Ika2/2, (pointing upward), and the force on the bottom is IaB =Ika2/2 (also upward). So the net force is

F=Ika2ˆz .

Problem 5.5

(a) K=I

2⇡a,because the length-perpendicular-to-ﬂow is the circumference.

(b) J=↵

s)I=Jda=↵1

ssdsd=2⇡↵ ds =2⇡↵a)↵=I

2⇡a;J=I

2⇡as.

Problem 5.6

(a) v=!r,so K=!r. (b) v=!rsin ✓ˆ

)J=⇢!rsin ✓ˆ

,where ⇢⌘Q/(4/3)⇡R3.

Problem 5.7

dt =d

dt V

⇢rd⌧=@⇢

@trd⌧=(r·J)rd⌧(by the continuity equation). Now product rule #5

says r·(xJ)=x(r·J)+J·(rx). But rx=ˆx ,sor·(xJ)=x(r·J)+Jx.Thus V(r·J)xd⌧=

Vr·(xJ)d⌧V

Jxd⌧. The ﬁrst term is SxJ·da(by the divergence theorem), and since Jis entirely

inside V, it is zero on the surface S. Therefore V(r·J)xd⌧=VJxd⌧, or, combining this with the yand

zcomponents, V(r·J)rd⌧=VJd⌧.Or, referring back to the ﬁrst line, dp

dt =Jd⌧.qed

Here’s a quicker method, if the distribution consists of a collection of point charges. Use Eqs. 5.30 and

3.100: Jd⌧=qivi=d

dt qiri=dp

dt .

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112 CHAPTER 5. MAGNETOSTATICS

Problem 5.8

(a) Use Eq. 5.37, with z=R, ✓2=✓1= 45,and four sides: B=p2µ0I

⇡R.

(b) z=R, ✓2=✓1=⇡

n, and nsides: B=nµ0I

2⇡Rsin(⇡/n).

2⇡R⇡

n=µ0I

2R(same as Eq. 5.41, with z= 0).

Problem 5.9

(a) The straight segments produce no ﬁeld at P. The two quarter-circles give B=µ0I

81

a1

b(out).

(b) The two half-lines are the same as one inﬁnite line: µ0I

2⇡R; the half-circle contributes µ0I

4R.

So B=µ0I

4R1+ 2

⇡(into the page).

Problem 5.10

(a) The forces on the two sides cancel. At the bottom, B=µ0I

2⇡s)F=µ0I

2⇡sIa =µ0I2a

2⇡s(up). At the

top, B=µ0I

2⇡(s+a))F=µ0I2a

2⇡(s+a)(down). The net force is µ0I2a2

2⇡s(s+a)(up).

(b) The force on the bottom is the same as before, µ0I2a/2⇡s(up). On the left side, B=µ0I

2⇡yˆz ;

dF=I(dl⇥B)=I(dx ˆx +dy ˆy +dz ˆz )⇥µ0I

2⇡yˆz =µ0I2

2⇡y(dx ˆy +dy ˆx ).But the xcomponent cancels the

corresponding term from the right side, and Fy=µ0I2

2⇡(s/p3+a/2)

s/p3

ydx. Here y=p3x, so

Fy=µ0I2

2p3⇡ln s/p3+a/2

s/p3=µ0I2

2p3⇡ln 1+p3a

2s.The force on the right side is the same, so the net

force on the triangle is µ0I2

2⇡a

s2

p3ln 1+p3a

2s.

CHAPTER 5. MAGNETOSTATICS 3

Problem 5.10

✲x

✻

✰

√3

! "# $

60◦

Problem 5.11

✲z

❄

! "# $

}

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CHAPTER 5. MAGNETOSTATICS 113

Problem 5.11

CHAPTER 5. MAGNETOSTATICS 3

Problem 5.10

✲x

✻

✰

√3

! "# $

60◦

Problem 5.11

✲z

❄

! "# $

}

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Use Eq. 5.41 for a ring of width dz, with I!nI dz:

B=µ0nI

2a2

(a2+z2)3/2dz. But z=acot ✓,

so dz =a

sin2✓d✓, and 1

(a2+z2)3/2=sin3✓

a3.

B=µ0nI

2a2sin3✓

a3sin2✓(ad✓)=µ0nI

2sin ✓d✓=µ0nI

2cos ✓✓2

✓1=µ0nI

2(cos ✓2cos ✓1).

For an inﬁnite solenoid, ✓2= 0, ✓1=⇡, so (cos ✓2cos ✓1)=1(1) = 2,and B=µ0nI. X

Problem 5.12

Rsinq

Rcosq

Field (at center of sphere) due to the ring at ✓(see ﬁgure) is (Eq. 5.41):

dB =µ0dI

(Rsin ✓)2

[(Rsin ✓)2+(Rcos ✓)2]3/2=µ0

2Rsin2✓dI.

dI =KRd✓,K=v, =Q

4⇡R2,v=!Rsin ✓,

dI =Q

4⇡R2!Rsin ✓Rd✓=Q!

4⇡sin ✓d✓.

B=µ0

4⇡⇡

sin3✓d✓=µ0Q!

8⇡R4

3.B=µ0Q!

6⇡Rˆz .

Problem 5.13

Magnetic attraction per unit length (Eqs. 5.40 and 5.13): fm=µ0

2⇡

2v2

Electric ﬁeld of one wire (Eq. 2.9): E=1

2⇡✏0



s. Electric repulsion per unit length on the other wire:

fe=1

2⇡✏0

2

d.They balance when µ0v2=1

✏0

, or v=1

p✏0µ0

.Putting in the numbers,

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114 CHAPTER 5. MAGNETOSTATICS

v=1

(8.85 ⇥1012)(4⇡⇥107)= 3.00 ⇥108m/s. This is precisely the speed of light(!), so in fact you could

never get the wires going fast enough; the electric force always dominates.

Problem 5.14

(a) B·dl=B2⇡s=µ0Ienc )B=0,for s<a;

µ0I

2⇡sˆ

,for s > a. 

(b) J=ks;I=a

Jda =a

ks(2⇡s)ds =2⇡ka3

3)k=3I

2⇡a3.I

enc =s

Jda =s

k¯s(2⇡¯s)d¯s=

2⇡ks3

3=Is3

a3, for s<a;Ienc =I, for s>a. So B=









µ0Is2

2⇡a3ˆ

,for s<a;

µ0I

2⇡sˆ

,for s > a.











Problem 5.15

By the right-hand-rule, the ﬁeld points in the ˆy direction for z>0, and in the +ˆy direction for z<0.

At z=0,B = 0. Use the amperian loop shown:

B·dl=Bl =µ0Ienc =µ0lzJ )B=µ0Jz ˆy (a<z<a). If z > a, Ienc =µ0laJ,

so B=µ0Jaˆy ,for z>+a;

+µ0Jaˆy ,for z>a. 

4CHAPTER 5. MAGNETOSTATICS

Problem 5.14

✲y

✻

✲

✛

! "# $

✠amperian loop

Problem 5.17

✲y

✰

✻

loop 1

loop 2

✿

ηη

η1

ηη

η2

❥

✣

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Problem 5.16

The ﬁeld inside a solenoid is µ0nI, and outside it is zero. The outer solenoid’s ﬁeld points to the right (ˆz ),

whereas the inner one points to the left (ˆz ). So: (i) B=µ0I(n2n1)ˆz ,(ii) B=µ0In2ˆz ,(iii) B=0.

Problem 5.17

From Ex. 5.8, the top plate produces a ﬁeld µ0K/2 (aiming out of the page, for points above it, and into

the page, for points below). The bottom plate produces a ﬁeld µ0K/2 (aiming into the page, for points above

it, and out of the page, for points below). Above and below both plates the two ﬁelds cancel; between the plates

they add up to µ0K, pointing in.

(a) B=µ0v(in) betweem the plates, B=0 elsewhere.

(b) The Lorentz force law says F=(K⇥B)da, so the force per unit area is f=K⇥B. Here K=v,

to the right, and B(the ﬁeld of the lower plate) is µ0v/2, into the page. So fm=µ02v2/2 (up).

fe=2/2✏0(down). They balance if µ0v2=1/✏0, or v=1/p✏0µ0=c(the speed of light), as in Prob. 5.13.

Problem 5.18

We might as well orient the axes so the ﬁeld point rlies on the yaxis: r=(0,y,0). Consider a source point

at (x0,y

0,z0) on loop #1:

=x0ˆx +(yy0)ˆy z0ˆz ;dl0=dx0ˆx +dy0ˆy ;

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CHAPTER 5. MAGNETOSTATICS 115

dl0⇥

=

ˆx ˆy ˆz

dx0dy00

x0(yy0)z0=(z0dy0)ˆx +(z0dx0)ˆy + [(yy0)dx0+x0dy0]ˆz .

dB1=µ0I

4⇡

dl0⇥

3=µ0I

4⇡

(z0dy0)ˆx +(z0dx0)ˆy + [(yy0)dx0+x0dy0]ˆz

[(x0)2+(yy0)2+(z0)2]3/2.

Now consider the symmetrically placed source element on

loop #2, at (x0,y

0,z0). Since z0changes sign, while every-

thing else is the same, the ˆx and ˆy components from dB1and

dB2cancel, leaving only a ˆz component. qed

With this, Amp´ere’s law yields immediately:

B=µ0nI ˆz ,inside the solenoid;

0,outside

(the same as for a circular solenoid—Ex. 5.9).

For the toroid, N/2⇡s=n(the number of turns per unit

length), so Eq. 5.60 yields B=µ0nI inside, and zero outside,

consistent with the solenoid. [Note: N/2⇡s=napplies only

if the toroid is large in circumference, so that sis essentially

constant over the cross-section.]

4CHAPTER 5. MAGNETOSTATICS

Problem 5.14

✲y

✻

✲

✛

! "# $

✠amperian loop

Problem 5.17

✲y

✰

✻

loop 1

loop 2

✿

❥

✣

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Problem 5.19

It doesn’t matter. According to Theorem 2, in Sect. 1.6.2, J·dais independent of surface, for any given

boundary line, provided that Jis divergenceless, which it is, for steady currents (Eq. 5.33).

Problem 5.20

(a) ⇢=charge

volume =charge

atom ·atoms

mole ·moles

gram ·grams

volume =(e)(N)1

M(d), where

e= charge of electron = 1.6⇥1019 C,

N= Avogadro0s number = 6.0⇥1023 mole,

M= atomic mass of copper = 64 gm/mole,

d= density of copper = 9.0 gm/cm3.

⇢=(1.6⇥1019)(6.0⇥1023)9.0

64 = 1.4⇥104C/cm3.

(b) J=I

⇡s2=⇢v)v=I

⇡s2⇢=1

⇡(2.5⇥103)(1.4⇥104)= 9.1⇥103cm/s, or about 33 cm/hr. This

is astonishingly small—literally slower than a snail’s pace.

2⇡I1I2

d=(4⇡⇥107)

2⇡= 2 ⇥107N/cm.

(d) E=1

2⇡✏0



d;fe=1

2⇡✏012

d=1

2⇡✏0I1I2

d=c2

v2µ0

2⇡I1I2

d=c2

v2fm,where

c⌘1/p✏0µ0=3.00 ⇥108m/s. Here fe

=c2

v2=3.0⇥1010

9.1⇥1032

= 1.1⇥1025.

fe=(1.1⇥1025)(2 ⇥107)= 2⇥1018 N/cm.

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116 CHAPTER 5. MAGNETOSTATICS

Problem 5.21

Amp´ere’s law says r⇥B=µ0J. Together with the continuity equation (5.29) this gives r·(r⇥B)=

µ0r·J=µ0@⇢/@t, which is inconsistent with div(curl)=0 unless ⇢is constant (magnetostatics). The other

Maxwell equations are OK: r⇥E=0)r·(r⇥E)=0(X), and as for the two divergence equations, there

is no relevant vanishing second derivative (the other one is curl(grad), which doesn’t involve the divergence).

Problem 5.22

At this stage I’d expect no changes in Gauss’s law or Amp´ere’s law. The divergence of Bwould take the

form r·B=↵0⇢m,where ⇢mis the density of magnetic charge, and ↵0is some constant (analogous to ✏0

and µ0). The curl of Ebecomes r⇥E=0Jm,where Jmis the magnetic current density (representing the

ﬂow of magnetic charge), and 0is another constant. Presumably magnetic charge is conserved, so ⇢mand Jm

satisfy a continuity equation: r·Jm=@⇢m/@t.

As for the Lorentz force law, one might guess something of the form qm[B+(v⇥E)] (where qmis the

magnetic charge). But this is dimensionally impossible, since Ehas the same units as vB. Evidently we

need to divide (v⇥E) by something with the dimensions of velocity-squared. The natural candidate is

c2=1/✏0µ0:F=qe[E+(v⇥B)] + qmB1

c2(v⇥E).In this form the magnetic analog to Coulomb’s

law reads F=↵0

4⇡

qm1qm2

r2ˆr , so to determine ↵0we would ﬁrst introduce (arbitrarily) a unit of magnetic charge,

then measure the force between unit charges at a given separation. [For further details, and an explanation of

the minus sign in the force law, see Prob. 7.38.]

Problem 5.23

A=µ0

4⇡Iˆz

dz =µ0I

4⇡ˆz z2

pz2+s2

=µ0I

4⇡ˆz ln z+z2+s2

=µ0I

4⇡ln z2+(z2)2+s2

z1+(z1)2+s2ˆz

CHAPTER 5. MAGNETOSTATICS 5

Problem 5.22

✲z

}

dl=dz

!"# $

Problem 5.26

✲y

✻

✠

✠ ✠ ✠ ✠ ✠ ✠ K

Problem 5.36

Problem 5.37

Problem 5.44

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B=r⇥A=@A

@sˆ

=µ0I

4⇡1

z2+(z2)2+s2

(z2)2+s21

z1+(z1)2+s2

(z1)2+s2ˆ



=µ0Is

4⇡z2(z2)2+s2

(z2)2[(z2)2+s2]

(z2)2+s2z1(z1)2+s2

1[(z1)2+s2]

(z1)2+s2ˆ



=µ0Is

4⇡1

s2z2

(z2)2+s21z1

(z1)2+s2+1

ˆ

=µ0I

4⇡sz2

(z2)2+s2z1

(z1)2+s2ˆ

,

or,since sin ✓1=z1

(z1)2+s2and sin ✓2=z2

(z2)2+s2,

=µ0I

4⇡s(sin ✓2sin ✓1)ˆ

(as in Eq.5.37).

Problem 5.24

A=k)B=r⇥A=1

@s(sk)ˆz =k

sˆz ;J=1

µ0

(r⇥B)= 1

µ0@

@sk

s ˆ

=k

µ0s2ˆ

.

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CHAPTER 5. MAGNETOSTATICS 117

Problem 5.25

r·A=1

2r·(r⇥B)=1

2[B·(r⇥r)r·(r⇥B)] = 0, since r⇥B=0(Bis uniform) and

r⇥r=0(Prob. 1.63). r⇥A=1

2r⇥(r⇥B)=1

2[(B·r)r(r·r)B+r(r·B)B(r·r)]. But

(r·r)B=0and r·B= 0 (since Bis uniform), and r·r=@x

@x+@y

@y+@z

@z= 1 + 1 + 1 = 3. Finally,

(B·r)r=Bx

@x+By

@y+Bz

@z(xˆx +yˆy +zˆz )=Bxˆx +Byˆy +Bzˆz =B.Sor⇥A=1

2(B3B)=B.

qed

Problem 5.26

(a) Ais parallel (or antiparallel) to I, and is a function only of s(the distance from the wire). In cylindrical

coordinates, then, A=A(s)ˆz ,soB=r⇥A=@A

@sˆ

=µ0I

2⇡sˆ

(the ﬁeld of an inﬁnite wire). Therefore

@s=µ0I

2⇡s, and A(r)=µ0I

2⇡ln(s/a)ˆz (the constant ais arbitrary; you could use 1, but then the units

look ﬁshy). r·A=@Az

@z=0.Xr⇥A=@Az

@sˆ

=µ0I

2⇡sˆ

=B.X

(b) Here Amp´ere’s law gives B·dl=B2⇡s=µ0Ienc =µ0J⇡s2=µ0

⇡R2⇡s2=µ0Is2

R2.

B=µ0

2⇡

R2ˆ

.@A

@s=µ0I

2⇡

R2)A=µ0I

4⇡R2(s2b2)ˆz . Here bis again arbitrary, except that since A

must be continuous at R,µ0I

2⇡ln(R/a)=µ0I

4⇡R2(R2b2), which means that we must pick aand bsuch that

2 ln(R/b)=1(b/R)2. I’ll use a=b=R. Then A=









µ0I

4⇡R2(s2R2)ˆz ,for sR;

µ0I

2⇡ln(s/R)ˆz ,for sR.











Problem 5.27

K=Kˆx )B=±µ0K

2ˆy (plus for z<0, minus for z>0).

Ais parallel (or antiparallel) to K, and depends only on z,soA=A(z)ˆx .

B=r⇥A=

ˆx ˆy ˆz

@/@x@/@y@/@z

A(z)0 0

=@A

@zˆy =±µ0K

2ˆy .

A=µ0K

2|z|ˆx will do the job—or this plus any constant.

CHAPTER 5. MAGNETOSTATICS 5

Problem 5.22

✲z

sη

}

dl=dz

!"# $

Problem 5.26

✲y

✻

✠

✠ ✠ ✠ ✠ ✠ ✠ K

Problem 5.36

Problem 5.37

Problem 5.44

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Problem 5.28

(a) r·A=µ0

4⇡r·J

d⌧0;r·J

=1

(r·J)+J·r1

. But the ﬁrst term is zero, because J(r0)

is a function of the source coordinates, not the ﬁeld coordinates. And since

=rr0,r1

=r01

.

So r·J

=J·r01

.Butr0·J

=1

(r0·J)+J·r01

, and r0·J= 0 in magnetostatics

(Eq. 5.33). So r·J

=r0·J

, and hence, by the divergence theorem, r·A=µ0

4⇡r0·J

d⌧0=

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118 CHAPTER 5. MAGNETOSTATICS

µ0

4⇡J

·da0, where the integral is now over the surface surrounding all the currents. But J=0on this

surface, so r·A=0.X

(b) r⇥A=µ0

4⇡r⇥J

d⌧0=µ0

4⇡1

(r⇥J)J⇥r1

d⌧0.Butr⇥J=0(since Jis

not a function of r), and r1

=ˆ

2(Eq. 1.101), so r⇥A=µ0

4⇡J⇥ˆ

2d⌧0=B.X

4⇡r2J

d⌧0. But r2J

=Jr21

(once again, Jis a constant, as far as

di↵erentiation with respect to ris concerned), and r21

=4⇡3(

) (Eq. 1.102). So

r2A=µ0

4⇡J(r0)4⇡3(

)d⌧0=µ0J(r).X

Problem 5.29

µ0I=B·dl=b

arU·dl=[U(b)U(a)] (by the gradient theorem), so U(b)6=U(a).qed

For an inﬁnite straight wire, B=µ0I

2⇡sˆ

. U =µ0I

2⇡would do the job, in the sense that

rU=µ0I

2⇡r()=µ0I

2⇡

@

@ ˆ

=B. But when advances by 2⇡, this function does not return to its initial

value; it works (say) for 0 <2⇡, but at 2⇡it “jumps” back to zero.

Problem 5.30

Use Eq. 5.69, with R!¯rand !⇢d¯r:

A=µ0!⇢

sin ✓

r2ˆ

r

¯r4d¯r+µ0!⇢

3rsin ✓ˆ

R

¯rd¯r

=µ0!⇢

3sin ✓1

r2r5

5+r

2R2r2ˆ

=µ0!⇢

2rsin ✓R2

3r2

5ˆ

.

B=r⇥A=µ0!⇢

21

rsin ✓

@✓ sin ✓rsin ✓R2

3r2

5ˆr 1

@rr2sin ✓R2

3r2

5ˆ

✓

=µ0!⇢ R2

3r2

5cos ✓ˆr R2

32r2

5sin ✓ˆ

✓.

Problem 5.31

(a) 









@Wz

@x=Fy)Wz(x, y, z)=x

0Fy(x0,y,z)dx0+C1(y, z).

@Wy

@x=Fz)Wy(x, y, z)=+x

0Fz(x0,y,z)dx0+C2(y, z).











These satisfy (ii) and (iii), for any C1and C2; it remains to choose these functions so as to satisfy (i):

x

@Fy(x0,y,z)

@ydx0+@C1

@yx

@Fz(x0,y,z)

@zdx0@C2

@z=Fx(x, y, z). But @Fx

@x+@Fy

@y+@Fz

@z= 0, so

x

@Fx(x0,y,z)

@x0dx0+@C1

@y@C2

@z=Fx(x, y, z). Now x

@Fx(x0,y,z)

@x0dx0=Fx(x, y, z)Fx(0,y,z), so

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CHAPTER 5. MAGNETOSTATICS 119

@C1

@y@C2

@z=Fx(0,y,z). We may as well pick C2=0,C

1(y, z)=y

Fx(0,y

0,z)dy0, and we’re done, with

Wx= 0; Wy=x

Fz(x0,y,z)dx0;Wz=y

Fx(0,y

0,z)dy0x

Fy(x0,y,z)dx0.

(b) r⇥W=@Wz

@y@Wy

@zˆx +@Wx

@z@Wz

@xˆy +@Wy

@x@Wx

@yˆz

=Fx(0,y,z)x

@Fy(x0,y,z)

@ydx0x

@Fz(x0,y,z)

@zdx0ˆx + [0 + Fy(x, y, z)] ˆy +[Fz(x, y, z)0] ˆz .

But r·F= 0, so the ˆx term is Fx(0,y,z)+x

@Fx(x0,y,z)

@x0dx0=Fx(0,y,z)+Fx(x, y, z)Fx(0,y,z),

so r⇥W=F.X

r·W=@Wx

@x+@Wy

@y+@Wz

@z=0+x

@Fz(x0,y,z)

@ydx0+y

@Fx(0,y

0,z)

@zdy0x

@Fy(x0,y,z)

@zdx06=0,

in general.

x0dx0=x2

2;Wz=y

y0dy0x

z dx0=y2

2zx.

W=x2

2ˆy +y2

2zxˆz .r⇥W=

ˆx ˆy ˆz

@/@x@/@y@/@z

0x2/2(y2/2zx)=yˆx +zˆy +xˆz =F.X

Problem 5.32

(a) At the surface of the solenoid, Babove =0,Bbelow =µ0nI ˆz =µ0Kˆz ;ˆn =ˆs ;soK⇥ˆn =Kˆz .

Evidently Eq. 5.76 holds. X

(b) In Eq. 5.69, both expressiions reduce to (µ0R2!/3) sin ✓ˆ

at the surface, so Eq. 5.77 is satisﬁed.

@rR+

=µ0R4!

32 sin ✓

r3ˆ

R

=2µ0R!

3sin ✓ˆ

;@A

@rR

=µ0R!

3sin ✓ˆ

. So the left side of

Eq. 5.78 is µ0R! sin ✓ˆ

. Meanwhile K=v=(!!!⇥r)=!Rsin ✓ˆ

, so the right side of Eq. 5.78 is

µ0!Rsin ✓ˆ

, and the equation is satisﬁed.

Problem 5.33

Because Aabove =Abelow at every point on the surface, it follows that @A

@xand @A

@yare the same above

and below; any discontinuity is conﬁned to the normal derivative.

Babove Bbelow =@Ayabove

@z+@Aybelow

@zˆx +@Axabove

@z@Axbelow

@zˆy . But Eq. 5.76 says this equals

µ0K(ˆy ). So @Ayabove

@z=@Aybelow

@z,and @Axabove

@z@Axbelow

@z=µ0K. Thus the normal derivative of the com-

ponent of Aparallel to Ksu↵ers a discontinuity µ0K, or, more compactly: @Aabove

@n@Abelow

@n=µ0K.

Problem 5.34

(Same idea as Prob. 3.36.) Write m=(m·ˆr )ˆr +(m·ˆ

✓)ˆ

✓=mcos ✓ˆr msin ✓ˆ

✓(Fig. 5.54). Then

3(m·ˆr )ˆr m=3mcos ✓ˆr mcos ✓ˆr +msin ✓ˆ

✓=2mcos ✓ˆr +msin ✓ˆ

✓, and Eq. 5.89 ,Eq. 5.88. qed

Problem 5.35

(a) m=Ia=I⇡R2ˆz .

(b) B⇡µ0

4⇡

I⇡R2

r32 cos ✓ˆr + sin ✓ˆ

✓.

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120 CHAPTER 5. MAGNETOSTATICS

2z3ˆz (for z<0, ✓=⇡,ˆr =ˆz , so the ﬁeld

is the same, with |z|3in place of z3). The exact answer (Eq. 5.41) reduces (for zR) to B⇡µ0IR2/2|z|3,

so they agree.

Problem 5.36

The ﬁeld of one side is given by Eq. 5.37, with s!

z2+(w/2)2and sin ✓2=sin ✓1=(w/2)

z2+w2/2;

B=µ0I

4⇡

z2+(w2/4)z2+(w2/2) . To pick o↵the vertical

component, multiply by sin =(w/2)

z2+(w/2)2; for all four

sides, multiply by 4: B=µ0I

2⇡

(z2+w2/4)z2+w2/2ˆz .

For zw,B⇡µ0Iw2

2⇡z3ˆz . The ﬁeld of a dipole m=Iw2,

for points on the zaxis (Eq. 5.88, with r!z,ˆr !ˆz ,✓=0)isB=µ0

2⇡

z3ˆz .X

Problem 5.37

(a) For a ring, m=I⇡r2. Here I!v dr =!rdr,som=R

0⇡r2!rdr =⇡!R4/4.

(b) The total charge on the shaded ring is dq =

(2⇡Rsin ✓)Rd✓. The time for one revolution is dt =2⇡/!.

So the current in the ring is I=dq

dt =!R2sin ✓d✓. The

area of the ring is ⇡(Rsin ✓)2, so the magnetic moment of the

ring is dm =(!R2sin ✓d✓)⇡R2sin2✓, and the total dipole

moment of the shell is

m=!⇡R4⇡

0sin3✓d✓=(4/3)!⇡R4, or m=4⇡

3!R4ˆz .

The dipole term in the multipole expansion for Ais therefore Adip =µ0

4⇡

3!R4sin ✓

r2ˆ

=µ0!R4

sin ✓

r2ˆ

,

which is also the exact potential (Eq. 5.69); evidently a spinning sphere produces a perfect dipole ﬁeld, with

no higher multipole contributions.

Problem 5.38

(a) Idl!Jd⌧,so A=µ0

4⇡



n=0

rn+1 (r0)nPn(cos ✓)Jd⌧.

(b) Amon =µ0

4⇡rJd⌧=µ0

4⇡r

dt (Prob. 5.7), where pis the total electric dipole moment. In mag-

netostatics,pis constant, so dp/dt =0, and hence Amon =0. qed

2I(r⇥dl)!m=1

2(r⇥J)d⌧.qed

Problem 5.39

(a) Yes. (Magnetic forces do no work.)

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CHAPTER 5. MAGNETOSTATICS 121

(b) B=µ0

2⇡sˆ

;v=˙sˆs +s˙

ˆ

+˙zˆz ;(v⇥B)=

ˆs ˆ

ˆz

˙ss

˙z

0µ0

2⇡s0=µ0

2⇡s(˙zˆs +˙sˆz ).

F=q(v⇥B)= qµ0

2⇡s(˙zˆr +˙sˆz ).

2)ˆs +(s¨

+2˙s˙

)ˆ

+¨zˆz (see, for example J. R. Taylor, Classical Mechanics, Eq. 1.47).

F=ma)(¨ss˙

2)ˆs +(s¨

+2˙s˙

)ˆ

+¨zˆz =↵

s(˙zˆs +˙sˆz ),

where ↵⌘qµ0

2⇡m. Thus

¨ss˙

2=↵˙z

s, s ¨

+2˙s˙

=0,¨z=↵˙s

(d) If ˙zis constant, then ¨z= 0, and it follows from the third equation that ˙s= 0, and hence sis constant.

Then the ﬁrst equation says ˙

2=↵(˙z/s2), so ˙

=±1

sp↵˙zis also constant (the second equation holds

automatically). The charge moves in a helix around the wire.

Problem 5.40

The mobile charges do pull in toward the axis, but the resulting concentration of (negative) charge sets up

an electric ﬁeld that repels away further accumulation. Equilibrium is reached when the electric repulsion on a

mobile charge qbalances the magnetic attraction: F=q[E+(v⇥B)] = 0)E=(v⇥B). Say the current

is in the zdirection: J=⇢vˆz (where ⇢and vare both negative).

B·dl=B2⇡s=µ0J⇡s2)B=µ0⇢vs

2ˆ

;

E·da=E2⇡sl =1

✏0

(⇢++⇢)⇡s2l)E=1

2✏0

(⇢++⇢)sˆs .

2✏0

(⇢++⇢)sˆs =(vˆz )⇥µ0⇢vs

2ˆ

=µ0

2⇢v2sˆs )⇢++⇢=⇢(✏0µ0v2)=⇢v2

c2.

Evidently ⇢+=⇢1v2

c2=⇢

2, or ⇢=2⇢+.In this naive model the mobile negative charges ﬁll a

smaller inner cylinder, leaving a shell of positive (stationary) charge at the outside. But since v⌧c, the e↵ect

is extremely small.

Problem 5.41

(a) If positive charges ﬂow to the right, they are deﬂected down, and the bottom plate acquires a positive

charge.

(b) qvB =qE )E=vB )V=Et =vBt, with the bottom at higher potential.

charge. The potential di↵erence is still the same, but this time the top plate is at the higher potential.

Problem 5.42

From Eq. 5.17, F=I(dl⇥B). But Bis constant, in this case, so it comes outside the integral: F=

Idl⇥B, and dl=w, the vector displacement from the point at which the wire ﬁrst enters the ﬁeld to

the point where it leaves. Since wand Bare perpendicular, F=IBw, and Fis perpendicular to w.

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122 CHAPTER 5. MAGNETOSTATICS

Problem 5.43

The angular momentum acquired by the particle as it moves out from the center to the edge is

L=dL

dt dt =Ndt =(r⇥F)dt =r⇥q(v⇥B)dt =qr⇥(dl⇥B)=q(r·B)dlB(r·dl).

But ris perpendicular to B, so r·B= 0, and r·dl=r·dr=1

2d(r·r)=1

2d(r2)=r dr = (1/2⇡)(2⇡rdr).

So L=q

2⇡R

B2⇡rdr =q

2⇡Bda. It follows that L=q

2⇡,where =Bda is the total ﬂux.

In particular, if = 0, then L= 0, and the charge emerges with zero angular momentum, which means it is

going along a radial line. qed

Problem 5.44

From Eq. 5.24, F=(K⇥Bave)da. Here K=v,v=!Rsin ✓ˆ

,da =R2sin ✓d✓d, and

Bave =1

2(Bin +Bout). From Eq. 5.70,

Bin =2

3µ0R!ˆz =2

3µ0R!(cos ✓ˆr sin ✓ˆ

✓).From Eq.5.69,

Bout =r⇥A=r⇥µ0R4!

sin ✓

r2ˆ

=µ0R4!

31

rsin ✓

@✓ sin2✓

r2ˆr 1

@rsin ✓

rˆ

✓

=µ0R4!

3r3(2 cos ✓ˆr + sin ✓ˆ

✓)=µ0R!

3(2 cos ✓ˆr + sin ✓ˆ

✓) (since r=R).

Bave =µ0R!

6(4 cos ✓ˆr sin ✓ˆ

✓).

K⇥Bave =(!Rsin ✓)µ0R!

6ˆ

⇥(4 cos ✓ˆr sin ✓ˆ

✓)=µ0

6(!R)2(4 cos ✓ˆ

✓+ sin ✓ˆr ) sin ✓.

Picking out the zcomponent of ˆ

✓(namely, sin ✓) and of ˆr (namely, cos ✓), we have

(K⇥Bave)z=µ0

2(!R)2sin2✓cos ✓,so

Fz=µ0

2(!R)2R2sin3✓cos ✓d✓d=µ0

2(!R2)22⇡sin4✓

4

⇡/2

,or F=µ0⇡

4(!R2)2ˆz .

Problem 5.45

(a) F=ma=qe(v⇥B)= µ0

4⇡

qeqm

r2(v⇥ˆr ); a=µ0

4⇡

qeqm

mr3(v⇥r).

(b) Because a?v,a·v= 0. But a·v=1

dt(v·v)= 1

dt(v2)=vdv

dt . So dv

dt =0.qed

dt =m(v⇥v)+m(r⇥a)µ0qeqm

4⇡

dt r

r=0+µ0qeqm

4⇡r3[r⇥(v⇥r]µ0qeqm

4⇡v

rr

dt 

=µ0qeqm

4⇡1

r3[r2v(r·v)r]v

r+r

dt(pr·r)=µ0qeqm

4⇡v

r(ˆr ·v)

rˆr v

r+ˆr

2(r·v)

r=0.X

(d) (i) Q·ˆ

=Q(ˆz ·ˆ

)=m(r⇥v)·ˆ

µ0qeqm

4⇡(ˆr ·ˆ

). But ˆz ·ˆ

=ˆr ·ˆ

= 0, so (r⇥v)·ˆ

= 0. But

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CHAPTER 5. MAGNETOSTATICS 123

r=rˆr , and v=dl

dt =˙rˆr +r˙

✓ˆ

✓+rsin ✓˙

ˆ

(where dots denote di↵erentiation with respect to time), so

r⇥v=

ˆr ˆ

✓ˆ



r00

˙rr

✓rsin ✓˙

=(r2sin ✓˙

)ˆ

✓+(r2˙

✓)ˆ

.

Therefore (r⇥v)·ˆ

=r2˙

✓= 0, so ✓is constant. qed

(ii) Q·ˆr =Q(ˆz ·ˆr )=m(r⇥v)·ˆr µ0qeqm

4⇡(ˆr ·ˆr ). But ˆz ·ˆr = cos ✓, and (r⇥v)?r)(r⇥v)·ˆr = 0, so

Qcos ✓=µ0qeqm

4⇡, or Q=µ0qeqm

4⇡cos ✓. And since ✓is constant, so too is Q.qed

(iii) Q·ˆ

✓=Q(ˆz ·ˆ

✓)=m(r⇥v)·ˆ

✓µ0qeqm

4⇡(ˆr ·ˆ

✓). But ˆz ·ˆ

✓=sin ✓,ˆr ·ˆ

✓= 0, and (r⇥v)·ˆ

✓=r2sin ✓˙



(from (i)), so Qsin ✓=mr2sin ✓˙

)˙

=Q

mr2=k

r2, with k⌘Q

m=µ0qeqm

4⇡mcos ✓.

(e) v2=˙r2+r2˙

✓2+r2sin2✓˙

2, but ˙

✓= 0 and ˙

=k

r2,so ˙r2=v2r2sin2✓k2

r4=v2k2sin2✓

r2.

dr

d2

=˙r2

2=v2(ksin ✓/r)2

(k2/r4)=r2vr

k2sin2✓;dr

d=rvr

k2sin2✓.

(f) dr

r(vr/k)2sin2✓

=d)0=1

sin ✓sec1vr

ksin ✓; sec[(0) sin ✓]= vr

ksin ✓,or

r()= A

cos[(0) sin ✓],where A⌘µ0qeqmtan ✓

4⇡mv .

Problem 5.46

Put the ﬁeld point on the xaxis, so r=(s, 0,0). Then B=

µ0

4⇡(K⇥ˆ

)

2da;da =Rddz;K=Kˆ

=K(sin ˆx +

cos ˆy );

=(sRcos )ˆx Rsin ˆy zˆz .

K⇥

=K

ˆx ˆy ˆz

sin cos 0

(sRcos )(Rsin )(z)=

K[(zcos )ˆx +(zsin )ˆy +(Rscos )ˆz ];

2=z2+R2+s22Rs cos .Thexand ycomponents

integrate to zero (zintegrand is odd, as in Prob. 5.18).

Bz=µ0

4⇡KR(Rscos )

(z2+R2+s22Rs cos )3/2ddz =µ0KR

4⇡2⇡

(Rscos )1

1

(z2+d2)3/2d,

where d2⌘R2+s22Rs cos .Now 1

1

(z2+d2)3/2=2z

d2pz2+d21

0=2

d2.

=µ0KR

2⇡2⇡

(Rscos )

(R2+s22Rs cos )d;(Rscos )= 1

2R(R2s2)+(R2+s22Rs cos ).

=µ0K

4⇡(R2s2)2⇡

d

(R2+s22Rs cos )+2⇡

d.

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124 CHAPTER 5. MAGNETOSTATICS

2⇡

d

a+bcos =2

⇡

d

a+bcos =4

pa2b2tan1pa2b2tan(/2)

a+b

⇡

pa2b2tan1pa2b2tan(⇡/2)

a+b=4

pa2b2⇡

2=2⇡

pa2b2.Here a=R2+s2,

b=2Rs, so a2b2=R4+2R2s2+s44R2s2=R42R2s2+s4=(R2s2)2;a2b2=|R2d2|.

Bz=µ0K

4⇡(R2s2)

|R2s2|2⇡+2⇡=µ0K

2R2s2

|R2s2|+1

.

Inside the solenoid, s<R, so Bz=µ0K

2(1+1) = µ0K.Outside the solenoid, s>R, so Bz=µ0K

2(1+1) = 0.

Here K=nI,so B=µ0nI ˆz (inside),and 0(outside) (as we found more easily using Amp´ere’s law, in Ex. 5.9).

Problem 5.47

(a) From Eq. 5.41, B=µ0IR2

21

[R2+(d/2+z)2]3/2+1

[R2+(d/2z)2]3/2.

@z=µ0IR2

2(3/2)2(d/2+z)

[R2+(d/2+z)2]5/2+(3/2)2(d/2z)(1)

[R2+(d/2z)2]5/2

=3µ0IR2

2(d/2+z)

[R2+(d/2+z)2]5/2+(d/2z)

[R2+(d/2z)2]5/2.

@zz=0

=3µ0IR2

2d/2

[R2+(d/2)2]5/2+d/2

[R2+(d/2)2]5/2=0.X

(b) Di↵erentiating again:

@2B

@z2=3µ0IR2

21

[R2+(d/2+z)2]5/2+(d/2+z)(5/2)2(d/2+z)

[R2+(d/2+z)2]7/2

+1

[R2+(d/2z)2]5/2+(d/2z)(5/2)2(d/2z)(1)

[R2+(d/2z)2]7/2.

@2B

@z2z=0

=3µ0IR2

22

[R2+(d/2)2]5/2+2(5/2)2(d/2)22

[R2+(d/2)2]7/2=3µ0IR2

[R2+(d/2)2]7/2R2d2

4+5d2

4

=3µ0IR2

[R2+(d/2)2]7/2d2R2.Zero if d=R, in which case

B(0) = µ0IR2

21

[R2+(R/2)2]3/2+1

[R2+(R/2)2]3/2=µ0IR21

(5R2/4)3/2=8µ0I

53/2R.

Problem 5.48

The total charge on the shaded ring is dq =(2⇡r)dr. The

time for one revolution is dt =2⇡/!. So the current in the

ring is I=dq

dt =!rdr. From Eq. 5.41, the magnetic ﬁeld of

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CHAPTER 5. MAGNETOSTATICS 125

this ring (for points on the axis) is dB=µ0

2!rr2

(r2+z2)3/2dr ˆz , and the total ﬁeld of the disk is

B=µ0!

2R

r3dr

(r2+z2)3/2ˆz .Let u⌘r2,so du =2r dr. Then

=µ0!

4R2

udu

(u+z2)3/2=µ0!

42u+2z2

pu+z2

=µ0!

2(R2+2z2)

pR2+z22zˆz .

When zR, the term in square brackets is

[]= 2z2(1 + R2/2z2)

z[1 + (R/z)2]1/22z⇡2z1+ R2

2z211

z2+3

z41

⇡2z1R2

2z2+3

z4+R2

2z2R4

4z41=2z1

z4=R4

4z4,

B⇡µ0!

4z4ˆz =µ0!R4

8z3ˆz .

Meanwhile, from Eq. 5.88, the dipole ﬁeld is

Bdip =µ0m

4⇡r32 cos ✓ˆr + sin ✓ˆ

✓,

and for points on the +zaxis ✓=0,r=z,ˆr =ˆz , so Bdip =µ0m

2⇡z3ˆz .In this case (Problem 5.37a) m=⇡!R4/4,

so Bdip =µ0!R4

8z3ˆz , in agreement with the approximation. X

Problem 5.49

B=µ0I

4⇡dl0⇥

=Rcos ˆx +(yRsin )ˆy +zˆz . (For simplicity I’ll drop the prime on

.)

2=R2cos2+y22Ry sin +R2sin2+z2=R2+y2+z22Ry sin . The source coordinates

(x0,y

0,z0) satisfy x0=Rcos )dx0=Rsin d;y0=Rsin )dy0=Rcos d;z0=0)dz0= 0. So

dl0=Rsin dˆx +Rcos dˆy .

dl0⇥

=

ˆx ˆy ˆz

Rsin dRcos d0

Rcos (yRsin )z=(Rz cos d)ˆx +(Rz sin d)ˆy +(Ry sin d+R2d)ˆz .

Bx=µ0IRz

4⇡2⇡

cos d

(R2+y2+z22Ry sin )3/2=µ0IRz

4⇡

R2+y2+z22Ry sin 

2⇡

=0,

since sin = 0 at both limits. The yand zcomponents are elliptic integrals, and cannot be expressed in terms

of elementary functions.

Bx= 0; By=µ0IRz

4⇡2⇡

sin d

(R2+y2+z22Ry sin )3/2;Bz=µ0IR

4⇡2⇡

(Rysin )d

(R2+y2+z22Ry sin )3/2.

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126 CHAPTER 5. MAGNETOSTATICS

Problem 5.50

From the Biot-Savart law, the ﬁeld of loop #1 is B=µ0I1

4⇡1

dl1⇥ˆ

2; the force on loop #2 is

F=I22

dl2⇥B=µ0

4⇡I1I212

dl2⇥(dl1⇥ˆ

)

2.Now dl2⇥(dl1⇥ˆ

)=dl1(dl2·ˆ

)ˆ

(dl1·dl2),so

F=µ0

4⇡I1I2 ˆ

2(dl1·dl2)dl1(dl2·ˆ

)

2

The ﬁrst term is what we want. It remains to show that the second term is zero:

=(x2x1)ˆx +(y2y1)ˆy +(z2z1)ˆz ,sor2(1/

)= @

@x2(x2x1)2+(y2y1)2+(z2z1)21/2ˆx

@y2(x2x1)2+(y2y1)2+(z2z1)21/2ˆy +@

@z2(x2x1)2+(y2y1)2+(z2z1)21/2ˆz

=(x2x1)

3ˆx (y2y1)

3ˆy (z2z1)

3ˆz =

3=ˆ

2. So ˆ

2·dl2=r21

·dl2= 0 (by

Corollary 2 in Sect. 1.3.3). qed

Problem 5.51

(a) B=µ0I

4⇡dl⇥ˆ

2=µ0I

4⇡dl⇥ˆr

r2,dl⇥ˆr =dl sin ˆz ,dl sin =rd✓(see diagram in the book),

so B=µ0I

4⇡d✓

r.X

(b) For a circular loop, r=Ris constant; B=µ0I

4⇡

Rd✓=µ0I

2R, which agrees with Eq. 5.41 (z= 0).

(c)

-0.5

0.5

1.0

1.5

-0.4

-0.2

0.2

0.4

0.6

0.8

B=µ0I

4⇡a2⇡

p✓d✓=µ0I

4⇡a2

3✓3/2

2⇡

=µ0Ip2⇡

3a.

(d) B=µ0I

4⇡p2⇡

(1 + ecos ✓)d✓=µ0I

4⇡p(2⇡)= µ0I

2p.X

Problem 5.52

(a)

B=µ0

4⇡J⇥ˆ

2d⌧)A=1

4⇡B⇥ˆ

2d⌧.

(b) Poisson’s equation (Eq. 2.24) says r2V=1

✏0

⇢. For dielectrics (with no free charge), ⇢b=r·P

(Eq. 4.12), and the resulting potential is V(r)= 1

4⇡✏0P(r0)·ˆ

2d⌧0. In general,⇢=✏0r·E(Gauss’s law),

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CHAPTER 5. MAGNETOSTATICS 127

so the analogy is P!✏0E, and hence V(r)=1

4⇡E(r0)·ˆ

2d⌧0.qed

[There are many other ways to obtain this result. For example, using Eq. 1.100:

r·ˆ

2=r0·ˆ

2=4⇡3(

)=4⇡3(rr0),

V(r)=V(r0)3(rr0)d⌧0=1

4⇡V(r0)r0·ˆ

2d⌧0=1

4⇡ˆ

2·r0V(r0)d⌧01

4⇡V(r0)ˆ

2·da0

(Eq. 1.59). But r0V(r0)=E(r0), and the surface integral !0 at 1, so V(r)=1

4⇡E(r0)·ˆ

2d⌧0, as

before. You can also check the result, by computing its gradient—but it’s not easy.]

Problem 5.53

(a) For uniform B,r

0(B⇥dl)=B⇥r

0dl=B⇥r6=A=1

2(B⇥r).

(b) B=µ0I

2⇡sˆ

,soB⇥dl=µ0I

2⇡aˆs µ0I

2⇡bˆs w=µ0Iw

2⇡1

a1

bˆs 6=0.

0d=1

2(r⇥B).

(d) B=µ0I

2⇡sˆ

;B(r)= µ0I

2⇡sˆ

;A=µ0I

2⇡s(r⇥ˆ

)1

1

d=µ0I

2⇡s(r⇥ˆ

). But rhere is the

vector from the origin—in cylindrical coordinates r=sˆs +zˆz .SoA=µ0I

2⇡ss(ˆs ⇥ˆ

)+z(ˆz ⇥ˆ

), and

(ˆs ⇥ˆ

)=ˆz ,(ˆz ⇥ˆ

)=ˆs . So A=µ0I

2⇡s(zˆs sˆz .

The examples in (c) and (d) happen to be divergenceless, but this is not in general the case. For (letting

L⌘1

0B(r)d, for short) r·A=r·(r⇥L)=[L·(r⇥r)r·(r⇥L)] = r·(r⇥L), and

r⇥L=1

0[r⇥B(r)] d=1

02[r⇥B(r)] d=µ01

02J(r)d,sor·A=µ0r·1

02J(r)d, and

it vanishes in regions where J=0(which is why the examples in (c) and (d) were divergenceless). To construct

an explicit counterexample, we need the ﬁeld at a point where J6=0—say, inside a wire with uniform current.

Here Amp´ere’s law gives B2⇡s=µ0Ienc =µ0J⇡s2)B=µ0J

2sˆ

,so

A=r⇥1

µ0J

2sˆ

d=µ0J

6s(r⇥ˆ

)=µ0Js

6(zˆs sˆz ).

r·A=µ0J

61

@s(s2z)+ @

@z(s2)=µ0J

61

s2sz=µ0Jz

36=0.

Conclusion: (ii) does not automatically yield r·A=0.

Problem 5.54

(a) Exploit the analogy with the electrical case:

E=1

4⇡✏0

r3[3(p·ˆr )ˆr p] (Eq.3.104) = rV, with V=1

4⇡✏0

p·ˆr

r2(Eq.3.102).

B=µ0

4⇡

r3[3(m·ˆr )ˆr m] (Eq.5.89) = rU, (Eq.5.67).

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128 CHAPTER 5. MAGNETOSTATICS

Evidently the prescription is p/✏0!µ0m:U(r)= µ0

4⇡

m·ˆr

r2.

(b) Comparing Eqs. 5.69 and 5.87, the dipole moment of the shell is m=(4⇡/3)!R4ˆz (which we also got

in Prob. 5.37). Using the result of (a), then, U(r)= µ0!R4

cos ✓

r2for r>R.

Inside the shell, the ﬁeld is uniform (Eq. 5.70): B=2

3µ0!Rˆz , so U(r)=2

3µ0!Rz + constant. We may

as well pick the constant to be zero, so U(r)=2

3µ0!Rr cos ✓for r<R.

[Notice that U(r) is not continuous at the surface (r=R): Uin(R)=2

3µ0!R2cos ✓6=Uout(R)=

3µ0!R2cos ✓. As I warned you on p. 245: if you insist on using magnetic scalar potentials, keep away from

places where there is current!]

(c)

B=µ0!Q

4⇡R13r2

5R2cos ✓ˆr 16r2

5R2sin ✓ˆ

✓=rU=@U

@rˆr 1

@✓ ˆ

✓1

rsin ✓

@ ˆ

.

@ =0)U(r, ✓,)=U(r, ✓).

@✓ =µ0!Q

4⇡R16r2

5R2sin ✓)U(r, ✓)=µ0!Q

4⇡R16r2

5R2rcos ✓+f(r).

@r=µ0!Q

4⇡R13r2

5R2cos ✓)U(r, ✓)=µ0!Q

4⇡Rrr3

5R2cos ✓+g(✓).

Equating the two expressions:

µ0!Q

4⇡R16r2

5R2rcos ✓+f(r)=µ0!Q

4⇡R1r2

5R2rcos ✓+g(✓),

or µ0!Q

4⇡R3r3cos ✓+f(r)=g(✓).

But there is no way to write r3cos ✓as the sum of a function of ✓and a function of r, so we’re stuck. The

reason is that you can’t have a scalar magnetic potential in a region where the current is nonzero.

Problem 5.55

(a) r·B=0,r⇥B=µ0J, and r·A=0,r⇥A=B)A=µ0

4⇡J

d⌧0,so

r·A=0,r⇥A=B, and r·W= 0 (we’ll choose it so), r⇥W=A)W=1

4⇡B

d⌧0.

(b) Wwill be proportional to Band to two factors of r(since di↵erentiating twice must recover B), so I’ll

try something of the form W=↵r(r·B)+r2B, and see if I can pick the constants ↵and in such a way

that r·W= 0 and r⇥W=A.

r·W=↵[(r·B)(r·r)+r·r(r·B)] + r2(r·B)+B·r(r2).rr=@x

@x+@y

@y+@z

@z= 1 + 1 + 1 = 3;

r(r·B)=r⇥(r⇥B)+B⇥(r⇥r)+(r·r)B+(B·r)r; but Bis constant, so all derivatives of Bvanish,

and r⇥r= 0 (Prob. 1.63), so

r(r·B)=(B·r)r=Bx

@x+By

@y+Bz

@z(xˆx +yˆy +zˆz )=Bxˆx +Byˆy +Bzˆz =B;

r(r2)=ˆx @

@x+ˆy @

@y+ˆz @

@z(x2+y2+z2)=2xˆx +2yˆy +2zˆz =2r. So

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CHAPTER 5. MAGNETOSTATICS 129

r·W=↵[3(r·B)+(r·B)] + [0 + 2(r·B)] = 2(r·B)(2↵+), which is zero if 2↵+= 0.

r⇥W=↵[(r·B)(r⇥r)r⇥r(r·B)] + r2(r⇥B)B⇥r(r2)=↵[0 (r⇥B)] + [0 2(B⇥r)]

=(r⇥B)(↵2)=1

2(r⇥B) (Prob. 5.25). So we want ↵2=1/2. Evidently ↵2(2↵)=5↵=1/2,

or ↵=1/10; =2↵=1/5. Conclusion: W=1

10 r(r·B)2r2B.(But this is certainly not unique.)

A·da. Integrate around the amperian loop shown, taking

Wto point parallel to the axis, and choosing W= 0 on the

axis:

Wl =s

0µ0nI

2l¯sd¯s=µ0nI

s2l

2(using Eq. 5.72 for A).

W=µ0nIs2

4ˆz (s<R).

For s>R,Wl =µ0nIR2l

4+s

Rµ0nI

2R2

¯sld¯s=µ0nIR2l

4+µ0nIR2l

2ln(s/R);

W=µ0nIR2

4[1 + 2 ln(s/R)] ˆz (s>R).

Problem 5.56

Apply the divergence theorem to the function [U⇥(r⇥V)], noting (from the product rule) that

r·[U⇥(r⇥V)] = (r⇥V)·(r⇥U)U·[r⇥(r⇥V)]:

r·[U⇥(r⇥V)] d⌧={(r⇥V)·(r⇥U)U·[r⇥(r⇥V)]}d⌧=[U⇥(r⇥V)] ·da.

As always, suppose we have two solutions, B1(and A1) and B2(and A2). Deﬁne B3⌘B2B1(and

A3⌘A2A1), so that r⇥A3=B3and r⇥B3=r⇥B1r⇥B2=µ0Jµ0J=0.SetU=V=A3

in the above identity:

{(r⇥A3)·(r⇥A3)A3·[r⇥(r⇥A3)]}d⌧={(B3)·(B3)A3·[r⇥B3]}d⌧=(B3)2d⌧

=[A3⇥(r⇥A3)] ·da=(A3⇥B3)·da.But either Ais speciﬁed (in which case A3=0), or else Bis

speciﬁed (in which case B3=0), at the surface. In either case (A3⇥B3)·da= 0. So (B3)2d⌧= 0, and

hence B1=B2. qed

Problem 5.57

From Eq. 5.88, Btot =B0ˆz µ0m0

4⇡r3(2 cos ✓ˆr +sin✓ˆ

✓). There-

fore B·ˆr =B0(ˆz ·ˆr )µ0m0

4⇡r32 cos ✓=B0µ0m0

2⇡r3cos ✓.

This is zero, for all ✓, when r=R, given by B0=µ0m0

2⇡R3, or

R=µ0m0

2⇡B01/3

.Evidently no ﬁeld lines cross this sphere.

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130 CHAPTER 5. MAGNETOSTATICS

Problem 5.58

(a) I=Q

(2⇡/!)=Q!

2⇡;a=⇡R2;m=Q!

2⇡⇡R2ˆz =Q

2!R2ˆz .L=RM v =M!R2;L=M!R2ˆz .

L=Q

!R2

M!R2=Q

2M.m=Q

2ML,and the gyromagnetic ratio is g=Q

2M.

(b) Because gis independent of R, the same ratio applies to all “donuts”, and hence to the entire sphere

(or any other ﬁgure of revolution): g=Q

2M.

2=e~

4m=(1.60 ⇥1019)(1.05 ⇥1034)

4(9.11 ⇥1031)= 4.61 ⇥1024 Am

Problem 5.59

(a) Bave =1

(3/4)⇡R3Bd⌧=3

4⇡R3(r⇥A)d⌧=

3

4⇡R3A⇥da=3

4⇡R3

µ0

4⇡J

d⌧0⇥da=

3µ0

(4⇡)2R3J⇥1

dad⌧0. Note that Jdepends on

the source point r0, not on the ﬁeld point r. To do the surface

integral, choose the (x, y, z) coordinates so that r0lies on the

zaxis (see diagram). Then

=R2+(z0)22Rz0cos ✓,

while da=R2sin ✓d✓dˆr . By symmetry, the xand ycom-

ponents must integrate to zero; since the zcomponent of ˆr is

cos ✓, we have

1

da=ˆz cos ✓

R2+(z0)22Rz0cos ✓R2sin ✓d✓d=2⇡R2ˆz ⇡

cos ✓sin ✓

R2+(z0)22Rz0cos ✓d✓.

Let u⌘cos ✓,so du =sin ✓d✓.

=2⇡R2ˆz 1

1

R2+(z0)22Rz0udu

=2⇡R2ˆz 22(R2+(z0)2)+2Rz0u

3(2Rz0)2R2+(z0)22Rz0u

1

=2⇡R2ˆz

3(Rz0)2R2+(z0)2+Rz0R2+(z0)22Rz0R2+(z0)2Rz0R2+(z0)2+2Rz0

=2⇡

3(z0)2ˆz R2+(z0)2+Rz0|Rz0|R2+(z0)2Rz0(R+z0)

=









4⇡

3z0ˆz =4⇡

3r0,(r0<R);

4⇡R3

3(z0)2ˆz =4⇡

(r0)3r0,(r0>R).











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CHAPTER 5. MAGNETOSTATICS 131

For now we want r0<R,soBave =3µ0

(4⇡)2R3

4⇡

3(J⇥r0)d⌧0=µ0

4⇡R3(J⇥r0)d⌧0. Now m=1

2(r⇥J)d⌧

(Eq. 5.90), so Bave =µ0

4⇡

R3. qed

(b) This time r0>R,soBave =3µ0

(4⇡)2R3

4⇡

3R3J⇥r0

(r0)3d⌧0=µ0

4⇡J⇥ˆ

2d⌧0, where

now

goes from the source point to the center (

=r0). Thus Bave =Bcen.qed

Problem 5.60

(a) Problem 5.53 gives the dipole moment of a shell: m=4⇡

3!R4ˆz . Let R!r, !⇢dr, and integrate:

m=4⇡

3!⇢ˆz R

r4dr =4⇡

3!⇢R5

5ˆz .But ⇢=Q

(4/3)⇡R3,so m=1

5Q!R2ˆz .

(b) Bave =µ0

4⇡

R3=µ0

4⇡

2Q!

5Rˆz .

=µ0

4⇡

msin ✓

r2ˆ

=µ0

4⇡

Q!R2

sin ✓

r2ˆ

.

(d) Use Eq. 5.69, with R!¯r, !⇢d¯r, and integrate:

A=µ0!⇢

sin ✓

r2ˆ

R

¯r4d¯r=µ0!

4⇡R3

sin ✓

5ˆ

=µ0

4⇡

Q!R2

sin ✓

r2ˆ

.

This is identical to (c); evidently the ﬁeld is pure dipole, for points outside the sphere.

(e) According to Prob. 5.30, the ﬁeld is B=µ0!Q

4⇡R13r2

5R2cos ✓ˆr 16r2

5R2sin ✓ˆ

✓. The average

obviously points in the zdirection, so take the zcomponent of ˆr (cos ✓) and ˆ

✓(sin ✓):

Bave =µ0!Q

4⇡R

(4/3)⇡R313r2

5R2cos2✓+16r2

5R2sin2✓r2sin ✓dr d✓d

=3µ0!Q

(4⇡R2)22⇡⇡

0r3

33

5R2cos2✓+R3

36

5R2sin2✓sin ✓d✓

=3µ0!Q

8⇡R4R3⇡

016

75 cos2✓+7

75 sin2✓sin ✓d✓=3µ0!Q

8⇡R

75 ⇡

07 + 9 cos2✓sin ✓d✓

=µ0!Q

200⇡R7 cos ✓3 cos3✓

⇡

0=µ0!Q

200⇡R(20) = µ0!Q

10⇡R(same as (b)).X

Problem 5.61

The issue (and the integral) is identical to the one in Prob. 3.48. The resolution (as before) is to regard

Eq. 5.89 as correct outside an inﬁnitesimal sphere centered at the dipole. Inside this sphere the ﬁeld is a

delta-function, A3(r), with Aselected so as to make the average ﬁeld consistent with Prob. 5.59:

Bave =1

(4/3)⇡R3A3(r)d⌧=3

4⇡R3A=µ0

4⇡

R3)A=2µ0m

3.The added term is 2µ0

3m3(r).

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132 CHAPTER 5. MAGNETOSTATICS

Problem 5.62

For a dipole at the origin and a ﬁeld point in the xz plane (= 0), we have

B=µ0

4⇡

r3(2 cos ✓ˆr + sin ✓ˆ

✓)= µ0

4⇡

r3[2 cos ✓(sin ✓ˆx + cos ✓ˆz ) + sin ✓(cos ✓ˆx sin ✓ˆz )]

=µ0

4⇡

r3[3 sin ✓cos ✓ˆx + (2 cos2✓sin2✓)ˆz ].

Here we have a stack of such dipoles, running from z=

L/2 to z=+L/2. Put the ﬁeld point at son the

xaxis. The ˆx components cancel (because of symmetri-

cally placed dipoles above and below z= 0), leaving B=

µ0

4⇡2Mˆz L/2

(3 cos2✓1)

r3dz,whereMis the dipole mo-

ment per unit length: m=I⇡R2=(vh)⇡R2=!R⇡R2h)

M=m

h=⇡!R3. Now sin ✓=s

r,so 1

r3=sin3✓

s3;z=

scot ✓)dz =s

sin2✓d✓. Therefore

B=µ0

2⇡(⇡!R3)ˆz ✓m

⇡/2

(3 cos2✓1)sin3✓

sin2✓d✓=µ0!R3

2s2ˆz ✓m

⇡/2

(3 cos2✓1) sin ✓d✓

=µ0!R3

2s2ˆz cos3✓+ cos ✓

✓m

⇡/2=µ0!R3

2s2cos ✓m1cos2✓mˆz =µ0!R3

2s2cos ✓msin2✓mˆz .

But sin ✓m=s

s2+(L/2)2,and cos ✓m=(L/2)

s2+(L/2)2,so B=µ0!R3L

4[s2+(L/2)2]3/2ˆz .

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CHAPTER 6. MAGNETIC FIELDS IN MATTER 133

Chapter 6

Magnetic Fields in Matter

Problem 6.1

N=m2⇥B1;B1=µ0

4⇡

r3[3(m1·ˆr )ˆr m1]; ˆr =ˆy ;m1=m1ˆz ;m2=m2ˆy .B1=µ0

4⇡

r3ˆz .

N=µ0

4⇡

m1m2

r3(ˆy ⇥ˆz )=µ0

4⇡

m1m2

r3ˆx . Here m1=⇡a2I,m2=b2I. So N=µ0

(abI)2

r3ˆx .Final orientation :

downward (ˆz ).

Problem 6.2

dF=Idl⇥B;dN=r⇥dF=Ir⇥(dl⇥B). Now (Prob. 1.6): r⇥(dl⇥B)+dl⇥(B⇥r)+B⇥(r⇥dl)=0.

But d[r⇥(r⇥B)] = dr⇥(r⇥B)+r⇥(dr⇥B) (since Bis constant), and dr=dl, so dl⇥(B⇥r)=r⇥(dl⇥B)

d[r⇥(r⇥B)]. Hence 2r⇥(dl⇥B)=d[r⇥(r⇥B)] B⇥(r⇥dl). dN=1

2I{d[r⇥(r⇥B)] B⇥(r⇥dl)}.

)N=1

2Id[r⇥(r⇥B)] B⇥(r⇥dl). But the ﬁrst term is zero (d(···)=0), and the second integral is

2a(Eq. 1.107). So N=I(B⇥a)=m⇥B.qed

Problem 6.3

(a)

Problem 6.3

✻

ˆz

R✲ˆy

✸B

ˆr

❥

Problem 6.5

✲

✻

❂

!J=J0ˆz

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According to Eq. 6.2, F=2⇡IRB cos ✓. But B=

µ0

4⇡

[3(m1·ˆr )ˆr m1]

r3, and Bcos ✓=B·ˆy ,soBcos ✓=

µ0

4⇡

r3[3(m1·ˆr )(ˆr ·ˆy )(m1·ˆy )]. But m1·ˆy = 0 and

ˆr ·ˆy = sin , while m1·ˆr =m1cos ✓.)Bcos ✓=

µ0

4⇡

r33m1sin cos .

F=2⇡IRµ0

4⇡

r33m1sin cos . Now sin =R

r,cos =pr2R2/r, so F=3

µ0

2m1IR2pr2R2

r5.

But IR2⇡=m2, so F=3µ0

2⇡m1m2pr2R2

r5, while for a dipole, R⌧r,so F=3µ0

2⇡

m1m2

r4.

(b) F=r(m2·B)=(m2·r)B=m2d

dz µ0

4⇡

z3(3(m1·ˆz )ˆz m1

  

2m1

)=µ0

2⇡m1m2ˆz d

dz 1

z3

  

31

or, since z=r:F=3µ0

2⇡

m1m2

r4ˆz .

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134 CHAPTER 6. MAGNETIC FIELDS IN MATTER

Problem 6.4

dF=I{(dy ˆy )⇥B(0,y,0) + (dz ˆz )⇥B(0,✏,z)(dy ˆy )⇥B(0,y,✏)(dz ˆz )⇥B(0,0,z)}

=I(dy ˆy )⇥[B(0,y,✏)B(0,y,0)]

  

⇡✏@B

+(dz ˆz )⇥[B(0,✏,z)B(0,0,z)]

  

⇡✏@B

@y

)I✏2ˆz ⇥@B

@yˆy ⇥@B

@z.Note that dy @B

@z0,y,0⇡✏@B

@z0,0,0and dz @B

@y0,0,z ⇡✏@B

@y0,0,0.

F=m

ˆx ˆy ˆz

001

@Bx

@By

@Bz



ˆx ˆy ˆz

010

@Bx

@By

@Bz

=mˆy @Bx

@yˆx @By

@yˆx @Bz

@z+ˆz @Bx

@z

=mˆx @Bx

@x+ˆy @Bx

@y+ˆz @Bx

@z

using r·B= 0 to write @By

@y+@Bz

@z=@Bx

@x.

But m·B=mBx(since m=mˆx , here), so r(m·B)=mr(Bx)=m@Bx

@xˆx +@Bx

@yˆy +@Bx

@zˆz .

Therefore F=r(m·B). qed

Problem 6.5

Problem 6.3

✻

ˆz

R✲ˆy

✸B

ˆr

❥

Problem 6.5

✲

✻

❂

!J=J0ˆz

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(a) B=µ0J0xˆy (Prob. 5.15).

m·B= 0, so Eq. 6.3 says F=0.

(b) m·B=m0µ0J0x, so F=m0µ0J0ˆx .

=p⇥(r⇥E)+E⇥(r⇥p)+(p·r)E+(E·r)p.

But pdoes not depend on (x, y, z), so the second

and fourth terms vanish, and r⇥E= 0, so the

ﬁrst term is zero. Hence r(p·E)=(p·r)E.qed

This argument does not apply to the magnetic analog,

since r⇥B6= 0. In fact, r(m·B)=(m·r)B+µ0(m⇥J).

(m·r)Ba=m0@

@x(B)=m0µ0J0ˆy ,(m·r)Bb=m0@

@y(µ0J0xˆy ) = 0.

Problem 6.6

Aluminum, copper, copper chloride, and sodium all have an odd number of electrons, so we expect them to

be paramagnetic. The rest (having an even number) should be diamagnetic.

Problem 6.7

✻

M✲ˆn

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Jb=r⇥M=0;Kb=M⇥ˆn =Mˆ

.

The ﬁeld is that of a surface current Kb=Mˆ

,

but that’s just a solenoid, so the ﬁeld

outside is zero, and inside B=µ0Kb=µ0M. Moreover, it points upward (in the drawing), so B=µ0M.

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CHAPTER 6. MAGNETIC FIELDS IN MATTER 135

Problem 6.8

r⇥M=Jb=1

@s(sks

2)ˆz =1

s(3ks2)ˆz =3ksˆz ,Kb=M⇥ˆn =ks2(ˆ

⇥ˆs )=kR2ˆz .

So the bound current ﬂows up the cylinder, and returns down the surface. [Incidentally, the total current should

be zero. . . is it? Yes, for Jbda =R

0(3ks)(2⇡sds)=2⇡kR3, while Kbdl =(kR2)(2⇡R)=2⇡kR3.] Since

these currents have cylindrical symmetry, we can get the ﬁeld by Amp`ere’s law:

B·2⇡s=µ0Ienc =µ0s

Jbda =2⇡kµ0s3)B=µ0ks2ˆ

=µ0M.

Outside the cylinder Ienc = 0, so B=0.

Problem 6.9

✿

③

▼

✌✎

✎ ✎ ✎ ✎ ✎ ✎K

✲

Kb=M×ˆn =Mˆ

φ.

(Essentially a long solenoid)

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Problem 6.7

✻

M✲ˆn

Problem 6.9

✿

③

▼

✌✎

✎ ✎ ✎ ✎ ✎ ✎K

✲

Kb=M×ˆn =Mˆ

φ.

(Essentially a long solenoid)

✶

✻

❄

✻

❄

(Essentially a physical dipole)

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✶

❖

✎

✛

✎ ✎

✲

M(Intermediate case)

Problem 6.12

}

✲

✻

❃

■Jb

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[The external ﬁelds are the same as in the electrical

case; the internal ﬁelds (inside the bar) are completely

di↵erent—in fact, opposite in direction.]

Problem 6.10

Kb=M, so the ﬁeld inside a complete ring would be µ0M. The ﬁeld of a square loop, at the center, is

given by Prob. 5.8: Bsq =p2µ0I/⇡R. Here I=Mw, and R=a/2, so

Bsq =p2µ0Mw

⇡(a/2) =2p2µ0Mw

⇡a; net ﬁeld in gap : B=µ0M12p2w

⇡a.

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136 CHAPTER 6. MAGNETIC FIELDS IN MATTER

Problem 6.11

As in Sec. 4.2.3, we want the average of B=Bout +Bin,whereBout is due to molecules outside a small

sphere around point P, and Bin is due to molecules inside the sphere. The average of Bout is same as ﬁeld at

center (Prob. 5.59b), and for this it is OK to use Eq. 6.10, since the center is “far” from all the molecules in

question:

Aout =µ0

4⇡

outside

M⇥ˆ

2d⌧

The average of Bin is µ0

4⇡2m

R3—Eq. 5.93—where m=4

3⇡R3M. Thus the average Bin is 2µ0M/3. But what is

left out of the integral Aout is the contribution of a uniformly magnetized sphere, to wit: 2µ0M/3 (Eq. 6.16),

and this is precisely what Bin puts back in. So we’ll get the correct macroscopic ﬁeld using Eq. 6.10. qed

Problem 6.12

}

✲

✻

❃

■Jb

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(a) M=ksˆz ;Jb=r⇥M=kˆ

;Kb=M⇥ˆn =kR ˆ

.

Bis in the zdirection (this is essentially a superposition of solenoids). So

B= 0 outside.Use the amperian loop shown (shaded)—inner side at radius s:

B·dl=Bl =µ0Ienc =µ0Jbda +Kbl=µ0[kl(Rs)+kRl]=µ0kls.

)B=µ0ksˆz inside.

(b) By symmetry, Hpoints in the zdirection. That same amperian loop gives H·dl=Hl =µ0Ifenc = 0, since

there is no free current here. So H=0, and hence B=µ0M.Outside M= 0, so B= 0; inside M=ksˆz ,

so B=µ0ksˆz .

Problem 6.13

(a) The ﬁeld of a magnetized sphere is 2

3µ0M(Eq. 6.16), so B=B02

3µ0M,with the sphere removed.

In the cavity, H=1

µ0B,soH=1

µ0B02

3µ0M=H0+M2

3M)H=H0+1

3M.

(b)

Problem 6.13

▼

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The ﬁeld inside a long solenoid is µ0K. Here K=M, so the ﬁeld of the bound current on

the inside surface of the cavity is µ0M, pointing down. Therefore

B=B0µ0M;

H=1

µ0

(B0µ0M)= 1

µ0

B0M)H=H0.

(c)

Problem 6.13

✰

✸Kb

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This time the bound currents are small, and far away from the center, so B=B0,

while H=1

µ0B0=H0+M)H=H0+M.

[Comment: In the wafer, Bis the ﬁeld in the medium; in the needle, His the Hin the medium; in the

sphere (intermediate case) both Band Hare modiﬁed.]

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CHAPTER 6. MAGNETIC FIELDS IN MATTER 137

Problem 6.14

✲

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;Bis the same as the ﬁeld of a short solenoid; H=1

µ0BM.

✲

✛

❄✻

✻❄

◆✍

✍◆

❯✕

✕❯

❥✯

✯❥

+✶

✶+

✲ ✲

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Problem 6.14

✛

❄✻

✻❄

◆✍

✍◆

❯✕

✕❯

❥✯

✯❥

*✶

✶*

✲ ✲

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Problem 6.15

“Potentials”:

Win(r, ✓)=

AlrlPl(cos ✓),(r<R);

Wout(r, ✓)=Bl

rl+1 Pl(cos ✓),(r>R).

Boundary Conditions:

(i) Win(R, ✓)=Wout(R, ✓),

(ii) @Wout

@rR+@Win

@rR=M?=Mˆz ·ˆr =Mcos ✓.

(The continuity of Wfollows from the gradient theorem: W(b)W(a)=b

arW·dl=b

aH·dl;

if the two points are inﬁnitesimally separated, this last integral !0.)

(i) )AlRl=Bl

Rl+1 )Bl=R2l+1Al,

(ii) )(l+ 1) Bl

Rl+2 Pl(cos ✓)+lAlRl2Pl(cos ✓)=Mcos ✓.

Combining these:

(2l+ 1)Rl1AlPl(cos ✓)=Mcos ✓,so Al=0(l6= 1),and 3A1=M)A1=M

Thus Win(r, ✓)= M

3rcos ✓=M

3z, and hence Hin =rWin =M

3ˆz =1

3M,so

B=µ0(H+M)=µ01

3M+M=2

3µ0M.X

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138 CHAPTER 6. MAGNETIC FIELDS IN MATTER

Problem 6.16

H·dl=Ifenc =I,soH=I

2⇡sˆ

.B=µ0(1 + m)H=µ0(1 + m)I

2⇡sˆ

.M=mH=mI

2⇡sˆ

.

Jb=r⇥M=1

@ssmI

2⇡sˆz = 0.Kb=M⇥ˆn =mI

2⇡aˆz ,at s=a;

mI

2⇡bˆz ,at s=b.

Total enclosed current, for an amperian loop between the cylinders:

I+mI

2⇡a2⇡a=(1+m)I, so B·dl=µ0Ienc =µ0(1 + m)I)B=µ0(1 + m)I

2⇡sˆ

.X

Problem 6.17

From Eq. 6.20: H·dl=H(2⇡s)=Ifenc =I(s2/a2),(s<a);

I(s>a).

H=Is

2⇡a2,(s<a)

2⇡s,(s>a),so B=µH =µ0(1+m)Is

2⇡a2,(s<a);

µ0I

2⇡s,(s>a).

Jb=mJf(Eq. 6.33), and Jf=I

⇡a2,so Jb=mI

⇡a2(same direction as I).

Kb=M⇥ˆn =mH⇥ˆn )Kb=mI

2⇡a(opposite direction to I).

Ib=Jb(⇡a2)+Kb(2⇡a)=mImI= 0 (as it should be, of course).

Problem 6.18

By the method of Prob. 6.15:

For large r, we want B(r, ✓)!B0=B0ˆz ,soH=1

µ0B!1

µ0B0ˆz , and hence W!1

µ0B0z=

1

µ0B0rcos ✓.

“Potentials”:

Win(r, ✓)=

AlrlPl(cos ✓),(r<R);

Wout(r, ✓)=1

µ0B0rcos ✓+Bl

rl+1 Pl(cos ✓),(r>R).

Boundary Conditions:

(i) Win(R, ✓)=Wout(R, ✓),

(ii) µ0@Wout

@rR+µ@Win

@rR=0.

(The latter follows from Eq. 6.26.)

(ii) )µ01

µ0

B0cos ✓+(l+ 1) Bl

Rl+2 Pl(cos ✓)+µlAlRl1Pl(cos ✓)=0.

For l6= 1, (i) )Bl=R2l+1Al,so[µ0(l+ 1) + µl]AlRl1= 0, and hence Al= 0.

For l= 1, (i) )A1R=1

µ0B0R+B1/R2, and (ii) )B0+2µ0B1/R3+µA1= 0, so A1=3B0/(2µ0+µ).

Win(r, ✓)=3B0

(2µ0+µ)rcos ✓=3B0z

(2µ0+µ).Hin =rWin =3B0

(2µ0+µ)ˆz =3B0

(2µ0+µ).

B=µH=3µB0

(2µ0+µ)=1+m

1+m/3B0.

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CHAPTER 6. MAGNETIC FIELDS IN MATTER 139

By the method of Prob. 4.23:

Step 1 :B0magnetizes the sphere: M0=mH0=m

µ0(1+m)B0. This magnetization sets up a ﬁeld within

the sphere given by Eq. 6.16:

B1=2

3µ0M0=2

m

1+m

B0=2

3B0(where ⌘m

1+m).

Step 2 :B1magnetizes the sphere an additional amount M1=

µ0B1. This sets up an additional ﬁeld in

the sphere:

B2=2

3µ0M1=2

3B1=2

32

B0,etc.

The total ﬁeld is:

B=B0+B1+B2+···=B0+ (2/3)B0+ (2/3)2B0+···=1 + (2/3) + (2/3)2+···B0=B0

(1 2/3).

12/3=3

32m/(1 + m)=3+3m

3+3m2m

=3(1 + m)

3+m

,so B=1+m

1+m/3B0.

Problem 6.19

m=e2r2

4meB;M=m

V=e2r2

4meVB, where Vis the volume per electron. M=mH(Eq. 6.29)

=m

µ0(1+m)B(Eq. 6.30). So m=e2r2

4meVµ0.[Note:m⌧1, so I won’t worry about the (1 + m)

term; for the same reason we need not distinguish Bfrom Belse, as we did in deriving the Clausius-Mossotti

equation in Prob. 4.41.] Let’s say V=4

3⇡r3. Then m=µ0

4⇡3e2

4mer. I’ll use 1 ˚

A= 1010 m for r.

Then m=(107)3(1.6⇥1019)2

4(9.1⇥1031)(1010 )=2⇥105,which is not bad—Table 6.1 says m=1⇥105.

However, I used only one electron per atom (copper has 29) and a very crude value for r. Since the orbital

radius is smaller for the inner electrons, they count for less (m⇠r2). I have also neglected competing

paramagnetic e↵ects. But never mind. . . this is in the right ball park.

Problem 6.20

Place the object in a region of zero magnetic ﬁeld, and heat it above the Curie point—or simply drop it on

a hard surface. If it’s delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it

back and forth many times; each time you reverse the direction, reduce the ﬁeld slightly.

Problem 6.21

(a) The magnetic force on the dipole is given by Eq. 6.3; to move the dipole in from inﬁnity we must exert an

opposite force, so the work done is

U=r

F·dl=r

1r(m·B)·dl=m·B(r)+m·B(1)

(I used the gradient theorem, Eq. 1.55). As long as the magnetic ﬁeld goes to zero at inﬁnity, then, U=m·B.

If the magnetic ﬁeld does not go to zero at inﬁnity, one must stipulate that the dipole starts out oriented

perpendicular to the ﬁeld.

(b) Identical to Prob. 4.8, but starting with Eq. 5.89 instead of 3.104.

4⇡

r3[3 cos ✓1cos ✓2cos(✓2✓1)]m1m2. Or, using cos(✓2✓1) = cos ✓1cos ✓2+ sin ✓1sin ✓2,

U=µ0

4⇡

m1m2

r3(sin ✓1sin ✓22 cos ✓1cos ✓2).

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140 CHAPTER 6. MAGNETIC FIELDS IN MATTER

Stable position occurs at minimum energy: @U

@✓1=@U

@✓2=0

@U

@✓1=µ0m1m2

4⇡r3(cos ✓1sin ✓2+ 2 sin ✓1cos ✓2)=0)2 sin ✓1cos ✓2=cos ✓1sin ✓2;

@✓2=µ0m1m2

4⇡r3(sin ✓1cos ✓2+ 2 cos ✓1sin ✓2)=0)2 sin ✓1cos ✓2=4 cos ✓1sin ✓2.

Thus sin ✓1cos ✓2= sin ✓2cos ✓1=0.





Either sin ✓1= sin ✓2=0: 1

! ! or 2

! 

or cos ✓1= cos ✓2=0: ""

3

or "#

4

Which of these is the stable minimum? Certainly not 2or 3—for these m2is not parallel to B1, whereas we

know m2will line up along B1. It remains to compare 1(with ✓1=✓2= 0) and 4(with ✓1=⇡/2, ✓2=⇡/2):

U1=µ0m1m2

4⇡r3(2); U2=µ0m1m2

4⇡r3(1). U1is the lower energy, hence the more stable conﬁguration.

Conclusion: They line up parallel, along the line joining them: ! !

(d) They’d line up the same way: ! ! ! ! ! !

Problem 6.22

F=Idl⇥B=Idl⇥B0+Idl⇥[(r·r0)B0]Idl⇥[(r0·r0)B0]=Idl⇥[(r·r0)B0]

(because dl=0). Now

(dl⇥B0)i=

j,k

✏ijkdlj(B0)k,and (r·r0)=

rl(r0)l,so

Fi=I

j,k,l

✏ijk rldlj[(r0)l(B0)k]Lemma 1 : rldlj=

✏ljmam(proof below).

=I

j,k,l,m

✏ijk✏ljmam(r0)l(B0)k



Lemma 2 : 

✏ijk✏ljm =ilkm imkl (proof below).





=I

k,l,m

(ilkm imkl)am(r0)l(B0)k=I

[ak(r0)i(B0)kai(r0)k(B0)k]

=I[(r0)i(a·B0)ai(r0·B0)] .

But r0·B0= 0 (Eq. 5.50), and m=Ia(Eq. 5.86), so F=r0(m·B0) (the subscript just reminds us to take

the derivatives at the point where mis located). qed

Proof of Lemma 1:

Eq. 1.108 says (c·r)dl=a⇥c=c⇥a. The jth component is pcprpdlj=p,m ✏jpmcpam. Pick

cp=pl (i.e. 1 for the lth component, zero for the others). Then rldlj=m✏jlmam=m✏ljmam.qed

Proof of Lemma 2:

✏ijk✏ljm = 0 unless ijk and ljm are both permutations of 123. In particular, imust either be lor m, and k

must be the other, so 

✏ijk✏ljm =Ailkm +Bimkl.

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CHAPTER 6. MAGNETIC FIELDS IN MATTER 141

To determine the constant A, pick i=l= 1, k=m= 3; the only contribution comes from j= 2:

✏123✏123 =1=A1133 +B1331 =A)A=1.

To determine B, pick i=m= 1, k=l= 3:

✏123✏321 =1=A1331 +B1133 =B)B=1.

So 

✏ijk✏ljm =ilkm imkl.qed

Problem 6.23

(a) B1=µ0

4⇡

z3ˆz (Eq. 5.88, with ✓= 0). So m2·B1=µ0

2⇡

z3.F=r(m·B) (Eq. 6.3) )F=@

@zµ0

2⇡

z3ˆz =

3µ0m2

2⇡z4ˆz . This is the magnetic force upward (on the upper magnet); it balances the gravitational force downward

(mdgˆz ):

3µ0m2

2⇡z4mdg=0)z=3µ0m2

2⇡mdg1/4

(b) The middle magnet is repelled upward by lower magnet and downward by upper magnet:

3µ0m2

2⇡x43µ0m2

2⇡y4mdg=0.

The top magnet is repelled upward by middle magnet, and attracted downward by lower magnet:

3µ0m2

2⇡y43µ0m2

2⇡(x+y)4mdg=0.

Subtracting: 3µ0m2

2⇡1

x41

y41

y4+1

(x+y)4mdg+mdg= 0, or 1

x42

y4+1

(x+y)4= 0, so: 2 = 1

(x/y)4+1

(x/y+1)4.

Let ↵⌘x/y; then 2 = 1

↵4+1

(↵+1)4. Mathematica gives the numerical solution ↵=x/y =0.850115 ...

Problem 6.24

(a) Forces on the upper charge:

Fq=1

4⇡✏0

z2ˆz ,Fm=r(m·B)=rm2µ0m

4⇡z3=µ0m2

2⇡3

z4ˆz .

At equilibrium,

4⇡✏0

z2=3µ0m2

2⇡z4)z2=6µ0✏0m2

q2)z=p6m

qc,

where 1/p✏0µ0=c, the speed of light.

(b) For electrons, q=1.6⇥1019 C (actually, it’s the magnitude of the charge we want in the expression

above), and m=9.22 ⇥1024 Am

2(the Bohr magneton—see Problem 5.58), so

z=p69.22 ⇥1024

(1.6⇥1019)(3 ⇥108)= 4.72 ⇥1013 m.

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142 CHAPTER 6. MAGNETIC FIELDS IN MATTER

(For comparison, the Bohr radius is 0.5⇥1010 m, so the equilibrium separation is about 1% of the size of a

hydrogen atom.)

could ﬁnd a way to maintain the orientation of the dipoles, and keep them on the zaxis, the structure would

fall apart.

Problem 6.25

(a) The electric ﬁeld inside a uniformly polarized sphere, E=1

3✏0P(Eq. 4.14) translates to H=1

3µ0(µ0M)=

1

3M.But B=µ0(H+M). So the magnetic ﬁeld inside a uniformly magnetized sphere is B=µ0(1

3M+M)=

3µ0M(same as Eq. 6.16).

(b) The electric ﬁeld inside a sphere of linear dielectric in an otherwise uniform electric ﬁeld is E=1

1+e/3E0

(Eq. 4.49). Now etranslates to m, for then Eq. 4.30 (P=✏0eE) goes to µ0M=µ0mH, or M=mH

(Eq. 6.29). So Eq. 4.49 )H=1

1+m/3H0.ButB=µ0(1 + m)H, and B0=µ0H0(Eqs. 6.31 and 6.32),

so the magnetic ﬁeld inside a sphere of linear magnetic material in an otherwise uniform magnetic ﬁeld is

µ0(1 + m)=1

(1 + m/3)

µ0

, or B=1+m

1+m/3B0(as in Prob. 6.18).

4⇡✏0

R3. Let’s pretend the charges

are all due to the frozen-in polarization of some medium (whatever ⇢might be, we can solve r·P=⇢to ﬁnd

the appropriate P). In this case there are no free charges, and p=Pd⌧, so Eave =1

4⇡✏0

R3Pd⌧, which

translates to

Have =1

4⇡µ0

R3µ0Md⌧=1

4⇡R3m.

But B=µ0(H+M), so Bave =µ0

4⇡

R3+µ0Mave, and Mave =m

3⇡R3, so Bave =µ0

4⇡

R3,in agreement

with Eq. 5.93. (We must assume for this argument that all the currents are bound, but again it doesn’t really

matter, since we can model any current conﬁguration by an appropriate frozen-in magnetization. See G. H.

Goedecke, Am. J. Phys. 66, 1010 (1998).)

Problem 6.26

Eq.2.15 : E=⇢1

4⇡✏0V

2d⌧0(for uniform charge density);

Eq.4.9: V=P·1

4⇡✏0V

2d⌧0(for uniform polarization);

Eq.6.11 : A=µ0✏0M⇥1

4⇡✏0V

2d⌧0(for uniform magnetization).

For a uniformly charged sphere (radius R): 





Ein =⇢1

3✏0r(Prob.2.12),

Eout =⇢1

3✏0

r2ˆr (Ex.2.3).

So the scalar potential of a uniformly polarized sphere is: Vin =1

3✏0(P·r),

Vout =1

3✏0

r2(P·ˆr ),

and the vector potential of a uniformly magnetized sphere is: Ain =µ0

3(M⇥r),

Aout =µ0

r2(M⇥ˆr ),

(conﬁrming the results of Ex. 4.2 and of Exs. 6.1 and 5.11).

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CHAPTER 6. MAGNETIC FIELDS IN MATTER 143

Problem 6.27

At the interface, the perpendicular component of Bis continuous (Eq. 6.26), and the parallel component of

His continuous (Eq. 6.25 with Kf= 0). So B?

1=B?

2,Hk

1=Hk

2. But B=µH(Eq. 6.31), so 1

µ1Bk

1=1

µ2Bk

Now tan ✓1=Bk

1/B?

1, and tan ✓2=Bk

2/B?

2, so

tan ✓2

tan ✓1

=Bk

=µ2

µ1

(the same form, though for di↵erent reasons, as Eq. 4.68).

Problem 6.28

In view of Eq. 6.33, there is a bound dipole at the center: mb=mm. So the net dipole moment at the

center is mcenter =m+mb= (1 + m)m=µ

µ0m. This produces a ﬁeld given by Eq. 5.89:

Bcenter

dipole =µ

4⇡

r3[3(m·ˆr )ˆr m].

This accounts for the ﬁrst term in the ﬁeld. The remainder must be due to the bound surface current (Kb) at

r=R(since there can be no volume bound current, according to Eq. 6.33). Let us make an educated guess

(based either on the answer provided or on the analogous electrical Prob. 4.37) that the ﬁeld due to the surface

bound current is (for interior points) of the form Bsurface

current =Am(i.e. a constant, proportional to m). In that

case the magnetization will be:

M=mH=m

µB=m

4⇡

r3[3(m·ˆr )ˆr m]+ m

µAm.

This will produce bound currents Jb=r⇥M=0, as it should, for 0 <r<R(no need to calculate this

curl—the second term is constant, and the ﬁrst is essentially the ﬁeld of a dipole, which we know is curl-less,

except at r= 0), and

Kb=M(R)⇥ˆr =m

4⇡R3(m⇥ˆr )+mA

µ(m⇥ˆr )=mm1

4⇡R3+A

µsin ✓ˆ

.

But this is exactly the surface current produced by a spinning sphere: K=v=!Rsin ✓ˆ

, with (!R)$

mmA

µ1

4⇡R3. So the ﬁeld it produces (for points inside) is (Eq. 5.70):

Bsurface

current =2

3µ0(!!

!R)=2

3µ0mmA

µ1

4⇡R3.

Everything is consistent, therefore, provided A=2

3µ0mA

µ1

4⇡R3, or A12µ0

3µm=2

µ0m

4⇡R3. But

m=µ

µ01, so A12

3+2

µ0

µ=2

(µµ0)

4⇡R3, or A1+2µ0

µ=2

(µ0µ)

4⇡R3;A=µ

4⇡

2(µ0µ)

R3(2µ0+µ), and hence

B=µ

4⇡1

r3[3(m·ˆr )ˆr m]+ 2(µ0µ)m

R3(2µ0+µ).qed

The exterior ﬁeld is that of the central dipole plus that of the surface current, which, according to Prob. 5.37,

is also a perfect dipole ﬁeld, of dipole moment

msurface

current =4

3⇡R3(!!

!R)=4

3⇡R33

2µ0

Bsurface

current=2⇡R3

µ0

4⇡

2(µ0µ)m

R3(2µ0+µ)=µ(µ0µ)m

µ0(2µ0+µ).

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144 CHAPTER 6. MAGNETIC FIELDS IN MATTER

So the total dipole moment is:

mtot =µ

µ0

m+µ

µ0

m(µ0µ)

(2µ0+µ)=3µm

(2µ0+µ),

and hence the ﬁeld (for r>R) is

B=µ0

4⇡3µ

2µ0+µ1

r3[3(m·ˆr )ˆr m].

Problem 6.29

The problem is that the ﬁeld inside a cavity is not the same as the ﬁeld in the material itself.

(a) Amp´ere type. The ﬁeld deep inside the magnet is that of a long solenoid, B0⇡µ0M. From Prob. 6.13:







Sphere : B=B02

3µ0M=1

3µ0M;

Needle : B=B0µ0M= 0;

Wafer : B=µ0M.

(b) Gilbert type. This is analogous to the electric case. The ﬁeld at the center is approximately that midway

between two distant point charges, B0⇡0. From Prob. 4.16 (with E!B,1/✏0!µ0,P!M):







Sphere : B=B0+µ0

3M=1

3µ0M;

Needle : B=B0= 0;

Wafer : B=B0+µ0M=µ0M.

In the cavities, then, the ﬁelds are the same for the two models, and this will be no test at all. Yes. Fund it

with $1 M from the Oﬃce of Alternative Medicine.

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CHAPTER 7. ELETRODYNAMICS 145

Chapter 7

Electrodynamics

Problem 7.1

(a) Let Qbe the charge on the inner shell. Then E=1

4⇡✏0

r2ˆr in the space between them, and (VaVb)=

a

bE·dr=1

4⇡✏0Qa

r2dr =Q

4⇡✏01

a1

b.

I=J·da=E·da=Q

✏0

=

✏0

4⇡✏0(VaVb)

(1/a 1/b)= 4⇡ (VaVb)

(1/a 1/b).

(b) R=VaVb

I=1

4⇡ 1

a1

b.

the region right around the inner sphere. Successive shells, as you go out, contribute less and less, because the

cross-sectional area (4⇡r2) gets larger and larger. For the two submerged spheres, R=2

4⇡a=1

2⇡a(one Ras

the current leaves the ﬁrst, one Ras it converges on the second). Therefore I=V /R = 2⇡aV.

Problem 7.2

(a) V=Q/C =IR. Because positive Imeans the charge on the capacitor is decreasing,

dt =I=1

RC Q, so Q(t)=Q0et/RC .ButQ0=Q(0) = CV0,so Q(t)=CV0et/RC .

Hence I(t)=dQ

dt =CV0

RC et/RC =V0

Ret/RC .

(b) W=1

2CV 2

0.The energy delivered to the resistor is 1

P dt =1

I2R dt =V2

R1

e2t/RC dt =

RRC

2e2t/RC 

2CV 2

0.X

dt =I=1

RC (CV0Q))dQ

QCV0

1

RC dt )ln(QCV0)=1

RC t+ constant )Q(t)=CV0+ket/RC . But Q(0) = 0 )k=CV0, so

Q(t)=CV01et/RC . I(t)= dQ

dt =CV01

RC et/RC =V0

Ret/RC .

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146 CHAPTER 7. ELETRODYNAMICS

(d) Energy from battery: 1

V0I dt =V2

R1

et/RC dt =V2

RRCet/RC 1

0=V2

RRC =CV 2

Since I(t) is the same as in (a), the energy delivered to the resistor is again 1

2CV 2

0.The ﬁnal energy in

the capacitor is also 1

2CV 2

0,so half the energy from the battery goes to the capacitor, and the other half

to the resistor.

Problem 7.3

(a) I=J·da, where the integral is taken over a surface enclosing the positively charged conductor. But

J=E, and Gauss’s law says E·da=1

✏0Q,soI=E·da=

✏0Q. But Q=CV , and V=IR,so

I=

✏0CIR, or R=✏0

C.qed

(b) Q=CV =CIR )dQ

dt =I=1

RC Q)Q(t)=Q0et/RC , or, since V=Q/C,V(t)=V0et/RC . The

time constant is ⌧=RC =✏0/.

Problem 7.4

I=J(s)2⇡sL )J(s)=I/2⇡sL. E =J/=I/2⇡sL=I/2⇡kL.

V=a

E·dl=I

2⇡kL(ab).So R=ba

2⇡kL.

Problem 7.5

I=E

r+R;P=I2R=E2R

(r+R)2;dP

dR =E21

(r+R)22R

(r+R)3=0)r+R=2R)R=r.

Problem 7.6

E=E·dl = zero for all electrostatic ﬁelds. It looks as though E=E·dl =(/✏0)h, as would indeed

be the case if the ﬁeld were really just /✏0inside and zero outside. But in fact there is always a “fringing

ﬁeld” at the edges (Fig. 4.31), and this is evidently just right to kill o↵the contribution from the left end of

the loop. The current is zero.

Problem 7.7

(a) E=d

dt =Bldx

dt =Blv;E=IR )I=Blv

R.(Never mind the minus sign—it just tells you the

direction of ﬂow: (v⇥B) is upward, in the bar, so downward through the resistor.)

(b) F=IlB =B2l2v

R,to the left.

dt =B2l2

Rv)dv

dt =B2l2

Rm v)v=v0eB2l2

mR t.

(d) The energy goes into heat in the resistor. The power delivered to resistor is I2R,so

dt =I2R=B2l2v2

R2R=B2l2

Rv2

0e2↵t,where ↵⌘B2l2

mR ;dW

dt =↵mv2

0e2↵t.

The total energy delivered to the resistor is W=↵mv2

01

e2↵tdt =↵mv2

e2↵t

2↵

=↵mv2

2↵=1

2mv2

0.X

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CHAPTER 7. ELETRODYNAMICS 147

Problem 7.8

(a) The ﬁeld of long wire is B=µ0I

2⇡sˆ

,so=B·da=µ0I

2⇡

s+a



s(ads)= µ0Ia

2⇡ln s+a

s.

(b) E=d

dt =µ0Ia

2⇡

dt ln s+a

s, and ds

dt =v, so µ0Ia

2⇡1

s+a

dt 1

dt =µ0Ia2v

2⇡s(s+a).

The ﬁeld points out of the page, so the force on a charge in the nearby side of the square is to the right.In

the far side it’s also to the right, but here the ﬁeld is weaker, so the current ﬂows counterclockwise.

Problem 7.9

Since r·B= 0, Theorem 2(c) (Sect. 1.6.2) guarantees that B·dais the same for all surfaces with a given

boundary line.

Problem 7.10

Contents

Problem 7.10

❘

✲

(view from above)

Φ=B·a=Ba2cos θ

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=B·a=Ba2cos ✓

Here ✓=!t,so

E=d

dt =Ba2(sin !t)!;

E=B!a2sin !t.

Problem 7.11

E=Blv =IR )I=Bl

Rv)upward magnetic force = IlB =B2l2

Rv. This opposes the gravitational force

downward:

mg B2l2

Rv=mdv

dt ;dv

dt =g↵v, where ↵⌘B2l2

mR .g↵vt=0)vt=g

↵=mgR

B2l2.

g↵v=dt )1

↵ln(g↵v)=t+ const. )g↵v=Ae↵t; at t=0,v=0,so A=g.

↵v=g(1 e↵t); v=g

↵(1 e↵t)= vt(1 e↵t).

At 90% of terminal velocity, v/vt=0.9=1e↵t)e↵t=10.9=0.1; ln(0.1) = ↵t; ln 10 = ↵t;

t=1

↵ln 10, or t90% =vt

gln 10.

Now the numbers: m=4⌘Al, where ⌘is the mass density of aluminum, Ais the cross-sectional area, and

lis the length of a side. R=4l/A, where is the conductivity of aluminum. So

vt=4⌘Alg4l

AB2l2=16⌘g

B2=16g⌘⇢

B2,and 









⇢=2.8⇥108⌦m

g=9.8m/s2

⌘=2.7⇥103kg/m3

B=1T











So vt=(16)(9.8)(2.7⇥103)(2.8⇥108)

1= 1.2 cm/s; t90% =1.2⇥102

9.8ln(10) = 2.8 ms.

If the loop were cut, it would fall freely, with acceleration g.

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148 CHAPTER 7. ELETRODYNAMICS

Problem 7.12

=⇡a

22B=⇡a2

4B0cos(!t); E=d

dt =⇡a2

4B0!sin(!t).I(t)= E

R=⇡a2!

4RB0sin(!t).

Problem 7.13

=B dx dy =kt2a

dx a

y3dy =1

4kt2a5.E=d

dt =1

2kta5.

Problem 7.14

②

Iind

✿

Iind

✲

ring

✲

pipe

✲

falling

magnet

✻

✗ ❖

✻❑ ✕

✿

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Suppose the current (I) in the magnet ﬂows counterclockwise (viewed from

above), as shown, so its ﬁeld, near the ends, points upward. A ring of

pipe below the magnet experiences an increasing upward ﬂux, as the magnet

approaches, and hence (by Lenz’s law) a current (Iind) will be induced in it

such as to produce a downward ﬂux. Thus Iind must ﬂow clockwise, which is

opposite to the current in the magnet. Since opposite currents repel, the force

on the magnet is upward. Meanwhile, a ring above the magnet experiences

adecreasing (upward) ﬂux, so its induced current is parallel to I, and it

attracts the magnet upward. And the ﬂux through rings next to the magnet

is constant, so no current is induced in them. Conclusion: the delay is due

to forces exerted on the magnet by induced eddy currents in the pipe.

Problem 7.15

In the quasistatic approximation, B=µ0nI ˆz ,(s<a);

0,(s>a).

Inside: for an “amperian loop” of radius s<a,

=B⇡s2=µ0nI⇡s2;E·dl=E2⇡s=d

dt =µ0n⇡s2dI

dt ;E=µ0ns

dt ˆ

.

Outside: for an “amperian loop” of radius s>a:

=B⇡a2=µ0nI⇡a2;E2⇡s=µ0n⇡a2dI

dt ;E=µ0na2

dt ˆ

.

Problem 7.16

(a) The magnetic ﬁeld (in the quasistatic approximation) is “circumferential”. This is analogous to the current

in a solenoid, and hence the ﬁeld is longitudinal.

(b) Use the “amperian loop” shown.

Outside, B=0,sohereE=0(like Boutside a solenoid).

So E·dl =El =d

dt =d

dt B·da=d

dt a

µ0I

2⇡s0l ds0

)E=µ0

2⇡

dt ln a

s.ButdI

dt =I0!sin !t,

so E=µ0I0!

2⇡sin(!t)lna

sˆz .

Problem 7.16

✲✲

✻

a✻

❄

✛ ✲

✻

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CHAPTER 7. ELETRODYNAMICS 149

Problem 7.17

(a) The ﬁeld inside the solenoid is B=µ0nI. So =⇡a2µ0nI )E=⇡a2µ0n(dI/dt).

In magnitude, then, E=⇡a2µ0nk. Now E=IrR,so Iresistor =⇡a2µ0nk

Bis to the right and increasing, so the ﬁeld of the loop is to the left, so the current is counterclockwise, or

to the right, through the resistor.

(b)  =2⇡a2µ0nI;I=dQ

dt =E

R=1

d

dt )Q=1

R, in magnitude. So Q=2⇡a2µ0nI

Problem 7.18

=B·da;B=µ0I

2⇡sˆ

;=µ0Ia

2⇡s+a

ds0

s0=µ0Ia

2⇡ln s+a

E=IloopR=dQ

dt R=d

dt =µ0a

2⇡ln(1 + a/s)dI

dt .

dQ =µ0a

2⇡Rln(1 + a/s)dI )Q=Iµ0a

2⇡Rln(1 + a/s).

The ﬁeld of the wire, at the square loop, is out of the page, and decreasing, so the ﬁeld of the induced

current must point out of page, within the loop, and hence the induced current ﬂows counterclockwise.

Problem 7.19

In the quasistatic approximation, B=µ0NI

2⇡sˆ

,(inside toroid);

0,(outside toroid)

(Eq. 5.60). The ﬂux around the toroid is therefore

=µ0NI

2⇡a+w

sh ds =µ0NIh

2⇡ln 1+w

a⇡µ0Nhw

2⇡aI. d

dt =µ0Nhw

2⇡a

dt =µ0Nhwk

2⇡a.

The electric ﬁeld is the same as the magnetic ﬁeld of a circular current (Eq. 5.41):

B=µ0I

(a2+z2)3/2ˆz ,

with (Eq. 7.19)

I!1

µ0

d

dt =Nhwk

2⇡a.So E=µ0

2Nhwk

2⇡aa2

(a2+z2)3/2ˆz =µ0

4⇡

Nhwka

(a2+z2)3/2ˆz .

Problem 7.20

@B/@tis nonzero along the left and right edges of the shaded rectangle:

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150 CHAPTER 7. ELETRODYNAMICS

The (inward) ﬂux through the strip on the left is increasing; the (inward) ﬂux through the strip on the right

is decreasing. This is analogous to two current sheets under Amp`ere’s law, with B!Eand µ0Ienc !d/dt

(Eq. 7.19). The one on the left is like a current ﬂowing out (taking account of the minus sign), so its ﬁeld is

counterclockwise and the one on the right is like a current ﬂowing in, so its ﬁeld is clockwise.

Problem 7.21

The answer is indeterminate, until some boundary conditions are supplied. We know that r⇥E=

(dB0/dt)ˆz (and r·E= 0), but this is insuﬃcient information to determine E. Ordinarily we would invoke

some symmetry of the conﬁguration, or require that the ﬁeld go to zero at inﬁnity, to resolve the ambiguity,

but neither is available in this case. E=1

dB0

dt (yˆx xˆy ) would do the job, in which case the force would be

zero, but we could add any constant vector to this, and make the force anything we like.

Problem 7.22

(a) From Eq. 5.41, the ﬁeld (on the axis) is B=µ0I

(b2+z2)3/2ˆz , so the ﬂux through the little loop (area ⇡a2)

is =µ0⇡Ia2b2

2(b2+z2)3/2.

(b) The ﬁeld (Eq. 5.88) is B=µ0

4⇡

r3(2 cos ✓ˆr + sin ✓ˆ

✓), where m=I⇡a2. Integrating over the spherical “cap”

(bounded by the big loop and centered at the little loop):

=B·da=µ0

4⇡

I⇡a2

r3(2 cos ✓)(r2sin ✓d✓d)=µ0Ia2

2r2⇡¯

✓

cos ✓sin ✓d✓

where r=pb2+z2and sin ¯

✓=b/r. Evidently =µ0I⇡a2

sin2✓

2

✓

0=µ0⇡Ia2b2

2(b2+z2)3/2,the same as in (a)!!

2(b2+z2)3/2.

Problem 7.23

E=d

dt =MdI

dt =Mk.

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CHAPTER 7. ELETRODYNAMICS 151

Problem 7.21

E=−

dΦ

dt =−MdI

dt =−Mk.

✻

❄

✻

❄

❖

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It’s hard to calculate Musing a current in the little loop, so, exploiting the equality of the mutual inductances,

I’ll ﬁnd the ﬂux through the little loop when a current Iﬂows in the big loop: =MI. The ﬁeld of one long

wire is B=µ0I

2⇡s)1=µ0I

2⇡2a

sads=µ0Ia

2⇡ln 2, so the total ﬂux is

=21=µ0Ialn 2

⇡)M=µ0aln 2

⇡)E=µ0ka ln 2

⇡,in magnitude.

Direction: The net ﬂux (through the big loop), due to Iin the little loop, is into the page. (Why? Field

lines point in, for the inside of the little loop, and out everywhere outside the little loop. The big loop encloses

all of the former, and only part of the latter, so net ﬂux is inward.) This ﬂux is increasing, so the induced

current in the big loop is such that its ﬁeld points out of the page: it ﬂows counterclockwise.

Problem 7.24

B=µ0nI )1=µ0nI⇡R2(ﬂux through a single turn). In a length lthere are nl such turns, so the

total ﬂux is =µ0n2⇡R2Il. The self-inductance is given by =LI, so the self-inductance per unit length is

L=µ0n2⇡R2.

Problem 7.25

The ﬁeld of one wire is B1=µ0

2⇡

s,so =2·µ0I

2⇡·l

d✏

✏

s=µ0Il

⇡ln d✏

✏. The ✏in the numerator is

negligible (compared to d), but in the denominator we cannot let ✏!0, else the ﬂux is inﬁnite.

L=µ0l

⇡ln(d/✏).Evidently the size of the wire itself is critical in determining L.

Problem 7.26

(a) In the quasistatic approximation B=µ0

2⇡sˆ

.So 1=µ0I

2⇡b

sh ds =µ0Ih

2⇡ln(b/a).

This is the ﬂux through one turn; the total ﬂux is Ntimes 1:=µ0Nh

2⇡ln(b/a)I0cos(!t).So

E=d

dt =µ0Nh

2⇡ln(b/a)I0!sin(!t)=(4⇡⇥107)(103)(102)

2⇡ln(2)(0.5)(2⇡60) sin(!t)

=2.61 ⇥104sin(!t) (in volts), where !=2⇡60 = 377/s. Ir=E

R=2.61 ⇥104

500 sin(!t)

= 5.22 ⇥107sin(!t) (amperes).

(b) Eb=LdIr

dt ; where (Eq. 7.28) L=µ0N2h

2⇡ln(b/a)=(4⇡⇥107)(106)(102)

2⇡ln(2) = 1.39 ⇥103(henries).

Therefore Eb=(1.39 ⇥103)(5.22 ⇥107!) cos(!t)= 2.74 ⇥107cos(!t) (volts).

Ratio of amplitudes: 2.74 ⇥107

2.61 ⇥104= 1.05 ⇥103=µ0N2h!

2⇡Rln(b/a).

Problem 7.27

With Ipositive clockwise, E=LdI

dt =Q/C, where Qis the charge on the capacitor; I=dQ

dt , so

d2Q

dt2=1

LC Q=!2Q, where !=1

pLC . The general solution is Q(t)=Acos !t+Bsin !t.Att= 0,

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152 CHAPTER 7. ELETRODYNAMICS

Q=CV ,soA=CV ;I(t)=dQ

dt =A!sin !t+B!sin !t.Att= 0, I= 0, so B= 0, and

I(t)=CV !sin !t=VC

Lsin t

pLC .

If you put in a resistor, the oscillation is “damped”. This time LdI

dt =Q

C+IR, so Ld2Q

dt2+RdQ

dt +1

CQ=0.

For an analysis of this case, see Purcell’s Electricity and Magnetism (Ch. 8) or any book on oscillations and

waves.

Problem 7.28

(a) W=1

2LI2.L=µ0n2⇡R2l(Prob. 7.24) W=1

2µ0n2⇡R2lI2.

(b) W=1

2(A·I)dl. A=(µ0nI/2)Rˆ

, at the surface (Eq. 5.72 or 5.73). So W1=1

µ0nI

2RI ·2⇡R, for one

turn. There are nl such turns in length l, so W=1

2µ0n2⇡R2lI2.X

2µ0B2d⌧.B=µ0nI, inside, and zero outside; d⌧=⇡R2l,soW=1

2µ0µ2

0n2I2⇡R2l=

2µ0n2⇡R2lI2.X

(d) W=1

2µ0B2d⌧(A⇥B)·da. This time B2d⌧=µ2

0n2I2⇡(R2a2)l. Meanwhile,

A⇥B=0outside (at s=b). Inside, A=µ0nI

2aˆ

(at s=a), while B=µ0nI ˆz .

A⇥B=1

2µ2

0n2I2a(ˆ

⇥ˆz



ˆs

)

Problem 7.26

☛

points inward (“out” of the volume)

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(A⇥B)·da=(1

2µ2

0n2I2aˆs )·[addz(ˆs )] = 1

2µ2

0n2I2a22⇡l.

W=1

2µ0µ2

0n2I2⇡(R2a2)l+µ2

0n2I2⇡a2l=1

2µ0n2I2R2⇡l. X

Contents

Problem 7.26

✲ˆz

❖

✕ˆs

✲ˆz

❄

✻

☛

✛ ✲

→I

✛

→

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Problem 7.29

B=µ0nI

2⇡s;W=1

2µ0B2d⌧=1

2µ0

µ2

0n2I2

4⇡21

s2hs dds =µ0n2I2

8⇡2h2⇡ln b

a=1

4⇡µ0n2I2hln(b/a).

L=µ0

2⇡n2hln (b/a) (same as Eq. 7.28).

Problem 7.30

B·dl=B(2⇡s)=µ0Ienc =µ0I(s2/R2))B=µ0Is

2⇡R2.

W=1

2µ0B2d⌧=1

2µ0

µ2

0I2

4⇡2R4R

s2(2⇡s)l ds =µ0I2l

4⇡R4s4

4

0=µ0l

16⇡I2=1

2LI2.

So L=µ0

8⇡l, and L=L/l =µ0/8⇡,independent of R!

Problem 7.31

(a) Initial current: I0=E0/R.SoLdI

dt =IR )dI

dt =R

LI)I=I0eRt/L, or I(t)= E0

ReRt/L.

(b) P=I2R=(E0/R)2e2Rt/LR=E2

Re2Rt/L =dW

dt .

W=E2

R1

e2Rt/L dt =E2

RL

2Re2Rt/L

=E2

R(0 + L/2R)= 1

2L(E0/R)2.

2LI2

0=1

2L(E0/R)2.X

Problem 7.32

(a) B1=µ0

4⇡

r3I1[3(a1·ˆ

)ˆ

a1], since m1=I1a1. The ﬂux through loop 2 is then

2=B1·a2=µ0

4⇡

3I1[3(a1·ˆ

)(a2·ˆ

)a1·a2]=MI1. M =µ0

4⇡

3[3(a1·ˆ

)(a2·ˆ

)a1·a2].

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CHAPTER 7. ELETRODYNAMICS 153

(b) E1=MdI2

dt ,dW

dt 1=E1I1=MI1dI2

dt . (This is the work done per unit time against the mutual emf in

loop 1—hence the minus sign.) So (since I1is constant) W1=MI1I2, where I2is the ﬁnal current in loop 2:

W=µ0

4⇡

3[3(m1·ˆ

)(m2·ˆ

)m1·m2].

Notice that this is opposite in sign to Eq. 6.35. In Prob. 6.21 we assumed that the magnitudes of the dipole

moments were ﬁxed, and we did not worry about the energy necessary to sustain the currents themselves—only

the energy required to move them into position and rotate them into their ﬁnal orientations. But in this

problem we are including it all, and it is a curious fact that this merely changes the sign of the answer. For

commentary on this subtle issue see R. H. Young, Am. J. Phys. 66, 1043 (1998), and the references cited

there.

Problem 7.33

(a) The (solenoid) magnetic ﬁeld is

B=µ0Kˆz =µ0!Rˆz (s<R),

0(s>R).

From Example 7.7, the electric ﬁeld is

E=









s

dt ˆ

=sR

2µ0˙!ˆ

(s<R),

R2

dt ˆ

=R3

2sµ0˙!ˆ

(s>R).

At the surface (s=R)E=1

2µ0R2˙!ˆ

,so the torque on a length `of the cylinder is

N=R(2⇡R`)1

2µ0R2˙!ˆz =⇡µ02R4˙!`ˆz ,

and the work done per unit length is W

`=⇡µ02R4d!

dt d.

But d=!dt, and the integral becomes

!f

!d!=1

2!2

f;)W

`=µ0⇡

2!fR22.

(This is the work done by the ﬁeld; the work you must do does not include the minus sign.)

(b) Because B=µ0Kˆz =µ0!fRˆz is uniform inside the solenoid (and zero outside), W=1

2µ0

B2⇡R2`.

`=1

2µ0

(µ0!fR)2⇡R2=µ0⇡

2!fR22.

Problem 7.34

The displacement current density (Sect. 7.3.2) is Jd=✏0@E

@t=I

A=I

⇡a2ˆz . Drawing an “amperian loop” at

radius s,

B·dl =B·2⇡s=µ0Idenc =µ0

⇡a2·⇡s2=µ0Is2

a2)B=µ0Is2

2⇡sa2;B=µ0Is

2⇡a2ˆ

.

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154 CHAPTER 7. ELETRODYNAMICS

Problem 7.35

(a) E=(t)

✏0

ˆz ;(t)=Q(t)

⇡a2=It

⇡a2;It

⇡✏0a2ˆz .

(b) Idenc =Jd⇡s2=✏0

dt ⇡s2=Is2

a2.B·dl=µ0Idenc )B2⇡s=µ0Is2

a2)B=µ0I

2⇡a2sˆ

.

of radius s. The charge density (at time t) is

(t)=[II(s)]t

⇡s2.

Since we are told this is independent of s, it must be that II(s)=s2, for some constant . But I(a) = 0,

so a2=I, or =I/a2. Therefore I(s)=I(1 s2/a2).

B2⇡s=µ0Ienc =µ0[II(s)] = µ0Is2

a2)B=µ0I

2⇡a2sˆ

.X

Problem 7.36

(a) Jd=✏0

µ0I0!2

2⇡cos(!t)ln(a/s)ˆz .ButI0cos(!t)=I. So Jd=µ0✏0

2⇡!2Iln(a/s)ˆz .

(b) Id=Jd·da=µ0✏0!2I

2⇡a

ln(a/s)(2⇡sds)=µ0✏0!2Ia

(sln asln s)ds

=µ0✏0!2I(ln a)s2

2s2

2ln s+s2

4

0=µ0✏0!2Ia2

2ln a

a2

2ln a

+a2

4=µ0✏0!2Ia2

I=µ0✏0!2a2

4.Since µ0✏0=1/c2,Id/I =(!a/2c)2. If a= 103m, and Id

I=1

100 , so that !a

2c=1

10 ,

!=2c

10a=3⇥108m/s

5⇥103m, or !=0.6⇥1011/s= 6⇥1010 /s; ⌫=!

2⇡⇡1010 Hz, or 104megahertz. (This is the

microwave region, way above radio frequencies.)

Problem 7.37

Physically, this is the ﬁeld of a point charge qat the origin, out to an expanding spherical shell of radius vt;

outside this shell the ﬁeld is zero. Evidently the shell carries the opposite charge, q.Mathematically, using

product rule #5 and Eq. 1.99:

r·E=✓(vt r)r·1

4⇡✏0

r2ˆr +1

4⇡✏0

r2ˆr ·r[✓(vt r)] = q

✏0

3(r)✓(vt r)+ 1

4⇡✏0

r2(ˆr ·ˆr )@

@r✓(vt r).

But 3(r)✓(vt r)=3(r)✓(t), and @

@r✓(vt r)=(vt r) (Prob. 1.46), so

⇢=✏0r·E=q3(r)✓(t)q

4⇡r2(vt r).

(For t<0 the ﬁeld and the charge density are zero everywhere.)

Clearly r·B= 0, and r⇥E=0(since Ehas only an rcomponent, and it is independent of ✓and ).

There remains only the Amp´ere/Maxwell law, r⇥B=0=µ0J+µ0✏0@E/@t. Evidently

J=✏0

@t=✏0q

4⇡✏0r2

@t[✓(vt r)]ˆr =q

4⇡r2v(vt r)ˆr .

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CHAPTER 7. ELETRODYNAMICS 155

(The stationary charge at the origin does not contribute to J, of course; for the expanding shell we have J=⇢v,

as expected—Eq. 5.26.)

Problem 7.38

From r·B=µ0⇢mit follows that the ﬁeld of a point monopole is B=µ0

4⇡

2ˆ

. The force law has the

form F/qmB1

c2v⇥E(see Prob. 5.22—the c2is needed on dimensional grounds). The proportionality

constant must be 1 to reproduce “Coulomb’s law” for point charges at rest. So F=qmB1

c2v⇥E.

Problem 7.39

Integrate the “generalized Faraday law” (Eq. 7.44iii), r⇥E=µ0Jm@B

@t, over the surface of the loop:

(r⇥E)·da=E·dl=E=µ0Jm·dad

dt B·da=µ0Imenc d

dt .

But E=LdI

dt ,so dI

dt =µ0

LImenc +1

d

dt ,or I=µ0

LQm+1

L,where Qmis the total magnetic charge

passing through the surface, and  is the change in ﬂux through the surface. If we use the ﬂat surface, then

Qm=qmand  = 0 (when the monopole is far away, = 0; the ﬂux builds up to µ0qm/2 just before

it passes through the loop; then it abruptly drops to µ0qm/2, and rises back up to zero as the monopole

disappears into the distance). If we use a huge balloon-shaped surface, so that qmremains inside it on the far

side, then Qm= 0, but rises monotonically from 0 to µ0qm. In either case,

I=µ0qm

[The analysis is slightly di↵erent for a superconducting loop, but the conclusion is the same.]

Problem 7.40

E=V

d)Jc=E=1

⇢E=V

⇢d.J

d=@D

@t=@

@t(✏E)=✏@

@tV0cos(2⇡⌫t)

d=✏V0

d[2⇡⌫ sin(2⇡⌫t)].

The ratio of the amplitudes is therefore:

=V0

⇢d

2⇡⌫✏V0

2⇡⌫✏⇢ =2⇡(4 ⇥108)(81)(8.85 ⇥1012)(0.23)1= 2.41.

Problem 7.41

Begin with a di↵erent problem: two parallel

wires carrying charges +and as shown.

Problem 7.39

✲y

✻

✰

+λ

−λ

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Field of one wire: E=

2⇡✏0sˆs ; potential: V=

2⇡✏0ln(s/a).

Potential of combination: V=

2⇡✏0ln(s/s+),

or V(y, z)= 

4⇡✏0ln (y+b)2+z2

(yb)2+z2.

Find the locus of points of ﬁxed V(i.e. equipotential surfaces):

e4⇡✏0V/⌘µ=(y+b)2+z2

(yb)2+z2=)µ(y22yb +b2+z2)=y2+2yb +b2+z2;

y2(µ1) + b2(µ1) + z2(µ1) 2yb(µ+ 1) = 0 =)y2+z2+b22yb=0 ⌘µ+1

µ1;

(yb)2+z2+b2b22=0=)(yb2)+z2=b2(21).

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156 CHAPTER 7. ELETRODYNAMICS

This is a circle, with center at y0=b=bµ+1

µ1and radius = b21=b(µ2+2µ+1)(µ22µ+1)

(µ1)2=2bpµ

µ1.

This suggests an image solution to the problem at hand. We want y0=d, radius = a, and V=V0. These

determine the parameters b,µ, and of the image solution:

a=y0

radius =bµ+1

µ1

2bpµ

µ1

=µ+1

2pµ.Call d

a⌘↵.

4↵2µ=(µ+ 1)2=µ2+2µ+1=)µ2+ (2 4↵2)µ+ 1 = 0;

µ=4↵22±4(1 2↵2)24

2=2↵21±14↵2+4↵41=2↵21±2↵↵21;

4⇡✏0V0

=lnµ=)=4⇡✏0V0

ln 2↵21±2↵p↵21.That’s the line charge in the image problem.

I=J·da=E·da=1

✏0

Qenc =

✏0

l.

The current per unit length is i=I

l=

✏0

=4⇡V0

ln 2↵21±2↵p↵21. Which sign do we want? Suppose

the cylinders are far apart, da, so that ↵1.

()=2↵21±2↵211/↵2=2↵21±2↵211

2↵21

8↵4+···

=2↵2(1 ±1) (1 ±1) ⌥1

4↵2±···=4↵221/2↵2+···⇡4↵2(+ sign),

1/4↵2(sign).

The current must surely decrease with increasing ↵, so evidently the + sign is correct:

i=4⇡V0

ln 2↵21+2↵p↵21,where ↵=d

Problem 7.42

From Prob. 3.24,











Vin(s, )=



k=1

skbksin(k),(s<a);

Vout(s, )= 1



k=1

skdksin(k),(s>a).

(We don’t need the cosine terms, because Vis clearly an odd function of .) At s=a, Vin =Vout =V0/2⇡.

Let’s start with Vin, and use Fourier’s trick to determine bk:



k=1

akbksin(k)=V0

2⇡)1



k=1

akbk⇡

⇡

sin(k) sin(k0)d=V0

2⇡⇡

⇡

sin(k0)d.But

⇡

⇡

sin(k) sin(k0)d=⇡kk0,and

⇡

⇡

sin(k0)d=1

(k0)2sin(k0)

k0cos(k0)

⇡

⇡

=2⇡

k0cos(k0)=2⇡

k0(1)k0.So

⇡akbk=V0

2⇡2⇡

k(1)k,or bk=V0

⇡k1

ak

,and hence Vin(s, )=V0

⇡



k=1

ks

aksin(k).

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CHAPTER 7. ELETRODYNAMICS 157

Similarly, Vout(s, )=V0

⇡



k=1

ka

sksin(k). Both sums are of the form S⌘1



k=1

k(x)ksin(k) (with

x=s/a for r<aand x=a/s for r>a). This series can be summed explicitly, using Euler’s formula

(ei✓= cos ✓+isin ✓): S= Im 1



k=1

k(x)keik= Im 1



k=1

kxeik.

But ln(1 + w)=w1

2w2+1

3w31

4w4···=1



k=1

k(w)k,so S=Im ln 1+xei.

Now ln Rei✓= ln R+i✓,so S=✓, where

tan ✓=Im 1+xei

Re (1 + xei)=

2i1+xei1+xei

2[(1 + xei)+(1+xei)] =xeiei

i[2 + x(ei+ei)] =xsin 

1+xcos .

Conclusion: 









Vin(s, )=

⇡tan1ssin 

a+scos ,(s<a);

Vout(s, )= V0

⇡tan1asin 

s+acos ,(s>a).

(b) From Eq. 2.36, ()=✏0@Vout

@ss=a@Vin

@ss=a.

@Vout

@s=V0

⇡









1+asin 

s+acos 2(asin )

(s+acos )2









=V0

⇡asin 

(s+acos )2+(asin )2

=V0

⇡asin 

s2+2as cos +a2;

@Vin

@s=V0

⇡









1+ssin 

a+scos 2[(a+scos ) sin ssin cos ]

(a+scos )2









=V0

⇡asin 

(a+scos )2+(ssin )2

=V0

⇡asin 

s2+2as cos +a2.

@Vin

@ss=a

=@Vout

@ss=a

=V0

2⇡asin 

1 + cos ,so ()=✏0V0

⇡a

sin 

(1 + cos )=✏0V0

⇡atan(/2).

Problem 7.43

(a) r2V=1

@ss@(zf)

@s+@2(zf)

@z2=z

ds sdf

ds=0)d

ds sdf

ds=0)sdf

ds =A(a constant) )

Ads

s=df )f=Aln(s/s0)(s0another constant).But (ii) )f(b) = 0, so ln(b/s0) = 0, so s0=b, and

V(s, z)=Az ln(s/b). But (i) )Az ln(a/b)=(I⇢z)/(⇡a2), so A=I⇢

⇡a2

ln(a/b);V(s, z)=I⇢z

⇡a2

ln(s/b)

ln(a/b).

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158 CHAPTER 7. ELETRODYNAMICS

(b) E=rV=@V

@sˆs @V

@zˆz =I⇢z

⇡a2

sln(a/b)ˆs +I⇢

⇡a2

ln(s/b)

ln(a/b)ˆz =I⇢

⇡a2ln(a/b)z

sˆs + ln s

bˆz .

⇡a2ln(a/b)z

a0=✏0I⇢z

⇡a3ln(a/b).

Problem 7.44

(a) Faraday’s law says r⇥E=@B

@t,soE=0)@B

@t=0)B(r) is independent of t.

(b) Faraday’s law in integral form (Eq. 7.19) says E·dl=d/dt. In the wire itself E=0,sothrough

the loop is constant.

@t, so E=0,B=0)J=0, and hence any current must be

at the surface.

(d) From Eq. 5.70, a rotating shell produces a uniform magnetic ﬁeld (inside): B=2

3µ0!aˆz . So to cancel

such a ﬁeld, we need !a=3

µ0

. Now K=v=!asin ✓ˆ

,so K=3B0

2µ0

sin ✓ˆ

.

Problem 7.45

(a) To make the ﬁeld parallel to the plane, we need image monopoles of the same sign (compare Figs. 2.13

and 2.14), so the image dipole points down (-z).

(b) From Prob. 6.3 (with r!2z):

F=3µ0

2⇡

(2z)4.3µ0

2⇡

(2h)4=Mg )h=1

23µ0m2

2⇡Mg 1/4

Problem 7.43

✻

❄

✲

r❯

✲ˆr

❯

ˆr 2

ˆr 1

✻

❄

−m

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B=µ0

4⇡

(r1)3{[3(mˆz ·ˆr 1)ˆr 1mˆz ] + [3(mˆz ·ˆr 2)ˆr 2+mˆz ]}

=3µ0m

4⇡(r1)3[(ˆz ·ˆr 1)ˆr 1(ˆz ·ˆr 2)ˆr 2].But ˆz ·ˆr 1=ˆz ·ˆr 2= cos ✓.

=3µ0m

4⇡(r1)3cos ✓(ˆr 1+ˆr 2).But ˆr 1+ˆr 2=2sin✓ˆr .

=3µ0m

2⇡(r1)3sin ✓cos ✓ˆr .But sin ✓=r

,cos ✓=h

,and r1=r2+h2.

=3µ0mh

2⇡

(r2+h2)5/2ˆr .

Now B=µ0(K⇥ˆz ))ˆz ⇥B=µ0ˆz ⇥(K⇥ˆz )=µ0[Kˆz (K·ˆz )] = µ0K. (I used the BAC-CAB rule,

and noted that K·ˆz = 0, because the surface current is in the xy plane.)

K=1

µ0

(ˆz ⇥B)=3mh

2⇡

(r2+h2)5/2(ˆz ⇥ˆr )=3mh

2⇡

(r2+h2)5/2ˆ

.qed

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CHAPTER 7. ELETRODYNAMICS 159

Problem 7.46

Say the angle between the dipole (m1) and the zaxis is ✓(see diagram).

Problem 7.44

✲x

✻

✒

−

❘

−

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The ﬁeld of the image dipole (m2) is

B(z)= µ0

4⇡

(h+z)3[3(m2·ˆz )ˆz m2]

for points on the zaxis (Eq. 5.89). The torque on m1is (Eq. 6.1)

N=m1⇥B=µ0

4⇡(2h)3[3(m2·ˆz )(m1⇥ˆz )(m1⇥m2)] .

But m1=m(sin ✓ˆx + cos ✓ˆz ),m2=m(sin ✓ˆx cos ✓ˆz ),so m2·ˆz =mcos ✓,m1⇥ˆz =msin ✓ˆy ,and

m1⇥m2=2m2sin ✓cos ✓ˆy .

N=µ0

4⇡(2h)33m2sin ✓cos ✓ˆy 2m2sin ✓cos ✓ˆy )=µ0m2

4⇡(2h)3sin ✓cos ✓ˆy .

Evidently the torque is zero for ✓= 0, ⇡/2, or ⇡. But 0 and ⇡are clearly unstable, since the nearby

ends of the dipoles (minus, in the ﬁgure) dominate, and they repel. The stable conﬁguration is ✓=⇡/2:

parallel to the surface (contrast Prob. 4.6).

In this orientation, B(z)=µ0m

4⇡(h+z)3ˆx , and the force on m1is (Eq. 6.3):

F=rµ0m2

4⇡(h+z)3z=h

=3µ0m2

4⇡(h+z)4ˆz z=h

=3µ0m2

4⇡(2h)4ˆz .

At equilibrium this force upward balances the weight Mg:

3µ0m2

4⇡(2h)4=Mg )h=1

23µ0m2

4⇡Mg 1/4

Incidentally, this is (1/2)1/4=0.84 times the height it would adopt in the orientation perpendicular to the

plane (Prob. 7.45b).

Problem 7.47

f=v⇥B;v=!asin ✓ˆ

;f=!aB0sin ✓(ˆ

⇥ˆz ). E=f·dl, and dl =ad✓ˆ

✓.

So E=!a2B0⇡/2

0sin ✓(ˆ

⇥ˆz )·ˆ

✓d✓. But ˆ

✓·(ˆ

⇥ˆz )=ˆz ·(ˆ

✓⇥ˆ

)=ˆz ·ˆr = cos ✓.

E=!a2B0⇡/2

sin ✓cos ✓d✓=!a2B0sin2✓

2⇡/2

0=1

2!a2B0(same as the rotating disk in Ex. 7.4).

Problem 7.48

Contents

✲x

✻

✲✛

}

l/2

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F=IBl;=2By

apa2x2dx (a= radius of circle).

E=d

dt =2Ba2y2dy

dt =2Bva2y2=IR.

I=2Bv

Ra2y2;l/2=a2y2. So F=4B2v

R(a2y2)=mg.

vcircle =mgR

4B2(a2y2);

tcircle =a

+ady

v=4B2

mgR a

a

(a2y2)dy =4B2

mgR (a2y1

3y3)

a

=4B2

mgR (4

3a3)= 16

B2a3

mgR .

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160 CHAPTER 7. ELETRODYNAMICS

Problem 7.49

(a) From Prob. 5.52a,

A(r,t)= 1

4⇡B(r0,t)⇥ˆ

2d⌧0,so E=@A

@t.

[Check: r⇥E=@

@t(r⇥A)=@B

@t, and we recover Faraday’s law.]

(b) The Coulomb ﬁeld is zero inside and 1

4⇡✏0

r2ˆr =1

4⇡✏0

4⇡R2

r2ˆr =R2

✏0r2ˆr outside. The Faraday ﬁeld is

@A

@t, where Ais given (in the quasistatic approximation) by Eq. 5.69, with !a function of time. Letting

˙!⌘d!/dt,

E(r, ✓,,t)= 









µ0R˙!

3rsin ✓ˆ

(r<R),

R2

✏0r2ˆr µ0R4˙!

sin ✓

r2ˆ

(r>R).

Problem 7.50

qBR =mv (Eq. 5.3). If Ris to stay ﬁxed, then qRdB

dt =mdv

dt =ma =F=qE, or E=RdB

dt . But

E·dl =d

dt ,so E2⇡R=d

dt ,so 1

2⇡R

d

dt =RdB

dt ,or B=1

21

⇡R2+ constant. If at time t=0

the ﬁeld is o↵, then the constant is zero, and B(R)=1

21

⇡R2(in magnitude). Evidently the ﬁeld at R

must be half the average ﬁeld over the cross-section of the orbit. qed

Problem 7.51

In the quasistatic approximation the magnetic ﬁeld of the wire is B=(µ0I/2⇡s)ˆ

, or, in Cartesian

coordinates,

B=µ0I

2⇡s(sin ˆx + cos ˆy )= µ0I

2⇡sy

sˆx +x

sˆy =µ0I

2⇡

(yˆx +xˆy )

x2+y2,

where xand yare measured from the wire. To convert to the stationary coordinates in the diagram, y!yvt:

B=µ0I

2⇡

[(yvt)ˆx +xˆy ]

x2+(yvt)2.

Faraday’s law says

r⇥E=@B

@t=µ0I

2⇡vˆx

x2+(yvt)2[(yvt)ˆx +xˆy ]

[x2+(yvt)2]2(2v)(yvt).

At t= 0, then,

r⇥E=µ0Iv

2⇡ˆx

x2+y2+2[y2ˆx +xy ˆy ]

[x2+y2]2=µ0Iv

2⇡(x2y2)ˆx +2xy ˆy ]

[x2+y2]2=µ0Iv

2⇡s2cos ˆs + sin ˆ

.

Our problem is to ﬁnd a vector function of sand (it obviously doesn’t depend on z) whose divergence is

zero, whose curl is as given above, that goes to zero at large s:

E(s, )=Es(s, )ˆs +E(s, )ˆ

+Ez(s, )ˆz ,

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CHAPTER 7. ELETRODYNAMICS 161

with

r·E=1

@s(sEs)+1

@E

@ =0,

(r⇥E)s=1

@Ez

@ =µ0Iv

2⇡s2cos 

(r⇥E)=@Ez

@s=µ0Iv

2⇡s2sin ,

(r⇥E)z=1

s@

@s(sE)@Es

@ =0.

The ﬁrst and last of these are satisﬁed if Es=E= 0, the middle two are satisﬁed by Ez=µ0Iv

2⇡ssin .

Evidently E=µ0Iv

2⇡ssin ˆz .The electric and magnetic ﬁelds ride along with the wire.

Problem 7.52

Initially, mv2

r=1

4⇡✏0

r2)T=1

2mv2=1

4⇡✏0

r. After the magnetic ﬁeld is on, the electron circles in a

new orbit, of radius r1and velocity v1:

mv2

4⇡✏0

+qv1B)T1=1

2mv2

1=1

4⇡✏0

2qv1r1B.

But r1=r+dr,so(r1)1=r11+ dr

r1⇠

=r11dr

r, while v1=v+dv,B=dB. To ﬁrst order, then,

T1=1

4⇡✏0

r1dr

r+1

2q(vr)dB, and hence dT =T1T=qvr

2dB 1

4⇡✏0

r2dr.

Now, the induced electric ﬁeld is E=r

dt (Ex. 7.7), so mdv

dt =qE =qr

dt , or m dv =qr

2dB. The increase in

kinetic energy is therefore dT =d(1

2mv2)=mv dv =qvr

2dB. Comparing the two expressions, I conclude that

dr = 0. qed

Problem 7.53

E=d

dt =↵. So the current in R1and R2is I=↵

R1+R2

; by Lenz’s law, it ﬂows counterclockwise. Now

the voltage across R1(which voltmeter #1 measures) is V1=IR1=↵R1

R1+R2

(Vbis the higher potential),

and V2=IR2=↵R2

R1+R2

(Vbis lower).

Problem 7.54

(a) E=d

dt =⇡r2dB

dt =↵⇡r2=IR )(in magnitude) I=⇡↵r2

R.If Bis out of the page, Lenz’s

law says the current is clockwise.

(b) Inside the shaded region, for a circle of radius s, apply Faraday’s law:

E·dl=E2⇡s=⇡s2↵)E=↵s

2ˆ

=↵s

2(sin ˆx + cos ˆy )=↵

2(ssin ˆx scos ˆy )=↵

2(yˆx xˆy ).

Along the line from Pto Q,dl =dx ˆx , and y=r/p2, so V=E·dl=↵

2y dx =↵

p2(rp2).Thus

Pis at the higher voltage, and the meter reads ↵r2

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162 CHAPTER 7. ELETRODYNAMICS

That’s the simplest way to do it. But you might instead regard the 3/4-circle+chord as a circuit, and use

Kirchho↵’s rule (the total emf around a closed loop is zero): V+IR1↵A1= 0, where R1=3/4Ris the

resistance of the curved portion, and A1= (3/4)⇡r2+r2/2=(r2/4)(3⇡+ 2) is the area (3/4 of the circle, plus

the triangle). Then

V=↵r2

4(3⇡+ 2) ⇡r2↵

4R=↵r2

4(3⇡+23⇡)=↵r2

2.X

Or we could do the same thing, using the small loop at the top: V+IR2↵A2= 0, where R2=(1/4)R

and A2= (1/4)⇡r2r2/2=(r2/4)(⇡2).

V=⇡↵r2

4↵r2

4(⇡2) = ↵r2

4(⇡⇡+ 2) = ↵r2

2.X

Problem 7.55

E=vBh =LdI

dt ;F=IhB =mdv

dt ;d2v

dt2=hB

dt =hB

mhB

Lv, d2v

dt2=!2v, with !=hB

pmL.

Problem 7.56

A point on the upper loop: r2=(acos 2,asin 2,z); a point on the lower loop: r1=(bcos 1,bsin 1,0).

2=(r2r1)2=(acos 2bcos 1)2+(asin 2bsin 1)2+z2

=a2cos222ab cos 2cos 1+b2cos21+a2sin222ab sin 1sin 2+b2sin21+z2

=a2+b2+z22ab(cos 2cos 1+ sin 2sin 1)=a2+b2+z22ab cos(21)

=(a2+b2+z2)12cos(21)=ab

12cos(21).

dl1=bd1ˆ

1=bd1[sin 1ˆx + cos 1ˆy ]; dl2=ad2ˆ

2=ad2[sin 2ˆx + cos 2ˆy ], so

dl1·dl2=ab d1d2[sin 1sin 2+ cos 1cos 2]=ab cos(21)d1d2.

M=µ0

4⇡ dl1·dl2

=µ0

4⇡

ab/ cos(21)

12cos(21)d2d1.

Both integrals run from 0 to 2⇡. Do the 2integral ﬁrst, letting u⌘21:

2⇡1



1

cos u

p12cos udu =

2⇡



cos u

p12cos udu

(since the integral runs over a complete cycle of cos u, we may as well change the limits to 0 !2⇡). Then the

1integral is just 2⇡, and

M=µ0

4⇡ab2⇡2⇡

cos u

p12cos udu =µ0

2ab2⇡

cos u

p12cos udu.

(a) If ais small, then ⌧1, so (using the binomial theorem)

p12cos u⇠

=1+cos u, and 2⇡

cos u

p12cos udu ⇠

=2⇡

cos udu+2⇡

cos2udu=0+⇡,

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CHAPTER 7. ELETRODYNAMICS 163

and hence M=(µ0⇡/2)ab3. Moreover, ⇠

=ab/(b2+z2), so M⇠

=µ0⇡a2b2

2(b2+z2)3/2(same as in Prob. 7.22).

(b) More generally,

(1 + ✏)1/2=11

2✏+3

8✏25

16✏3+···=)1

p12cos u=1+cos u+3

22cos2u+5

23cos3u+···,

M=µ0

2ab2⇡

cos udu+2⇡

cos2udu+3

222⇡

cos3udu+5

232⇡

cos4udu+···

=µ0

2ab0+(⇡)+3

22(0) + 5

23(3

4⇡)+···=µ0⇡

2ab31+15

82+()4+···.qed

Problem 7.57

Let be the ﬂux of Bthrough a single loop of either coil, so that 1=N1and 2=N2.Then

E1=N1

d

dt ,E2=N2

d

dt ,so E2

=N2

.qed

Problem 7.58

(a) Suppose current I1ﬂows in coil 1, and I2in coil 2. Then (if is the ﬂux through one turn):

1=I1L1+MI2=N1;2=I2L2+MI1=N2,or =I1

+I2

=I2

+I1

In case I1= 0, we have M

N1=L2

N2; if I2=0,wehave L1

N1=M

N2. Dividing: M

L1=L2

M, or L1L2=M2. qed

(b) E1=d1

dt =L1dI1

dt +MdI2

dt =V1cos(!t); E2=d2

dt =L2dI2

dt +MdI1

dt =I2R.qed

dt +L2dI2

dt M=L2V1cos !t. Plug in L2dI2

dt =I2RMdI1

dt .

M2dI1

dt MRI2M2dI1

dt =L2V1cos !t)I2(t)=L2V1

MR cos !t. L1dI1

dt +ML2V1

MR !sin !t=V1cos !t.

dI1

dt =V1

L1cos !tL2

R!sin !t)I1(t)= V1

L11

!sin !t+L2

Rcos !t.

(d) Vout

Vin

=I2R

V1cos !t=L2V1

MR cos !tR

V1cos !t=L2

M=N2

. The ratio of the amplitudes is N2

. qed

(e) Pin =VinI1=(V1cos !t)V1

L11

!sin !t+L2

Rcos !t=(V1)2

L11

!sin !tcos !t+L2

Rcos2!t.

Pout =VoutI2=(I2)2R=(L2V1)2

M2Rcos2!t. Average of cos2!tis 1/2; average of sin !tcos !tis zero.

So hPini=1

2(V1)2L2

L1R;hPouti=1

2(V1)2(L2)2

M2R=1

2(V1)2(L2)2

L1L2R;hPini=hPouti=(V1)2L2

2L1R.

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164 CHAPTER 7. ELETRODYNAMICS

Problem 7.59

(a) The charge ﬂowing into dz in time dt is

dq =I(z)dt I(z+dz)dt =dI

dt dz dt =(t+dt)dz (t)dz =d

dt dt dz )d

dt =dI

dz .X

Since the left side is a function only of t, and the right side is a function only of z, they must both be constant;

call it k:d

dt =k)(t)=kt +C1;dI

dz =k)I(z)=kz +C2.

If (0) = 0 and I(0) = 0 the constants C1and C2must both be zero: (t)=kt, I(z)=kz. X

(b) In the quasistatic approximation, E=

2⇡✏0sˆs =kt

2⇡✏0sˆs ;B=µ0I

2⇡sˆ

=µ0kz

2⇡sˆ

.

r·E=1

@sskt

2⇡✏0s=0 X

r·B=1

@ µ0kz

2⇡s=0 X

r⇥E=@E

@zˆ

1

@ ˆz =0=@B

@tX

r⇥B=@B

@zˆ

+1

@(sB)

@sˆz =µ0k

2⇡sˆz =µ0✏0

@tX

For a gaussian cylinder of radius sabout the zaxis, at height zand with width dz:

E·da=kt

2⇡✏0s(2⇡s)dz =kt

✏0

dz =dz

✏0

=Qenc

✏0

.XB·da=0.X

For a circular amperian loop of radius sabout the zaxis, at height z:

E·dl=0=d

dt B·da,XB·dl=µ0kz

2⇡s(2⇡s)=µ0I=µ0Ienc +µ0✏0

dt E·da.X

(Note that B·daand E·daare zero through this loop.)

Problem 7.60

(a) The continuity equation says @⇢

@t=r·J. Here the right side is independent of t, so we can integrate:

⇢(t)=(r·J)t+ constant. The “constant” may be a function of r—it’s only constant with respect to t. So,

putting in the rdependence explicitly, and noting that r·J=˙⇢(r,0), ⇢(r,t)= ˙⇢(r,0)t+⇢(r,0). qed

(b) Suppose E=1

4⇡✏0⇢ˆ

2d⌧and B=µ0

4⇡J⇥ˆ

2d⌧. We want to show that r·B= 0, r⇥B=

µ0J+µ0✏0@E

@t;r·E=1

✏0⇢, and r⇥E=@B

@t, provided that Jis independent of t.

We know from Ch. 2 that Coulomb’s law E=1

4⇡✏0⇢ˆ

2d⌧satisﬁes r·E=1

✏0⇢and r⇥E=0. Since

Bis constant (in time), the r·Eand r⇥Eequations are satisﬁed. From Chapter 5 (speciﬁcally, Eqs. 5.47-

5.50) we know that the Biot-Savart law satisﬁes r·B= 0. It remains only to check r⇥B. The argument in

Sect. 5.3.2 carries through until the equation following Eq. 5.54, where I invoked r0·J= 0. In its place we

now put r0·J=˙⇢:

r⇥B=µ0Jµ0

4⇡(J·r)ˆ

  

(J·r0)ˆ

d⌧(Eqs. 5.51-5.53)

(Eq. 5.54)

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CHAPTER 7. ELETRODYNAMICS 165

Integration by parts yields two terms, one of which becomes a surface integral, and goes to zero. The other is

3r0·J=ˆ

2(˙⇢). So:

r⇥B=µ0Jµ0

4⇡ˆ

2(˙⇢)d⌧=µ0J+µ0✏0

@t1

4⇡✏0⇢ˆ

3d⌧=µ0J+µ0✏0

@t.qed

Problem 7.61

Problem 7.56

vt −ϵvt

!"# $

−z

✲z

✻

✼

θdE

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(a) dEz=1

4⇡✏0

()dz

2sin ✓

sin ✓=z

;

=pz2+s2

Ez=

4⇡✏0z dz

(z2+s2)3/2=

4⇡✏01

pz2+s2

vt✏

Ez=

4⇡✏01

(vt ✏)2+s21

(vt)2+s2.

(b)

E=

4⇡✏0a

01

(vt ✏)2+s21

(vt)2+s22⇡sds=

2✏0(vt ✏)2+s2(vt)2+s2

=

2✏0(vt ✏)2+a2(vt)2+a2(✏vt)+(vt).

dE

dt =

2v(vt ✏)

(vt ✏)2+a2v(vt)

(vt)2+a2+2v.

As ✏!0, vt < ✏also !0, so Id!

2(2v)=v=I. With an inﬁnitesimal gap we attribute the magnetic ﬁeld

to displacement current, instead of real current, but we get the same answer. qed

Problem 7.62

Contents

Problem 7.26

✲ˆz

❖

✕ˆs

✲ˆz

❄

✻

☛

✛ ✲

→I

✛

→

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(a) Parallel-plate capacitor: E=1

✏0

;V=Eh =1

✏0

wlh)C=Q

V=✏0wl

h)C=✏0w

(b) B=µ0K=µ0

w;=Bhl =µ0I

whl =LI )L=µ0h

wl)L=µ0h

(Propagation speed 1/pLC =1/pµ0✏0=2.999 ⇥108m/s=c.)

(d) D=,E=D/✏=/✏,so just replace ✏0by ✏;

H=K, B =µH =µK, so just replace µ0by µ.LC =✏µ;v=1/p✏µ.

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166 CHAPTER 7. ELETRODYNAMICS

Problem 7.63

(a) J=(E+v⇥B); Jﬁnite, =1)E+(v⇥B)=0. Take the curl: r⇥E+r⇥(v⇥B)=0. But

Faraday’s law says r⇥E=@B

@t.So @B

@t=r⇥(v⇥B). qed

(b) r·B=0)B·da= 0 for any closed surface. Apply this at time (t+dt) to the surface consisting of

S,S0, and R:S0

B(t+dt)·da+R

B(t+dt)·daS

B(t+dt)·da=0

(the sign change in the third term comes from switching outward dato inward da).

d=S0

B(t+dt)·daS

B(t)·da=SB(t+dt)B(t)

  

@tdt ·daR

B(t+dt)·da

(for inﬁnitesimal dt)

d=S

@t·dadt R

B(t+dt)·(dl⇥v)dt(Figure 7.13).

Since the second term is already ﬁrst order in dt, we can replace B(t+dt)byB(t) (the distinction would be

second order):

d=dt S

@t·dadt C

B·(dl⇥v)

  

(v⇥B)·dl

=dt S@B

@t·daSr⇥(v⇥B)·da.

d

dt =S@B

@tr⇥(v⇥B)·da=0.qed

Problem 7.64

(a)

r·E0=(r·E) cos ↵+c(r·B) sin ↵=1

✏0

⇢ecos ↵+cµ0⇢msin ↵

✏0

(⇢ecos ↵+cµ0✏0⇢msin ↵)= 1

✏0

(⇢ecos ↵+1

c⇢msin ↵)= 1

✏0

⇢0

e.X

r·B0=(r·B) cos ↵1

c(r·E) sin ↵=µ0⇢mcos ↵1

c✏0

⇢esin ↵

=µ0(⇢mcos ↵1

cµ0✏0

⇢esin ↵)=µ0(⇢mcos ↵c⇢esin ↵)=µ0⇢0

m.X

r⇥E0=(r⇥E) cos ↵+c(r⇥B) sin ↵=µ0Jm@B

@tcos ↵+cµ0Je+µ0✏0

@tsin ↵

=µ0(Jmcos ↵cJesin ↵)@

@tBcos ↵1

cEsin ↵=µ0J0

m@B0

@t.X

r⇥B0=(r⇥B) cos ↵1

c(r⇥E) sin ↵=µ0Je+µ0✏0

@tcos ↵1

cµ0Jm@B

@tsin ↵

=µ0(Jecos ↵+1

cJmsin ↵)+µ0✏0

@t(Ecos ↵+cBsin ↵)=µ0J0

e+µ0✏0

@E0

@t.X

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CHAPTER 7. ELETRODYNAMICS 167

(b)

F0=q0

e(E0+v⇥B0)+q0

m(B01

c2v⇥E0)

=qecos ↵+1

cqmsin ↵(Ecos ↵+cBsin ↵)+v⇥Bcos ↵1

cEsin ↵

+(qmcos ↵cqesin ↵)Bcos ↵1

cEsin ↵1

c2v⇥(Ecos ↵+cBsin ↵)

=qeEcos2↵+cBsin ↵cos ↵cBsin ↵cos ↵+Esin2↵

+v⇥Bcos2↵1

cEsin ↵cos ↵+1

cEsin ↵cos ↵+Bsin2↵

+qm1

cEsin ↵cos ↵+Bsin2↵+Bcos2↵1

cEsin ↵cos ↵

+v⇥1

cBsin ↵cos ↵1

c2Esin2↵1

c2Ecos2↵1

cBsin ↵cos ↵

=qe(E+v⇥B)+qmB1

c2v⇥E=F.qed

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168 CHAPTER 8. CONSERVATION LAWS

Chapter 8

Conservation Laws

Problem 8.1

Example 7.13.

E=

2⇡✏0

sˆs

B=µ0I

2⇡

sˆ













S=1

µ0

(E⇥B)= I

4⇡2✏0

s2ˆz ;

P=S·da=



S2⇡sds=I

2⇡✏0



sds =I

2⇡✏0

ln(b/a).

But V=



E·dl=

2⇡✏0



sds =

2⇡✏0

ln(b/a),so P=IV.

Problem 7.62.

E=

✏0

ˆz

B=µ0Kˆx =µ0I

wˆx











S=1

µ0

(E⇥B)= I

✏0wˆy ;

P=S·da=Swh =Ih

✏0

,but V=E·dl=

✏0

h, so P=IV.

Problem 8.2

(a) E=

✏0

ˆz ;=Q

⇡a2;Q(t)=It )E(t)= It

⇡✏0a2ˆz .

B2⇡s=µ0✏0

@t⇡s2=µ0✏0

I⇡s2

⇡✏0a2)B(s, t)= µ0Is

2⇡a2ˆ

.

(b) uem =1

2✏0E2+1

µ0

B2=1

2✏0It

⇡✏0a22

µ0µ0Is

2⇡a22=µ0I2

2⇡2a4(ct)2+(s/2)2.

S=1

µ0

(E⇥B)= 1

µ0It

⇡✏0a2µ0Is

2⇡a2(ˆs )= I2t

2⇡2✏0a4sˆs .

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CHAPTER 8. CONSERVATION LAWS 169

@uem

@t=µ0I2

2⇡2a42c2t=I2t

⇡2✏0a4;r·S=I2t

2⇡2✏0a4r·(sˆs )= I2t

⇡2✏0a4=@uem

@t.X

2⇡2a4b

[(ct)2+(s/2)2]sds=µ0wI2

⇡a4(ct)2s2

2+1

4

=µ0wI2b2

2⇡a4(ct)2+b2

8.Over a surface at radius b:Pin =S·da=I2t

2⇡2✏0a4[bˆs ·(2⇡bw ˆs )] = I2wtb2

⇡✏0a4.

dUem

dt =µ0wI2b2

2⇡a42c2t=I2wtb2

⇡✏0a4=Pin.X(Set b=afor total.)

Problem 8.3

The force is clearly in the zdirection, so we need

T·da)z=Tzx dax+Tzy day+Tzz daz=1

µ0BzBxdax+BzByday+BzBzdaz1

2B2daz

µ0Bz(B·da)1

2B2daz.

Now B=2

3µ0R!ˆz (inside) and B=µ0m

4⇡r3(2 cos ✓ˆr + sin ✓ˆ

✓) (outside), where m=4

3⇡R3(!R). (From

Eq. 5.70, Prob. 5.37, and Eq. 5.88.) We want a surface that encloses the entire upper hemisphere—say a

hemispherical cap just outside r=Rplus the equatorial circular disk.

Hemisphere:

Bz=µ0m

4⇡R32 cos ✓(ˆr )z+ sin ✓(ˆ

✓)z=µ0m

4⇡R32 cos2✓sin2✓=µ0m

4⇡R33 cos2✓1.

da=R2sin ✓d✓dˆr ;B·da=µ0m

4⇡R3(2 cos ✓)R2sin ✓d✓d;daz=R2sin ✓d✓dcos ✓;

B2=µ0m

4⇡R324 cos2✓+ sin2✓=µ0m

4⇡R323 cos2✓+1

.

T·da)z=1

µ0µ0m

4⇡R323 cos2✓12 cos ✓R2sin ✓d✓d1

23 cos2✓+1

R2sin ✓cos ✓d✓d

=µ0!R

321

2R2sin ✓cos ✓d✓d12 cos2✓43 cos2✓1

=µ0

2!R2

329 cos2✓5sin ✓cos ✓d✓d.

(Fhemi)z=µ0

2!R2

32

2⇡

⇡/2



09 cos3✓5 cos ✓sin ✓d✓=µ0⇡!R2

329

4cos4✓+5

2cos2✓

⇡/2

=µ0⇡!R2

320+9

45

2=µ0⇡

4wR2

32

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170 CHAPTER 8. CONSERVATION LAWS

Disk:

Bz=2

3µ0R!;da=r dr dˆ

=r dr dˆz ;

B·da=2

3µ0R!rdrd;B2=2

3µ0R!2

;daz=r dr d.

T·da)z=1

µ02

3µ0R!2r dr d+1

2r dr d=1

2µ02

3µ0R!2

r dr d.

(Fdisk)z=2µ0!R

32

2⇡



r dr =2⇡µ0!R2

32

Total:

F=⇡µ0!R2

322+1

4ˆz =⇡µ0!R2

22

ˆz (agrees with Prob.5.44).

Alternatively, we could use a surface consisting of the entire equatorial plane, closing it with a hemispherical

surface “at inﬁnity,” where (since the ﬁeld is zero out there) the contribution is zero. We have already done

the integral over the disk; it remains to do rest of the integral over the plane, from Rto 1. On the plane,

✓= 0, and (for r>R)B=µ0m

4⇡r3ˆ

✓=µ0m

4⇡r3ˆz ,so

T·da)z=1

µ0Bz(Bzdaz)1

2B2

zdaz)=1

2µ0

zdaz=1

2µ0µ0m

4⇡r32(r dr d).

The contribution from the rest of the plane is therefore

(Frest)z=1

2µ0µ0m

4⇡22⇡1

r5dr =µ0

16⇡4

3⇡R4!21

4r4

=µ0⇡R4!

321

4R4

=µ0⇡

4!R2

32

This is the same as (Fhemi)z, so—when added to (Fdisk)z—it will yield the same total force as before.

Problem 8.4

(a)

T·da)z=Tzx dax+Tzy day+Tzz daz.

But for the xy plane dax=day= 0, and daz=

r dr d(I’ll calculate the force on the upper charge).

T·da)z=✏0EzEz1

2E2(r dr d).

Now E=1

4⇡✏0

2cos ✓ˆr , and cos ✓=r

,soEz=

0,E

2=q

2⇡✏02r2

(r2+a2)3. Therefore

Hello

✲y

✻

❂

✇

✼θ

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Fz=1

2✏0q

2⇡✏02

2⇡



r3dr

(r2+a2)3=q2

4⇡✏0



udu

(u+a2)3(letting u⌘r2)

=q2

4⇡✏0

21

(u+a2)+a2

2(u+a2)3

=q2

4⇡✏0

20+ 1

a2a2

2a4=q2

4⇡✏0

(2a)2.X

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CHAPTER 8. CONSERVATION LAWS 171

(b) In this case E=1

4⇡✏0

2sin ✓ˆz , and sin ✓=a

, so

E2=E2

z=qa

2⇡✏021

(r2+a2)3,and hence ($

T·da)z=✏0

2qa

2⇡✏02r dr d

(r2+a2)3.Therefore

Fz=✏0

2qa

2⇡✏02

2⇡



r dr

(r2+a2)3=q2a2

4⇡✏01

(r2+a2)21

=q2a2

4⇡✏00+ 1

4a4=q2

4⇡✏0

(2a)2.X

Problem 8.5

(a) E=

✏0

ˆz ,B=µ0vˆx ,g=✏0(E⇥B)=µ02vˆy ,p=(dA)g=dAµ02vˆy .

(b) (i) There is a magnetic force, due to the (average) magnetic ﬁeld at the upper plate:

F=q(u⇥B)=A[(uˆz )⇥(1

2µ0vˆx )] = 1

2µ02Avu ˆy ,

I1=Fdt =1

2µ02Av ˆy udt=1

2dµ02Av ˆy .

[The velocity of the patch (of area A) is actually v+u=vˆy uˆz , but the ycomponent produces a magnetic

force in the zdirection (a repulsion of the plates) which reduces their (electrical) attraction but does not deliver

(horizontal) momentum to the plates.]

(ii) Meanwhile, in the space immediately above the upper plate the magnetic ﬁeld drops abruptly to

zero (as the plate moves past), inducing an electric ﬁeld by Faraday’s law. The magnetic ﬁeld in the vicinity

of the top plate (at d(t)=d0ut) can be written, using Problem 1.46(b),

B(z,t)=µ0v✓(dz)ˆx ,)@B

@t=µ0vu (dz)ˆx .

In the analogy at the beginning of Section 7.2.2, the Faraday-induced electric ﬁeld is just like the magnetostatic

ﬁeld of a surface current K=vu ˆx .Referring to Eq. 5.58, then,

Eind =1

2µ0vu ˆy ,for z<d,

2µ0vu ˆy ,for z>d.

This induced electric ﬁeld exerts a force on area Aof the bottom plate, F=(A)(1

2µ0vu ˆy ), and delivers

an impulse

I2=1

2µ02Av ˆy udt=1

2µ02Avd ˆy .

(I dropped the subscript on d0, reverting to the original notation: dis the initial separation of the plates.)

The total impulse is thus I=I1+I2=dAµ02vˆy ,matching the momentum initially stored in the ﬁelds,

from part (a). [I thank Michael Ligare for untangling this surprisingly subtle problem. Incidentally, there is

also “hidden momentum” in the original conﬁguration. It is not relevant here; it is (relativistic) mechanical

momentum (see Example 12.13), and is delivered to the plates as they come together, so it does not a↵ect the

overall conservation of momentum.]

Problem 8.6

(a) gem =✏0(E⇥B)=✏0EB ˆy ;pem =✏0EBAd ˆy .

(b) II

I=1

Fdt =1

I(l⇥B)dt =1

IBd(ˆz ⇥ˆx )dt =(Bd ˆy )1

0dQ

dt dt

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172 CHAPTER 8. CONSERVATION LAWS

=(Bdˆy )[Q(1)Q(0)] = BQd ˆy .But the original ﬁeld was E=/✏0=Q/✏0A, so Q=✏0EA, and hence

III=✏0EBAd ˆy .

Problem 8.7

(a) Ex=Ey=0,E

z=/✏0.Therefore

Txy =Txz =Tyz =···= 0; Txx =Tyy =✏0

2E2=2

2✏0

;Tzz =✏0E2

z1

2E2=✏0

2E2=2

2✏0

T=2

2✏0

10 0

010

00+1



.

(b) F=$

T·da(S=0, since B=0); integrate over the xy plane: da=dx dy ˆz (negative because

outward with respect to a surface enclosing the upper plate). Therefore

Fz=Tzz daz=2

2✏0

A, and the force per unit area is f=F

A=2

2✏0

ˆz .

area, per unit time.

(d) The recoil force is the momentum delivered per unit time, so the force per unit area on the top plate is

f=2

2✏0

ˆz (same as (b)).

Problem 8.8

B=µ0nI ˆz (for s<R; outside the solenoid B= 0). The force on a segment ds of spoke is

dF=I0dl⇥B=I0µ0nI ds(ˆs ⇥ˆz )=I0µ0nI ds ˆ

.

The torque on the spoke is

N=r⇥dF=I0µ0nI



sds(ˆs ⇥ˆ

)=I0µ0nI 1

2R2a2(ˆz ).

Therefore the angular momentum of the cylinders is L=Ndt =1

2µ0nI(R2a2)ˆz I0dt. But I0dt =Q,

L=1

2µ0nIQ(R2a2)ˆz (in agreement with Eq.8.34).

Problem 8.9

(a) Between the shells, E=Q

4⇡✏0

r2ˆr ,g=✏0(E⇥B)= QB0

4⇡r2(ˆr ⇥ˆz ).

L=(r⇥g)d⌧=QB0

4⇡1

r2[r⇥(ˆr ⇥ˆz )]r2sin ✓dr d✓d=QB0

4⇡r[ˆr ⇥(ˆr ⇥ˆz )] sin ✓dr d✓d.

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CHAPTER 8. CONSERVATION LAWS 173

Now ˆr ⇥(ˆr ⇥ˆz )=ˆr (ˆr ·ˆz )ˆz (ˆr ·ˆr )=ˆr cos ✓=ˆz , but Lhas to be along the zdirection, so we pick o↵the z

component of ˆr :[ˆr ⇥(ˆr ⇥ˆz )]z= cos2✓1=sin2✓.

Lz=QB0

4⇡rsin3✓dr d✓d=QB0

4⇡2⇡⇡

sin3✓d✓b

r dr =QB0

24

3b2a2

2=1

3QB0(b2a2).

L=1

3QB0(b2a2)ˆz .

(b) The changing magnetic ﬁeld induces an electric ﬁeld (Example 7.7). Assuming symmetry about the z

axis, E=s

dt ˆ

.The force on a patch of area da is dF=Eda, and the torque on this patch is dN=s⇥dF.

The net torque on the sphere at radius ais (using s=asin ✓ˆs and da =a2sin ✓d✓d):

Na=Q

4⇡a21

dt ˆz s2da =Qa2

8⇡

dt ˆz 2⇡sin3✓d✓=Qa2

dt ˆz .

Similarly, Nb=Qb2

dt ˆz ,so the total torque is N=Q

dt (b2a2)ˆz ,and the angular momentum delivered

to the spheres is

L=Ndt =Q

3(b2a2)ˆz 0

dt dt =Q

3(b2a2)B0ˆz .X

Problem 8.10

(a)

E=









0,(r<R)

4⇡✏0

r2ˆr ,(r>R)







;B=









3µ0Mˆz ,(r<R)

µ0

4⇡

r32 cos ✓ˆr + sin ✓ˆ

✓,(r>R)







(Ex.6.1)

(where m=4

3⇡R3M); g=✏0(E⇥B)= µ0

(4⇡)2

r5(ˆr ⇥ˆ

✓) sin ✓, and (ˆr ⇥ˆ

✓)= ˆ

,so

`=r⇥g=µ0

(4⇡)2

r4sin ✓(ˆr ⇥ˆ

).

But (ˆr ⇥ˆ

)=ˆ

✓, and only the zcomponent will survive integration, so (since (ˆ

✓)z=sin ✓):

L=µ0mQ

(4⇡)2ˆz sin2✓

r4r2sin ✓dr d✓d.

2⇡



d=2⇡;

⇡



sin3✓d✓=4



r2dr =1

r

L=µ0mQ

(4⇡)2ˆz (2⇡)4

31

R=2

9µ0MQR2ˆz .

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174 CHAPTER 8. CONSERVATION LAWS

✲x

✻

❂

✇

✼

−q

✻

✲x

✻

✸

✶

r′

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(b) Apply Faraday’s law to the ring shown:

E·dl=E(2⇡rsin ✓)=d

dt =⇡(rsin ✓)22

3µ0

dt 

)E=µ0

dt (rsin ✓)ˆ

.

The force on a patch of surface (da) is dF=Eda =µ0

dt (rsin ✓)da ˆ

=Q

4⇡R2.

The torque on the patch is dN=r⇥dF=µ0

dt r2sin ✓da (ˆr ⇥ˆ

). But (ˆr ⇥ˆ

)=ˆ

✓, and we want

only the zcomponent (ˆ

✓z=sin ✓):

N=µ0

dt ˆz r2sin2✓r2sin ✓d✓d.

Here r=R;

⇡



sin3✓d✓=4

2⇡



d=2⇡, so N=µ0

dt ˆz R44

3(2⇡)= 2µ0

9QR2dM

dt ˆz .

L=Ndt =2µ0

9QR2ˆz



dM =2µ0

9MQR2ˆz (same as (a)).

4⇡R2. The charge below

(“south of”) the ring in the ﬁgure is

qs=2⇡R2⇡



sin ✓0d✓0=q

2(cos ✓0)|⇡

✓=q

2(1 + cos ✓).

So the total current crossing the ring (ﬂowing “north”) is I(t)=1

dt (1 + cos ✓), and hence

K(t)= I

2⇡Rsin ✓(ˆ

✓)= 1

4⇡R

(1 + cos ✓)

sin ✓ˆ

✓. The force on a patch of area da is dF=(K⇥B)da.

Bave =2

3µ0Mˆz +µ0

4⇡

3⇡R3M

R3(2 cos ✓ˆr + sin ✓ˆ

✓)1

2=µ0M

6[2 ˆz + 2 cos ✓ˆr + sin ✓ˆ

✓];

K⇥B=1

4⇡R

µ0M

(1 + cos ✓)

sin ✓[2(ˆ

✓⇥ˆz ) + 2 cos ✓(ˆ

✓⇥ˆr )

  

ˆ



dN=Rˆr ⇥dF=µ0M

24⇡dq

dt (1 + cos ✓)

sin ✓2[ ˆr ⇥(ˆ

✓⇥ˆz )

  

✓(ˆr ·ˆz )ˆz (ˆr ·ˆ

✓)

cos ✓(ˆr ⇥ˆ

)

  

ˆ

✓

]R2sin ✓d✓d

=µ0M

12⇡dq

dt (1 + cos ✓)R2[cos ✓ˆ

✓+ cos ✓ˆ

✓]d✓d=µ0MR2

6⇡dq

dt (1 + cos ✓) cos ✓d✓dˆ

✓.

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CHAPTER 8. CONSERVATION LAWS 175

The xand ycomponents integrate to zero; (ˆ

✓)z=sin ✓, so (using

2⇡



d=2⇡):

Nz=µ0MR2

6⇡dq

dt (2⇡)

⇡



(1 + cos ✓) cos ✓sin ✓d✓=µ0MR2

3dq

dt sin2✓

2cos3✓

3

⇡

=µ0MR2

3dq

dt 2

3=2µ0

9MR2dq

dt .N=2µ0

9MR2dq

dt ˆz .

Therefore

L=Ndt =2µ0

9MR2ˆz



dq =2µ0

9MR2Qˆz (same as (a)).

(I used the average ﬁeld at the discontinuity—which is the correct thing to do—but in this case you’d get the

same answer using either the inside ﬁeld or the outside ﬁeld.)

Problem 8.11 The magnetic ﬁeld of the upper loop is given by Eq. 5.41: B=µ0Ib

(b2+z2)3/2ˆz .(I put the

origin at the center of the upper loop, with the zaxis pointing up, so z=hat the location of the lower loop.)

Treat the lower loop as a magnetic dipole, with moment m=Ia⇡a2ˆz . The force on mis given by Eq. 6.3:

F=r(m·B)=rIa⇡a2µ0Ib

(b2+z2)3/2=µ0⇡a2b2IaIb

@zb2+z23/2ˆz

=µ0⇡a2b2IaIb

23

2b2+z25/2(2z)ˆz =3⇡

2µ0IaIb

a2b2h

(b2+h2)5/2ˆz .X

Problem 8.12 Following the method of Section 7.2.4 (leading up to Eq. 7.30), the power delivered to the two

loops is

dt =EaIaEbIb

(where these are the changing values as the currents are turned on, not the ﬁnal values). Now

Ea=La

dIa

dt MdIb

dt ,Eb=Lb

dIb

dt MdIa

dt ,

dt =LaIa

dIa

dt +MIa

dIb

dt +LbIb

dIb

dt +MIb

dIa

dt =d

dt 1

2LaI2

a+1

2LbI2

b+MIaIb.

The total work done, then, to increase the currents from zero to their ﬁnal values, is

W=1

2LaI2

a+1

2LbI2

b+MIaIb.qed

Problem 8.13

(a) E=d

dt ;=⇡a2B;B=µ0nIs;E=IrR. So Ir=1

Rµ0⇡a2ndIs

dt .

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176 CHAPTER 8. CONSERVATION LAWS

(b) E·dl=d

dt )E(2⇡a)=µ0⇡a2ndIs

dt )E=1

2µ0andIs

dt ˆ

.B=µ0Ir

(b2+z2)3/2ˆz (Eq. 5.41).

S=1

µ0

(E⇥B)= 1

µ0µ0an

dIs

dt µ0Ir

(b2+z2)3/2(ˆ

⇥ˆz )= 1

4µ0Ir

dIs

ab2n

(b2+z2)3/2ˆs .

Power:

P=S·da=



1

(S)(2⇡a)dz =1

2⇡µ0a2b2nIr

dIs



1

(b2+z2)3/2dz

The integral is z

b2pz2+b21

1 =1

b21

b2=2

b2.

=⇡µ0a2ndIs

dt Ir=(RIr)Ir=Ir2R. qed

Problem 8.14 The ﬁelds are zero for s<aand s>b; between the cylinders,

E=1

2⇡✏0



sˆs ,B=µ0

2⇡

sˆ

,

where I=v.

(a) From Eq. 8.5:

u=1

2✏0E2+1

µ0

B2=1

2✏01

2⇡✏022

s2+1

µ0µ0

2⇡22v2

s2=2

8⇡2✏0

(1 + ✏0µ0v2)1

s2.

Writing ✏0µ0=1/c2, and integrating over the volume between the cylinders:

`=2

8⇡2✏01+v2

c2b

s22⇡sds=2

4⇡✏01+ v2

c2ln(b/a).

(b) From Eq. 8.29:

g=✏0(E⇥B)=✏0

2⇡✏0sµ0v

2⇡sˆz =µ02v

4⇡2s2ˆz .

`=µ02v

4⇡2ˆz b

s22⇡sds=µ02v

2⇡ln(b/a)ˆz .

S=1

µ0

(E⇥B)= 1

✏0µ0

g=c2g.dW

dt =S·da=µ02c2v

2⇡ln(b/a)ˆz .

Problem 8.15

(a) The ﬁelds are E=1

4⇡✏0

r2ˆr ,B=µ0

2⇡

sˆ

(for points inside the toroid—see Eq. 5.60).

g=✏0(E⇥B)=✏0

4⇡✏0a2

µ0NI

2⇡aˆz =µ0

8⇡2

qNI

a3ˆz .

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CHAPTER 8. CONSERVATION LAWS 177

(Note that inside the toroid r⇡aˆs and s⇡a.)

p=µ0

8⇡2

qNI

a3ˆz (2⇡awh)= µ0

4⇡

qNIwh

a2ˆz .

(b) The changing magnetic ﬁeld induces an electric ﬁeld, as given by Problem 7.19 (with z= 0):

E=µ0

4⇡a2NhwdI

dt ˆz .The impulse delivered to qis therefore

I=Fdt =qEdt =qµ0

4⇡a2Nhwˆz dI

dt dt =µ0q

4⇡a2NhwI ˆz .X

Problem 8.16

According to Eqs. 3.104, 4.14, 5.89, and 6.16, the ﬁelds are

E=









1

3✏0

P,(r<R),

4⇡✏0

r3[3(p·ˆr )ˆr p],(r>R),











B=









3µ0M,(r<R),

µ0

4⇡

r3[3(m·ˆr )ˆr m],(r>R),











where p= (4/3)⇡R3P,and m= (4/3)⇡R3M. Now p=✏0(E⇥B)d⌧, and there are two contributions, one

from inside the sphere and one from outside.

Inside:

pin =✏01

3✏0

P⇥2

3µ0Md⌧=2

9µ0(P⇥M)d⌧=2

9µ0(P⇥M)4

3⇡R3=8

27µ0⇡R3(M⇥P).

Outside:

pout =✏0

4⇡✏0

µ0

4⇡1

r6{[3(p·ˆr )ˆr p]⇥[3(m·ˆr )ˆr m]}d⌧.

Now ˆr ⇥(p⇥m)=p(ˆr ·m)m(ˆr ·p), so ˆr ⇥[ˆr ⇥(p⇥m)] = (ˆr ·m)(ˆr ⇥p)(ˆr ·p)(ˆr ⇥m), whereas using the BAC-

CAB rule directly gives ˆr ⇥[ˆr ⇥(p⇥m)] = ˆr [ˆr ·(p⇥m)](p⇥m)(ˆr ·ˆr ). So {[3(p·ˆr )ˆr p]⇥[3(m·ˆr )ˆr m]}=

3(p·ˆr )(ˆr ⇥m)+3(m·ˆr )(ˆr ⇥p)+(p⇥m)=3{ˆr [ˆr ·(p⇥m)] (p⇥m)}+(p⇥m)=2(p⇥m)+3ˆr [ˆr ·(p⇥m)].

pout =µ0

16⇡21

r6{2(p⇥m)+3ˆr [ˆr ·(p⇥m)]}r2sin ✓dr d✓d.

To evaluate the integral, set the zaxis along (p⇥m); then ˆr ·(p⇥m)=|p⇥m|cos ✓. Meanwhile, ˆr =

sin ✓cos ˆx +sin✓sin ˆy +cos ✓ˆz . But sin and cos integrate to zero, so the ˆx and ˆy terms drop out, leaving

pout =µ0

16⇡21

r4dr2(p⇥m)sin ✓d✓d+3|p⇥m|ˆz cos2✓sin ✓d✓d

=µ0

16⇡21

3r3

R2(p⇥m)4⇡+ 3(p⇥m)4⇡

3=µ0

12⇡R3(p⇥m)

=µ0

12⇡R34

3⇡R3P⇥4

3⇡R3M=4µ0

27 R3(M⇥P).

ptot =8

27 +4

27µ0R3(M⇥P)= 4

9µ0R3(M⇥P).

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178 CHAPTER 8. CONSERVATION LAWS

Problem 8.17

(a) From Eq. 5.70 and Prob. 5.37,











r<R:E=0,B=2

3µ0R!ˆz ,with =e

4⇡R2;

r>R:E=1

4⇡✏0

r2ˆr ,B=µ0

4⇡

r3(2 cos ✓ˆr + sin ✓ˆ

✓),with m=4

3⇡!R4.

The energy stored in the electric ﬁeld is (Ex. 2.9):

WE=1

8⇡✏0

The energy density of the internal magnetic ﬁeld is:

uB=1

2µ0

B2=1

2µ02

3µ0R!e

4⇡R22

=µ0!2e2

72⇡2R2,so WBin =µ0!2e2

72⇡2R2

3⇡R3=µ0e2!2R

54⇡.

The energy density in the external magnetic ﬁeld is:

uB=1

2µ0

µ2

16⇡2

r64 cos2✓+ sin2✓=e2!2R4µ0

18(16⇡2)

r63 cos2✓+1

,so

WBout =µ0e2!2R4

(18)(16)⇡2



r6r2dr

⇡



03 cos2✓+1

sin ✓d✓

2⇡



d=µ0e2!2R4

(18)(16)⇡21

3R3(4)(2⇡)=µ0e2!2R

108⇡.

WB=WBin +Wbout =µ0e2!2R

108⇡(2 + 1) = µ0e2!2R

36⇡;W=WE+WB=1

8⇡✏0

R+µ0e2!2R

36⇡.

(b) Same as Prob. 8.10(a), with Q!eand m!1

3e!R2:L=µ0e2!R

18⇡ˆz .

18⇡!R=~

2)!R=9⇡~

µ0e2=(9)(⇡)(1.05 ⇥1034)

(4⇡⇥107)(1.60 ⇥1019)2= 9.23 ⇥1010 m/s.

8⇡✏0

R1+2

9!R

c2=mc2;1+2

9!R

c2=1+2

99.23 ⇥1010

3⇥1082

=2.10 ⇥104;

R=(2.10 ⇥104)(1.6⇥1019)2

8⇡(8.85 ⇥1012)(9.11 ⇥1031)(3 ⇥108)2= 2.95 ⇥1011 m; !=9.23 ⇥1010

2.95 ⇥1011 = 3.13 ⇥1021 rad/s.

Since !R, the speed of a point on the equator, is 300 times the speed of light, this “classical” model is clearly

unrealistic.

Problem 8.18

Maxwell’s equations with magnetic charge (Eq. 7.44):

(i) r·E=1

✏0

⇢e,(iii) r⇥E=µ0Jm@B

@t,

(ii) r·B=µ0⇢m,(iv) r⇥B=µ0Je+µ0✏0

@t.

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CHAPTER 8. CONSERVATION LAWS 179

The Lorentz force law becomes (Eq. 8.44)

F=qe(E+v⇥B)+qmB1

c2v⇥E.

Following the argument in Section 8.1.2:

F·dl=qe(E+v⇥B)+qmB1

c2v⇥E·vdt =(qeE+qmB)·vdt,

dt =(E·Je+B·Jm)d⌧

(which generalizes Eq. 8.6). Use (iii) and (iv) to eliminate Jeand Jm:

(E·Je+B·Jm)= 1

µ0

E·(r⇥B)✏0E·@E

@t1

µ0

B·(r⇥E)1

µ0

B·@B

@t,

but r·(E⇥B)=B·(r⇥E)E·(r⇥B), so

(E·Je+B·Jm)=1

µ0r·(E⇥B)1

@t✏0E2+1

µ0

B2

(generalizing Eq. 8.8). Thus

dt =d

dt 1

2✏0E2+1

µ0

B2d⌧1

µ0(E⇥B)·da,

which is identical to Eq. 8.9. Evidently Poynting’s theorem is unchanged(!), and

u=1

2✏0E2+1

µ0

B2,S=1

µ0

(E⇥B),

the same as before.

To construct the stress tensor we begin with the generalization of Eq. 8.13:

F=(E+v⇥B)⇢e+B1

c2v⇥E⇢md⌧.

The force per unit volume (Eq. 8.14) becomes

f=(⇢eE+Je⇥B)+⇢mB1

c2Jm⇥E

=✏0(r·E)E+1

µ0r⇥B✏0

@t⇥B+1

µ0

(r·B)Bµ0✏01

µ0r⇥E1

µ0

@t⇥E

=✏0[(r·E)EE⇥(r⇥E)] + 1

µ0

[(r·B)BB⇥(r⇥B)] ✏0

@t(E⇥B)

=✏0(r·E)E1

2r(E2)+(E·r)E+1

µ0(r·B)B1

2r(B2)+(B·r)B✏0

@t(E⇥B).

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180 CHAPTER 8. CONSERVATION LAWS

This is identical to Eq. 8.16, so the stress tensor is the same as before:

Tij ⌘✏0EiEj1

2ij E2+1

µ0BiBj1

2ij B2.

Likewise, Eq. 8.20 is still valid. In fact, this argument is more straightforward when you include magnetic

charge, since you don’t need artiﬁcially to insert the (r·B)Bterm (after Eq. 8.15).

The electromagnetic momentum density (Eq. 8.29) also stays the same, since the argument in Section 8.2.3

is formulated entirely in terms of the ﬁelds:

g=✏0(E⇥B).

Problem 8.19

E=qe

4⇡✏0

r3;

B=µ0qm

4⇡

r03=µ0qm

4⇡

(rdˆz )

(r2+d22rd cos ✓)3/2.

✲x

✻

❂

✇

✼

−q

✻

✲x

✻

✸

✶

r′

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Momentum density (Eq. 8.32):

g=✏0(E⇥B)=µ0qeqm

(4⇡)2

(d)(r⇥ˆz )

r3(r2+d22rd cos ✓)3/2.

Angular momentum density (Eq. 8.33):

`=(r⇥g)=µ0qeqmd

(4⇡)2

r⇥(r⇥ˆz )

r3(r2+d22rd cos ✓)3/2.But r⇥(r⇥ˆz )=r(r·ˆz )r2ˆz =r2cos ✓ˆr r2ˆz .

The xand ycomponents will integrate to zero; using (ˆr )z= cos ✓, we have:

L=µ0qeqmd

(4⇡)2ˆz r2(cos2✓1)

r3(r2+d22rd cos ✓)3/2r2sin ✓dr d✓d.Let u⌘cos ✓:

=µ0qeqmd

(4⇡)2ˆz (2⇡)



1



r1u2

(r2+d22rdu)3/2du dr.

Do the rintegral ﬁrst:



r dr

(r2+d22rdu)3/2=(ru d)

d(1 u2)pr2+d22rdu

d(1 u2)+d

d(1 u2)d=u+1

d(1 u2)=1

d(1 u).

Then

L=µ0qeqmd

8⇡ˆz 1



1

(1 u2)

(1 u)du =µ0qeqm

8⇡ˆz



1

(1 + u)du =µ0qeqm

8⇡ˆz u+u2

2

1

=µ0qeqm

4⇡ˆz .

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CHAPTER 8. CONSERVATION LAWS 181

Problem 8.20

p=✏0V

(E⇥B)d⌧=✏0V

(rV)⇥Bd⌧=✏0V

[r⇥(VB)Vr⇥B]d⌧

=✏0S

VB⇥da+✏0µ0V

VJd⌧=1

c2V

VJd⌧.

(I used Problem 1.61(b) in the penultimate step. Here Vis all of space, and Sis its surface at inﬁnity, where

B=0, so the surface integral vanishes.) Using V(r)⇡V(0)+(rV)·r=V(0)E(0)·r,

p=1

c2V(0)Jd⌧1

c2[E(0)·r]Jd⌧.

For a current loop, Jd⌧!Idl =Idl=0,and (Eq. 1.108):

[E(0)·r]Jd⌧![E(0)·r]Idl =I[E(0)·r]dl=Ia⇥E(0)=m⇥E.

p=1

c2(m⇥E).X

Problem 8.21

(a) The rotating shell at radius bproduces a solenoidal magnetic ﬁeld:

B=µ0Kˆz ,where K=b!bb, and b=Q

2⇡bl .So B=µ0!bQ

2⇡lˆz (a<s<b).

(Note that if angular velocity is deﬁned with respect to the zaxis, then !bis a negative number.) The shell at

aalso produces a magnetic ﬁeld (µ0!aQ/2⇡l)ˆz , in the region s<a, so the total ﬁeld inside the inner shell is

B=µ0Q

2⇡l(!a!b)ˆz ,(s<a).

Meanwhile, the electric ﬁeld is

E=1

2⇡✏0



sˆs =Q

2⇡✏0ls ˆs ,(a<s<b).

g=✏0(E⇥B)=✏0Q

2⇡✏0lsµ0!bQ

2⇡l(ˆs ⇥ˆz )= µ0!bQ2

4⇡2l2sˆ

;`=r⇥g=µ0!bQ2

4⇡2l2s(r⇥ˆ

).

Now r⇥ˆ

=(sˆs +zˆz )⇥ˆ

=sˆz zˆs , and the ˆs term integrates to zero, so

L=µ0!bQ2

4⇡2l2ˆz d⌧=µ0!bQ2

4⇡2l2⇡(b2a2)lˆz =µ0!bQ2(b2a2)

4⇡lˆz .

(b) The extra electric ﬁeld induced by the changing magnetic ﬁeld due to the rotating shells is given by

E2⇡s=d

dt )E=1

2⇡s

d

dt ˆ

, and in the region a<s<b

=µ0Q

2⇡l(!a!b)⇡a2µ0Q!b

2⇡l⇡s2a2=µ0Q

2l!aa2!bs2;E(s)=1

2⇡s

µ0Q

2la2d!a

dt s2d!b

dt ˆ

.

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182 CHAPTER 8. CONSERVATION LAWS

In particular,

E(a)=µ0Qa

4⇡ld!a

dt d!b

dt ˆ

,and E(b)=µ0Q

4⇡lb a2d!a

dt b2d!b

dt ˆ

.

The torque on a shell is N=r⇥qE=qsE ˆz ,so

Na=Qa µ0Qa

4⇡ld!a

dt d!b

dt ˆz ;La=1

Nadt =µ0Q2a2

4⇡l(!a!b)ˆz .

Nb=Qb µ0Q

4⇡lba2d!a

dt b2d!b

dt ˆz ;Lb=1

Nbdt =µ0Q2

4⇡la2!ab2!bˆz .

Ltot =La+Lb=µ0Q2

4⇡la2!ab2!ba2!a+a2!bˆz =µ0Q2!b

4⇡l(b2a2)ˆz .

Thus the reduction in the ﬁnal mechanical angular momentum (b) is equal to the residual angular momentum

in the ﬁelds (a). X

Problem 8.22

B=µ0nI ˆz ,(s<R); E=q

4⇡✏0

3, where

=(xa, y, z).

g=✏0(E⇥B)=✏0(µ0nI)q

4⇡✏01

⇥ˆz )= µ0qnI

4⇡

3[yˆx (xa)ˆy ].

Linear Momentum.

p=gd⌧=µ0qnI

4⇡yˆx (xa)ˆy

[(xa)2+y2+z2]3/2dx dy dz. The ˆx term is odd in y; it integrates to zero.

=µ0qnI

4⇡ˆy (xa)

[(xa)2+y2+z2]3/2dx dy dz. Do the zintegral ﬁrst :

[(xa)2+y2](xa)2+y2+z2

1

[(xa)2+y2].

=µ0qnI

2⇡ˆy (xa)

[(xa)2+y2]dx dy. Switch to polar coordinates :

x=scos ,y=ssin , dx dy )sdsd; [(xa)2+y2]=s2+a22sa cos .

=µ0qnI

2⇡ˆy (scos a)

(s2+a22sa cos )sdsd

Now 2⇡

cos d

(A+Bcos )=2⇡

B1A

pA2B2;2⇡

d

(A+Bcos )=2⇡

pA2B2.

Here A2B2=(s2+a2)24s2a2=s4+2s2a2+a44s2a2=(s2a2)2;A2B2=a2s2.

=µ0qnI

2aˆy 1a2+s2

a2s2+2a2

(a2s2)sds=µ0qnI

aˆy R

sds=µ0qnIR2

2aˆy .

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CHAPTER 8. CONSERVATION LAWS 183

Angular Momentum.

`=r⇥g=µ0qnI

4⇡

3r⇥[yˆx (xa)ˆy ]= µ0qnI

4⇡

3z(xa)ˆx +zy ˆy [x(xa)+y2]ˆz .

The ˆx and ˆy terms are odd in z, and integrate to zero,so

L=µ0qnI

4⇡ˆz x2+y2xa

[(xa)2+y2+z2]3/2dx dy dz. The zintegral is the same as before.

=µ0qnI

2⇡ˆz x2+y2xa

[(xa)2+y2]dx dy =µ0qnI

2⇡ˆz sacos 

(s2+a22sa cos )s2ds d

=µ0qnI ˆz s2

a2s2+1a2+s2

a2s2sds=µ0qnI ˆz R

s2s2

a2s2sds= zero.

Problem 8.23

(a) If we’re only interested in the work done on free charges and currents, Eq. 8.6 becomes

dt =V

(E·Jf)d⌧. But Jf=r⇥H@D

@t(Eq. 7.56), so E·Jf=E·(r⇥H)E·@D

@t. From product

rule #6, r·(E⇥H)=H·(r⇥E)E·(r⇥H), while r⇥E=@B

@t,so

E·(r⇥H)=H·@B

@tr·(E⇥H).Therefore E·Jf=H·@B

@tE·@D

@tr·(E⇥H), and hence

dt =VE·@D

@t+H·@B

@td⌧S

(E⇥H)·da.

This is Poynting’s theorem for the ﬁelds in matter. Evidently the Poynting vector, representing the power per

unit area transported by the ﬁelds, is S=E⇥H, and the rate of change of the electromagnetic energy density

is @uem

@t=E·@D

@t+H·@B

@t.

For linear media, D=✏Eand H=1

µB, with ✏and µconstant (in time); then

@uem

@t=✏E·@E

@t+1

µB·@B

@t=1

2✏@

@t(E·E)+ 1

2µ

@t(B·B)=1

@t(E·D+B·H),

so uem =1

2(E·D+B·H).qed

(b) If we’re only interested in the force on free charges and currents, Eq. 8.13 becomes f=⇢fE+Jf⇥B.

But ⇢f=r·D, and Jf=r⇥H@D

@t,sof=E(r·D)+(r⇥H)⇥B@D

@t⇥B. Now

@t(D⇥B)=@D

@t⇥B+D⇥@B

@t, and @B

@t=r⇥E, so @D

@t⇥B=@

@t(D⇥B)+D⇥(r⇥E), and

hence f=E(r·D)D⇥(r⇥E)B⇥(r⇥H)@

@t(D⇥B). As before, we can with impunity add the

term H(r·B), so

f={[E(r·D)D⇥(r⇥E)] + [H(r·B)B⇥(r⇥H)]}@

@t(D⇥B).

The term in curly brackets can be written as the divergence of a stress tensor (as in Eq. 8.19), and the last

term is (minus) the rate of change of the momentum density, g=D⇥B.

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184 CHAPTER 8. CONSERVATION LAWS

Problem 8.24

(a) Initially, the disk will rise like a helicopter. The force on one charge (velocity v=!Rˆ

+vzˆz ) is

Fi=q(v⇥B)=qk 

ˆs ˆ

ˆz

0!Rv

R02z=qk 2!Rz ˆs Rvzˆ

+!R2ˆz .

The net force on all the charges is



i=1

Fi=nqkR2!ˆz =Md2z

dt2ˆz ;d2z

dt2=nqkR2

M!.[1]

The net torque on the disk is



i=1

(ri⇥Fi)=n(Rˆs )⇥(qkRvzˆ

)=nqkR2vzˆz =Id!

dt ˆz

where Iis the moment of inertia of the disk. So

dt =nqkR2

Idz

dt .[2]

Di↵erentiate [2], and combine with [1]:

d2z

dt2=I

nqkR2d2!

dt2=nqkR2

M!)d2!

dt2=↵!,where ↵⌘nqkR2

pMI .

The solution (with initial angular velocity !0and initial angular acceleration 0) is

!(t)=!0cos ↵t.

Meanwhile,

dt =I

nqkR2d!

dt =I

nqkR2!0↵sin ↵t=!0I

Msin ↵t.

z(t)=!0I

Mt

sin ↵tdt=!0

↵I

M(1 cos ↵t).

[The problem is a little cleaner if you make the disk massless, and assign a mass mto each of the charges.

Then M!nm and I!nmR2,so↵=qkR/m and z(t)!(!0R/↵)(1 cos ↵t).]

(b) The disk rises and falls harmonically, as its rotation slows down and speeds up. The total energy is

E=1

2Mv2

z+1

2I!2=1

2M!2

Msin2↵t+1

2I!2

0cos2↵t=1

2I!2

0(sin2↵t+ cos2↵t)=1

2I!2

which is constant (and equal to the initial energy). [Of course, if you didn’t notice that the rotation rate is

changing, you might think the magnetic force is doing work, as the disk oscillates up and down.]

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CHAPTER 9. ELECTROMAGNETIC WAVES 185

Chapter 9

Electromagnetic Waves

Problem 9.1

@f1

@z=2Ab(zvt)eb(zvt)2;@2f1

@z2=2Ab eb(zvt)22b(zvt)2eb(zvt)2;

@f1

@t=2Abv(zvt)eb(zvt)2;@2f1

@t2=2Abv veb(zvt)2+2bv(zvt)2eb(zvt)2=v2@2f1

@z2.X

@f2

@z=Ab cos[b(zvt)]; @2f2

@z2=Ab2sin[b(zvt)];

@f2

@t=Abv cos[b(zvt)]; @2f2

@t2=Ab2v2sin[b(zvt)] = v2@2f2

@z2.X

@f3

@z=2Ab(zvt)

[b(zvt)2+ 1]2;@2f3

@z2=2Ab

[b(zvt)2+ 1]2+8Ab2(zvt)2

[b(zvt)2+ 1]3;

@f3

@t=2Abv(zvt)

[b(zvt)2+ 1]2;@2f3

@t2=2Abv2

[b(zvt)2+ 1]2+8Ab2v2(zvt)2

[b(zvt)2+ 1]3=v2@2f3

@z2.X

@f4

@z=2Ab2zeb(bz2+vt);@2f4

@z2=2Ab2eb(bz2+vt)2b2z2eb(bz2+vt);

@f4

@t=Abveb(bz2+vt);@2f4

@t2=Ab2v2eb(bz2+vt)6=v2@2f4

@z2.

@f5

@z=Ab cos(bz) cos(bvt)3;@2f5

@z2=Ab2sin(bz) cos(bvt)3;@f5

@t=3Ab3v3t2sin(bz) sin(bvt)3;

@2f5

@t2=6Ab3v3tsin(bz) sin(bvt)39Ab6v6t4sin(bz) cos(bvt)36=v2@2f5

@z2.

Problem 9.2

@z=Ak cos(kz) cos(kvt); @2f

@z2=Ak2sin(kz) cos(kvt);

@t=Akv sin(kz) sin(kvt); @2f

@t2=Ak2v2sin(kz) cos(kvt)=v2@2f

@z2.X

Use the trig identity sin ↵cos =1

2[sin(↵+) + sin(↵)] to write

f=A

2{sin[k(z+vt)] + sin[k(zvt)]},

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186 CHAPTER 9. ELECTROMAGNETIC WAVES

which is of the form 9.6, with g=(A/2) sin[k(zvt)] and h=(A/2) sin[k(z+vt)].

Problem 9.3

(A3)2=A3ei3A3ei3=A1ei1+A2ei2A1ei1+A2ei2

=(A1)2+(A2)2+A1A2ei1ei2+ei1ei2=(A1)2+(A2)2+A1A22 cos(12);

A3=(A1)2+(A2)2+2A1A2cos(12).

A3ei3=A3(cos 3+isin 3)=A1(cos 1+isin 1)+A2(cos 2+isin 2)

=(A1cos 1+A2cos 2)+i(A1sin 1+A2sin 2).tan 3=A3sin 3

A3cos 3

=A1sin 1+A2sin 2

A1cos 1+A2cos 2

;

3= tan1A1sin 1+A2sin 2

A1cos 1+A2cos 2.

Problem 9.4

The wave equation (Eq. 9.2) says @2f

@z2=1

@2f

@t2. Look for solutions of the form f(z,t)=Z(z)T(t). Plug

this in: Td2Z

dz2=1

v2Zd2T

dt2. Divide by ZT :1

d2Z

dz2=1

v2T

d2T

dt2. The left side depends only on z, and the

right side only on t, so both must be constant. Call the constant k2.











d2Z

dz2=k2Z)Z(z)=Aeikz +Beikz,

d2T

dt2=(kv)2T)T(t)=Ceikvt +Deikvt.











(Note that kmust be real,elseZand Tblow up; with no loss of generality we can assume kis positive.)

f(z,t)=Aeikz +BeikzCeikvt +Deikvt=A1ei(kz+kvt)+A2ei(kzkvt)+A3ei(kz+kvt)+A4ei(kzkvt).

The general linear combination of separable solutions is therefore

f(z,t)=1

0A1(k)ei(kz+!t)+A2(k)ei(kz!t)+A3(k)ei(kz+!t)+A4(k)ei(kz!t)dk,

where !⌘kv. But we can combine the third term with the ﬁrst, by allowing kto run negative (!=|k|v

remains positive); likewise the second and the fourth:

f(z,t)=1

1 A1(k)ei(kz+!t)+A2(k)ei(kz!t)dk.

Because (in the end) we shall only want the the real part of f, it suﬃces to keep only one of these terms (since

kgoes negative, both terms include waves traveling in both directions); the second is traditional (though either

would do). Speciﬁcally,

Re(f)=1

1

[Re(A1) cos(kz +!t)Im(A1) sin(kz +!t) + Re(A2) cos(kz !t)Im(A2) sin(kz !t)] dk.

The ﬁrst term, cos(kz +!t) = cos(kz !t), combines with the third, cos(kz !t), since the negative kis

picked up in the other half of the range of integration, and the second, sin(kz+!t)=sin(kz!t), combines

with the fourth for the same reason. So the general solution, for our purposes, can be written in the form

f(z,t)=1

1

A(k)ei(kz!t)dk qed (the tilde0s remind us that we want the real part).

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CHAPTER 9. ELECTROMAGNETIC WAVES 187

Problem 9.5

Equation 9.26 )gI(v1t)+hR(v1t)=gT(v2t). Now @gI

@z=1

@gI

@t;@hR

@z=1

@hR

@t;@gT

@z=1

@gT

@t.

Equation 9.27 )1

@gI(v1t)

@t+1

@hR(v1t)

@t=1

@gT(v2t)

@t)gI(v1t)hR(v1t)=v1

gT(v2t)+

(where is a constant).

Adding these equations, we get 2gI(v1t)=1+v1

v2gT(v2t)+, or gT(v2t)=2v2

v1+v2gI(v1t)+0

(where 0⌘v2

v1+v2

). Now gI(z, t), gT(z,t), and hR(z,t) are each functions of a single variable u(in the

ﬁrst case u=zv1t, in the second u=zv2t, and in the third u=z+v1t). Thus

gT(u)=2v2

v1+v2gI(v1u/v2)+0.

Multiplying the ﬁrst equation by v1/v2and subtracting, 1v1

v2gI(v1t)1+v1

v2hR(v1t)=)

hR(v1t)=v2v1

v1+v2gI(v1t)v2

v1+v2, or hR(u)=v2v1

v1+v2gI(u)+0.

[The notation is tricky, so here’s an example: for a sinusoidal wave,







gI=AIcos(k1z!t)=AIcos[k1(zv1t)] )gI(u)=AIcos(k1u).

gT=ATcos(k2z!t)=ATcos[k2(zv2t)] )gT(u)=ATcos(k2u).

hR=ARcos(k1z!t)=ARcos[k1(z+v1t)] )hR(u)=ARcos(k1u).

Here 0= 0, and the boundary conditions say AT

=2v2

v1+v2

,AR

=v2v1

v1+v2

(same as Eq. 9.32), and v1

v2k1=k2

(consistent with Eq. 9.24).]

Problem 9.6

(a) Tsin ✓+Tsin ✓=ma )T@f

@z0+@f

@z0=m@2f

@t20

(b) ˜

AI+˜

AR=˜

AT;T[ik2˜

ATik1(˜

AI˜

AR)] = m(!2˜

AT),or k1(˜

AI˜

AR)=k2im!2

T˜

AT.

Multiply ﬁrst equation by k1and add: 2k1˜

AI=k1+k2im!2

T˜

AT,or ˜

AT=2k1

k1+k2im!2/T ˜

AI.

AR=˜

AT˜

AI=2k1(k1+k2im!2/T )

k1+k2im!2/T ˜

AI=k1k2+im!2/T

k1+k2im!2/T ˜

AI.

If the second string is massless, so v2=T/µ2=1, then k2/k1= 0, and we have ˜

AT=2

1i˜

AI,

AR=1+i

1i˜

AI, where ⌘m!2

k1T=m(k1v1)2

k1T=mk1

µ1

, or =mk1

µ1

.Now 1+i

1i=Aei,with

A2=1+i

1i1i

1+i=1)A=1,and ei=(1 + i)2

(1 i)(1 + i)=1+2i2

1+2)

tan =2

12.Thus AReiR=eiAIeiI)AR=AI,R=I+ tan12

12.

Similarly, 2

1i=Aei)A2=2

1i 2

1+i=4

1+2)A=2

1+2.

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188 CHAPTER 9. ELECTROMAGNETIC WAVES

Aei=2(1 + i)

(1 i)(1 + i)=2(1 + i)

(1 + 2))tan =.So ATeiT=2

1+2eiAIeiI;

AT=2

1+2AI;T=I+ tan1.

Problem 9.7

(a) F=T@2f

@z2z@f

@tz=µz@2f

@t2, or T@2f

@z2=µ@2f

@t2+@f

@t.

(b) Let ˜

f(z,t)= ˜

F(z)ei!t; then Te

i!td2˜

dz2=µ(!2)˜

i!t+(i!)˜

i!t)

Td2˜

dz2=!(µ!+i)˜

F, d2˜

dz2=˜

k2˜

F, where ˜

k2⌘!

T(µ!+i).Solution : ˜

F(z)= ˜

Aei˜

kz +˜

Bei˜

kz.

Resolve ˜

kinto its real and imaginary parts: ˜

k=k+i)˜

k2=k22+2ik=!

T(µ!+i).

2k=!

T)=!

2kT ;k22=k2!

2T21

k2=µ!2

T; or k4k2(µ!2/T )(!/2T)2=0)

k2=1

2(µ!2/T )±(µ!2/T )2+ 4(!/2T)2=µ!2

2T1±1+(/µ!)2. But kis real, so k2is positive, so

we need the plus sign: k=!µ

2T1+1+(/µ!)2.=!

2kT =

p2Tµ 1+1+(/µ!)21/2.

Plugging this in, ˜

F=Aei(k+i)z+Bei(k+i)z=Aezeikz +Bezeikz . But the Bterm gives an expo-

nentially increasing function, which we don’t want (I assume the waves are propagating in the +zdirection),

so B= 0, and the solution is ˜

f(z,t)= ˜

Aezei(kz!t).(The actual displacement of the string is the real part

of this, of course.)

z=1

=p2Tµ

1+1+(/µ!)2; this is the characteristic penetration depth.

(d) This is the same as before, except that k2!k+i. From Eq. 9.29, ˜

AR=k1ki

k1+k+i˜

AI;

AR

AI2

=k1ki

k1+k+ik1k+i

k1+ki=(k1k)2+2

(k1+k)2+2. AR=(k1k)2+2

(k1+k)2+2AI

(where k1=!/v1=!µ1/T , while kand are deﬁned in part b). Meanwhile

k1ki

k1+k+i=(k1ki)(k1+k+i)

(k1+k)2+2=(k1)2k222ik1

(k1+k)2+2)R= tan12k1

(k1)2k22.

Problem 9.8

(a) fv(z,t)=Acos(kz !t)ˆx ;fh(z,t)=Acos(kz !t+ 90)ˆy =

Asin(kz!t)ˆy .Since f2

v+f2

h=A2, the vector sum f=fv+fhlies

on a circle of radius A. At time t=0,f=Acos(kz)ˆx Asin(kz)ˆy .

At time t=⇡/2!,f=Acos(kz 90)ˆx Asin(kz 90)ˆy =

Asin(kz)ˆx +Acos(kz)ˆy . Evidently it circles counterclockwise .

To make a wave circling the other way, use h=90.

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CHAPTER 9. ELECTROMAGNETIC WAVES 189

(b)

Problem 9.9

(a) k=!

cˆx ;ˆn =ˆz .k·r=!

cˆx ·(xˆx +yˆy +zˆz )=!

cx;k⇥ˆn =ˆx ⇥ˆz =ˆy .

E(x, t)=E0cos !

cx+!tˆz ;B(x, t)=E0

ccos !

cx+!tˆy .

(b) k=!

cˆx +ˆy +ˆz

p3;ˆn =ˆx ˆz

p2.(Since ˆn is parallel to the xz plane, it must have the form ↵ˆx +ˆz ;

since ˆn ·k=0,=↵; and since it is a unit vector, ↵=1/p2.)

k·r=!

p3c(ˆx +ˆy +ˆz )·(xˆx +yˆy +zˆz )= !

p3c(x+y+z); ˆ

k⇥ˆn =1

p6

ˆx ˆy ˆz

11 1

101=1

p6(ˆx +2ˆy ˆz ).

E(x, y, z, t)=E0cos !

p3c(x+y+z)!tˆx ˆz

p2;

B(x, y, z, t)=E0

ccos !

p3c(x+y+z)!tˆx +2ˆy ˆz

p6.

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190 CHAPTER 9. ELECTROMAGNETIC WAVES

Problem 9.10

P=I

c=1.3⇥103

3.0⇥108= 4.3⇥106N/m2. For a perfect reﬂector the pressure is twice as great:

8.6⇥106N/m2. Atmospheric pressure is 1.03 ⇥105N/m2, so the pressure of light on a reﬂector is

(8.6⇥106)/(1.03 ⇥105)= 8.3⇥1011 atmospheres.

Problem 9.11 The ﬁelds are E(z,t)=E0cos(kz !t)ˆx ,B(z, t)= 1

cE0cos(kz !t)ˆy , with !=ck.

(a) The electric force is Fe=qE=qE0cos(kz !t)ˆx =ma=mdv

dt , so

v=qE0

mˆx cos(kz !t)dt =qE0

m!sin(kz !t)ˆx +C.

But vave =C=0,so v=qE0

m!sin(kz !t)ˆx .

(b) The magnetic force is

Fm=q(v⇥B)=qqE0

m!E0

csin(kz !t) cos(kz !t)(ˆx ⇥ˆy )= q2E2

m!csin(kz !t) cos(kz !t)ˆz .

m!cˆz T

sin(kz !t) cos(kz !t)dt, where T=2⇡/!is the

period. The integral is 1

2!sin2(kz !t)

0=1

2!sin2(kz 2⇡)sin2(kz)= 0, so (Fm)ave =0.

(d) Adding in the damping term,

F=qEmv=qE0cos(kz !t)ˆx mv=mdv

dt )dv

dt +v=qE0

mcos(kz !t)ˆx .

The steady state solution has the form v=Acos(kz !t+✓)ˆx ,dv

dt =A!sin(kz !t+✓)ˆx .Putting this

in, and using the trig identity cos u= cos ✓cos(u+✓) + sin ✓sin(u+✓),

A!sin(kz !t+✓)+Acos(kz !t+✓)=qE0

m[cos ✓cos(kz !t+✓) + sin ✓sin(kz !t+✓)] .

Equating like terms:

A!=qE0

msin ✓,A=qE0

mcos ✓)tan ✓=!

,A

2(!2+2)=qE0

m2

)A=qE0

m!2+2.

v=qE0

m!2+2cos(kz !t+✓)ˆx ,✓⌘tan1(!/); Fm=q2E2

mc!2+2cos(kz !t+✓) cos(kz !t)ˆz .

To calculate the time average, write cos(kz !t+✓) = cos ✓cos(kz !t)sin ✓sin(kz !t). We already

know that the average of cos(kz !t) sin(kz !t) is zero, so

(Fm)ave =q2E2

mc!2+2ˆz cos ✓T

cos2(kz !t)dt.

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CHAPTER 9. ELECTROMAGNETIC WAVES 191

The integral is T/2=⇡/!, and cos ✓=/!2+2(see ﬁgure), so (Fm)ave =⇡q2E2

m!c(!2+2)ˆz .

Problem 9.12

hfgi=1

TT

acos(k·r!t+a)bcos(k·r!t+b)dt

=ab

2TT

[cos(2k·r2!t+a+b)+cos(ab)] dt =ab

2Tcos(ab)T=1

2ab cos(ab).

Meanwhile, in the complex notation: ˜

f=˜aeik·r!t),˜g=˜

beik·r!t),where ˜a=aeia,˜

b=beib. So

2˜

f˜g⇤=1

2˜aei(k·r!t)˜

b⇤ei(k·r!t)=1

2˜a˜

b⇤=1

2abei(ab),Re 1

2˜

f˜g⇤=1

2ab cos(ab)=hfgi.qed

Problem 9.13

Tij =✏0EiEj1

2ij E2+1

µ0BiBj1

2ij B2.

With the ﬁelds in Eq. 9.48, Ehas only an xcomponent, and Bonly a ycomponent. So all the “o↵-diagonal”

(i6=j) terms are zero. As for the “diagonal” elements:

Txx =✏0ExEx1

2E2+1

µ01

2B2=1

2✏0E21

µ0

B2=0.

Tyy =✏01

2E2+1

µ0ByBy1

2B2=1

2✏0E2+1

µ0

B2=0.

Tzz =✏01

2E2+1

µ01

2B2=u.

So Tzz =✏0E2

0cos2(kz !t+) (all other elements zero).

The momentum of these ﬁelds is in the zdirection, and it is

being transported in the zdirection, so yes, it does make sense

that Tzz should be the only nonzero element in Tij . According

to Sect. 8.2.3, $

T·dais the rate at which momentum crosses

an area da. Here we have no momentum crossing areas ori-

ented in the xor ydirection; the momentum per unit time

per unit area ﬂowing across a surface oriented in the zdirec-

tion is Tzz =u=gc(Eq. 9.59), so p=gcAt, and hence

p/t=gcA = momentum per unit time crossing area A.

Evidently momentum ﬂux density = energy density. X

Problem 9.14

R=E0R

E0I2

(Eq.9.86) )R=1

1+2

(Eq.9.82),where ⌘µ1v1

µ2v2

.T=✏2v2

✏1v1E0T

E0I2

(Eq.9.87)

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192 CHAPTER 9. ELECTROMAGNETIC WAVES

)T=2

1+2

(Eq.9.82).[Note that ✏2v2

✏1v1

=µ1

µ2

✏2µ2

✏1µ1

=µ1

µ2v1

v22v2

=µ1v1

µ2v2

=.]

T+R=1

(1 + )24+ (1 )2=1

(1 + )2(4+12+2)= 1

(1 + )2(1 + 2+2)=1.X

Problem 9.15

Equation 9.78 is replaced by ˜

E0Iˆx +˜

E0Rˆn R=˜

E0Tˆn T, and Eq. 9.80 becomes ˜

E0Iˆy ˜

E0R(ˆz ⇥ˆn R)=

˜

E0T(ˆz ⇥ˆn T). The ycomponent of the ﬁrst equation is ˜

E0Rsin ✓R=˜

E0Tsin ✓T; the xcomponent of the

second is ˜

E0Rsin ✓R=˜

E0Tsin ✓T. Comparing these two, we conclude that sin ✓R= sin ✓T= 0, and hence

✓R=✓T= 0. qed

Problem 9.16

Aeiax +Beibx =Ceicx for all x, so (using x= 0), A+B=C.

Di↵erentiate: iaAeiax +ibBeibx =icCeicx, so (using x= 0), aA +bB =cC.

Di↵erentiate again: a2Aeiax b2Beibx =c2Ceicx, so (using x= 0), a2A+b2B=c2C.

a2A+b2B=c(cC)=c(aA +bB); (A+B)(a2A+b2B)=(A+B)c(aA +bB)=cC(aA +bB);

a2A2+b2AB +a2AB +b2B2=(aA +bB)2=a2A2+2abAB +b2B2,or (a2+b22ab)AB = 0, or

(ab)2AB = 0. But Aand Bare nonzero, so a=b. Therefore (A+B)eiax =Ceicx.

a(A+B)=cC, or aC =cC, so (since C6= 0) a=c.Conclusion: a=b=c. qed

Problem 9.17







EI=˜

E0Iei(kI·r!t)ˆy ,

BI=1

E0Iei(kI·r!t)(cos ✓1ˆx + sin ✓1ˆz ); 











ER=˜

E0Rei(kR·r!t)ˆy ,

BR=1

E0Rei(kR·r!t)(cos ✓1ˆx + sin ✓1ˆz ); 











ET=˜

E0Tei(kT·r!t)ˆy ,

BT=1

E0Tei(kT·r!t)(cos ✓2ˆx + sin ✓2ˆz ); 





Boundary conditions: 









(i) ✏1E?

1=✏2E?

2,(iii) Ek

1=Ek

(ii) B?

1=B?

2,(iv) 1

µ1Bk

1=1

µ2Bk

Law of refraction: sin ✓2

sin ✓1

=v2

. [Note: kI·r!t=kR·r!t=kT·r!t, at z= 0, so we can drop all

exponential factors in applying the boundary conditions.]

Boundary condition (i): 0 = 0 (trivial). Boundary condition (iii): ˜

E0I+˜

E0R=˜

E0T.

Boundary condition (ii): 1

E0Isin ✓1+1

E0Rsin ✓1=1

E0Tsin ✓2)˜

E0I+˜

E0R=v1sin ✓2

v2sin ✓1˜

E0T.

But the term in parentheses is 1, by the law of refraction, so this is the same as (iii).

Boundary condition (iv): 1

µ11

E0I(cos ✓1)+ 1

E0Rcos ✓1=1

µ2v2

E0T(cos ✓2))

E0I˜

E0R=µ1v1cos ✓2

µ2v2cos ✓1˜

E0T. Let ↵⌘cos ✓2

cos ✓1

;⌘µ1v1

µ2v2

.Then ˜

E0I˜

E0R=↵ ˜

E0T.

Solving for ˜

E0Rand ˜

E0T:2

E0I=(1+↵)˜

E0T)˜

E0T=2

1+↵ ˜

E0I;

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CHAPTER 9. ELECTROMAGNETIC WAVES 193

E0R=˜

E0T˜

E0I=2

1+↵ 1+↵

1+↵ ˜

E0I)˜

E0R=1↵

1+↵ ˜

E0I.

Since ↵and are positive, it follows that 2/(1 + ↵) is positive, and hence the transmitted wave is in phase

with the incident wave, and the (real) amplitudes are related by E0T=2

1+↵ E0I.The reﬂected wave is

in phase if ↵ <1 and 180out of phase if ↵ >1; the (real) amplitudes are related by E0R=

1↵

1+↵ E0I.

These are the Fresnel equations for polarization perpendicular to the plane of incidence.

To construct the graphs, note that ↵ =1sin2✓/2

cos ✓=2sin2✓

cos ✓, where ✓is the angle of incidence,

so, for =1.5, ↵ =2.25 sin2✓

cos ✓.[In the ﬁgure, the minus signs on the vertical axis should be decimal

points.]

Is there a Brewster’s angle? Well, E0R= 0 would mean that ↵ = 1, and hence that

↵=1(v2/v1)2sin2✓

cos ✓=1

=µ2v2

µ1v1

,or 1 v2

v12

sin2✓=µ2v2

µ1v12

cos2✓,so

1=v2

v12sin2✓+(µ2/µ1)2cos2✓. Since µ1⇡µ2, this means 1 ⇡(v2/v1)2, which is only true for optically

indistinguishable media, in which case there is of course no reﬂection—but that would be true at any angle,

not just at a special “Brewster’s angle”. [If µ2were substantially di↵erent from µ1, and the relative velocities

were just right, it would be possible to get a Brewster’s angle for this case, at

v1

v22

=1cos2✓+µ2

µ12

cos2✓)cos2✓=(v1/v2)21

(µ2/µ1)21=(µ2✏2/µ1✏1)1

(µ2/µ1)21=(✏2/✏1)(µ1/µ2)

(µ2/µ1)(µ1/µ2).

But the media would be very peculiar.]

By the same token, Ris either always 0, or always ⇡, for a given interface—it does not switch over as you

change ✓, the way it does for polarization in the plane of incidence. In particular, if =3/2, then ↵ >1, for

↵ =2.25 sin2✓

cos ✓>1 if 2.25 sin2✓>cos2✓,or 2.25 >sin2✓+ cos2✓=1.X

In general, for >1,↵ >1, and hence R=⇡. For <1,↵ <1,and R= 0.

At normal incidence, ↵= 1, so Fresnel’s equations reduce to E0T=2

1+E0I;E0R=

1

1+E0I,

consistent with Eq. 9.82.

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194 CHAPTER 9. ELECTROMAGNETIC WAVES

Reﬂection and Transmission coeﬃcients: R=E0R

E0I2

=1↵

1+↵ 2

.Referring to Eq. 9.116,

T=✏2v2

✏1v1

↵E0T

E0I2

=↵ 2

1+↵ 2

R+T=(1 ↵)2+4↵

(1 + ↵)2=12↵ +↵22+4↵

(1 + ↵)2=(1 + ↵)2

(1 + ↵)2=1.X

Problem 9.18

Equation 9.106 )=2.42; Eq. 9.110 )

↵=1(sin ✓/2.42)2

cos ✓.

(a) ✓=0)↵= 1. Eq. 9.109 )E0R

E0I=↵

↵+

=12.42

1+2.42 =1.42

3.42 =0.415;

E0T

E0I=2

↵+=2

3.42 = 0.585.

(b) Equation 9.112 )✓B= tan1(2.42) = 67.5.

(4.42)2cos2✓=1sin2✓/(2.42)2;

(4.42)2(1 sin2✓) = (4.42)2(4.42)2sin2✓

=10.171 sin2✓; 19.51 = (19.50.17) sin2✓;

18.5 = 19.3 sin2✓; sin2✓= 18.5/19.3=0.959;

sin ✓=0.979; ✓= 78.3.

Problem 9.19

(a) Equation 9.120 )⌧=✏/. Now ✏=✏0✏r(Eq. 4.34), ✏r⇠

=n2(Eq. 9.70), and for glass the index

of refraction is typically around 1.5, so ✏⇡(1.5)2⇥8.85 ⇥1012 =2⇥1011 C2/N m2, while =1/⇢⇡

1012 (⌦m)1(Table 7.1). Then ⌧= (2⇥1011)/1012 = 20 s. (But the resistivity of glass varies enormously

from one type to another, so this answer could be o↵by a factor of 100 in either direction.)

(b) For silver, ⇢=1.59 ⇥108(Table 7.1), and ✏⇡✏0,so!✏ =2⇡⇥1010 ⇥8.85 ⇥1012 =0.56.

Since =1/⇢=6.25 ⇥107!✏, the skin depth (Eq. 9.128) is

d=1

⇠

=2

!µ=2

2⇡⇥1010 ⇥6.25 ⇥107⇥4⇡⇥107=6.4⇥107m=6.4⇥104mm.

I’d plate silver to a depth of about 0.001 mm; there’s no point in making it any thicker, since the ﬁelds don’t

penetrate much beyond this anyway.

Since !✏, Eq. 9.126 )k⇡!µ

2, so (Eq. 9.129)

=2⇡2

!µ0

=2⇡2

2⇡⇥106⇥6⇥107⇥4⇡⇥107=4⇥104m= 0.4 mm.

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CHAPTER 9. ELECTROMAGNETIC WAVES 195

From Eq. 9.129, the propagation speed is v=!

k=!

2⇡=⌫ =(4⇥104)⇥106= 400 m/s. In vacuum,

=c

⌫=3⇥108

106= 300 m; v=c= 3 ⇥108m/s. (But really, in a good conductor the skin depth is so small,

compared to the wavelength, that the notions of “wavelength” and “propagation speed” lose their meaning.)

Problem 9.20

(a) Use the binomial expansion for the square root in Eq. 9.126:

⇠

=!✏µ

21+1

2

✏! 211/2

=!✏µ



✏! =

2µ

✏.

So (Eq. 9.128) d=1

⇠

✏

µ. qed

For pure water, 





✏=✏r✏0= 80.1✏0(Table 4.2),

µ=µ0(1 + m)=µ0(1 9.0⇥106)⇠

=µ0(Table 6.1),

=1/(2.5⇥105) (Table 7.1).

So d= (2)(2.5⇥105)(80.1)(8.85 ⇥1012)

4⇡⇥107= 1.19 ⇥104m.

(b) In this case (/✏!)2dominates, so (Eq. 9.126) k⇠

=, and hence (Eqs. 9.128 and 9.129)

=2⇡

k⇠

=2⇡

=2⇡d, or d=

2⇡. qed

Meanwhile ⇠

=!✏µ

2

✏! =!µ

2=(1015)(4⇡⇥107)(107)

2=8⇥107;d=1

=1

8⇥107=

1.3⇥108= 13 nm. So the ﬁelds do not penetrate far into a metal—which is what accounts for their opacity.

=, as we found in (b), Eq. 9.134 says = tan1(1) = 45. qed

Meanwhile, Eq. 9.137 says B0

E0⇠

=✏µ

✏! =µ

!.For a typical metal, then, B0

=(107)(4⇡⇥107)

1015 =

107s/m.(In vacuum, the ratio is 1/c =1/(3 ⇥108)=3⇥109s/m, so the magnetic ﬁeld is comparatively

about 100 times larger in a metal.)

Problem 9.21

(a) u=1

2✏E2+1

µB2=1

2e2z✏E2

0cos2(kz !t+E)+ 1

µB2

0cos2(kz !t+E+). Averaging

over a full cycle, using hcos2i=1

2and Eq. 9.137:

hui=1

2e2z✏

2E2

0+1

2µB2

0=1

4e2z✏E2

0+1

µE2

0✏µ1+

✏! 2=1

4e2z✏E2

01+1+

✏! 2.

But Eq. 9.126 )1+1+

✏! 2=2

✏µ

!2, so hui=1

4e2z✏E2

✏µ

!2=k2

2µ!2E2

0e2z.So the ratio of the

magnetic contribution to the electric contribution is

humagi

hueleci=B2

0/µ

0✏=1

µ✏µ✏1+

✏! 2=1+

✏! 2>1.qed

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196 CHAPTER 9. ELECTROMAGNETIC WAVES

(b) S=1

µ(E⇥B)= 1

µE0B0e2zcos(kz!t+E) cos(kz!t+E+)ˆz ;hSi=1

2µE0B0e2zcos ˆz . [The

average of the product of the cosines is (1/2⇡)2⇡

0cos ✓cos(✓+)d✓= (1/2) cos .] So I=1

2µE0B0e2zcos =

2µE2

0e2zK

!cos , while, from Eqs. 9.133 and 9.134, Kcos =k,so I=k

2µ!E2

0e2z.qed

Problem 9.22

According to Eq. 9.147, R=

E0R

E0I

=

1˜



1+ ˜



=1˜



1+ ˜

1˜

⇤

1+ ˜

⇤,where ˜

=µ1v1

µ2!˜

=µ1v1

µ2!(k2+i2) (Eqs. 9.125 and 9.146). Since silver is a good conductor (✏!), Eq. 9.126 reduces to

2⇠

=k2⇠

=!✏2µ2

2

✏2!=!µ2

2,so ˜

=µ1v1

µ2!!µ2

2(1 + i)=µ1v1

2µ2!(1 + i).

Let ⌘µ1v1

2µ2!=µ0c

2µ0!=cµ0

2!=(3⇥108)(6 ⇥107)(4⇡⇥107)

(2)(4 ⇥1015)= 29.Then

R=1i

1++i1+i

1+i=(1 )2+2

(1 + )2+2= 0.93. Evidently 93% of the light is reﬂected.

Problem 9.23

(a) We are told that v=↵p, where ↵is a constant. But =2⇡/k and v=!/k,so

!=↵k2⇡/k =↵p2⇡k. From Eq. 9.150, vg=d!

dk =↵p2⇡1

2pk=1

2↵2⇡

k=1

2↵p=1

2v, or v=2vg.

(b) i(px Et)

~=i(kx !t))k=p

~,!=E

~=p2

2m~=~k2

2m.Therefore v=!

k=E

p=p

2m=~k

2m;

vg=d!

dk =2~k

2m=~k

m=p

m.So v=1

2vg.Since p=mvc(where vcis the classical speed of the particle), it

follows that vg(not v) corresponds to the classical veloctity.

Problem 9.24

E=1

4⇡✏0

a3)F=qE =1

4⇡✏0

a3x=kspringx=m!2

0x(Eq. 9.151). So !0=q2

4⇡✏0ma3.

⌫0=!0

2⇡=1

2⇡(1.6⇥1019)2

4⇡(8.85 ⇥1012)(9.11 ⇥1031)(0.5⇥1010)3= 7.16 ⇥1015 Hz. This is ultraviolet.

From Eqs. 9.173 and 9.174,

A=nq2

2m✏0

,N= # of molecules per unit volume = Avogadro0s#

22.4 liters =6.02⇥1023

22.4⇥103=2.69 ⇥1025,

f= # of electrons per molecule = 2 (for H2).

=(2.69 ⇥1025)(1.6⇥1019)2

(9.11 ⇥1031)(8.85 ⇥1012)(4.5⇥1016)2= 4.2⇥105(which is about 1/3 the actual value);

B=2⇡c

!02

=2⇡⇥3⇥108

4.5⇥1016 2

= 1.8⇥1015 m2(which is about 1/4 the actual value).

So even this extremely crude model is in the right ball park.

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CHAPTER 9. ELECTROMAGNETIC WAVES 197

Problem 9.25

Equation 9.170 )n=1+ Nq2

2m✏0

(!2

0!2)

[(!2

0!2)2+2!2]. Let the denominator ⌘D. Then

d!=Nq2

2m✏02!

D(!2

0!2)

D22(!2

0!2)(2!)+22!=0)2!D=(!2

0!2)2(!2

0!2)22!;

(!2

0!2)2+2!2= 2(!2

0!2)22(!2

0!2),or (!2

0!2)2=2(!2+!2

0!2)=2!2

0)(!2

0!2)=±!0;

!2=!2

0⌥!0,!=!01⌥/!0⇠

=!0(1 ⌥/2!0)=!0⌥/2. So !2=!0+/2,!1=!0/2, and the

width of the anomalous region is !=!2!1=.

From Eq. 9.171, ↵=Nq2!2

m✏0c



(!2

0!2)2+2!2, so at the maximum (!=!0), ↵max =Nq2

m✏0c.

At !1and !2,!2=!2

0⌥!0,so↵=Nq2!2

m✏0c



2!2

0+2!2=↵max !2

!2+!2

0.But

!2+!2

=!2

0⌥!0

2!2

0⌥!0=1

(1 ⌥/!0)

(1 ⌥/2!0)⇠

21⌥

!01±

2!0⇠

21⌥

2!0⇠

So ↵⇠

2↵max at !1and !2. qed

Problem 9.26

Equation 9.170 for a single resonance:

!=1+Nq2

2m✏0(!2

0!2)

(!2

0!2)2+2!2)ck =!1+ a!2

0(!2

0!2)

(!2

0!2)2+2!2where a⌘Nq2

2m✏0!2

From Eq. 9.150, the group velocity is vg=d!/dk, so

=cdk

=1+ a!2

0(!2

0!2)

(!2

0!2)2+2!2+!a!2

[(!2

0!2)2+2!2](2!)(!2

0!2)[2(!2

0!2)(2!)+2(2!)]

[(!2

0!2)2+2!2]2

=1+a!2

(!2

0!2)[(!2

0!2)2+2!2]2!2[(!2

0!2)2+2!2]+4!2(!2

0!2)222!2(!2

0!2)

[(!2

0!2)2+2!2]2

=1+a!2

(!2

0!2)3+2!2(!2

0!2)22!2(!2

0!2)22!4

[(!2

0!2)2+2!2]2

=1+a!2

0(!2

0+!2)(!2

0!2)22!2

[(!2

0!2)2+2!2]2.

vg=c1+a!2

0(!2

0+!2)(!2

0!2)22!2

[(!2

0!2)2+2!2]21

(a) For = 0 and a=0.003:

c=1+a!2

(!2

0+!2)

(!2

0!2)21

)y=1+0.003 1+x

(1 x)21

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198 CHAPTER 9. ELECTROMAGNETIC WAVES

0.0

0.5

1.0

1.5

2.0

0.2

0.4

0.6

0.8

1.0

(b) For = (0.1)!0and a=0.003:

y=1+0.003(1 + x)(1 x)20.01x

[(1 x)2+0.01x]21

0.0

0.5

1.0

1.5

2.0

1.0

1.5

2.0

2.5

Problem 9.27

(a) From Eqs. 9.176 and 9.177, r⇥˜

E=@˜

@t=i!˜

B0ei(kz!t);r⇥˜

B=1

@˜

@t=i!

c2˜

E0ei(kz!t).

In the terminology of Eq. 9.178:

(r⇥˜

E)x=@˜

@y@˜

@z=@˜

E0z

@yik ˜

E0yei(kz!t). So (ii) @Ez

@yikEy=i!Bx.

(r⇥˜

E)y=@˜

@z@˜

@x=ik ˜

E0x@˜

E0z

@xei(kz!t). So (iii) ikEx@Ez

@x=i!By.

(r⇥˜

E)z=@˜

@x@˜

@y=@˜

E0y

@x@˜

E0x

@yei(kz!t). So (i) @Ey

@x@Ex

@y=i!Bz.

(r⇥˜

B)x=@˜

@y@˜

@z=@˜

B0z

@yik ˜

B0yei(kz!t). So (v) @Bz

@yikBy=i!

c2Ex.

(r⇥˜

B)y=@˜

@z@˜

@x=ik ˜

B0x@˜

B0z

@xei(kz!t). So (vi) ikBx@Bz

@x=i!

c2Ey.

(r⇥˜

B)z=@˜

@x@˜

@y=@˜

B0y

@x@˜

B0x

@yei(kz!t). So (iv) @By

@x@Bx

@y=i!

c2Ez.

This conﬁrms Eq. 9.179. Now multiply (iii) by k, (v) by !, and subtract: ik2Exk@Ez

@x!@Bz

@y+i!kBy=

ik!By+i!2

c2Ex)ik2!2

c2Ex=k@Ez

@x+!@Bz

@y,or (i) Ex=i

(!/c)2k2k@Ez

@x+!@Bz

@y.

Multiply (ii) by k, (vi) by !, and add: k@Ez

@yik2Ey+i!kBx!@Bz

@x=i!kBxi!2

c2Ey)i!2

c2k2Ey=

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CHAPTER 9. ELECTROMAGNETIC WAVES 199

k@Ez

@y+!@Bz

@x,or (ii) Ey=i

(!/c)2k2k@Ez

@y!@Bz

@x.

Multiply (ii) by !/c2, (vi) by k, and add: !

@Ez

@yi!k

c2Ey+ik2Bxk@Bz

@x=i!2

c2Bxi!k

c2Ey)

ik2!2

c2Bx=k@Bz

@x!

@Ez

@y,or (iii) Bx=i

(!/c)2k2k@Bz

@x!

@Ez

@y.

Multiply (iii) by !/c2, (v) by k, and subtract: i!k

c2Ex!

@Ez

@xk@Bz

@y+ik2By=i!2

c2By+i!k

c2Ex)

ik2!2

c2By=!

@Ez

@x+k@Bz

@y,or (iv) By=i

(!/c)2k2k@Bz

@y+!

@Ez

@x.

This completes the conﬁrmation of Eq. 9.180.

(b) r·˜

E=@˜

@x+@˜

@y+@˜

@z=@˜

E0x

@x+@˜

E0y

@y+ik ˜

E0zei(kz!t)=0)@Ex

@x+@Ey

@y+ikEz= 0.

Using Eq. 9.180, i

(!/c)2k2k@2Ez

@x2+!@2Bz

@x@y+i

(!/c)2k2k@2Ez

@2y!@2Bz

@x@y+ikEz=0,

or @2Ez

@x2+@2Ez

@2y+(!/c)2k2Ez=0.

Likewise, r·˜

B=0)@Bx

@x+@By

@y+ikBz=0)

(!/c)2k2k@2Bz

@x2!

@2Ez

@x@y+i

(!/c)2k2k@2Bz

@y2+!

@Ez

@x@y+ikBz=0)

@2Bz

@x2+@2Bz

@2y+(!/c)2k2Bz= 0.

This conﬁrms Eqs. 9.181. [You can also do it by putting Eq. 9.180 into Eq. 9.179 (i) and (iv).]

Problem 9.28

Here Ez= 0 (TE) and !/c =k(n=m= 0), so Eq. 9.179(ii) )Ey=cBx, Eq. 9.179(iii) )Ex=cBy,

Eq. 9.179(v) )@Bz

@y=ikBy!

c2Ex=ikBy!

cBy= 0, Eq. 9.179(vi) )@Bz

@x=ikBx+!

c2Ey=

ikBx!

cBx= 0. So @Bz

@x=@Bz

@y= 0, and since Bzis a function only of xand y, this says Bzis in fact

aconstant (as Eq. 9.186 also suggests). Now Faraday’s law (in integral form) says E·dl=@B

@t·da,

and Eq. 9.176 )@B

@t=i!B,soE·dl=i!B·da. Applied to a cross-section of the waveguide this gives

E·dl=i!ei(kz!t)Bzda =i!Bzei(kz!t)(ab) (since Bzis constant, it comes outside the integral). But

if the boundary is just inside the metal, where E= 0, it follows that Bz=0.So this would be a TEM mode,

which we already know cannot exist for this guide.

Problem 9.29

Here a=2.28 cm and b=1.01 cm, so ⌫10 =1

2⇡!10 =c

2a=0.66 ⇥1010 Hz; ⌫20 =2c

2a=1.32 ⇥1010 Hz;

⌫30 =3c

2a=1.97 ⇥1010 Hz; ⌫01 =c

2b=1.49 ⇥1010 Hz; ⌫02 =2c

2b=2.97 ⇥1010 Hz; ⌫11 =c

21

a2+1

b2=

1.62 ⇥1010 Hz. Evidently just four modes occur: 10, 20, 01, and 11.

To get only one mode you must drive the waveguide at a frequency between ⌫10 and ⌫20:

0.66 ⇥1010 <⌫<1.32 ⇥1010 Hz. =c

⌫,so 10 =2a;20 =a. 2.28 cm <<4.56 cm.

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200 CHAPTER 9. ELECTROMAGNETIC WAVES

Problem 9.30

From Prob. 9.11, hSi=1

2µ0

Re( ˜

E⇥˜

B⇤). Here (Eq. 9.176) ˜

E=˜

E0ei(kz!t),˜

B⇤=˜

B⇤

0ei(kz!t), and, for

the TEmn mode (Eqs. 9.180 and 9.186)

B⇤

x=ik

(!/c)2k2m⇡

aB0sin m⇡x

acos n⇡y

b;

B⇤

y=ik

(!/c)2k2n⇡

bB0cos m⇡x

asin n⇡y

b;

B⇤

z=B0cos m⇡x

acos n⇡y

b;

Ex=i!

(!/c)2k2n⇡

bB0cos m⇡x

asin n⇡y

b;

Ey=i!

(!/c)2k2m⇡

aB0sin m⇡x

acos n⇡y

b;

Ez=0.

hSi=!k⇡2B2

[(!/c)2k2]2n

b2cos2m⇡x

asin2n⇡y

b+m

a2sin2m⇡x

acos2n⇡y

bˆz .

hSi·da=1

8µ0

!k⇡2B2

[(!/c)2k2]2ab m

a2+n

b2.[In the last step I used

a

0sin2(m⇡x/a)dx =a

0cos2(m⇡x/a)dx =a/2; b

0sin2(n⇡y/b)dy =b

0cos2(n⇡y/b)dy =b/2.]

Similarly,

hui=1

4✏0˜

E·˜

E⇤+1

µ0

B·˜

B⇤

=✏0

!2⇡2B2

[(!/c)2k2]2n

b2cos2m⇡x

asin2n⇡y

b+m

a2sin2m⇡x

acos2n⇡y

b

4µ0B2

0cos2m⇡x

acos2n⇡y

b

+k2⇡2B2

[(!/c)2k2]2n

b2cos2m⇡x

asin2n⇡y

b+m

a2sin2m⇡x

acos2n⇡y

b.

huida =ab

4✏0

!2⇡2B2

[(!/c)2k2]2n

b2+m

a2+B2

4µ0

k2⇡2B2

[(!/c)2k2]2n

b2+m

a2.

These results can be simpliﬁed, using Eq. 9.190 to write (!/c)2k2=(!mn/c)2,✏0µ0=1/c2to eliminate ✏0,

and Eq. 9.188 to write (m/a)2+(n/b)2=(!mn/⇡c)2:

hSi·da=!kabc2

8µ0!2

0;huida =!2ab

8µ0!2

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CHAPTER 9. ELECTROMAGNETIC WAVES 201

Evidently

energy per unit time

energy per unit length =hSi·da

huida =kc2

!=c

!!2!2

mn =vg(Eq.9.192).qed

Problem 9.31

Following Sect. 9.5.2, the problem is to solve Eq. 9.181 with Ez6=0,B

z= 0, subject to the boundary

conditions 9.175. Let Ez(x, y)=X(x)Y(y); as before, we obtain X(x)=Asin(kxx)+Bcos(kxx). But the

boundary condition requires Ez= 0 (and hence X= 0) when x= 0 and x=a, so B= 0 and kx=m⇡/a.

But this time m=1,2,3,..., but not zero, since m= 0 would kill Xentirely. The same goes for Y(y). Thus

Ez=E0sin m⇡x

asin n⇡y

bwith n, m =1,2,3,....

The rest is the same as for TE waves: !mn =c⇡(m/a)2+(n/b)2is the cuto↵frequency, the wave

velocity is v=c/1(!mn/!)2, and the group velocity is vg=c1(!mn/!)2. The lowest TM mode is

11, with cuto↵frequency !11 =c⇡(1/a)2+ (1/b)2. So the ratio of the lowest TM frequency to the lowest

TE frequency is c⇡(1/a)2+ (1/b)2

(c⇡/a)=1+(a/b)2.

Problem 9.32

(a) r·E=1

@s(sEs)=0X;r·B=1

@(B)=0X;r⇥E=@Es

@zˆ

1

@Es

@ ˆz =E0ksin(kz !t)

sˆ

?

@B

@t=E0!

sin(kz !t)

sˆ

X(since k=!/c); r⇥B=@B

@zˆs +1

@s(sB)ˆz =E0k

sin(kz !t)

sˆs ?

@t=E0!

sin(kz !t)

sˆs X.Boundary conditions: Ek=Ez=0X;B?=Bs=0X.

(b) To determine , use Gauss’s law for a cylinder of radius sand length dz:

E·da=E0

cos(kz !t)

s(2⇡s)dz =1

✏0

Qenc =1

✏0

dz )=2⇡✏0E0cos(kz !t).

To determine I, use Amp´ere’s law for a circle of radius s(note that the displacement current through this

loop is zero, since Eis in the ˆs direction): B·dl=E0

cos(kz !t)

s(2⇡s)=µ0Ienc )I=2⇡E0

µ0ccos(kz !t).

The charge and current on the outer conductor are precisely the opposite of these, since E=B=0inside

the metal, and hence the total enclosed charge and current must be zero.

Problem 9.33

f(z,0) = 1

1

A(k)eikz dk )˜

f(z,0)⇤=1

1

A(k)⇤eikz dk. Let l⌘k; then ˜

f(z,0)⇤=

1

A(l)⇤eilz(dl)=1

1

A(l)⇤eilz dl =1

1

A(k)⇤eikz dk (renaming the dummy variable l!k).

f(z,0) = Re ˜

f(z,0)=1

2˜

f(z,0) + ˜

f(z,0)⇤=1

1

2˜

A(k)+ ˜

A(k)⇤eikz dk. Therefore

2˜

A(k)+ ˜

A(k)⇤=1

2⇡1

1

f(z,0)eikz dz.

Meanwhile, ˙

f(z,t)=1

1

A(k)(i!)ei(kz!t)dk )˙

f(z,0) = 1

1

[i!˜

A(k)]eikz dk.

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202 CHAPTER 9. ELECTROMAGNETIC WAVES

(Note that !=|k|v, here, so it does not come outside the integral.)

f(z,0)⇤=1

1

[i!˜

A(k)⇤]eikz dk =1

1

[i|k|v˜

A(k)⇤]eikz dk =1

[i|l|v˜

A(l)⇤]eilz(dl)

=1

1

[i|k|v˜

A(k)⇤]eikz dk =1

1

[i!˜

A(k)⇤]eikz dk.

f(z,0) = Re ˙

f(z,0)=1

2˙

f(z,0) + ˙

f(z,0)⇤=1

1

2[i!˜

A(k)+i!˜

A(k)⇤]eikz dk.

i!

2˜

A(k)˜

A(k)⇤=1

2⇡1

1

f(z,0)eikz dz, or 1

2˜

A(k)˜

A(k)⇤=1

2⇡1

1 i

!˙

f(z,0)eikz dz.

Adding these two results, we get ˜

A(k)= 1

2⇡1

1 f(z,0) + i

!˙

f(z,0)eikz dz. qed

Problem 9.34

(a) Since (E⇥B) points in the direction of propagation, B=E0

c[cos(kz !t)+cos(kz +!t)] ˆy .

(b) From Eq. 7.63, K⇥(ˆz )= 1

µ0

B=E0

µ0c[2 cos(!t)] ˆy ,K=2E0

µ0ccos(!t)ˆx .

µ0c2[cos(!t)ˆx ]⇥[cos(!t)ˆy ]= 2✏0E2

0cos2(!t)ˆz .The time

average of cos2(!t) is 1/2, so

fave =✏0E2

This is twice the pressure in Eq. 9.64, but that was for a perfect absorber, whereas this is a perfect reﬂector.

Problem 9.35

(a) (i) Gauss’s law: r·E=1

rsin ✓

@E

@ =0.X

(ii) Faraday’s law:

@B

@t=r⇥E=1

rsin ✓

@✓(sin ✓E)ˆr 1

@r(rE)ˆ

✓

rsin ✓

@✓ E0

sin2✓

rcos u1

kr sin uˆr 1

@rE0sin ✓cos u1

kr sin uˆ

✓.

But @

@rcos u=ksin u;@

@rsin u=kcos u.

rsin ✓

r2 sin ✓cos ✓cos u1

kr sin uˆr 1

rE0sin ✓ksin u+1

kr2sin u1

rcos uˆ

✓.

Integrating with respect to t, and noting that cos udt=1

!sin uand sin udt=1

!cos u, we obtain

B=2E0cos ✓

!r2sin u+1

kr cos uˆr +E0sin ✓

!rkcos u+1

kr2cos u+1

rsin uˆ

✓.

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CHAPTER 9. ELECTROMAGNETIC WAVES 203

(iii) Divergence of B:

r·B=1

@rr2Br+1

rsin ✓

@✓(sin ✓B✓)

@r2E0cos ✓

!sin u+1

kr cos u+1

rsin ✓

@✓ E0sin2✓

!rkcos u+1

kr2cos u+1

rsin u

2E0cos ✓

!kcos u1

kr2cos u1

rsin u

rsin ✓

2E0sin ✓cos ✓

!rkcos u+1

kr2cos u+1

rsin u

=2E0cos ✓

!r2kcos u1

kr2cos u1

rsin ukcos u+1

kr2cos u+1

rsin u=0.X

(iv) Amp´ere/Maxwell:

r⇥B=1

r@

@r(rB✓)@Br

@✓ ˆ



r@

@rE0sin ✓

!kcos u+1

kr2cos u+1

rsin u@

@✓ 2E0cos ✓

!r2sin u+1

kr cos uˆ



=E0sin ✓

!rk2sin u2

kr3cos u1

r2sin u1

r2sin u+k

rcos u+2

r2sin u+2

kr3cos uˆ



E0sin ✓

rksin u+1

rcos uˆ

=1

E0sin ✓

rksin u+1

rcos uˆ

.

@t=1

E0sin ✓

r!sin u+!

kr cos uˆ

=1

E0sin ✓

rksin u+1

rcos uˆ



E0sin ✓

rksin u+1

rcos uˆ

=r⇥B.X

(b) Poynting Vector:

S=1

µ0

(E⇥B)=E0sin ✓

µ0rcos u1

kr sin u2E0cos ✓

!r2sin u+1

kr cos uˆ

✓

+E0sin ✓

!rkcos u+1

kr2cos u+1

rsin u(ˆr )

=E2

0sin ✓

µ0!r22 cos ✓

rsin ucos u+1

kr(cos2usin2u)1

k2r2sin ucos uˆ

✓

sin ✓kcos2u+1

kr2cos2u+1

rsin ucos u+1

rsin ucos u1

k2r3sin ucos u1

kr2sin2uˆr 

0sin ✓

µ0!r22 cos ✓

r11

k2r2sin ucos u+1

kr(cos2usin2u)ˆ

✓

+ sin ✓2

r+1

k2r3sin ucos u+kcos2u+1

kr2(sin2ucos2u)ˆr .

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204 CHAPTER 9. ELECTROMAGNETIC WAVES

Averaging over a full cycle, using hsin ucos ui=0,hsin2ui=hcos2ui=1

2, we get the intensity:

I=hSi=E2

0sin ✓

µ0!r2k

2sin ✓ˆr =E2

0sin2✓

2µ0cr2ˆr .

It points in the ˆr direction, and falls o↵as 1/r2, as we would expect for a spherical wave.

2µ0csin2✓

r2r2sin ✓d✓d=E2

2µ0c2⇡⇡

sin3✓d✓=4⇡

µ0c.

Problem 9.36

z<0: ˜

EI(z,t)= ˜

EIei(k1z!t)ˆx ,˜

BI(z,t)= 1

EIei(k1z!t)ˆy

ER(z,t)= ˜

ERei(k1z!t)ˆx ,˜

BR(z,t)=1

ERei(k1z!t)ˆy .

0<z<d:˜

Er(z,t)= ˜

Erei(k2z!t)ˆx ,˜

Br(z,t)= 1

Erei(k2z!t)ˆy

El(z,t)= ˜

Elei(k2z!t)ˆx ,˜

Bl(z,t)=1

Elei(k2z!t)ˆy .

z>d:˜

ET(z,t)= ˜

ETei(k3z!t)ˆx ,˜

BT(z,t)= 1

ETei(k3z!t)ˆy .

Boundary conditions: Ek

1=Ek

2,Bk

1=Bk

2, at each boundary (assuming µ1=µ2=µ3=µ0):

z=0: 









EI+˜

ER=˜

Er+˜

El;

EI1

ER=1

Er1

El)˜

EI˜

ER=(˜

Er˜

El),where ⌘v1/v2.

z=d:









Ereik2d+˜

Eleik2d=˜

ETeik3d;

Ereik2d1

Eleik2d=1

ETeik3d)˜

Ereik2d˜

Eleik2d=↵˜

ETeik3d,where ↵⌘v2/v3.

We have here four equations; the problem is to eliminate ˜

ER,˜

Er, and ˜

El, to obtain a single equation for

ETin terms of ˜

EI.

Add the ﬁrst two to eliminate ˜

ER:2

EI= (1 + )˜

Er+ (1 )˜

El;

Add the last two to eliminate ˜

El:2

Ereik2d=(1+↵)˜

ETeik3d;

Subtract the last two to eliminate ˜

Er:2

Eleik2d=(1↵)˜

ETeik3d.

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CHAPTER 9. ELECTROMAGNETIC WAVES 205

Plug the last two of these into the ﬁrst:

2˜

EI= (1 + )1

2eik2d(1 + ↵)˜

ETeik3d+ (1 )1

2eik2d(1 ↵)˜

ETeik3d

4˜

EI=(1 + ↵)(1 + )eik2d+ (1 ↵)(1 )eik2d˜

ETeik3d

=(1 + ↵)eik2d+eik2d+(↵+)eik2deik2d˜

ETeik3d

= 2 [(1 + ↵) cos(k2d)i(↵+) sin(k2d)] ˜

ETeik3d.

Now the transmission coeﬃcient is T=v3✏3E2

v1✏1E2

=v3

v1µ0✏3

µ0✏1|˜

ET|2

|˜

EI|2=v1

|˜

ET|2

|˜

EI|2=↵ |˜

ET|2

|˜

EI|2,so

T1=1

↵ |˜

EI|2

|˜

ET|2=1

↵ 

2[(1 + ↵) cos(k2d)i(↵+) sin(k2d)] eik3d

4↵ (1 + ↵)2cos2(k2d)+(↵+)2sin2(k2d).But cos2(k2d)=1sin2(k2d).

4↵ (1 + ↵)2+(↵2+2↵ +212↵ ↵22) sin2(k2d)

4↵ (1 + ↵)2(1 ↵2)(1 2) sin2(k2d).

But n1=c

2=c

3=c

,so ↵=n3

,=n2

4n1n3(n1+n3)2+(n2

1n2

2)(n2

3n2

sin2(k2d).

Problem 9.37

T=1)sin kd =0)kd =0,⇡,2⇡....Theminimum (nonzero) thickness is d=⇡/k.Butk=!/v =

2⇡⌫/v =2⇡⌫n/c, and n=✏µ/✏0µ0(Eq. 9.69), where (presumably) µ⇡µ0. So n=✏/✏0=p✏r, and hence

d=⇡c

2⇡⌫p✏r

2⌫p✏r

=3⇥108

2(10 ⇥109)p2.5=9.49 ⇥103m, or 9.5 mm.

Problem 9.38

From Eq. 9.199,

T1=1

4(4/3)(1) [(4/3) + 1]2+[(16/9) (9/4)][1 (9/4)]

(9/4) sin2(3!d/2c)

16 49

9+(17/36)(5/4)

(9/4) sin2(3!d/2c)=49

48 +85

(48)(36) sin2(3!d/2c).

T=48

49 + (85/36) sin2(3!d/2c).

Since sin2(3!d/2c) ranges from 0 to 1, Tmin =48

49 + (85/36) = 0.935; Tmax =48

49 = 0.980. Not much

variation, and the transmission is good (over 90%) for all frequencies. Since Eq. 9.199 is unchanged when you

switch 1 and 3, the transmission is the same either direction, and the ﬁsh sees you just as well as you see it.

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206 CHAPTER 9. ELECTROMAGNETIC WAVES

Problem 9.39

(a) Equation 9.91 )˜

ET(r,t)= ˜

E0Tei(kT·r!t);kT·r=kT(sin ✓Tˆx + cos ✓Tˆz )·(xˆx +yˆy +zˆz )=

kT(xsin ✓T+zcos ✓T)=xkTsin ✓T+izkTsin2✓T1=kx +iz, where

k⌘kTsin ✓T=!n2

cn1

sin ✓I=!n1

csin ✓I,

⌘kTsin2✓T1=!n2

c(n1/n2)2sin2✓I1=!

cn2

1sin2✓In2

2.So

ET(r,t)= ˜

E0Tezei(kx!t).qed

(b) R=

E0R

E0I

=

↵

↵+

. Here is real (Eq. 9.106) and ↵is purely imaginary (Eq. 9.108); write ↵=ia,

with areal: R=ia 

ia +ia 

ia +=a2+2

a2+2= 1.

1↵

1+↵ E0I,so R=

1↵

1+↵ 

=

1ia

1+ia

=(1 ia)(1 + ia)

(1 + ia)(1 ia)= 1.

(d) From the solution to Prob. 9.17, the transmitted wave is

E(r,t)= ˜

E0Tei(kT·r!t)ˆy ,˜

B(r,t)= 1

E0Tei(kT·r!t)(cos ✓Tˆx + sin ✓Tˆz ).

Using the results in (a): kT·r=kx +iz, sin ✓T=ck

!n2

,cos ✓T=ic

!n2

E(r,t)= ˜

E0Tezei(kx!t)ˆy ,˜

B(r,t)= 1

E0Tezei(kx!t)ic

!n2

ˆx +ck

!n2

ˆz .

We may as well choose the phase constant so that ˜

E0Tis real. Then

E(r,t)=E0ezcos(kx !t)ˆy ;

B(r,t)= 1

E0ezc

!n2

Re {[cos(kx !t)+isin(kx !t)] [iˆx +kˆz ]}

!E0ez[sin(kx !t)ˆx +kcos(kx !t)ˆz ].qed

(I used v2=c/n2to simplfy B.)

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CHAPTER 9. ELECTROMAGNETIC WAVES 207

(e) (i) r·E=@

@yE0ezcos(kx !t)=0.X

(ii) r·B=@

@xE0

!ezsin(kx !t)+@

@zE0

!ezkcos(kx !t)

=E0

!ezkcos(kx !t)ezkcos(kx !t)=0.X

(iii) r⇥E=

ˆx ˆy ˆz

@/@x@/@y@/@z

0Ey0=@Ey

@zˆx +@Ey

@xˆz

=E0ezcos(kx !t)ˆx E0ezksin(kx !t)ˆz .

@B

@t=E0

!ez[! cos(kx !t)ˆx +k!sin(kx !t)ˆz ]

=E0ezcos(kx !t)ˆx kE0ezsin(kx !t)ˆz =r⇥E.X

(iv) r⇥B=

ˆx ˆy ˆz

@/@x@/@y@/@z

Bx0Bz=@Bx

@z@Bz

@xˆy

=E0

!2ezsin(kx !t)+E0

!ezk2sin(kx !t)ˆy =(k22)E0

!ezsin(kx !t)ˆy .

Eq.9.202 )k22=!

c2n2

1sin2✓I(n1sin ✓I)2+(n2)2=n2!

c2=!2✏2µ2.

=✏2µ2!E0ezsin(kx !t)ˆy .

µ2✏2

@t=µ2✏2E0ez!sin(kx !t)ˆy =r⇥BX.

(f)

S=1

µ2

(E⇥B)= 1

µ2

!e2z

ˆx ˆy ˆz

0 cos(kx !t)0

sin(kx !t)0kcos(kx !t)

=E2

µ2!e2zkcos2(kx !t)ˆx sin(kx !t) cos(kx !t)ˆz .

Averaging over a complete cycle, using hcos2i=1/2 and hsin cosi= 0, hSi=E2

2µ2!e2zˆx . On average,

then, no energy is transmitted in the zdirection, only in the xdirection (parallel to the interface). qed

Problem 9.40

Look for solutions of the form E=E0(x, y, z)ei!t,B=B0(x, y, z)ei!t, subject to the boundary condi-

tions Ek=0,B

?= 0 at all surfaces. Maxwell’s equations, in the form of Eq. 9.177, give

r·E=0 )r·E0= 0; r⇥E=@B

@t)r⇥E0=i!B0;

r·B=0)r·B0= 0; r⇥B=1

@t)r⇥B0=i!

c2E0.

From now on I’ll leave o↵the subscript (0). The problem is to solve the (time independent) equations

r·E= 0; r⇥E=i!B;

r·B= 0; r⇥B=i!

c2E.

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208 CHAPTER 9. ELECTROMAGNETIC WAVES

From r⇥E=i!Bit follows that I can get Bonce I know E, so I’ll concentrate on the latter for the moment.

r⇥(r⇥E)=r(r·E)r

2E=r2E=r⇥(i!B)=i!i!

c2E=!2

c2E. So

r2Ex=!

c2Ex;r2Ey=!

c2Ey;r2Ez=!

c2Ez. Solve each of these by separation of variables:

Ex(x, y, z)=X(x)Y(y)Z(z))YZd2X

dx2+ZX d2Y

dy2+XY d2Z

dz2=!

c2XY Z, or 1

d2X

dx2+1

d2Y

dy2+1

d2Z

dz2=

(!/c)2. Each term must be a constant, so d2X

dx2=k2

xX, d2Y

dy2=k2

yY, d2Z

dz2=k2

zZ, with

x+k2

y+k2

z=(!/c)2. The solution is

Ex(x, y, z)=[Asin(kxx)+Bcos(kxx)][Csin(kyy)+Dcos(kyy)][Esin(kzz)+Fcos(kzz)].

But Ek= 0 at the boundaries )Ex= 0 at y= 0 and z= 0, so D=F= 0, and Ex= 0 at y=band z=d, so

ky=n⇡/b and kz=l⇡/d, where nand lare integers. A similar argument applies to Eyand Ez.Conclusion:

Ex(x, y, z)=[Asin(kxx)+Bcos(kxx)] sin(kyy) sin(kzz),

Ey(x, y, z) = sin(kxx)[Csin(kyy)+Dcos(kyy)] sin(kzz),

Ez(x, y, z) = sin(kxx) sin(kyy)[Esin(kzz)+Fcos(kzz)],

where kx=m⇡/a. (Actually, there is no reason at this stage to assume that kx,ky, and kzare the same for

all three components, and I should really aﬃx a second subscript (xfor Ex,yfor Ey, and zfor Ez), but in a

moment we shall see that in fact they do have to be the same, so to avoid cumbersome notation I’ll assume

they are from the start.)

Now r·E=0)kx[Acos(kxx)Bsin(kxx)] sin(kyy) sin(kzz)+kysin(kxx)[Ccos(kyy)Dsin(kyy)] sin(kzz)+

kzsin(kxx) sin(kyy)[Ecos(kzz)Fsin(kzz)] = 0.In particular, putting in x= 0, kxAsin(kyy) sin(kzz) = 0,

and hence A= 0. Likewise y=0)C= 0 and z=0)E= 0. (Moreover, if the k’s were not equal for di↵erent

components, then by Fourier analysis this equation could not be satisﬁed (for all x,y, and z) unless the other

three constants were also zero, and we’d be left with no ﬁeld at all.) It follows that (Bkx+Dky+Fk

z)=0

(in order that r·E= 0), and we are left with

E=Bcos(kxx) sin(kyy) sin(kzz)ˆx +Dsin(kxx) cos(kyy) sin(kzz)ˆy +Fsin(kxx) sin(kyy) cos(kzz)ˆz ,

with kx=(m⇡/a),k

y=(n⇡/b),k

z=(l⇡/d)(l, m, n all integers),and Bkx+Dky+Fk

z=0.

The corresponding magnetic ﬁeld is given by B=(i/!)r⇥E:

Bx=i

!@Ez

@y@Ey

@z=i

![Fk

ysin(kxx) cos(kyy) cos(kzz)Dkzsin(kxx) cos(kyy) cos(kzz)] ,

By=i

!@Ex

@z@Ez

@x=i

![Bkzcos(kxx) sin(kyy) cos(kzz)Fk

xcos(kxx) sin(kyy) cos(kzz)] ,

Bz=i

!@Ey

@x@Ex

@y=i

![Dkxcos(kxx) cos(kyy) sin(kzz)Bkycos(kxx) cos(kyy) sin(kzz)] .

Or:

B=i

!(Fk

yDkz) sin(kxx) cos(kyy) cos(kzz)ˆx i

!(BkzFk

x) cos(kxx) sin(kyy) cos(kzz)ˆy

i

!(DkxBky) cos(kxx) cos(kyy) sin(kzz)ˆz .

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CHAPTER 9. ELECTROMAGNETIC WAVES 209

These automatically satisfy the boundary condition B?=0(Bx= 0 at x= 0 and x=a,By= 0 at y= 0 and

y=b, and Bz= 0 at z= 0 and z=d).

As a check, let’s see if r·B=0:

r·B=i

!(Fk

yDkz)kxcos(kxx) cos(kyy) cos(kzz)i

!(BkzFk

x)kycos(kxx) cos(kyy) cos(kzz)

i

!(DkxBky)kzcos(kxx) cos(kyy) cos(kzz)

=i

!(Fk

xkyDkxkz+BkzkyFk

xky+DkxkzBkykz) cos(kxx) cos(kyy) cos(kzz)=0.X

The boxed equations satisfy all of Maxwell’s equations, and they meet the boundary conditions. For TE

modes, we pick Ez= 0, so F= 0 (and hence Bkx+Dky= 0, leaving only the overall amplitude undetermined,

for given l,m, and n); for TM modes we want Bz= 0 (so DkxBky= 0, again leaving only one amplitude

undetermined, since Bkx+Dky+Fk

z= 0). In either case (TElmn or TMlmn), the frequency is given by

!2=c2(k2

x+k2

y+k2

z)=c2(m⇡/a)2+(n⇡/b)2+(l⇡/d)2, or !=c⇡(m/a)2+(n/b)2+(l/d)2.

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210 CHAPTER 10. POTENTIALS AND FIELDS

Chapter 10

Potentials and Fields

Problem 10.1

22V+@L

@t=r2Vµ0✏0

@2V

@t2+@

@t(r·A)+µ0✏0

@2V

@t2=r2V+@

@t(r·A)=1

✏0

⇢.X

22ArL=r2Aµ0✏0

@2A

@t2rr·A+µ0✏0

@t=µ0J.X

Problem 10.2

(a) W=1

2✏0E2+1

µ0

B2d⌧.At t1=d/c, x d=ct1,so E=0,B=0,and hence W(t1)=0.

At T2=(d+h)/c,ct2=d+h:

E=µ0k

2(d+hx)ˆz ,B=1

µ0k

2(d+hx)ˆy ,

so B2=1

c2E2, and ✏0E2+1

µ0

B2=✏0E2+1

µ0✏0

c2E2=2✏0E2.

Therefore

W(t2)=1

2(2✏0)µ2

0k2

(d+h)



(d+hx)2dx (lw)= ✏0µ2

0k2lw

4(d+hx)3

3d+h

=✏0µ2

0k2lwh3

12 .

(b) S(x)= 1

µ0

(E⇥B)= 1

µ0cE2[ˆz ⇥(±ˆy )] = ±1

µ0cE2ˆx =±µ0k2

4c(ct |x|)2ˆx

(plus sign for x>0, as here). For |x|> ct, S= 0.

So the energy per unit time entering the box in this time interval is

dt =P=S(d)·da=µ0k2lw

4c(ct d)2.

Note that no energy ﬂows out the top, since S(d+h) = 0.

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CHAPTER 10. POTENTIALS AND FIELDS 211



P dt =µ0k2lw

(d+h)/c



d/c

(ct d)2dt =µ0k2lw

4c(ct d)3

3c(d+h)/c

d/c

=µ0k2lwh3

12c2.

Since 1/c2=µ0✏0, this agrees with the answer to (a).

Problem 10.3

(a)

E=rV@A

@t=1

4⇡✏0

r2ˆr .B=r⇥A=0.

This is a funny set of potentials for a stationary point charge qat the origin. (V=1

4⇡✏0

r,A=0would, of

course, be the customary choice.) Evidently ⇢=q3(r); J=0.

(b)

V0=V@

@t=01

4⇡✏0

r=1

4⇡✏0

r;A0=A+r=1

4⇡✏0

r2ˆr +1

4⇡✏0

qt1

r2ˆr =0.

This gauge function transforms the “funny” potentials into the “ordinary” potentials of a stationary point

charge.

Problem 10.4

E=rV@A

@t=A0cos(kx !t)ˆy (!)= A0!cos(kx !t)ˆy ,

B=r⇥A=ˆz @

@x[A0sin(kx !t)] = A0kcos(kx !t)ˆz .

Hence r·E=0X,r·B=0X.

r⇥E=ˆz @

@x[A0!cos(kx !t)] = A0!ksin(kx !t)ˆz ,@B

@t=A0!ksin(kx !t)ˆz ,

so r⇥E=@B

@tX.

r⇥B=ˆy @

@x[A0kcos(kx !t)] = A0k2sin(kx !t)ˆy ,@E

@t=A0!2sin(kx !t)ˆy .

So r⇥B=µ0✏0

@tprovided k2=µ0✏0!2,or, since c2=1/µ0✏0,!=ck.

Problem 10.5

Ex. 10.1: r·A= 0; @V

@t=0.Both Coulomb and Lorentz.

Prob. 10.3: r·A=qt

4⇡✏0r·ˆr

r2=qt

✏0

3(r); @V

@t=0. Neither.

Prob. 10.4: r·A= 0; @V

@t=0.Both.

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212 CHAPTER 10. POTENTIALS AND FIELDS

Problem 10.6

Suppose r·A6=µ0✏0

@t. (Let r·A+µ0✏0

@t=—some known function.) We want to pick such

that A0and V0(Eq. 10.7) do obey r·A0=µ0✏0

@V0

@t.

r·A0+µ0✏0

@V0

@t=r·A+r2+µ0✏0

@tµ0✏0

@2

@t2=+22.

This will be zero provided we pick for the solution to 22=, which by hypothesis (and in fact) we know

how to solve.

We could always ﬁnd a gauge in which V0= 0, simply by picking =t

0Vdt

0.Wecannot in general pick

A=0—this would make B=0. [Finding such a gauge function would amount to expressing Aas r, and

we know that vector functions cannot in general be written as gradients—only if they happen to have curl

zero, which A(ordinarily) does not.]

Problem 10.7

(a) Using Eq. 1.99,

r·J=˙q

4⇡r·ˆr

r2=˙q

4⇡4⇡3(r)=˙q3(r)=@⇢

@t.X

(b) From Eq. 10.10,

V(r,t)= q

4⇡✏03(r0)

d⌧0=1

4⇡✏0

q(t)

By symmetry, B=0(what direction could it point?), so r⇥A=0,r·A= 0, and A!0at inﬁnity.

Evidently A=0.

@t=1

4⇡✏0

q(t)

r2ˆr .B=0. Checking Maxwell’s equations:

r·E=q

4⇡✏0r·ˆr

r2=q

4⇡✏0

4⇡3(r)=q3(r)

✏0

=⇢

✏0

r·B=0.X

r⇥E=0=@B

@t.X

r⇥B=0;µ0J+µ0✏0

@t=µ01

4⇡

˙q

r2ˆr +µ0✏0˙q

4⇡✏0

r2ˆr =0.X

[Note that the displacement current exactly cancels the conduction current. Physically, this conﬁguration is a

point charge at the origin that is changing with time as current ﬂows in symmetrically (from inﬁnity).]

Problem 10.8

Noting the Ais independent of tand Bis independent of r, use Eq. 10.19:

dt =@A

@t+(v·r)A=(v·r)1

2(r⇥B)

=1

2vx

@x+vy

@y+vz

@z[(yBzzBy)ˆx +(zBxxBz)ˆy +(xByyBx)ˆz ]

=1

2[vx(Bzˆy +Byˆz )+vy(Bzˆx Bxˆz )+vz(Byˆx +Bxˆy )]

=1

2[(vyBzvzBy)ˆx +(vzBxvxBz)ˆy +(vxByvyBx)ˆz ]=1

2(v⇥B).X

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CHAPTER 10. POTENTIALS AND FIELDS 213

Equation 10.20 says

dt(p+qA)= dp

dt q

2(v⇥B)=qr(v·A)=q

2r[v·(r⇥B)],

or dp

dt =q

2(v⇥B)q

2r[r·(B⇥v)].

Now, for a vector cthat is independent of position,

r(r·c)=ˆx @

@x+ˆy @

@y+ˆz @

@z(xcx+ycy+zcz)=cxˆx +cyˆy +czˆz )=c.

In this case c=(B⇥v)=(v⇥B), so

dt =q

2(v⇥B)+ q

2(v⇥B)=q(v⇥B).X

Problem 10.9

dt(T+qV )= d

dt 1

2mv2+qdV

dt =mv·dv

dt +q@V

@t+(v·r)V=v·F+q@V

@t+(v·r)V.

But Eq. 10.17 says

v·F=qv·rV@A

@t+v⇥(r⇥A)=qv·rV+v·@A

@t,

dt(T+qV )=q(v·r)V+v·@A

@t+q@V

@t+(v·r)V=q@V

@tv·@A

@t=@

@t[q(Vv·A)] .

Problem 10.10

From the product rule:

r·J

=1

(r·J)+J·r1

,r0·J

=1

(r0·J)+J·r01

.

But r1

=r01

, since

=rr0. So

r·J

=1

(r·J)J·r01

=1

(r·J)+ 1

(r0·J)r0·J

.

But

r·J=@Jx

@x+@Jy

@y+@Jz

@z=@Jx

@tr

@x+@Jy

@tr

@y+@Jz

@tr

@z,

and @tr

@x=1

@x,@tr

@y=1

@y,@tr

@z=1

@z,

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214 CHAPTER 10. POTENTIALS AND FIELDS

r·J=1

c@Jx

@tr

@x+@Jy

@tr

@y+@Jz

@tr

@z=1

@tr·(r

Similarly,

r0·J=@⇢

@t1

@tr·(r0

[The ﬁrst term arises when we di↵erentiate with respect to the explicit r0, and use the continuity equation.]

thus

r·J

=1

1

@tr·(r0

)+1

@⇢

@t1

@tr·(r0

)r·J

=1

@⇢

@tr0·J



(the other two terms cancel, since r

=r0

). Therefore:

r·A=µ0

4⇡@

@t⇢

d⌧r0·J

d⌧=µ0✏0

@t1

4⇡✏0⇢

d⌧µ0

4⇡J

·da.

The last term is over the suface at “inﬁnity”, where J= 0, so it’s zero. Therefore r·A=µ0✏0

@t.X

Problem 10.11

(a) As in Ex. 10.2, for t < s/c, A=0; for t > s/c,

A(s, t)=µ0

4⇡ˆz 2

p(ct)2s2



k(tps2+z2/c)

ps2+z2dz =µ0k

2⇡ˆz 









p(ct)2s2



ps2+z21

p(ct)2s2



dz









=µ0k

2⇡ˆz tln ct +(ct)2s2

s1

c(ct)2s2.Accordingly,

E(s, t)=@A

@t=µ0k

2⇡ˆz ln ct +(ct)2s2

s+

ts

ct +(ct)2s21

sc+1

2c2t

(ct)2s21

2c2t

(ct)2s2

=µ0k

2⇡ˆz ln ct +(ct)2s2

s+ct

(ct)2s2ct

(ct)2s2

=µ0k

2⇡ln ct +(ct)2s2

sˆz (or zero, for t < s/c).

B(s, t)=@Az

@sˆ



=µ0k

2⇡









ts

ct +(ct)2s2s1

(2s)

p(ct)2s2ct (ct)2s2

s21

(2s)

(ct)2s2











=µ0k

2⇡ct2

s(ct)2s2+s

c(ct)2s2ˆ

=µ0k

2⇡

(c2t2+s2)

sc(ct)2s2

=µ0k

2⇡sc(ct)2s2ˆ

.

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CHAPTER 10. POTENTIALS AND FIELDS 215

(b) A(s, t)= µ0

4⇡ˆz 1

1

q0(t

/c)

dz. But

=ps2+z2, so the integrand is even in z:

A(s, t)=µ0q0

4⇡ˆz 21

(t

/c)

dz.

Now z=

2s2)dz =1

2s2=

2s2, and z=0)

=s, z =1)

=1. So:

A(s, t)=µ0q0

2⇡ˆz 1

t

c

2s2.

Now (t

/c)=c(

ct) (Ex. 1.15); therefore A=µ0q0

2⇡ˆz c1

(

ct)

2s2d

, so

A(s, t)=µ0q0c

2⇡

(ct)2s2ˆz (or zero, if ct < s);

E(s, t)=@A

@t=µ0q0c

2⇡1

22c2t

[(ct)2s2]3/2ˆz =µ0q0c3t

2⇡[(ct)2s2]3/2ˆz (or zero, for t < s/c);

B(s, t)=@Az

@tˆ

=µ0q0c

2⇡1

22s

[(ct)2s2]3/2ˆ

=µ0q0cs

2⇡[(ct)2s2]3/2ˆ

(or zero, for t < s/c).

Problem 10.12

A=µ0

4⇡I(tr)

dl =µ0k

4⇡(t

/c)

dl=µ0k

4⇡tdl

1

cdl.

But for the complete loop, dl= 0, so A=µ0kt

4⇡1

a1

dl+1

b2

dl+2ˆx b

x. Here 1dl=2aˆx (inner

circle), 2dl=2bˆx (outer circle), so

A=µ0kt

4⇡1

a(2a)+1

b(2b) + 2 ln(b/a)ˆx )A=µ0kt

2⇡ln(b/a)ˆx ,E=@A

@t=µ0k

2⇡ln(b/a)ˆx .

The changing magnetic ﬁeld induces the electric ﬁeld. Since we only know Aat one point (the center), we

can’t compute r⇥Ato get B.

Problem 10.13

In this case ˙⇢(r,t)= ˙⇢(r,0) and ˙

J(r,t) = 0, so Eq. 10.36 )

E(r,t)= 1

4⇡✏0⇢(r0,0) + ˙⇢(r0,0)tr

2+˙⇢(r0,0)

ˆ

d⌧0,but tr=t

c(Eq.10.18),so

4⇡✏0⇢(r0,0) + ˙⇢(r0,0)t

2˙⇢(r0,0)(

/c)

2+˙⇢(r0,0)

ˆ

d⌧0=1

4⇡✏0⇢(r0,t)

2ˆ

d⌧0.qed

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216 CHAPTER 10. POTENTIALS AND FIELDS

Problem 10.14

In this approximation we’re dropping the higher derivatives of J,so ˙

J(tr)= ˙

J(t), and Eq. 10.38 )

B(r,t)= µ0

4⇡1

2J(r0,t)+(trt)˙

J(r0,t)+

c˙

J(r0,t)⇥ˆ

d⌧0,but trt=

c(Eq.10.25),so

=µ0

4⇡J(r0,t)⇥ˆ

2d⌧0.qed

Problem 10.15

At time tthe charge is at r(t)=a[cos(!t)ˆx + sin(!t)ˆy ], so v(t)=!a[sin(!t)ˆx + cos(!t)ˆy ]. Therefore

=zˆz a[cos(!tr)ˆx + sin(!tr)ˆy ], and hence

2=z2+a2(of course), and

=pz2+a2.

·v=1

(

·v)= 1

!a2[sin(!tr) cos(!tr) + sin(!tr) cos(!tr)]=0,so 1ˆ

·v

c=1.

Therefore

V(z,t)= 1

4⇡✏0

pz2+a2;A(z,t)= q!a

4⇡✏0c2pz2+a2[sin(!tr)ˆx + cos(!tr)ˆy ),where tr=tpz2+a2

Problem 10.16

Term under square root in (Eq. 10.49) is:

I=c4t22c2t(r·v)+(r·v)2+c2r2c4t2v2r2+v2c2t2

=(r·v)2+(c2v2)r2+c2(vt)22c2(r·vt).put in vt=rR2.

=(r·v)2+(c2v2)r2+c2(r2+R22r·R)2c2(r2r·R)=(r·v)2r2v2+c2R2.

but

(r·v)2r2v2= ((R+vt)·v)2(R+vt)2v2

=(R·v)2+v4t2+ 2(R·v)v2tR2v22(R·v)tv2v2t2v2

=(R·v)2R2v2=R2v2cos2✓R2v2=R2v21cos2✓

=R2v2sin2✓.

Therefore

I=R2v2sin2✓+c2R2=c2R21v2

c2sin2✓.

Hence

V(r,t)= 1

4⇡✏0

R1v2

c2sin2✓

.qed

Problem 10.17

Once seen, from a given point x, the particle will forever remain in view—to disappear it would have to

travel faster than light.

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CHAPTER 10. POTENTIALS AND FIELDS 217

Problem 10.18

First calculate tr:tr=t|rw(tr)|/c )

c(trt)=xb2+c2t2

r)c(trt)+x=b2+c2t2

c2t2

r2c2trt+c2t2+2xctr2xct +x2=b2+c2t2

2ctr(xct)+(x22xct +c2t2)=b2;

2ctr(xct)=b2(xct)2,or tr=b2(xct)2

2c(xct).

Now V(x, t)= 1

4⇡✏0

(

c

·v),and

c

·v=

(cv);

=c(ttr).

v=1

b2+c2t2

2c2tr=c2tr

c(trt)+x=c2tr

ctr+(xct);(cv)=c2tr+c(xct)c2tr

ctr+(xct)=c(xct)

ctr+(xct);

c

·v=c(ttr)c(xct)

ctr+(xct)=c2(ttr)(xct)

ctr+(xct);ctr+(xct)=b2(xct)2

2(xct)+(xct)= b2+(xct)2

2(xct);

ttr=2ct(xct)b2+(xct)2

2c(xct)=(xct)(x+ct)b2

2c(xct)=(x2c2t2b2)

2c(xct).Therefore

c

·v=b2+(xct)2

2(xct)1

c2(xct)

2c(xct)

[2ct(xct)b2+(xct)2]=b2+(xct)2

c(xct)[2ct(xct)b2+(xct)2].

The term in square brackets simpliﬁes to (2ct +xct)(xct)b2=(x+ct)(xct)b2=x2c2t2b2.

So V(x, t)= q

4⇡✏0

b2+(xct)2

(xct)(x2c2t2b2).

Meanwhile

A=V

c2v=c2tr

ctr+(xct)

c2ˆx =b2(xct)2

2c(xct)2(xct)

b2+(xct)2

4⇡✏0

b2+(xct)2

(xct)(x2c2t2b2)ˆx

4⇡✏0c

b2(xct)2

(xct)(x2c2t2b2)ˆx .

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218 CHAPTER 10. POTENTIALS AND FIELDS

Problem 10.19

From Eq. 10.44, c(ttr)=

)c2(ttr)2=

. Di↵erentiate with respect to t:

2c2(ttr)1@tr

@t=2

·@

@t,or c

1@tr

@t=

·@

@t. Now

=rw(tr),so

@t=@w

@tr

@t=v@tr

@t;c

1@tr

@t=

·v@tr

@t;c

=@tr

@t(c



·v)=

@tr

@t(

·u) (Eq.10.71),and hence @tr

@t=c

·u.qed

Now Eq. 10.47 says A(r,t)= v

c2V(r,t), so

@t=1

c2@v

@tV+v@V

@t=1

c2@v

@tr

@tV+v@V

@t

c2a@tr

4⇡✏0

·u+v1

4⇡✏0

qc

(

·u)2

@t(

c

·v)

4⇡✏0a

·u

@tr

@tv

(

·u)2c@

@t@

@t·v

·@v

@t.

But

=c(ttr))@

@t=c1@tr

@t,

=rw(tr))@

@t=v@tr

@t(as above),and

@t=@v

@tr

@t=a@tr

@t.

4⇡✏0c(

·u)2a(

·u)@tr

@tvc21@tr

@t+v2@tr

@t

·a@tr

@t

4⇡✏0c(

·u)2c2v+(

·u)a+(c2v2+

·a)v@tr

@t

4⇡✏0c(

·u)2c2v+(

·u)a+(c2v2+

·a)vc

·u

4⇡✏0c(

·u)3c2v(

·u)+c

(

·u)a+c

(c2v2+

·a)v

=qc

4⇡✏0

(

c

·v)3(

c

·v)v+

ca+

c(c2v2+

·a)v.qed

Problem 10.20

E=q

4⇡✏0

(

·u)3(c2v2)u+

⇥(u⇥a). Here

v=vˆx ,a=aˆx ,and,for points to the right,ˆ

=ˆx .

So u=(cv)ˆx ,u⇥a=0,and

·u=

(cv).

E=q

4⇡✏0

3(cv)3(c2v2)(cv)ˆx =q

4⇡✏0

(c+v)(cv)2

(cv)3ˆx =q

4⇡✏0

2c+v

cvˆx ;

B=1

cˆ

⇥E=0.qed

For ﬁeld points to the left,ˆ

=ˆx and u=(c+v)ˆx ,so

·u=

(c+v), and

E=q

4⇡✏0

3(c+v)3(c2v2)(c+v)ˆx =q

4⇡✏0

2cv

c+vˆx ;B=0.

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CHAPTER 10. POTENTIALS AND FIELDS 219

Problem 10.21

By Gauss’s law (in integral form) the answer has to be q/✏0.

E=1

4⇡✏0

q(1 v2/c2)

(1 v2

c2sin2✓)3/2

R2(Eq. 10.75), so

E·da=q(1 v2/c2)

4⇡✏0R2sin ✓d✓d

R2(1 v2

c2sin2✓)3/2=q(1 v2/c2)

4⇡✏0

2⇡⇡

sin ✓d✓

(1 v2

c2sin2✓)3/2.

Let u⌘cos ✓,sodu =sin ✓d✓, sin2✓=1u2.

E·da=q(1 v2/c2)

2✏01

1

[1 v2

c2+v2

c2u2]3/2=q(1 v2/c2)

2✏0c

v31

1

c2

v21+u23/2.

The integral is: u

c2

v21c2

v21+u2

1

c2

v21c

=v

c32

(1 v2/c2).So

E·da=q(1 v2/c2)

2✏0c

v3v

c32

(1 v2/c2)=q

✏0

Problem 10.22

(a) E=

4⇡✏0

(1 v2/c2)ˆ

1(v/c)2sin2✓3/2.

The horizontal components cancel; the vertical com-

ponent of ˆ

Ris sin ✓(see diagram). Here d=Rsin ✓, so

R2=sin2✓

d2;x

d= cot ✓,sodx =d(csc2✓)d✓=d

sin2✓d✓;

R2dx =d

sin2✓

d2d✓=d✓

d.Thus

E=

4⇡✏0

(1 v2/c2)ˆy

d⇡

sin ✓

1(v/c)2sin2✓3/2d✓.Let z⌘cos ✓,so sin2✓=1z2.

=(1 v2/c2)ˆy

4⇡✏0d1

1

[1 (v/c)2+(v/c)2z2]3/2dz

=(1 v2/c2)ˆy

4⇡✏0d1

(v/c)3

(c2/v21)(c/v)21+z2

1

=(1 v2/c2)

4⇡✏0d

(1 v2/c2)

(c/v)21+1ˆy =1

4⇡✏0

2

dˆy (same as for a line charge at rest).

(b) B=1

c2(v⇥E) for each segment dq =dx. Since vis constant, it comes outside the integral, and the

same formula holds for the total ﬁeld:

B=1

c2(v⇥E)= 1

c2v1

4⇡✏0

2

d(ˆx ⇥ˆy )=µ0✏0v1

4⇡✏0

2

dˆz =µ0

4⇡

2v

dˆz .

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220 CHAPTER 10. POTENTIALS AND FIELDS

But v=I, so B=µ0

4⇡

dˆ

(the same as we got in magnetostatics, Eq. 5.39 and Ex. 5.7).

Problem 10.23

w(t)=R[cos(!t)ˆx + sin(!t)ˆy ];

v(t)=R![sin(!t)ˆx + cos(!t)ˆy ];

a(t)=R!2[cos(!t)ˆx + sin(!t)ˆy ]=!2w(t);

=w(tr);

=R;

tr=tR/c;

=[cos(!tr)ˆx + sin(!tr)ˆy ];

u=cˆ

v(tr)=c[cos(!tr)ˆx + sin(!tr)ˆy ]!R[sin(!tr)ˆx + cos(!tr)ˆy ]

={[ccos(!tr)!Rsin(!tr)] ˆx +[csin(!tr)+!Rcos(!tr)] ˆy };

⇥(u⇥a)=(

·a)u(

·u)a;

·a=w·(!2w)=!2R2;

·u=Rccos2(!tr)!Rsin(!tr) cos(!tr)+csin2(!tr)+!Rsin(!tr) cos(!tr)=Rc;

v2=(!R)2. So (Eq. 10.72):

E=q

4⇡✏0

(Rc)3uc2!2R2+u(!R)2a(Rc)=q

4⇡✏0

cuRa

(Rc)2

4⇡✏0

(Rc)2[c2cos(!tr)!Rc sin(!tr)] ˆx [c2sin(!tr)+!Rc cos(!tr)] ˆy

+R2!2cos(!tr)ˆx +R2!2sin(!tr)ˆy 

4⇡✏0

(Rc)2!2R2c2cos(!tr)+!Rc sin(!tr)ˆx +!2R2c2sin(!tr)!Rc cos(!tr)ˆy .

B=1

cˆ

⇥E=1

cˆ

xEyˆ

yExˆz

=1

4⇡✏0

(Rc)2cos(!tr)!2R2c2sin(!tr)!Rc cos(!tr)

sin(!tr)!2R2c2cos(!tr)+!Rc sin(!tr)ˆz

=q

4⇡✏0

R2c3!Rc cos2(!tr)!Rc sin2(!tr)ˆz =q

4⇡✏0

R2c3!Rc ˆz =q

4⇡✏0

Rc2ˆz .

Notice that Bis constant in time.

To obtain the ﬁeld at the center of a circular ring of charge, let q!(2⇡R); for this ring to carry current

I, we need I=v=!R,so=I/!R, and hence q!(I/!R)(2⇡R)=2⇡I/!. Thus B=2⇡I

4⇡✏0

Rc2ˆz , or,

since 1/c2=✏0µ0,B=µ0I

2Rˆz ,the same as Eq. 5.41, in the case z= 0.

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CHAPTER 10. POTENTIALS AND FIELDS 221

Problem 10.24

(,t)=0|sin(✓/2)|,where ✓=!t. So the (retarded) scalar potential at the center is (Eq. 10.26)

V(t)= 1

4⇡✏0

dl0=1

4⇡✏02⇡

0|sin[(!tr)/2]|

aad

=0

4⇡✏02⇡

sin(✓/2) d✓=0

4⇡✏0

[2 cos(✓/2)]

2⇡

=0

4⇡✏0

[2 (2)] = 0

⇡✏0

(Note: at ﬁxed tr,d=d✓, and it goes through one full cycle of or ✓.)

Meanwhile I(,t)=v=0!a|sin[(!t)/2]|ˆ

. From Eq. 10.26 (again)

A(t)= µ0

4⇡I

dl0=µ0

4⇡2⇡

0!a|sin[(!tr)/2]|ˆ



aad.

But tr=ta/c is again constant,for the integration,and ˆ

=sin ˆx + cos ˆy .

=µ00!a

4⇡2⇡

0|sin[(!tr)/2]|(sin ˆx + cos ˆy )d.Again,switch variables to ✓=!tr,

and integrate from ✓= 0 to ✓=2⇡(so we don0t have to worry about the absolute value).

=µ00!a

4⇡2⇡

sin(✓/2) [sin(✓+!tr)ˆx + cos(✓+!tr)ˆy ]d✓.Now

2⇡

sin (✓/2) sin(✓+!tr)d✓=1

22⇡

[cos (✓/2+!tr)cos (3✓/2+!tr)] d✓

22 sin (✓/2+!tr)2

3sin (3✓/2+!tr)

2⇡

= sin(⇡+!tr)sin(!tr)1

3sin(3⇡+!tr)+1

3sin(!tr)

=2 sin(!tr)+2

3sin(!tr)=4

3sin(!tr).

2⇡

sin (✓/2) cos(✓+!tr)d✓=1

22⇡

[sin (✓/2+!tr) + sin (3✓/2+!tr)] d✓

22 cos (✓/2+!tr)2

3cos (3✓/2+!tr)

2⇡

= cos(⇡+!tr)cos(!tr)1

3cos(3⇡+!tr)+1

3cos(!tr)

=2 cos(!tr)+2

3cos(!tr)=4

3cos(!tr).So

A(t)=µ00!a

4⇡4

3[sin(!tr)ˆx cos(!tr)ˆy ]= µ00!a

3⇡{sin[!(ta/c)] ˆx cos[!(ta/c)] ˆy }.

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222 CHAPTER 10. POTENTIALS AND FIELDS

Problem 10.25

Hello

r·A=1

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Problem 10.26

⇢(r,t)=





(4/3)⇡R3=3Q

4⇡v3t3,(r<R=vt),

0,(otherwise).

For a point at the center, tr=tr/c,so

⇢(r,t

r)=





4⇡v3(tr/c)3=3Q

4⇡(v/c)3(ct r)3,r<v(tr/c))r< vt

1+v/c,

0,(otherwise).

Let a⌘vt/(1 + v/c); then

Qe↵=3Q

4⇡(v/c)3a

(ct r)34⇡r2dr =3Q

(v/c)3a

(ct r)3dr

=3Q

(v/c)3ln(ct r)+ 2ct

(ct r)(ct)2

2(ct r)2

=3Q

(v/c)3ln(ct a)+ 2ct

(ct a)(ct)2

2(ct a)2ln(ct)2+1

2

=3Q

(v/c)33

2+ ln ct

ct a2ct

ct a+1

2ct

ct a2.

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CHAPTER 10. POTENTIALS AND FIELDS 223

Now, ct a=ct vt

1+v/c =ct

1+v/c 1+v

cv

c=ct

1+v/c, so ct

ct a=1+v

c,and hence

Qe↵=3Q

(v/c)33

2+ ln 1+ v

c21+v

c+1

21+v

c2

=3Q

(v/c)33

2+ ln 1+ v

c22v

c+1

2+v

c+1

2v

c2=3Q

(v/c)3ln 1+v

cv

c+1

2v

c2.

If ✏⌧1, then ln(1 + ✏)=✏1

2✏2+1

3✏31

4✏4+..., so for v⌧c,

Qe↵⇡3Q

(v/c)3v

c1

2v

c2+1

3v

c31

4v

c4v

c+1

2v

c2=Q13v

4c.X

Problem 10.27

Using Product Rule #5, Eq. 10.50 )

r·A=µ0

4⇡qcv·r(c2tr·v)2+(c2v2)(r2c2t2)1/2

=µ0qc

4⇡v·1

2(c2tr·v)2+(c2v2)(r2c2t2)3/2r(c2tr·v)2+(c2v2)(r2c2t2)

=µ0qc

8⇡(c2tr·v)2+(c2v2)(r2c2t2)3/2v·2(c2tr·v)r(r·v)+(c2v2)r(r2).

Product Rule #4 )

r(r·v)=v⇥(r⇥r)+(v·r)r,but r⇥r=0,

(v·r)r=vx

@x+vy

@y+vz

@z(xˆx +yˆy +zˆz )=vxˆx +vyˆy +vzˆz =v,and

r(r2)=r(r·r)=2r⇥(r⇥r) + 2(r·r)r=2r.So

r·A=µ0qc

8⇡(c2tr·v)2+(c2v2)(r2c2t2)3/2v·2(c2tr·v)v+(c2v2)2r

=µ0qc

4⇡(c2tr·v)2+(c2v2)(r2c2t2)3/2(c2tr·v)v2(c2v2)(r·v).

But the term in curly brackets is : c2tv2v2(r·v)c2(r·v)+v2(r·v)=c2(v2tr·v).

=µ0qc3

4⇡

(v2tr·v)

[(c2tr·v)2+(c2v2)(r2c2t2)]3/2.

Meanwhile, from Eq. 10.49,

µ0✏0

@t=µ0✏0

4⇡✏0

qc 1

2(c2tr·v)2+(c2v2)(r2c2t2)3/2⇥

@t(c2tr·v)2+(c2v2)(r2c2t2)

=µ0qc

8⇡(c2tr·v)2+(c2v2)(r2c2t2)3/22(c2tr·v)c2+(c2v2)(2c2t)

=µ0qc3

4⇡

(c2tr·vc2t+v2t)

[(c2tr·v)2+(c2v2)(r2c2t2)]3/2=r·A.X

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224 CHAPTER 10. POTENTIALS AND FIELDS

Problem 10.28

(a) F2=q1q2

4⇡✏0

(b2+c2t2)ˆx .

(This is just Coulomb’s law, since q1is at rest.)

(b) I2=q1q2

4⇡✏01

1

(b2+c2t2)dt =q1q2

4⇡✏01

bc tan1(ct/b)

1

=q1q2

4⇡✏0bc tan1(1)tan1(1)

=q1q2

4⇡✏0bc ⇡

2⇡

2=q1q2

4⇡✏0

⇡

bc.

4⇡✏0

x2cv

c+vˆx . Here x

and vare to be evaluated at the retarded time tr, which is

given by c(ttr)=x(tr)=b2+c2t2

r)c2t22cttr+

c2t2

r=b2+c2t2

r)tr=c2t2b2

2c2t.Note: As we found

in Prob. 10.17, q2ﬁrst “comes into view” (for q1) at time

t= 0. Before that it can exert no force on q1, and there is

no retarded time. From the graph of trversus twe see that

trranges all the way from 1 to 1while t>0.

x(tr)=c(ttr)=2c2t2c2t2+b2

2ct =b2+c2t2

2ct (for t>0).v(t)=1

2c2t

pb2+c2t2=c2t

x,so

v(tr)=c2t2b2

2t 2ct

b2+c2t2=cc2t2b2

c2t2+b2(for t>0).Therefore

cv

c+v=(c2t2+b2)(c2t2b2)

(c2t2+b2)+(c2t2b2)=2b2

2c2t2=b2

c2t2(for t>0).E=q2

4⇡✏0

4c2t2

(b2+c2t2)2

c2t2ˆx )

F1=





0,t<0;

q1q2

4⇡✏0

4b2

(b2+c2t2)2ˆx ,t>0.

(d) I1=q1q2

4⇡✏0

4b21

(b2+c2t2)2dt. The integral is

c41

[(b/c)2+t2]2dt =1

c4c2

2b2 t

(b/c)2+t2

+1

[(b/c)2+t2)] dt=1

2c2b2⇡c

2b=⇡

4cb3.

So I1=q1q2

4⇡✏0

⇡

bc.

(e) F16=F2, so Newton’s third law is not obeyed. On the other hand, I1=I2in this instance, which

suggests that the net momentum delivered from (1) to (2) is equal and opposite to the net momentum delivered

from (2) to (1), and hence that the total mechanical momentum is conserved. (In general the ﬁelds might carry

o↵some momentum, leaving the mechanical momentum altered; but that doesn’t happen in the present case.)

Problem 10.29

The electric ﬁeld of q1at q2[Eq. 10.75, with ✓=45and R=(vt ˆx +vt ˆy )] is

E1=q1

4⇡✏0

1v2/c2

(1 v2/2c2)3/2

2p2(vt)2(ˆx +ˆy ).

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CHAPTER 10. POTENTIALS AND FIELDS 225

The magnetic ﬁeld [Eq. 10.76, with v1=vˆx ] is

B1=1

c2(v1⇥E)=v

c2(ˆx ⇥E)=v

4⇡✏0

1v2/c2

(1 v2/2c2)3/2

2p2(vt)2ˆz .

The force on q2is therefore [Lorentz force law with v2=vˆy ]

F2=q2(E1+v2⇥B1)= q1q2

4⇡✏0

1v2/c2

(1 v2/2c2)3/2

2p2(vt)2ˆx +ˆy +v2

c2ˆx .

The electric ﬁeld of q2at q1is reversed, E2=E1; so is the magnetic ﬁeld B2=B1. The electric force

is also reversed, but the magnetic force now points in the ydirection instead of the xdirection. So the force

on q1is

F1=q1q2

4⇡✏0

1v2/c2

(1 v2/2c2)3/2

2p2(vt)2ˆx ˆy +v2

c2ˆy .

The forces are equal in magnitude, but not opposite in direction. No, Newton’s third law is not obeyed.

Problem 10.30

c(tt1)=vt1,t=t1(1 + v/c), t1=t

1+v/c. c(tt2)=vt2+L, t L/c =t2(1 + v/c), t2=tL/c

1+v/c .

x1=vt1=vt

1+v/c. x2=vt2+L=v(tL/c)

1+v/c +L=vt vL/c +L+vL/c

1+v/c =vt +L

1+v/c.

V(0,t)= 1

4⇡✏0x2



xdx =

4⇡✏0

ln x2

x1=

4⇡✏0

ln vt +L

vt .

If L⌧vt,V=

4⇡✏0

ln 1+ L

vt⇡

4⇡✏0

vt =q

4⇡✏01

vt.The Li´enard-Wiechert potential is

V=q

4⇡✏0(



·v/c).Here (



·v/c)=vtr+v2tr/c =v(1 + v/c)tr=v(1 + v/c)t

1+v/c =vt.

(In the limit L!0, tr=t1=t2.) So V=q

4⇡✏01

vt.X

Problem 10.31

S=1

µ0

(E⇥B); B=1

c2(v⇥E) (Eq.10.76).

So S=1

µ0c2[E⇥(v⇥E)] = ✏0E2v(v·E)E.

The power crossing the plane is P=S·da,

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226 CHAPTER 10. POTENTIALS AND FIELDS

and da=2⇡rdrˆx (see diagram). So

P=✏0(E2vE2

xv)2⇡rdr;Ex=Ecos ✓,so E2E2

x=E2sin2✓.

=2⇡✏0vE2sin2✓rdr. From Eq.10.75,E=q

4⇡✏0

2

R21(v/c)2sin2✓3/2where ⌘1

1v2/c2.

=2⇡✏0vq

4⇡✏021

21

rsin2✓

R41(v/c)2sin2✓3dr. Now r=atan ✓)dr =a1

cos2✓d✓;1

R=cos ✓

24

4⇡✏0

a2⇡/2

sin3✓cos ✓

1(v/c)2sin2✓3d✓.Let u⌘sin2✓,so du =2sin✓cos ✓d✓.

=vq2

16⇡✏0a241

[1 (v/c)2u]3du =vq2

16⇡✏0a244

2=vq2

32⇡✏0a2.

Problem 10.32

(a) F12(t)= 1

4⇡✏0

q1q2

(vt)2ˆz .

(b) From Eq. 10.75, with ✓= 180,R=vt, and ˆ

R=ˆz :

F21(t)=1

4⇡✏0

q1q2(1 v2/c2)

(vt)2ˆz .

No, Newton’s third law does not hold: F12 6=F21,

because of the extra factor (1 v2/c2).

But the latter, when integrated over all space, is independent of time. We want only the time-dependent part:

p(t)=✏0(E1⇥B2)d⌧.Now E1=1

4⇡✏0

r2ˆr , while, from Eq. 10.76, B2=1

c2(v⇥E2), and (Eq. 10.75)

E2=q2

4⇡✏0

(1 v2/c2)

(1 v2sin2✓0/c2)3/2

R2.But R=rvt;R2=r2+v2t22rvt cos ✓; sin ✓0=rsin ✓

R.So

E2=q2

4⇡✏0

(1 v2/c2)

[1 (vr sin ✓/Rc)2]3/2

(rvt)

R3. Finally, noting that v⇥(rvt)=v⇥r=vr sin ✓ˆ

, we get

B2=q2(1 v2/c2)

4⇡✏0c2

vr sin ✓

[R2(vr sin ✓/c)2]3/2ˆ

.So p(t)=✏0

4⇡✏0

q2(1 v2/c2)v

4⇡✏0c21

rsin ✓(ˆr ⇥ˆ

)

[R2(vr sin ✓/c)2]3/2.

But ˆr ⇥ˆ

=ˆ

✓=(cos ✓cos ˆx + cos ✓sin ˆy sin ✓ˆz ), and the xand ycomponents integrate to zero, so:

p(t)=q1q2v(1 v2/c2)ˆz

(4⇡c)2✏0sin2✓

r[r2+(vt)22rvt cos ✓(vr sin ✓/c)2]3/2r2sin ✓dr d✓d

=q1q2v(1 v2/c2)ˆz

8⇡c2✏0rsin3✓

[r2+(vt)22rvt cos ✓(vr sin ✓/c)2]3/2dr d✓.

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CHAPTER 10. POTENTIALS AND FIELDS 227

I’ll do the rintegral ﬁrst. According to the CRC Tables,

1

(a+bx +cx2)3/2dx =2(bx +2a)

(4ac b2)pa+bx +cx2

=2

4ac b2b

pc2a

pa

=2

pc(4ac b2)b2pac=2

(2pac b)

(2pac b) (2pac +b)=2

pc2pac +b1.

In this case x=r, a =(vt)2,b=2vt cos ✓,and c=1(v/c)2sin2✓. So the rintegral is

1(v/c)2sin2✓2vt1(v/c)2sin2✓2vt cos ✓=1

vt1(v/c)2sin2✓1(v/c)2sin2✓cos ✓

=1(v/c)2sin2✓+ cos ✓

vt1(v/c)2sin2✓1(v/c)2sin2✓cos2✓=1

vt sin2✓(1 v2/c2)

1+ cos ✓

1(v/c)2sin2✓

.

p(t)=q1q2v(1 v2/c2)ˆz

8⇡c2✏0

vt(1 v2/c2)⇡

sin2✓

1+ cos ✓

1(v/c)2sin2✓

sin3✓d✓

=q1q2ˆz

8⇡c2✏0t



⇡

sin ✓d✓+c

v⇡

cos ✓sin ✓

(c/v)2sin2✓

d✓



.

But ⇡

0sin ✓d✓= 2. In the second integral let u⌘cos ✓,sodu =sin ✓d✓:

⇡

cos ✓sin ✓

(c/v)2sin2✓

d✓=1

1

(c/v)21+u2du = 0 (the integrand is odd, and the interval is even).

Conclusion: p(t)=µ0q1q2

4⇡tˆz (plus a term constant in time).

(d)

F12 +F21 =1

4⇡✏0

q1q2

v2t2ˆz 1

4⇡✏0

q1q2(1 v2/c2)

v2t2ˆz =q1q2

4⇡✏0v2t211+ v2

c2ˆz =q1q2

4⇡✏0c2t2ˆz =µ0q1q2

4⇡t2ˆz .

dp

dt =µ0q1q2

4⇡t2ˆz =F12 +F21.qed

Since q1is at rest, and q2is moving at constant velocity, there must be another force (Fmech) acting on

them, to balance F12 +F21; what we have found is that Fmech =dpem/dt, which means that the impulse

imparted to the system by the external force ends up as momentum in the ﬁelds. [For further discussion of

this problem see J. J. G. Scanio, Am. J. Phys. 43, 258 (1975).]

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228 CHAPTER 10. POTENTIALS AND FIELDS

Problem 10.33

Maxwell’s equations with magnetic charge read:

(i) r·E=1

✏0⇢e,(iii) r⇥E=µ0Jm@B

@t,

(ii) r·B=µ0⇢m,(iv) r⇥B=µ0Je+µ0✏0

@t,

where ⇢eis the electric charge density, ⇢mis the magnetic charge density, Jeis the electric current density,

and Jmis the magnetic current density.

If there are only electric charges and currents, then the usual potential formulation applies (Section 10.1.1),

with E=rVe@Ae/@t, B=r⇥Ae. If there are only magnetic charges and currents,

(i) r·E=0,(iii) r⇥E=µ0Jm@B

@t,

(ii) r·B=µ0⇢m,(iv) r⇥B=µ0✏0

@t.

This time equation (i) says that Ecan be expressed as the curl of a vector potential: E=r⇥Am, and

plugging this into (iv) yields r⇥(B+µ0✏0@Am/@t)=0, which tells us that the quantity in parentheses can

be represented as the gradient of a scalar: B=rVmµ0✏0@Am/@t. [The signs of Vmand Amare arbitrary,

but I think this choice yields the most symmetrical formulation.] Putting these into (ii) and (iii) yields

22Vm=µ0⇢mµ0✏0

@tr·Am+@Vm

@t,22Am=µ0Jm+rr·Am+@Vm

@t.

The “Lorenz gauge” for the magnetic potentials, r·Am=@Vm/@t, kills the terms in parentheses.

In general, if there are both electric and magnetic charges and currents present, the total ﬁelds are the sums

(by the superposition principle):

E=rVe@Ae

@tr⇥Am,B=rVmµ0✏0

@Am

@t+r⇥Ae.

If we choose to work in the Lorenz gauge:

r·Ae=µ0✏0

@Ve

@t,r·Am=@Vm

@t,

Maxwell’s equations become (in terms of the potentials)

22Ve=1

✏0

⇢e,22Vm=µ0⇢m,22Ae=µ0Je,22Am=µ0Jm,

and the retarded solutions are

Ve(r,t)= 1

4⇡✏0⇢e(r0,t

d⌧0,V

m(r,t)= µ0

4⇡⇢m(r0,t

d⌧0,

Ae(r,t)= µ0

4⇡Je(r0,t

d⌧0,Am(r,t)= µ0

4⇡Jm(r0,t

d⌧0.

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CHAPTER 10. POTENTIALS AND FIELDS 229

Problem 10.34

The retarded potentials are

V(r,t)= 1

4⇡✏0⇢(r0,t

d⌧0,A(r,t)= µ0

4⇡J(r0,t

d⌧0.

We want a source localized at the origin, so we expand in powers of r0, keeping terms up to ﬁrst order:

2=(rr0)·(rr0)=r22r·r0+(r0)2,

⇡r1r·r0

r2,

⇡1

r1+r·r0

r2,

t

c⇡tr

c+r·r0

rc =t0+r·r0

rc ,

where t0⌘tr/c is the retarded time for a source at the origin. Thus (Taylor expanding about t0)

⇢(r0,t

r)=⇢r0,t

c⇡⇢(r0,t

0)+r·r0

rc ˙⇢(r0,t

0),

⇢(r0,t

r1+r·r0

r2⇢(r0,t

0)+r·r0

rc ˙⇢(r0,t

0)⇡1

r⇢(r0,t

0)+r·r0

r3⇢(r0,t

0)+r·r0

r2c˙⇢(r0,t

0),

and hence

V(r,t)= 1

4⇡✏01

r⇢(r0,t

0)d⌧0+r

r3·r0⇢(r0,t

0)d⌧0+r

r2c·r0˙⇢(r0,t

0)d⌧0.

The ﬁrst integral is the total charge of the dipole, which is zero; the second integral is the dipole moment, and

the third is the time derivative of the dipole moment:

V(r,t)= 1

4⇡✏0

ˆr

r2·p(t0)+r

c˙

p(t0).

By the same reasoning, the vector potential is

A(r,t)= µ0

4⇡1

rJ(r0,t

0)d⌧0+1

r3(r·r0)J(r0,t

0)d⌧0+1

r2c(r·r0)˙

J(r0,t

0)d⌧0.

From Eq. 5.31 J(r0,t)d⌧0=˙

p(t),and J=⇢v0=⇢(dr0/dt) is already ﬁrst order in r0, so the second and

third integrals are second order, and we drop them.

A(r,t)= µ0

4⇡˙

p(t0)

r.

In calculating the ﬁelds, remember that the dipole moments themselves depend on r, through their argument

(t0=tr/c). For a function of t0alone,

rf(t0)= df

dt0rt0=˙

f1

crr=1

c˙

fˆr ,

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230 CHAPTER 10. POTENTIALS AND FIELDS

so you simply replace rby ˆr

dt. For a structure of the form f(t0)+ r

c˙

f(t0)

rf(t0)+r

c˙

f(t0)=ˆr

c˙

f+r

cˆr

c¨

f+˙

cˆr =r

c2¨

Using the product rule for r(A·B):

rV=1

4⇡✏0ˆr

r2⇥r⇥p+r

c˙

p+p+r

c˙

p⇥r⇥ˆr

r2+ˆr

r2·rp+r

c˙

p+p+r

c˙

p·rˆr

r2

4⇡✏0ˆr

r2⇥r

c2⇥¨

p+0+ˆr

r2·r

c2¨

p+1

r3p+r

c˙

p3ˆr ˆr ·p+r

c˙

p

4⇡✏0ˆr (ˆr ·¨

rc2+[p+(r/c)˙

p]3ˆr (ˆr ·[p+(r/c)˙

p])

r3.

The others are easier:

r⇥A=µ0

4⇡1

r(r⇥˙

p)˙

p⇥r1

r=µ0

4⇡1

rˆr

c⇥¨

p+˙

p⇥ˆr

r2=µ0

4⇡ˆr ⇥[( ˙

p+(r/c)¨

r2,

@t=µ0

4⇡¨

r.

The ﬁelds are therefore

E(r,t)=rV@A

@t=µ0

4⇡¨

pˆr (ˆr ·¨

r+c2[p+(r/c)˙

p]3ˆr (ˆr ·[p+(r/c)˙

p])

r3,

B(r,t)=r⇥A=µ0

4⇡ˆr ⇥[˙

p+(r/c)¨

r2

(with all dipole moments evaluated at the retarded time t0=tr/c).

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CHAPTER 11. RADIATION 231

Chapter 11

Radiation

Problem 11.1

From Eq. 11.17, A=µ0p0!

4⇡

rsin[!(tr/c)](cos ✓ˆr sin ✓ˆ

✓), so

r·A=µ0p0!

4⇡1

@rr21

rsin[!(tr/c)] cos ✓+1

rsin ✓

@✓ sin2✓1

rsin[!(tr/c)]

=µ0p0!

4⇡1

r2sin[!(tr/c)] !r

ccos[!(tr/c)]cos ✓2 sin ✓cos ✓

r2sin ✓sin[!(tr/c)]

=µ0✏0p0!

4⇡✏01

r2sin[!(tr/c)] + !

rc cos[!(tr/c)]cos ✓.

Meanwhile, from Eq. 11.12,

@t=p0cos ✓

4⇡✏0r!2

ccos[!(tr/c)] !

rsin[!(tr/c)]

=p0!

4⇡✏01

r2sin[!(tr/c)] + !

rc cos[!(tr/c)]cos ✓.So r·A=µ0✏0

@t.qed

Problem 11.2

Eq. 11.14: V(r,t)=!

4⇡✏0c

p0·ˆr

rsin[!(tr/c)].Eq. 11.17: A(r,t)=µ0!

4⇡

rsin[!(tr/c)].

Now p0⇥ˆr =p0sin ✓ˆ

and ˆr ⇥(p0⇥ˆr )=p0sin ✓(ˆr ⇥ˆ

)=p0sin ✓ˆ

✓, so

Eq. 11.18: E(r,t)=µ0!2

4⇡

ˆr ⇥(p0⇥ˆr )

rcos[!(tr/c)].Eq. 11.19: B(r,t)=µ0!2

4⇡c

(p0⇥ˆr )

rcos[!(tr/c)].

Eq. 11.21: hSi=µ0!4

32⇡2c

(p0⇥ˆr )2

r2ˆr .

Problem 11.3

P=I2R=q2

0!2sin2(!t)R(Eq.11.15) )hPi=1

2q2

0!2R. Equate this to Eq. 11.22:

2q2

0!2R=µ0q2

0d2!4

12⇡c)R=µ0d2!2

6⇡c; or, since !=2⇡c

,

R=µ0d2

6⇡c

4⇡2c2

2=2

3⇡µ0cd

2

3⇡(4⇡⇥107)(3 ⇥108)d

2

= 80⇡2d

2

⌦= 789.6(d/)2⌦.

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232 CHAPTER 11. RADIATION

For the wires in an ordinary radio, with d=5⇥102m and (say) = 103m, R= 790(5 ⇥105)2=2⇥106⌦,

which is negligible compared to the Ohmic resistance.

Problem 11.4

By the superposition principle, we can add the potentials of the two dipoles. Let’s ﬁrst express V(Eq. 11.14)

in Cartesian coordinates: V(x, y, z, t)=p0!

4⇡✏0cz

x2+y2+z2sin[!(tr/c)]. That’s for an oscillating dipole

along the zaxis. For one along xor y, we just change zto xor y. In the present case,

p=p0[cos(!t)ˆx + cos(!t⇡/2) ˆy ], so the one along yis delayed by a phase angle ⇡/2:

sin[!(tr/c)] !sin[!(tr/c)⇡/2] = cos[!(tr/c)] (just let !t!!t⇡/2). Thus

V=p0!

4⇡✏0cx

x2+y2+z2sin[!(tr/c)] y

x2+y2+z2cos[!(tr/c)]

=p0!

4⇡✏0c

sin ✓

r{cos sin[!(tr/c)] sin cos[!(tr/c)]}.Similarly,

A=µ0p0!

4⇡r{sin[!(tr/c)] ˆx cos[!(tr/c)] ˆy }.

We could get the ﬁelds by di↵erentiating these potentials, but I prefer to work with Eqs. 11.18 and 11.19,

using superposition. Since ˆz = cos ✓ˆr sin ✓ˆ

✓, and cos ✓=z/r, Eq. 11.18 can be written

E=µ0p0!2

4⇡rcos[!(tr/c)] ˆz z

rˆr . In the case of the rotating dipole, therefore,

E=µ0p0!2

4⇡rcos[!(tr/c)] ˆx x

rˆr + sin[!(tr/c)] ˆy y

rˆr ,

B=1

c(ˆr ⇥E).

S=1

µ0

(E⇥B)= 1

µ0c[E⇥(ˆr ⇥E)] = 1

µ0cE2ˆr (E·ˆr )E=E2

µ0cˆr (notice that E·ˆr = 0). Now

E2=µ0p0!2

4⇡r2a2cos2[!(tr/c)] + b2sin2[!(tr/c)] + 2(a·b) sin[!(tr/c)] cos[!(tr/c)],

where a⌘ˆx (x/r)ˆr and b⌘ˆy (y/r)ˆr . Noting that ˆx ·r=xand ˆy ·r=y, we have

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CHAPTER 11. RADIATION 233

a2=1+x2

r22x2

r2=1x2

r2;b2=1y2

r2;a·b=y

rx

r+xy

r2=xy

r2.

E2=µ0p0!2

4⇡r21x2

r2cos2[!(tr/c)] + 1y2

r2sin2[!(tr/c)]

2xy

r2sin[!(tr/c)] cos[!(tr/c)]

=µ0p0!2

4⇡r211

r2x2cos2[!(tr/c)] + 2xy sin[!(tr/c)] cos[!(tr/c)] + y2sin2[!(tr/c)]

=µ0p0!2

4⇡r211

r2(xcos[!(tr/c)] + ysin[!(tr/c)])2

But x=rsin ✓cos and y=rsin ✓sin .

=µ0p0!2

4⇡r21sin2✓(cos cos[!(tr/c)] + sin sin[!(tr/c)])2

=µ0p0!2

4⇡r21(sin ✓cos[!(tr/c)])2.

S=µ0

cp0!2

4⇡r21(sin ✓cos[!(tr/c)])2ˆr .

hSi=µ0

cp0!2

4⇡r211

2sin2✓ˆr .

P=hSi·da=µ0

cp0!2

4⇡21

r211

2sin2✓r2sin ✓d✓d

=µ0p2

0!4

16⇡2c2⇡⇡

sin ✓d✓1

2⇡

sin3✓d✓=µ0p2

0!4

8⇡c21

2·4

3=µ0p2

0!4

6⇡c.

This is twice the power radiated by either oscillating dipole alone (Eq. 11.22). In general, S=1

µ0

(E⇥B)=

µ0

[(E1+E2)⇥(B1+B2)] = 1

µ0

[(E1⇥B1)+(E2⇥B2)+(E1⇥B2)+(E2⇥B1)] = S1+S2+ cross terms.

In this particular case the ﬁelds of 1 and 2 are 90out of phase, so the cross terms go to zero in the time

averaging, and the total power radiated is just the sum of the two individual powers.

Problem 11.5

Go back to Eq. 11.33:

A=µ0m0

4⇡sin ✓

r1

rcos[!(tr/c)] !

csin[!(tr/c)]ˆ

.

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234 CHAPTER 11. RADIATION

Since V=0here,

E=@A

@t=µ0m0

4⇡sin ✓

r1

r(!) sin[!(tr/c)] !

c!cos[!(tr/c)]ˆ



=µ0m0!

4⇡sin ✓

r1

rsin[!(tr/c)] + !

ccos[!(tr/c)]ˆ

.

B=r⇥A=1

rsin ✓

@✓ (Asin ✓)ˆr 1

@r(rA)ˆ

✓

=µ0m0

4⇡1

rsin ✓

2 sin ✓cos ✓

r1

rcos[!(tr/c)] !

csin[!(tr/c)]ˆr

sin ✓

r1

r2cos[!(tr/c)] + !

rc sin[!(tr/c)] !

c!

ccos[!(tr/c)]ˆ

✓

µ0m0

4⇡2 cos ✓

r21

rcos[!(tr/c)] !

csin[!(tr/c)]ˆr

sin ✓

r1

r2cos[!(tr/c)] + !

rc sin[!(tr/c)] + !

c2cos[!(tr/c)]ˆ

✓.

These are precisely the ﬁelds we studied in Prob. 9.35, with A!µ0m0!2

4⇡c. The Poynting vector (quoting

the solution to that problem) is

S=µ0m2

0!3

16⇡2c2sin ✓

r22 cos ✓

r1c2

!2r2sin ucos u+c

!rcos2usin2uˆ

✓

sin ✓2

r+c2

!2r3sin ucos u+!

ccos2u+c

!r2sin2ucos2uˆr ,

where u⌘!(tr/c). The intensity is hSi=µ0m2

0!4

32⇡2c3

sin2✓

r2ˆr ,the same as Eq. 11.39.

Problem 11.6

I2R=I2

0Rcos2(!t))hPi=1

2I2

0R=µ0m2

0!4

12⇡c3=µ0⇡2b4I2

0!4

12⇡c3, so R=µ0⇡b4!4

6c3; or, since !=2⇡c

,

R=µ0⇡b4

6c3

16⇡4c4

4=8

3⇡5µ0cb

4

3(⇡5)(4⇡⇥107)(3 ⇥108)(b/)4= 3.08 ⇥105(b/)4⌦.

Because b⌧, and Rgoes like the fourth power of this small number, Ris typically much smaller than the

electric radiative resistance (Prob. 11.3). For the dimensions we used in Prob. 11.3 (b= 5 cm and = 103m),

R=3⇥105(5 ⇥105)4=2⇥1012 ⌦, which is a millionth of the comparable electrical radiative resistance.

Problem 11.7

With ↵= 90, Eq. 7.68 )E0=cB,B0=E/c, q0

m=cqe)m0⌘q0

md=cqed=cp0. So

E0=cµ0(m0/c)!2

4⇡csin ✓

rcos[!(tr/c)] ˆ

=µ0m0!2

4⇡csin ✓

rcos[!(tr/c)] ˆ

.

B0=1

cµ0(m0/c)!2

4⇡sin ✓

rcos[!(tr/c)] ˆ

✓=µ0m0!2

4⇡c2sin ✓

rcos[!(tr/c)] ˆ

✓.

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CHAPTER 11. RADIATION 235

These are identical to the ﬁelds of an Amp´ere dipole (Eqs. 11.36 and 11.37), which is consistent with our

general experience that the two models generate identical ﬁelds except right at the dipole (not relevant here,

since we’re in the radiation zone).

Problem 11.8

(a) The power radiated (Eq. 11.60) is dWr

dt =µ0

6⇡c¨p2.In this case p=Qd, so ¨p=¨

Qd =Q01

RC 2

et/RC d,

so dWr

dt =µ0

6⇡c

(Q0d)2

(RC)4e2t/RC , and the total energy radiated is

Wr=µ0

6⇡c

(Q0d)2

(RC)41

e2t/RC dt =µ0

6⇡c

(Q0d)2

(RC)4RC

2e2t/RC 

=µ0

6⇡c

(Q0d)2

(RC)4

2=µ0

12⇡c

(Q0d)2

(RC)3.

The fraction of the original energy that is radiated is therefore

f=Wr

=µ0

12⇡c

(Q0d)2

(RC)3

=µ0

6⇡c

R3C2.

[Technically, ˙

Q(t) is discontinuous at t= 0, and ¨

Qpicks up a delta function. But any real circuit has some

(self-)inductance, which smoothes out the sudden change in ˙

Q.]

(b) With the parameters given,

f=4⇡⇥107

6⇡(3 ⇥108)

108

(106)(1024)= 2.22 ⇥106.

About a millionth of the total energy is radiated away, so yes, this is negligible.

Problem 11.9

p(t)=p0[cos(!t)ˆx + sin(!t)ˆy ])¨

p(t)=!2p0[cos(!t)ˆx + sin(!t)ˆy ])

[¨

p(t)]2=!4p2

0[cos2(!t) + sin2(!t)] = p2

0!4.So Eq. 11.59 says S=µ0p2

0!4

16⇡2c

sin2✓

r2ˆr .(This appears to disagree

with the answer to Prob. 11.4. The reason is that in Eq. 11.59 the polar axis is along the direction of ¨

p(t0);

as the dipole rotates, so do the axes. Thus the angle ✓here is not the same as in Prob. 11.4.) Meanwhile,

Eq. 11.60 says P=µ0p2

0!4

6⇡c.(This does agree with Prob. 11.4, because we have now integrated over all angles,

and the orientation of the polar axis irrelevant.)

Problem 11.10

At t= 0 the dipole moment of the ring is

p0=rdl =(0sin )(bsin ˆy +bcos ˆx )bd=0b2ˆy 2⇡

sin2d+ˆx 2⇡

sin cos d

=b2(⇡ˆy +0ˆx )=⇡b20ˆy .

As it rotates (counterclockwise, say) p(t)=p0[cos(!t)ˆy sin(!t)ˆx ], so ¨

p=!2p, and hence (¨

p)2=!4p2

Therefore (Eq. 11.60) P=µ0

6⇡c!4(⇡b20)2=⇡µ0!4b42

6c.

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236 CHAPTER 11. RADIATION

Problem 11.11

Here V= 0 (since the ring is neutral), and the current depends only on t(not on position), so the retarded

vector potential (Eq. 11.52) is A(r,t)= µ0

4⇡I(t

/c)

dl0. But in this case it does not suﬃce to replace

by rin the denominator—that would lead to Eq. 11.54, and hence to A=0(since p=0). Instead, use

Eq. 11.30: 1

⇠

r1+ b

rsin ✓cos 0. Meanwhile, dl0=bd0ˆ

=b(sin 0ˆx + cos 0ˆy )d0, and

I(t

/c)⇠

=I(tr/c +(b/c) sin ✓cos 0)=I(t0+(b/c) sin ✓cos 0)⇠

=I(t0)+ ˙

I(t0)b

csin ✓cos 0

(carrying all terms to ﬁrst order in b). As always, t0=tr/c. (From now on I’ll suppress the argument: I,

I, etc. are all to be evaluated at t0.) Then

A(r,t)= µ0

4⇡1

r1+ b

rsin ✓cos 0I+˙

csin ✓cos 0b(sin 0ˆx + cos 0ˆy )d0

⇠

=µ0b

4⇡r2⇡

0I+˙

csin ✓cos 0+Ib

rsin ✓cos 0(sin 0ˆx + cos 0ˆy )d0.

But 2⇡

sin 0d0=2⇡

cos 0d0=2⇡

sin 0cos 0d0=0,while 2⇡

cos20d0=⇡.

=µ0b

4⇡r(⇡ˆy )˙

csin ✓+Ib

rsin ✓=µ0b2

4r2sin ✓I+r

c˙

Iˆy .

In general (i.e. for points not on the xz plane) ˆy !ˆ

; moreover, in the radiation zone we are not interested

in terms that go like 1/r2,so A(r,t)=µ0b2

4c˙

I(tr/c)sin ✓

rˆ

.

E(r,t)=@A

@t=µ0b2

4c¨

I(tr/c)sin ✓

rˆ

.

B(r,t)=r⇥A=1

rsin ✓

@✓ (Asin ✓)ˆr 1

@r(rA)ˆ

✓

=µ0b2

4c˙

rsin ✓

r2 sin ✓cos ✓ˆr 1

r¨

I1

csin ✓ˆ

✓=µ0b2

4c2¨

Isin ✓

rˆ

✓.

S=1

µ0

(E⇥B)= 1

µ0cµ0b2

4c¨

Isin ✓

r2ˆ

⇥ˆ

✓=µ0

16c3b2¨

I2sin2✓

r2ˆr .

P=S·da=µ0

16c3b2¨

I2sin2✓

r2r2sin ✓d✓d=µ0

16c3b2¨

I2(2⇡)4

3=µ0⇡

6c3b2¨

I2

=µ0¨m2

6⇡c3.(Note that m=I⇡b2,so ¨m=¨

I⇡b2.)

Problem 11.12

p=ey ˆy ,y=1

2gt2,so p=1

2get2ˆy ;¨

p=ge ˆy .Therefore (Eq.11.60) :P=µ0

6⇡c(ge)2. Now, the time

it takes to fall a distance his given by h=1

2gt2)t=2h/g, so the energy radiated in falling a distance h

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CHAPTER 11. RADIATION 237

is Urad =Pt =µ0(ge)2

6⇡c2h/g. Meanwhile, the potential energy lost is Upot =mgh. So the fraction is

f=Urad

Upot

=µ0g2e2

6⇡c2h

mgh =µ0e2

6⇡mc2g

h.=(4⇡⇥107)(1.6⇥1019)2

6⇡(9.11 ⇥1031)(3 ⇥108)(2)(9.8)

(0.01) = 2.76 ⇥1022.

Evidently almost all the energy goes into kinetic form (as indeed I assumed in saying y=1

2gt2).

Problem 11.13

The power radiated (Eq. 11.70) is dWr

dt =µ0q2a2

6⇡c.Here F=1

4⇡✏0

x2=ma,soa=k

x2, where k⌘qQ

4⇡✏0m.

The energy radiated (twice that radiated on the way out) is

Wr=µ0q2

6⇡ck2

x4dt =µ0q2

6⇡c2k21

vdx,

where x0is the distance of closest approach. Conservation of energy (ignoring radiative losses, as suggested)

says 1

2mv2

0=1

2mv2+1

4⇡✏0

x)v2=v2

02qQ

4⇡✏0m

x=v2

02k

x,and x0=2k

(note that qis at x0when v= 0). So

Wr=µ0q2

6⇡c2k21

x4v2

02k/x dx =µ0q2

6⇡c

2k2

p2k1

x4(1/x0)(1/x)dx =µ0q2

6⇡c

2k2

p2k

15x5/2

(I used Mathematica to do the integral.) Simplifying,

Wr=2µ0q2v5

45⇡ck =2µ0q2v5

45⇡c

4⇡✏0m

qQ =8qmv5

45c3Q)f=Wr

=8qmv5

45c3Q

mv2

=16q

45Qv0

c3.

Problem 11.14

F=1

4⇡✏0

r2=ma =mv2

r)v=1

4⇡✏0

mr .At the beginning (r0=0.5˚

A),

c=(1.6⇥1019)2

4⇡(8.85 ⇥1012)(9.11 ⇥1031)(5 ⇥1011)1/21

3⇥108=0.0075,

and when the radius is one hundredth of this v/c is only 10 times greater (0.075), so for most of the trip the

velocity is safely nonrelativistic.

From the Larmor formula, P=µ0q2

6⇡cv2

r2

=µ0q2

6⇡c1

4⇡✏0

mr22

(since a=v2/r), and P=dU/dt,

where Uis the (total) energy of the electron:

U=Ukin +Upot =1

2mv21

4⇡✏0

r=1

21

4⇡✏0

r1

4⇡✏0

r=1

8⇡✏0

So dU

dt =1

8⇡✏0

dt =P=q2

6⇡✏0c31

4⇡✏0

mr22

,and hence dr

dt =1

3cq2

2⇡✏0mc 21

r2,or

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238 CHAPTER 11. RADIATION

dt =3c2⇡✏0mc

q22

r2dr )t=3c2⇡✏0mc

q220

r2dr =c2⇡✏0mc

q22

= (3 ⇥108)2⇡(8.85 ⇥1012)(9.11 ⇥1031)(3 ⇥108)

(1.6⇥1019)22

(5 ⇥1011)3= 1.3⇥1011 s. (Not very long!)

Problem 11.15

According to Eq. 11.74, the maximum occurs at d

d✓sin2✓

(1 cos ✓)5= 0. Thus

2 sin ✓cos ✓

(1 cos ✓)5 5 sin2✓(sin ✓)

(1 cos ✓)6=0)2 cos ✓(1 cos ✓)=5sin2✓=5(1 cos2✓);

2 cos ✓2cos2✓=55cos2✓,or 3cos2✓+ 2 cos ✓5=0.So

cos ✓=2±4 + 602

6=1

3±1 + 1521. We want the plus sign, since ✓m!90(cos ✓m= 0) when

!0 (Fig. 11.11): ✓max = cos11 + 1521

3.

For v⇡c, ⇡1; write =1✏(where ✏⌧1), and expand to ﬁrst order in ✏:

1 + 1521

3=1

3(1 ✏)1 + 15(1 ✏)21⇠

3(1 + ✏)1 + 15(1 2✏)1

3(1 + ✏)p16 30✏1=1

3(1 + ✏)41(15✏/8) 1=1

3(1 + ✏)4115

16✏1

3(1 + ✏)315

4✏= (1 + ✏)(1 5

4✏)⇠

=1+✏5

4✏=11

4✏.

Evidently ✓max ⇡0,so cos ✓max ⇠

=11

2✓2

max =11

4✏)✓2

max =1

2✏,or ✓max ⇠

=✏/2= (1 )/2.

Let f⌘(dP/d⌦|✓m)ur

(dP/d⌦|✓m)rest

=sin2✓max

(1 cos ✓max)5ur

. Now sin2✓max ⇠

=✏/2, and

(1 cos ✓max)⇠

=1(1 ✏)(1 1

4✏)⇠

=1(1 ✏1

4✏)=5

4✏. So f=✏/2

(5✏/4)5=4

551

2✏4.But

=1

12=1

1(1 ✏)2⇠

1(1 2✏)=1

p2✏)✏=1

22. Therefore

f=4

551

2(22)4=1

48

55

8=2.628.

Problem 11.16

Equation 11.72 says dP

d⌦=q2

16⇡2✏0ˆ

⇥(u⇥a)2

(ˆ

·u)5. Let ⌘v/c.

u=cˆ

v=cˆ

vˆz )ˆ

·u=cv(ˆ

·ˆz )=cvcos ✓=c1v

ccos ✓=c(1 cos ✓);

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CHAPTER 11. RADIATION 239

a·u=ac(ˆx ·ˆ

)av(ˆx ·ˆz )=ac sin ✓cos ;u2=u·u=c22cv(ˆ

·ˆz )+v2=c2+v22cv cos ✓.

⇥(u⇥a)=(ˆ

·a)u(ˆ

·u)a;

|ˆ

⇥(u⇥a)|2=(ˆ

·a)2u22(u·a)( ˆ

·a)( ˆ

·u)+(ˆ

·u)2a2

=(c2+v22cv cos ✓)(asin ✓cos )22(ac sin ✓cos )(asin ✓cos )(cvcos ✓)+a2c2(1 cos ✓)2

=a2c2(1 cos ✓)2+ (sin2✓cos2)(c2+v22cv cos ✓2c2+2cv cos ✓

=a2c2(1 cos ✓)2(1 2)(sin ✓cos )2.

d⌦=µ0q2a2

16⇡2c(1 cos ✓)2(1 2) sin2✓cos2

(1 cos ✓)5.

The total power radiated (in all directions) is:

P=dP

d⌦d⌦=dP

d⌦sin ✓d✓d=µ0q2a2

16⇡2c(1 cos ✓)2(1 2) sin2✓cos2

(1 cos ✓)5sin ✓d✓d.

But 2⇡

d=2⇡and 2⇡

cos2d=⇡.

=µ0q2a2

16⇡2c⇡⇡

02(1 cos ✓)2(1 2) sin2✓

(1 cos ✓)5sin ✓d✓.

Let w⌘(1 cos ✓). Then (1 w)/= cos ✓; sin2✓=2(1 w)2/2, and the numerator becomes

2w2(1 2)

2(21+2ww2)= 1

22w22+ (1 2)22(1 2)w+w2(1 2)

2(1 2)22(1 2)w+ (1 + 2)w2;

dw =sin ✓d✓)sin ✓d✓=1

dw. When ✓=0,w=(1); when ✓=⇡,w=(1+).

P=µ0q2a2

16⇡c

3(1+)

(1)

w5(1 2)22(1 2)w+ (1 + 2)w2dw. The integral is

Int = (1 2)21

w5dw 2(1 2)1

w4dw + (1 + 2)1

w3dw

=(1 2)21

4w42(1 2)1

3w3+ (1 + 2)1

2w2

1+

1

w2

1+

1=1

(1 + )21

(1 )2=(1 2+2)(1 + 2+2)

(1 + )2(1 )2=4

(1 2)2.

w3

1+

1=1

(1 + )31

(1 )3=(1 3+323)(1 + 3+32+3)

(1 + )3(1 )3=2(3 + 2)

(1 2)3.

w4

1+

1=1

(1 + )41

(1 )4=(1 4+6243+4)(1 + 4+62+43+4)

(1 + )4(1 )4=8(1 + 2)

(1 2)4.

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240 CHAPTER 11. RADIATION

Int = (1 2)21

48(1 + 2)

(1 2)42(1 2)1

32(3 + 2)

(1 2)3+ (1 + 2)1

24

(1 2)2

=2

(1 2)2(1 + 2)2

3(3 + 2)+(1+2)=8

3

(1 2)2.

P=µ0q2a2

16⇡c

3

(1 2)2=µ0q2a24

6⇡c,where =1

12.

Is this consistent with the Li´enard formula (Eq. 11.73)? Here v⇥a=va(ˆz ⇥ˆx )=va ˆy ,so

a2v

c⇥a2=a21v2

c2= (1 2)a2=1

2a2, so the Li´enard formula says P=µ0q26

6⇡c

2.X

Problem 11.17

(a) To counteract the radiation reaction (Eq. 11.80), you must exert a force Fe=µ0q2

6⇡c˙

For circular motion, r(t)=R[cos(!t)ˆx + sin(!t)ˆy ],v(t)=˙

r=R![sin(!t)ˆx + cos(!t)ˆy ];

a(t)= ˙

v=R!2[cos(!t)ˆx + sin(!t)ˆy ]=!2r;˙

a=!2˙

r=!2v. So Fe=µ0q2

6⇡c!2v.

Pe=Fe·v=µ0q2

6⇡c!2v2.This is the power you must supply.

Meanwhile, the power radiated is (Eq. 11.70) Prad =µ0q2a2

6⇡c, and a2=!4r2=!4R2=!2v2,so

Prad =µ0q2

6⇡c!2v2, and the two expressions agree.

(b) For simple harmonic motion, r(t)=Acos(!t)ˆz ;v=˙

r=A!sin(!t)ˆz ;a=˙

v=A!2cos(!t)ˆz =

!2r;˙

a=!2˙

r=!2v.SoFe=µ0q2

6⇡c!2v;Pe=µ0q2

6⇡c!2v2.But this time a2=!4r2=!4A2cos2(!t),

whereas !2v2=!4A2sin2(!t), so

Prad =µ0q2

6⇡c!4A2cos2(!t)6=Pe=µ0q2

6⇡c!4A2sin2(!t);

the power you deliver is not equal to the power radiated. However, since the time averages of sin2(!t) and

cos2(!t) are equal (to wit: 1/2), over a full cycle the energy radiated is the same as the energy input. (In the

mean time energy is evidently being stored temporarily in the nearby ﬁelds.)

2gt2ˆy ;v=gt ˆy ;a=gˆy ;˙

a= 0. So Fe=0; the radiation reaction is zero, and

hence Pe=0.But there is radiation: Prad =µ0q2

6⇡cg2.Evidently energy is being continuously extracted from

the nearby ﬁelds. This paradox persists even in the exact solution (where we do not assume v⌧c, as in the

Larmor formula and the Abraham-Lorentz formula)—see Prob. 11.34.

Problem 11.18

(a) From Eq. 11.80, Frad =µ0q2

6⇡c˙a=m⌧...

x(Eq. 11.82). The equation of motion is

F=m¨x=Fspring +Frad =kx +m⌧...

x, or ¨x+!2

0x⌧...

x=0,with !0=k/m.

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CHAPTER 11. RADIATION 241

Since the damping is small, it oscillates at the natural frequency !0, and hence ...

x=!2

0˙x, so ¨x+!2

0⌧˙x+!2

0x=0,

or ¨x+˙x+!2

0x= 0, with =!2

0⌧.The solution with x(0) = 0 is (for ⌧!0)

x(t)=Aet/2sin(!0t); v(t)=

2Aet/2sin(!0t)+!0Aet/2cos(!0t),so v(0) = A!0=v0)A=v0

and x(t)= v0

et/2sin(!0t).

According to the Larmor formula (Eq. 11.70), the power radiated is P=µ0q2a2

6⇡c.In this case (still assuming

⌧!0)

a=!2

0x=!2

et/2sin(!0t),P=µ0q2

6⇡c(!0v0)2etsin2(!0t).

Averaging over a full cycle (holding etconstant) hPi=µ0q2!2

0v2

6⇡cet1

2, and the total energy radiated is

1

0hPidt =µ0q2!2

0v2

12⇡c1

et

=µ0q2!2

0v2

12⇡c=µ0q2!2

0v2

12⇡c!2

0⌧=µ0q2v2

12⇡c6⇡mc

µ0q2=1

2mv2

0.X

(b) This “equivalent” single oscillator has twice the charge, and twice the mass, so ⌧(and hence ) is

doubled. Since hPidt goes like q2/, it also doubles. The power radiated is indeed four times as great, but

the oscillations die away faster, and the total energy radiated is just twice as much as for one oscillator.

Problem 11.19

(a) a=⌧˙a+F

m)dv

dt =⌧da

dt +F

m)dv

dt dt =⌧da

dt dt +1

mF dt.

[v(t0+✏)v(t0✏)] = ⌧[a(t0+✏)a(t0✏)] + 2✏

mFave, where Fave is the average force during the inter-

val. But vis continuous, so as long as Fis not a delta function, we are left (in the limit ✏!0) with

[a(t0+✏)a(t0✏)] = 0. Thus a, too, is continuous. qed

(b) (i) a=⌧˙a=⌧da

dt )da

a=1

⌧dt )da

a=1

⌧dt )ln a=t

⌧+ constant )a(t)=Aet/⌧,where A

is a constant.

(ii) a=⌧˙a+F

m)⌧da

dt =aF

m)da

aF/m =1

⌧dt )ln(aF/m)= t

⌧+ constant )aF

m=Bet/⌧)

a(t)= F

m+Bet/⌧,where Bis some other constant.

(iii) Same as (i): a(t)=Cet/⌧,where Cis a third constant.

a(t)=











[(F/m)+B]et/⌧,t0;

(F/m)+Bet/⌧,0tT;

(F/m)eT/⌧+Bet/⌧,tT.

To eliminate the runaway in region (iii), we’d need B=(F/m)eT/⌧; to avoid preacceleration in region

(i), we’d need B=(F/m). Obviously, we cannot do both at once.

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242 CHAPTER 11. RADIATION

(d) If we choose to eliminate the runaway, then

a(t)=











(F/m)1eT/⌧et/⌧,t0;

(F/m)1e(tT)/⌧,0tT;

0,tT.

(i) v=(F/m)1eT/⌧et/⌧dt =(F⌧/m)1eT/⌧et/⌧+D, where Dis a constant determined by the

condition v(1)=0)D=0.

(ii) v=(F/m)t⌧e(tT)/⌧+E, where Eis a constant determined by the continuity of vat t= 0:

(F⌧/m)1eT/⌧=(F/m)⌧eT/⌧+E)E=(F⌧/m).

(iii) vis a constant determined by the continuity of vat t=T:v=(F/m)[T+⌧⌧]=(F/m)T.

v(t)=











(F⌧/m)1eT/⌧et/⌧,t0;

(F/m)t+⌧⌧e(tT)/⌧,0tT;

(F/m)T, t T.

(e)

Problem 11.20

(a) From Eq.11.80, Fend

rad =µ0(q/2)2

6⇡c˙a, so Frad =Fint

rad +2Fend

rad =µ0q2

6⇡c˙a1

2+21

4=µ0q2

6⇡c˙a. X

(b) Following the suggested method:

F(q)=Fint(q)+21

4F(q))1

2F(q)=Fint(q))F(q)=2Fint(q)=µ0q2˙a

6⇡c.X

12⇡c˙aL

0y1

2dy22dy1. (Running the y2

integral up to y1insures that y1y2, so we don’t count the

same pair twice. Alternatively, run both integrals from 0 to

L—intentionally double-counting—and divide the result by 2.)

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CHAPTER 11. RADIATION 243

Frad =µ0˙a

12⇡c(42)L

y1dy1=µ0˙a

12⇡c(42)L2

2=µ0

6⇡c(L)2˙a=µ0q2

6⇡c˙a. X

Problem 11.21

(a) The total torque is twice the torque on +q; we might as well calculate it at time t= 0. First we

need the electric ﬁeld at +q, due to qwhen it was at the retarded point P(Eq. 10.72). From the ﬁgure,

= cos ✓ˆx sin ✓ˆy ,

=2Rcos ✓.The velocity of q(at P) was v=!R(sin 2✓ˆx + cos 2✓ˆy ), and its

acceleration was a=!2R(cos 2✓ˆx sin 2✓ˆy ).Quantities we will need in Eq. 10.72 are:

u=cˆ

v=(ccos ✓+!Rsin 2✓)ˆx (csin ✓!Rcos 2✓)ˆy ,

·u=2Rcos ✓(c+!Rsin ✓),

·a=2(!Rcos ✓)2.

-q R

E=q

4⇡✏0

2Rcos ✓

(2Rcos ✓)3(c+!Rsin ✓)3[c2(!R)2+ 2(!Rcos ✓)2][(ccos ✓+!Rsin 2✓)ˆx (csin ✓!Rcos 2✓)ˆy ]

2Rcos ✓(c+!Rsin ✓)!2R(cos 2✓ˆx sin 2✓ˆy )

The total torque (about the origin) is

N= 2(Rˆx )⇥(qE)=2q2R

4⇡✏0

ˆz

(2Rcos ✓)2(c+!Rsin ✓)3c2(!R)2+ 2(!Rcos ✓)2(csin ✓!Rcos 2✓)

+ 2(!R)2cos ✓(c+!Rsin ✓) sin 2✓

=q2

4⇡✏0

ˆz

2Rcos2✓(c+!Rsin ✓)3c3sin ✓+c2!R(2 cos2✓1) + c(!R)2(2 cos2✓+ 1) sin ✓+(!R)3

=q2

4⇡✏0

2Rcos2✓(1 + sin ✓)3sin ✓+(2 cos2✓1) + 2(2 cos2✓+ 1) sin ✓+3ˆz ,

where ⌘!R/c. [Since Eand

both lie in the xy plane, B=(1/c)ˆ

⇥Eis along the zdirection, v⇥Bis

radial, and hence the magnetic contribution to the torque is zero.]

The angle ✓is determined by the retarded time condition,

=ctr(note that tris negative, here), and

2✓is the angle through which the dipole rotates in time tr,so2Rcos ✓=ctr=c(2✓/!), or ✓=cos ✓.We

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244 CHAPTER 11. RADIATION

can use this to eliminate the trig functions:

N=q2

4⇡✏0



2R✓21+2✓23(42+2✓2)+2✓2(2+2✓21)ˆz .

Meanwhile, expanding in powers of ✓:

=✓sec ✓=✓+1

2✓3+5

24✓5+61

720✓7+....

This can be “solved” (for ✓as a function of ) by reverting the series:

✓=1

23+13

245541

7207+....

Then

✓2=212+4

34+...

,2✓2=212

32+4

54+...

,

✓2=1

21+2+...

,1

1+2✓23=132+...,

(42+2✓2)+2✓2(2+2✓21)=8

3414

52+...

.

To leading order in , then,

N=q2

4⇡✏0



2

84

3ˆz =q2

4⇡✏0

43

3Rˆz .

(b) The radiation reaction force on +qis (Eq. 11.80) F=µ0q2

6⇡c˙

a.In this case ˙

a=!2v=!3Rˆy , so the

net torque (counting both ends) is

N=2Rµ0q2

6⇡c!3Rˆz =q2

4⇡✏0

43

3Rˆz .

Adding this to the interaction torque from (a), the total is

N=q2

4⇡✏0

83

3Rˆz =µ0p2!3

6⇡cˆz .

p=2q¨

r=2q(!2)r=!2p, so Eq. 11.60 says the power radiated is P=µ0

6⇡c!4p2.The power

associated with the torque in (b) is N!=µ0p2!3

6⇡c!, so they are in agreement. X

Problem 11.22

(a) This is an oscillating electric dipole, with amplitude p0=qd and frequency !=k/m. The (averaged)

Poynting vector is given by Eq. 11.21: hSi=µ0p2

0!4

32⇡2csin2✓

r2ˆr , so the power per unit area of ﬂoor is

If=hSi·ˆz =µ0p2

0!4

32⇡2csin2✓cos ✓

r2.But sin ✓=R

r,cos ✓=h

r,and r2=R2+h2.

=µ0q2d2!4

32⇡2cR2h

(R2+h2)5/2.

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CHAPTER 11. RADIATION 245

dIf

dR =0)d

dR R2

(R2+h2)5/2=0)2R

(R2+h2)5/25

(R2+h2)7/22R=0)

(R2+h2)5

2R2=0)h2=3

2R2)R=2/3h, for maximum intensity.

(b)

P=If(R)da =If(R)2⇡R dR =2⇡µ0(qd)2!4

32⇡2ch1

(R2+h2)5/2dR. Let x⌘R2:

1

(R2+h2)5/2dR =1

21

(x+h2)5/2dx =1

(2)(1/2)

(5/2) =2

3h.

=2⇡µ0q2d2!4

32⇡2ch2

3h=µ0q2d2!4

24⇡c,

which should be (and is) half the total radiated power (Eq. 11.22)—the rest hits the ceiling, of course.

2kx2

0is the energy, at time t, and dU/dt =2Pis the power radiated:

2kd

dt(x2

0)=µ0!4

12⇡cq2x2

0)d

dt(x2

0)=µ0!4q2

6⇡kc (x2

0)=x2

0)x2

0=d2etor x0(t)=det/2.

⌧=2

=12⇡kc

µ0q2k2m2=12⇡cm2

µ0q2k.

Problem 11.23

(a) From Eq. 11.39, hSi=µ0m2

0!4

32⇡2c3sin2✓

r2ˆr . Here sin ✓=

R/r, r =pR2+h2, and the total radiated power (Eq. 11.40) is

P=µ0m2

0!4

12⇡c3. So the intensity is I(R)=12P

32⇡R2

(R2+h2)2=

8⇡

(R2+h2)2.

(b) The intensity directly below the antenna (R= 0) would (ideally) have been zero. The engineer should

have measured it at the position of maximum intensity:

dR =3P

8⇡2R

(R2+h2)22R2

(R2+h2)32R=3P

8⇡

(R2+h2)3R2+h22R2=0)R=h.

At this location the intensity is I(h)=3P

8⇡

(2h2)2=3P

32⇡h2.

32⇡(200)2=0.026 W/m2= 2.6 µW/cm2. Yes, KRUD is in compliance.

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246 CHAPTER 11. RADIATION

Problem 11.24

(a)

V±=⌥p0!

4⇡✏0ccos ✓±

r±sin[!(tr±/c)].V

tot =V++V.

r±=r2+(d/2)2⌥2r(d/2) cos ✓⇠

=r1⌥(d/r) cos ✓⇠

=r1⌥d

2rcos ✓.

r±⇠

r1±d

2rcos ✓.

cos ✓±=rcos ✓⌥(d/2)

r±

=rcos ✓⌥d

2r1

r1±d

2rcos ✓= cos ✓±d

2rcos2✓⌥d

= cos ✓⌥d

2r(1 cos2✓) = cos ✓⌥d

2rsin2✓.

sin[!(tr±/c)] = sin !tr

c1⌥d

2rcos ✓= sin !t0±!d

2ccos ✓,where t0⌘tr/c.

= sin(!t0) cos !d

2ccos ✓±cos(!t0) sin !d

2ccos ✓⇠

=sin(!t0)±!d

2ccos ✓cos(!t0).

V±=⌥p0!

4⇡✏0cr 1±d

2rcos ✓cos ✓⌥d

2rsin2✓sin(!t0)±!d

2ccos ✓cos(!t0)

=⌥p0!

4⇡✏0cr cos ✓⌥d

2rsin2✓±d

2rcos2✓sin(!t0)±!d

2ccos ✓cos(!t0)

=⌥p0!

4⇡✏0cr cos ✓sin(!t0)±!d

2ccos2✓cos(!t0)±d

2rcos2✓sin2✓sin(!t0.

Vtot =p0!

4⇡✏0cr !d

ccos2✓cos(!t0)+d

rcos2✓sin2✓sin(!t0)

=p0!2d

4⇡✏0c2rcos2✓cos(!t0)+ c

!rcos2✓sin2✓sin(!t0.

In the radiation zone (r!/c) the second term is negligible, so V=p0!2d

4⇡✏0c2rcos2✓cos[!(tr/c)].

Meanwhile

A±=⌥µ0p0!

4⇡r±

sin[!(tr±/c)] ˆz

=⌥µ0p0!

4⇡r1±d

2rcos ✓sin(!t0)±!d

2ccos ✓cos(!t0)ˆz

=⌥µ0p0!

4⇡rsin(!t0)±!d

2ccos ✓cos(!t0)±d

2rcos ✓sin(!t0)ˆz .

Atot =A++A=µ0p0!

4⇡r!d

ccos ✓cos(!t0)+d

rcos ✓sin(!t0)ˆz

=µ0p0!2d

4⇡cr cos ✓cos(!t0)+ c

!rsin(!t0)ˆz .

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CHAPTER 11. RADIATION 247

In the radiation zone, A=µ0p0!2d

4⇡cr cos ✓cos[!(tr/c)] ˆz .

(b) To simplify the notation, let ↵⌘µ0p0!2d

4⇡. Then

V=↵cos2✓

rcos[!(tr/c)];

rV=@V

@rˆr +1

@✓ ˆ

✓=↵cos2✓1

r2cos[!(tr/c)] + !

rc sin[!(tr/c)]ˆr

+↵2 cos ✓sin ✓

r2cos[!(tr/c)] ˆ

✓=↵!

cos2✓

rsin[!(tr/c)]ˆr (in the radiation zone).

A=↵

cos ✓

rcos[!(tr/c)] cos ✓ˆr sin ✓ˆ

✓.@A

@t=↵!

cos ✓

rsin[!(tr/c)] cos ✓ˆr sin ✓ˆ

✓.

E=rV@A

@t=↵!

cr sin[!(tr/c)] cos2✓ˆr cos2✓ˆr + sin ✓cos ✓ˆ

✓

=↵!

cr sin ✓cos ✓sin[!(tr/c)] ˆ

✓.

B=r⇥A=1

r@

@r(rA✓)@Ar

@✓ ˆ



=↵

cr @

@r(cos ✓cos[!(tr/c)](sin ✓)) @

@✓ cos2✓

rcos[!(tr/c)]ˆ



=↵

cr (sin ✓cos ✓)!

csin[!(tr/c)] ˆ

(in the radiation zone) = ↵!

c2rsin ✓cos ✓sin[!(tr/c)] ˆ

.

Notice that B=1

c(ˆr ⇥E) and E·ˆr =0.

S=1

µ0

(E⇥B)= 1

µ0cE⇥(ˆr ⇥E)= 1

µ0cE2ˆr (E·ˆr )E=E2

µ0cˆr

µ0c↵!

rc sin ✓cos ✓sin[!(tr/c)]2ˆr . I =1

2µ0c↵!

rc sin ✓cos ✓2.

P=hSi·da=1

µ0c↵!

c2sin2✓cos2✓sin ✓d✓d=1

2µ0c↵!

c22⇡⇡

(1 cos2✓) cos2✓sin ✓d✓.

The integral is : cos3✓

3

⇡

0+cos5✓

5

⇡

0=2

32

5=4

15.

2µ0c

µ2

16⇡2(p0d)2!42⇡4

15 =µ0

60⇡c3(p0d)2!6.

Notice that it goes like !6, whereas dipole radiation goes like !4.

Problem 11.25

(a) m(t)=Mcos ˆz +Msin [cos(!t)ˆx + sin(!t)ˆy ]. As in Prob. 11.4, the power radiated will be twice

that of an oscillating magnetic dipole with dipole moment of amplitude m0=Msin . Therefore (quoting

Eq. 11.40): P=µ0M2!4sin2

6⇡c3.(Alternatively, you can get this from the answer to Prob. 11.11.)

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248 CHAPTER 11. RADIATION

(b) From Eq. 5.88, with r!R,m!M, and ✓=⇡/2: B=µ0

4⇡

R3, so

M=4⇡R3

µ0

B=4⇡(6.4⇥106)3(5 ⇥105)

4⇡⇥107= 1.3⇥1023 Am

6⇡(3 ⇥108)32⇡

24 ⇥60 ⇥604

= 4 ⇥105W (not much).

(d) P=µ0(4⇡R3B/µ0)2!4sin2

6⇡c3=8⇡

3µ0c3!2R3Bsin 2. Using the average value (1/2) for sin2 ,

P=8⇡

3(4⇡⇥107)(3 ⇥108)32⇡

1032

(104)3(108)21

2= 2 ⇥1036 W (a lot).

Problem 11.26

(a) Write p(t)=q(t)dˆz , with q(t)=kt2, where kd =(1/2)¨p0.As in Eq. 11.5,

V(r,t)= 1

4⇡✏0k(t

+/c)2

+k(t

/c)2



4⇡✏0t2(2t/c)

+/c2

+t2(2t/c)

+

+/c2



4⇡✏0t21

+1

+1

c2(

+

).

From Eqs. 11.8 and 11.9,

±=r1⌥d

2rcos ✓,1

r1±d

2rcos ✓,

V(r,t)= k

4⇡✏0t2

rd

rcos ✓r

c2d

rcos ✓=k

4⇡✏0c2dcos ✓ct

r2

1=µ0¨p0

8⇡cos ✓ct

r2

1.

As in Eq. 11.15, I(t)=(dq/dt)ˆz =2kt ˆz , so (following Eqs. 11.16, and 11.17),

A(r,t)= µ0

4⇡ˆz d/2

d/2

2k(t

/c)

dz =µ0

4⇡2k(tr/c)

rdˆz =µ0¨p0

4⇡cct

r1ˆz .

E=rV@A

@t=µ0¨p0

8⇡cos ✓2(ct)2

r3ˆr 1

rsin ✓ct

r2

1ˆ

✓µ0¨p0

4⇡cc

rˆz

=µ0¨p0

4⇡rcos ✓ct

r2

ˆr +1

2sin ✓ct

r2

✓1

2sin ✓ˆ

✓(cos ✓ˆr sin ✓ˆ

✓)

=µ0¨p0

4⇡rct

r2

1cos ✓ˆr +1

2ct

r2

sin ✓ˆ

✓.

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CHAPTER 11. RADIATION 249

B=r⇥A=µ0¨p0

4⇡cr⇥ct

r1(cos ✓ˆr sin ✓ˆ

✓)

=µ0¨p0

4⇡cr @

@rrct

r1sin ✓@

@✓ ct

r1cos ✓ ˆ



=µ0¨p0

4⇡cr sin ✓+ct

r1sin ✓ˆ

=µ0¨p0t

4⇡r2sin ✓ˆ

.

(b) The Poynting vector is

S=1

µ0

(E⇥B)= µ0¨p2

16⇡2r3sin ✓ct

r2

1cos ✓(ˆ

✓)+1

2ct

r2

sin ✓ˆr .

S·da=µ0¨p2

32⇡2ct

r2

sin2✓

r3(r2sin ✓d✓d).

P(r, t)= µ0¨p2

32⇡2rct

r2

2⇡⇡

sin3✓d✓=µ0¨p2

12⇡rct

r2

.

12⇡rt0+r

cc2

r2t2

0+2t0

c+r2

c2+1

=µ0¨p2

12⇡c1+ct0

r2+2ct0

r+ct0

r2.

Prad(t0) = lim

r!1 P(r, t0+r/c)= µ0¨p2

6⇡c,

in agreement with Eq. 11.60.

Problem 11.27

The momentum ﬂux density is (minus) the Maxwell stress tensor (Section 8.2.3),

Tij =✏0EiEj1

2ij E2+1

µ0BiBj1

2ij B2,

(Eq. 8.17) so the momentum radiated per unit (retarded) time is

dtr

=1

(@tr/@t)$

T·dA,

where (Eq. 10.78)

@tr

@t=

·u=1ˆ

·v

c1

(This factor is 1, when v=0; it can be ignored in deriving the Larmor formula, but it does contribute to the

momentum radiated.) The integration is over a large spherical surface centered on the charge:

dA=

2sin ✓d✓dˆ

As in the case of the power radiated, only the radiation ﬁelds contribute (Eq. 11.66):

Erad =q

4⇡✏0

(

·u)3[

⇥(u⇥a)],Brad =1

cˆ

⇥Erad.

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250 CHAPTER 11. RADIATION

Thus

T·dA=✏0E(E·dA)1

2E2dA+1

µ0B(B·dA)1

2B2dA=1

2✏0E2+1

µ0

B2dA=✏0E2dA

(note that B·dA= 0, and, for radiation ﬁelds, E·dA= 0, (1/µ0)B2= (1/µ0c2)( ˆ

⇥E)·(ˆ

⇥E)=

✏0ˆ

·[E⇥(ˆ

⇥E)] = ✏0ˆ

·[ˆ

E2E(ˆ

·E)] = ✏0E2). So

dtr

=✏0

·u

cE2dA=q2

16⇡2✏0c

(

·u)5[

⇥(u⇥a)]2dA.

We expand the integrand to ﬁrst order in ⌘v/c:u=cˆ

v=cˆ

,

·u=c

(1 ˆ

·),

(

·u)5=1

5c5[1 + 5( ˆ

·)],

⇥(u⇥a)=u(

·a)a(

·u)=

cˆ

(ˆ

·a)a

c(1 ˆ

·)

cˆ

(ˆ

·a)a+a(ˆ

·)(ˆ

·a).

[

⇥(u⇥a)]2=

2c2(ˆ

·a)22( ˆ

·a)2+a2+ 2( ˆ

·a)2(ˆ

·)2( ˆ

·)( ˆ

·a)22a2(ˆ

·)

+ 2(a·)( ˆ

·a)=

2c2a2(ˆ

·a)22a2(ˆ

·) + 2( ˆ

·a)(a·).

(

·u)5[

⇥(u⇥a)]2=1

2c3[1 + 5( ˆ

·)] a2(ˆ

·a)22a2(ˆ

·) + 2( ˆ

·a)(a·)

2c3a2(ˆ

·a)22a2(ˆ

·) + 2( ˆ

·a)(a·) + 5( ˆ

·)a25( ˆ

·)( ˆ

·a)2

2c3a2(ˆ

·a)2+3a2(ˆ

·) + 2( ˆ

·a)(a·)5( ˆ

·)( ˆ

·a)2

2c3(ˆ

⇥a)2+ˆ

·[3a2+ 2(a·)a]5( ˆ

·)( ˆ

·a)2.

To integrate the ﬁrst term we set the polar axis along a,so(ˆ

⇥a)2=a2sin2✓, while ˆ

= sin ✓cos ˆ

sin ✓sin ˆ

y+ cos ✓ˆ

z. The integral kills the ˆ

xand ˆ

ycomponents, leaving

1

2(ˆ

⇥a)2dA=2⇡a2ˆ

z⇡

sin3✓cos ✓d✓=0.

[Note that if v=0then dp/dtr= 0—a particle instantaneously at rest radiates no momentum. That’s why

we had to carry the expansion to ﬁrst order in , whereas in deriving the Larmor formula we could a↵ord to

set v= 0.] The second term is of the form ( ˆ

·g), for a constant vector g. Setting the polar axis along g, so

·g=gcos ✓, the xand ycomponents again vanish, leaving

1

2(ˆ

·g)dA=2⇡gˆ

z⇡

cos2✓sin ✓d✓=4⇡

3g.

The last term involves ( ˆ

·)( ˆ

·a)2; this time we orient the polar axis along aand let vlie in the xz plane:

=xˆ

x+zˆ

z, so ( ˆ

·)=xsin ✓cos +zcos ✓. Then

1

2(ˆ

·)( ˆ

·a)2dA=a2(xsin ✓cos +zcos ✓) cos2✓sin ✓cos ˆ

x+ sin ✓sin ˆ

y+ cos ✓ˆ

zsin ✓d✓d

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CHAPTER 11. RADIATION 251

=a2xˆ

xsin3✓cos2✓cos2d✓d+zˆ

zcos4✓sin ✓d✓d=4⇡

15 a2(xˆ

x+3zˆ

z)=4⇡

15 a2+ 2(a·)a.

Putting all this together,

dtr

=q2

16⇡2✏0

c44⇡

32a(a·)+3a254⇡

15 a2+ 2(a·)a=µ0q2

6⇡c3a2v.

The angular momentum radiated is

dtr

=1

(@tr/@t)

⇥$

T·dA.

Because of the “extra”

in the integrand, it seems at ﬁrst glance that the radiation ﬁelds alone will produce a

result that grows without limit (as

!1); however, the coeﬃcient of this term is precisely zero: 

⇥$

T·

dA=✏0E2(

⇥dA) = 0 (for radiation ﬁelds). The ﬁnite contribution comes from the cross terms, in which

one ﬁeld (in $

T) is a radiation ﬁeld and the other a Coulomb ﬁeld Ecoul ⌘q

4⇡✏0

(

·u)3(c2v2)u:

Tij =✏0E(r)

iE(c)

j+E(c)

iE(r)

jij E(r)·E(c)+1

µ0B(r)

iB(c)

j+B(c)

iB(r)

jij B(r)·B(c).

This time $

T·dA=✏0E(r)(E(c)·dA)E(r)·E(c)dA1

µ0B(r)·B(c)dA,

(

⇥$

T)·dA=✏0

⇥E(r)E(c)·dA.

Thus dL

dtr

=q2c

16⇡2✏02

(

·u)5

⇥[

⇥(u⇥a)] u·dA,where ⌘1

1(v/c)2.

Now, {

⇥[

⇥(u⇥a)]}=

[

·(u⇥a)](u⇥a)

2,and (u⇥a)=c(ˆ

)⇥a=c(ˆ

⇥a)(⇥a),

·(u⇥a)=cˆ

·(⇥a), so {

⇥[

⇥(u⇥a)]}=

2c(ˆ

⇥a)(⇥a)+ ˆ

ˆ

·(⇥a).

Meanwhile u·dA=(u·ˆ

)

2d⌦, where d⌦⌘sin ✓d✓d. Expanding to ﬁrst order in ,

dtr

=µ0q2c3

16⇡2

(

·u)5(

2c)(ˆ

⇥a)(⇥a)+ ˆ

ˆ

·(⇥a)

(

·u)d⌦

=µ0q2

16⇡21 + 4( ˆ

·)(ˆ

⇥a)(⇥a)+ ˆ

ˆ

·(⇥a)d⌦

=µ0q2

16⇡2(ˆ

⇥a)(⇥a)+ ˆ

[ˆ

·(⇥a)] + 4( ˆ

·)( ˆ

⇥a)d⌦

The ﬁrst integral is

a⇥ˆ

sin ✓d✓d=a⇥(sin ✓cos ˆ

x+ sin ✓sin ˆ

y+ cos ✓ˆ

z) sin ✓d✓d=0,

the second is

(⇥a)sin ✓d✓d=4⇡(⇥a),

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252 CHAPTER 11. RADIATION

for the third we set the polar axis along g⌘⇥aand get

gˆ

cos ✓sin ✓d✓d=2⇡gˆ

zcos2✓sin ✓d✓=4⇡

3g=4⇡

3(⇥a),

and for the last we put the polar axis along aand let =xˆ

x+zˆ

zlie in the xz plane:

4a(xsin ✓cos +zcos ✓) sin ✓(ˆ

) sin ✓d✓d

=4a(xsin ✓cos +zcos ✓)(sin ˆx cos ˆy ) sin2✓d✓d

=4axˆ

ycos2sin3✓d✓d=16⇡

3axˆ

y=16⇡

3(⇥a).

Putting this all together, we conclude

dtr

=µ0q2

16⇡24⇡(⇥a)+4⇡

3(⇥a)+16⇡

3(⇥a)=µ0q2

6⇡c(v⇥a).

Problem 11.28

(a) A(x, t)= µ0

4⇡K(tr)

=µ0ˆz

4⇡K(tr)

pr2+x22⇡rdr

=µ0ˆz

2K(tpr2+x2/c)

pr2+x2r dr.

The maximum ris given by tpr2+x2/c = 0;

rmax =c2t2x2(since K(t) = 0 for t<0).

(i)

A(x, t)=µ0K0ˆz

2rm

pr2+x2dr =µ0K0ˆz

2r2+x2

0=µ0K0ˆz

2r2

mx2x=µ0K0(ct x)

2ˆz .

E(x, t)=@A

@t=µ0K0c

2ˆz ,for ct > x, and 0,for ct < x.

B(x, t)=r⇥A=@Az

@xˆy =µ0K0

2ˆy ,for ct > x, and 0,for ct < x.

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CHAPTER 11. RADIATION 253

(ii)

A(x, t)=µ0↵ˆz

2rm

0tpr2+x2/c

pr2+x2r dr =µ0↵ˆz

2trm

pr2+x2dr 1

crm

r dr

=µ0↵ˆz

2t(ct x)1

2c(c2t2x2)=µ0↵ˆz

4c(x22ctx +c2t2)=µ0↵(xct)2

4cˆz .

E(x, t)=@A

@t=µ0↵(xct)

2ˆz ,for ct > x, and 0,for ct < x.

B(x, t)=r⇥A=@Az

@xˆy =µ0↵

2c(xct)ˆy ,for ct > x, and 0,for ct < x.

(b) Let u⌘1

cr2+x2x,sodu =1

c1

pr2+x22r dr=1

pr2+x2dr, and

tpr2+x2

c=tx

cu, and as r:0!1,u:0!1. Then A(x, t)= µ0cˆz

21

Ktx

cudu. qed

E(x, t)=@A

@t=µ0cˆz

21

@tKtx

cudu. But @

@tKtx

cu=@

@uKtx

cu.

=µ0c

2ˆz 1

@uKtx

cudu =µ0c

2ˆz Ktx

cu1

0=µ0c

2[K(tx/c)K(1)] ˆz

=µ0c

2K(tx/c)ˆz ,[if K(1) = 0].

Note that (i) and (ii) are consistent with this result. Meanwhile

B(x, t)=@Az

@xˆy =µ0c

cˆy 1

@xKtx

cudu. But @

@xKtx

cu=1

@uKtx

cu.

=µ0

2ˆy 1

@uKtx

cudu =µ0

2ˆy Ktx

cu1

0=µ0

2[K(tx/c)K(1)] ˆy

=µ0

2K(tx/c)ˆy ,[if K(1) = 0].

S=1

µ0

(E⇥B)= 1

µ0µ0c

2µ0

2K(tx/c)[ˆz ⇥ˆy ]= µ0c

4[K(tx/c)]2ˆx .

This is the power per unit area that reaches xat time t; it left the surface at time (tx/c). Moreover, an

equal amount of energy is radiated downward, so the total power leaving the surface at time tis µ0c

2[K(t)]2.

Problem 11.29

With ↵= 90, Eq. 7.68 gives E0=cB,B0=1

cE,q

m=cqe. Use this to “translate” Eqs. 10.72, 10.73,

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254 CHAPTER 11. RADIATION

and 11.70:

E0=c1

cˆ

⇥E=ˆ

⇥(cB0)=c(ˆ

⇥B0).

B0=1

cE=1

4⇡✏0

(

·u)3(c2v2)u+

⇥(u⇥a)

=1

(q0

m/c)

4⇡✏0

(

·u)3(c2v2)u+

⇥(u⇥a)=µ0q0

4⇡

(

·u)3(c2v2)u+

⇥(u⇥a).

P=µ0a2

6⇡cq2

e=µ0a2

6⇡c1

cq0

m2

=µ0a2

6⇡c3(q0

m)2.

Or, dropping the primes,

B(r,t)=µ0qm

4⇡

(

·u)3(c2v2)u+

⇥(u⇥a).

E(r,t)=c(ˆ

⇥B).

P=µ0q2

ma2

6⇡c3.

Problem 11.30

(a) Wext =F dx =FT

v(t)dt. From Prob. 11.19, v(t)= F

mt+⌧⌧e(tT)/⌧. So

Wext =F2

mT

t dt +⌧T

dt ⌧eT/⌧T

et/⌧dt=F2

mt2

2+⌧t⌧eT/⌧⌧et/⌧

=F2

m1

2T2+⌧T⌧2eT/⌧eT/⌧1=F2

m1

2T2+⌧T⌧2+⌧2eT/⌧.

(b) From Prob. 11.19, the ﬁnal velocity is vf=(F/m)T,soWkin =1

2mv2

f=1

2mF2

m2T2=F2T2

2m.

6⇡c, and (again from Prob. 11.19)

a(t)=





(F/m)1eT/⌧et/⌧,(t0);

(F/m)1e(tT)/⌧,(0 tT).

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CHAPTER 11. RADIATION 255

Wrad =µ0q2

6⇡c

m21eT/⌧20

1

e2t/⌧dt +T

01e(tT)/⌧2dt

=⌧F2

m1eT/⌧2⌧

2e2t/⌧

1

+T

dt 2eT/⌧T

et/⌧dt +e2T/⌧T

e2t/⌧dt

=⌧F2

m⌧

21eT/⌧2+T2eT/⌧⌧et/⌧

0+e2T/⌧⌧

2e2t/⌧

0

=⌧F2

m⌧

212eT/⌧+e2T/⌧+T2⌧eT/⌧eT/⌧1+⌧

2e2T/⌧e2T/⌧1

=⌧F2

m⌧

2⌧eT/⌧+⌧

2e2T/⌧+T2⌧+2⌧eT/⌧+⌧

2⌧

2e2T/⌧=⌧F2

mT⌧+⌧eT/⌧.

Energy conservation requires that the work done by the external force equal the ﬁnal kinetic energy plus

the energy radiated:

Wkin +Wrad =F2T2

2m+⌧F2

mT⌧+⌧eT/⌧=F2

m1

2T2+⌧T⌧2+⌧2eT/⌧=Wext.X

Problem 11.31

(a) a=⌧˙a+k

m(t))✏

✏

a(t)dt =v(✏)v(✏)=⌧✏

✏

dt dt +k

m✏

✏

(t)dt =⌧[a(✏)a(✏)] + k

If the velocity is continuous, so v(✏)=v(✏), then a(✏)a(✏)=k

m⌧.

When t<0,a=⌧˙a)a(t)=Aet/⌧; when t>0,a=⌧˙a)a(t)=Bet/⌧;a=BA=k

m⌧

)B=Ak

m⌧, so the general solution is a(t)=Aet/⌧,(t<0);

[A(k/m⌧)] et/⌧,(t>0).

To eliminate the runaway we’d need A=k/m⌧; to eliminate preacceleration we’d need A= 0. Obviously,

you can’t do both. If you choose to eliminate the runaway, then a(t)=(k/m⌧)et/⌧,(t<0);

0,(t>0).

v(t)=t

1

a(t)dt =k

m⌧t

1

et/⌧dt =k

m⌧⌧et/⌧

1 =k

met/⌧(for t<0);

for t>0,v(t)=v(0) + t

a(t)dt =v(0) = k

m. So v(t)=(k/m)et/⌧,(t<0);

(k/m),(t>0).

For an uncharged particle we would have a(t)= k

m(t),v(t)=t

1

a(t)dt =0,(t<0);

(k/m),(t>0).

The graphs:

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256 CHAPTER 11. RADIATION

(b)

Wext =F dx =Fvdt =k(t)v(t)dt =kv(0) = k2

Wkin =1

2mv2

f=1

2mk

m2

=k2

2m.

Wext =Prad dt =µ0q2

6⇡c[a(t)]2dt =⌧mk

m⌧20

1

e2t/⌧dt =k2

m⌧⌧

2e2t/⌧

1

=k2

m⌧

⌧

2=k2

2m.

Clearly, Wext =Wkin +Wrad.X

Problem 11.32

Our task is to solve the equation a=⌧˙a+U0

m[(x)+(xL)], subject to the boundary conditions

(1) xcontinuous at x= 0 and x=L;

(2) vcontinuous at x= 0 and x=L;

(3) a=±U0/m⌧v(plus at x= 0, minus at x=L).

The third of these follows from integrating the equation of motion:

dv

dt dt =⌧da

dt dt +U0

m[(x)+(xL)] dt,

v=⌧a+U0

m[(x)+(xL)] dt

dx dx =0,

a== U0

m⌧1

v[(x)+(xL)] dx =±U0

m⌧v.

In each of the three regions the force is zero (it acts only at x= 0 and x=L), and the general solution is

a(t)=Aet/⌧;v(t)=A⌧et/⌧+B;x(t)=A⌧2et/⌧+Bt +C.

(I’ll put subscripts on the constants A,B, and C, to distinguish the three regions.)

Region iii (x>L): To avoid the runaway we pick A3= 0; then a(t)=0,v(t)=B3,x(t)=B3t+C3. Let

the ﬁnal velocity be vf(= B3), set the clock so that t= 0 when the particle is at x= 0, and let Tbe the time

it takes to traverse the barrier, so x(T)=L=vfT+C3, and hence C3=LvfT. Then

a(t) = 0; v(t)=vf,x(t)=L+vf(tT),(t<T).

Region ii (0 <x<L): a=A2et/⌧,v=A2⌧et/⌧+B2,x=A2⌧2et/⌧+B2t+C2.

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CHAPTER 11. RADIATION 257

(3) )0A2eT/⌧=U0

m⌧vf)A2=U0

m⌧vf

eT/⌧.

(2) )vf=A2⌧eT/⌧+B2=U0

mvf

+B2)B2=vfU0

mvf

(1) )L=A2⌧2eT/⌧+B2T+C2=U0⌧

mvf

+vfTU0T

mvf

+C2=vfT+U0

mvf

(⌧T)+C2)

C2=LvfT+U0

mvf

(T⌧).

a(t)= U0

m⌧vf

e(tT)/⌧;

v(t)=vf+U0

mvfe(tT)/⌧1;

x(t)=L+vf(tT)+ U0

mvf⌧e(tT)/⌧t+T⌧;

(0 <t<T).

[Note: if the barrier is suﬃciently wide (or high) the particle may turn around before reaching L, but we’re

interested here in the r´egime where it does tunnel through.]

In particular, for t= 0 (when x= 0):

0=LvfT+U0

mvf⌧eT/⌧+T⌧)L=vfTU0

mvf⌧eT/⌧+T⌧.qed

Region i(x<0): a=A1et/⌧,v=A1⌧et/⌧+B1,x=A1⌧2et/⌧+B1t+C1. Let vibe the incident velocity

(at t!1); then B1=vi. Condition (3) says

m⌧vf

eT/⌧A1=U0

m⌧v0

where v0is the speed of the particle as it passes x= 0. From the solution in region (ii) it follows that

v0=vf+U0

mvfeT/⌧1. But we can also express it in terms of the solution in region (i): v0=A1⌧+vi.

Therefore

vi=vf+U0

mvfeT/⌧1A1⌧=vf+U0

mvfeT/⌧1+U0

mv0U0

mvf

eT/⌧

=vfU0

mvf

+U0

mv0

=vfU0

mvf1vf

v0=vfU0

mvf1vf

vf+(U0/mvf)eT/⌧1

=vfU0

mvf11

1+(U0/mv2

f)eT/⌧1.qed

If 1

2mv2

f=1

2U0, then

L=vfTvf⌧eT/⌧+T⌧=vfT⌧eT/⌧T+⌧=⌧vf1eT/⌧;

vi=vfvf11

1+eT/⌧1=vf11+eT/⌧=vfeT/⌧.

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258 CHAPTER 11. RADIATION

Putting these together,

⌧vf

=1eT/⌧)eT/⌧=1L

⌧vf)eT/⌧=1

1(L/⌧vf))vi=vf

1(L/vf⌧).qed

In particular, for L=vf⌧/4, vi=vf

11/4=4

3vf, so KEi

KEf

2mv2

=vi

vf2

=16

KEi=16

9KEf=16

2U0=8

9U0.

Problem 11.33

(a) From Eq. 10.72, E1=(q/2)

4⇡✏0

(

·u)3(c2v2)u+(

·a)u(

·u)a. Here u=cˆ

v,

lˆx +dˆy ,v=vˆx ,a=aˆx , so

·v=lv,

·a=la,

·u=c



·v=c

lv. We want only the x

component. Noting that ux=(c/

)lv=(cl v

, we have:

E1x=q

8⇡✏0

lv)31

(cl v

)(c2v2+la)a(c

lv)

8⇡✏0

lv)3(cl v

)(c2v2)+cl2av

la ac

2+alv

.But

2=l2+d2.

8⇡✏0

lv)3(cl v

)(c2v2)acd2.

Fself =q2

8⇡✏0

lv)3(cl v

)(c2v2)acd2ˆx .(This generalizes Eq.11.90.)

Now x(t)x(tr)=l=vT +1

2aT 2+1

6˙aT 3+···, where T=ttr, and v,a, and ˙aare all evaluated at the

retarded time tr.

(cT )2=

2=l2+d2=d2+(vT +1

2aT 2+1

6˙aT 3)2=d2+v2T2+vaT3+1

3v˙aT 4+1

4a2T4;

c2T2(1 v2/c2)=c2T2/2=d2+vaT3+1

3v˙a+1

4a2T4. Solve for Tas a power series in d:

T=d

c1+Ad +Bd2+···)c2

2

2d2

c21+2Ad +2Bd2+A2d2=d2+va3d3

c3(1+3Ad)+v˙a

3+a2

44

c4d4.

Comparing like powers of d:A=1

2va3

c3;2B+A2=3va3

c3A+v˙a

3+a2

44

c4.

2B=3va3

2va3

c31

4v2a26

c6+v˙a

4

c4+a24

4c4=v˙a

4

c4+6a2

4c41

2v2

c2+3

v2a26

=4

c4v˙a

3+a22

41v2

c2v2

c2+6v2

c2)B=4

2c4v˙a

3+2a2

41+4v2

c2.

T=d

c1+va

3

c3d+4

2c4v˙a

3+2a2

41+4v2

c2d2+()d4+··· (generalizing Eq.11.93).

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CHAPTER 11. RADIATION 259

l=vT +1

2aT 2+1

6˙aT 3+···

=vd

c1+va

3

c3d+4

2c4v˙a

3+2a2

41+4v2

c2d2+1

2a2d2

c21+va3

c3d+1

6˙a3

c3d3

=v

cd+a

4

c21v2

c2+v2

c2d2+v

4

c4v˙a

3+2a2

41+4v2

c2+1

2a2

c2va3

c3+1

6˙a3

c3d3

=v

cd+a4

2c2d2+3

2c3˙a

31+2v2

c2+v4a2

c21

4+v2

c2+1v2

c2d3

=v

cd+a4

2c2d2+5

2c3˙a

3+5

v2a2

c2d3+()d4+···

=cT =d1+va

3

c3d+4

2c4v˙a

3+2a21

4+v2

c2d2+()d4+···

lv =cd+va4

2c2d2+5

2c3v˙a

3+2a21

4+v2

c2d3v2

cdav4

2c2d25v

2c3˙a

3+5

v2a2

c2d3+···

=cd1v2

c2+5

2c3v˙a

3+2a21

4+v2

c2v˙a

35

v22a2

c2d3+···

d+5a2

8c3d3+()d4+···

cl v

=vd+a4

2cd2+5

2c2˙a

3+5

v2a2

c2d3vdv2a

4

c3d2v5

2c4v˙a

3+2a21

4+v2

c2d3

=a4

2c1v2

c2d2+5

2c2˙a

3+5

v2a2

c2v2

˙a

3v2a2

c21

4+v2

c2d3+()d4+···

=a2

2cd2+5

2c2˙a

32+v2a2

c25

41

4v2

c2d3+()d4+···

=a2

2cd2+3

2c2˙a

3+v2a2

c2d3+()d4+···

lv)3=cd

1+6a2

8c4d23

=

cd3136a2

8c4d2+···

Fself =q2

8⇡✏0

cd3136a2

8c4d2a2

2cd2+3

2c2˙a

3+v2a2

c2d3c2

2acd2ˆx

=q2

8⇡✏0

3

c3d13

6a2

c4d2ac

2+

2˙a

3+v2a2

c2dˆx

=q2

8⇡✏0

3

c3d

2ac +˙a

3+v2a2

c2d+()d2+···ˆx

=q2

4⇡✏03a

4c2d+4

4c3˙a

3+v2a2

c2+()d+···ˆx (generalizing Eq.11.95).

Switching to t:v(tr)=v(t)+ ˙v(t)(trt)+···=v(t)a(t)T=v(t)ad/c. (When multiplied by d, it

doesn’t matter—to this order—whether we evaluate at tor at tr.)

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260 CHAPTER 11. RADIATION

1v(tr)

c2

=1[v(t)22vad/c]

c2=1v(t)2

c21+2av3d

c3,so

=1v(tr)

c21/2

=(t)1va3

c3d;a(tr)=a(t)T˙a=a(t)˙a

cd.

Evaluating everything now at time t:

Fself =q2

4⇡✏0313va3d/c3(a˙ad/c)

4c2d+4

4c3˙a

3+v2a2

c2+()d2+···ˆx

=q2

4⇡✏03a

4c2d+3

4c2˙a

c+3va22

c3+4

4c3˙a

3+v2a2

c2+()d+···ˆx

=q2

4⇡✏03a

4c2d+4

4c3˙a+˙a

3+3va22

c2+v2a2

c2+()d+···ˆx

=q2

4⇡✏03a

4c2d+4

3c3˙a+3va22

c2+()d+···ˆx (generalizing Eq.11.96).

The ﬁrst term is the electromagnetic mass; the radiation reaction itself is the second term:

Fint

rad =µ0q2

12⇡c4˙a+3va22

c2(generalizing Eq. 11.99), so the generalization of Eq. 11.100 is

Frad =µ0q2

6⇡c4˙a+3va22

c2.

(b) Frad =A4˙a+32a2v

c2, where A⌘µ0q2

6⇡c.P=Aa26(Eq. 11.75). What we must show is that

t2

Fradvdt=t2

P dt, or t2

4˙av +3v2a22

c2dt =t2

a26dt

(except for boundary terms—see Sect. 11.2.2).

Rewrite the ﬁrst term: t2

4˙av dt =t2

(4v)da

dt dt =4va

t1t2

dt(4v)a dt.

Now d

dt(4v)=43d

dt v+4a;d

dt =d

dt 1

1v2/c2=1

(1 v2/c2)3/22va

c2=va3

c2.So

dt(4v)=43vva3

c2+4a=6a1v2

c2+4v2

c2=6a1+3v2

c2.

t2

4˙av dt =4va

t1t2

6a21+3v2

c2dt, and hence

t2

4˙av +32a2v2

c2dt =4va

+t2

t16a21+3v2

c2+36a2v2

c2dt =4va

t1t2

6a2dt. qed

Problem 11.34

(a) P=µ0q2a26

6⇡c(Eq.11.75).w=b2+c2t2(Eq.10.52); v=˙w=c2t

pb2+c2t2;

a=˙v=c2

pb2+c2t2c2t(c2t)

(b2+c2t2)3/2=c2

(b2+c2t2)3/2b2+c2t2c2t2=b2c2

(b2+c2t2)3/2;

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CHAPTER 11. RADIATION 261

2=1

1v2/c2=1

1[c2t2/(b2+c2t2)] =b2+c2t2

b2+c2t2c2t2=1

b2b2+c2t2.So

P=µ0q2

6⇡c

b4c4

(b2+c2t2)3

b6=q2c

6⇡✏0b2.Yes, it radiates (in fact, at a constant rate).

(b) Frad =µ0q24

6⇡c˙a+32a2v

c2;˙a=3

b2c2(2c2t)

(b2+c2t2)5/2=3b2c4t

(b2+c2t2)5/2;˙a+32a2v

c2=

3b2c4t

(b2+c2t2)5/2+3

(b2+c2t2)

b4c4

(b2+c2t2)3

c2t

pb2+c2t2=0. Frad =0.No, the radiation reaction is zero.

Problem 11.35

The ﬁelds of a nonstatic ideal dipole (Problem 10.34) contain terms that go like 1/r,1/r2, and 1/r3, for

ﬁxed t0⌘tr/c; radiation comes from the ﬁrst of these:

Er=µ0

4⇡r[¨

pˆr (ˆr ·¨

p)],Br=µ0

4⇡rc(ˆr ⇥¨

where the dipole moments are evaluated at time t0. The 1/r2term in the Poynting vector is

Sr=1

µ0

(Er⇥Br)= µ0

16⇡2cr2[¨

pˆr (ˆr ·¨

p)] ⇥(ˆr ⇥¨

p)= µ0

16⇡2cr2{¨

p⇥(ˆr ⇥¨

p)(ˆr ·¨

p)[ˆr ⇥(ˆr ⇥¨

p)]}

=µ0

16⇡2cr2ˆr ¨p2¨

p(ˆr ·¨

p)(ˆr ·¨

p)[ˆr (ˆr ·¨

p)¨

p]=µ0

16⇡2cr2¨p2(ˆr ·¨

p)2ˆr .

Setting the zaxis along ¨

p,¨p2(ˆr ·¨

p)2=¨p2sin2✓, and the power radiated is

P=Sr·da=µ0

16⇡2c¨p21

r2sin2✓r2sin ✓d✓d=µ0

8⇡c¨p2⇡

sin3✓d✓=µ0

6⇡c¨p2.

For the sinusoidal case, p=p0cos !t, we have ¨

p=!2p0cos !t,¨p2=!4p2

0cos2!t, and its time average

is (1/2)p2

0!4, so

hPi=µ0

12⇡c¨p2

0!4,

in agreement with Eq. 11.22. And in the case of quadratic time dependence our result agrees with the answer

to Problem 11.26.

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262 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Chapter 12

Electrodynamics and Relativity

Problem 12.1

Let ube the velocity of a particle in S,¯

uits velocity in ¯

S, and vthe velocity of ¯

Swith respect to S.

Galileo’s velocity addition rule says that u=¯

u+v. For a free particle, uis constant (that’s Newton’s ﬁrst

law in S.

(a) If vis constant, then ¯

u=uvis also constant, so Newton’s ﬁrst law holds in ¯

S, and hence ¯

Sis inertial.

(b) If ¯

Sis inertial, then ¯

uis also constant, so v=u¯

uis constant.

Problem 12.2

(a) mAuA+mBuB=mCuC+mDuD;ui=¯

ui+v.

mA(¯

uA+v)+mB(¯

uB+v)=mC(¯

uC+v)+mD(¯

uD+v),

mA¯

uA+mB¯

uB+(mA+mB)v=mC¯

uC+mD¯

uD+(mC+mD)v.

Assuming mass is conserved, (mA+mB)=(mC+mD), it follows that

mA¯

uA+mB¯

uB=mC¯

uC+mD¯

uD, so momentum is conserved in ¯

(b)1

2mAu2

A+1

2mBu2

B=1

2mCu2

C+1

2mDu2

2mA(¯u2

A+2

uA·v+v2)+1

2mB(¯u2

B+2

uB·v+v2)=1

2mC(¯u2

C+2

uC·v+v2)+1

2mD(¯u2

D+2

uD·v+v2)

2mA¯u2

A+1

2mB¯u2

B+v·(mA¯

uA+mB¯

uB)+1

2v2(mA+mB)

2mC¯u2

C+1

2mD¯u2

D+v·(mC¯

uC+mD¯

uD)+1

2v2(mC+mD).

But the middle terms are equal by conservation of momentum, and the last terms are equal by conservation

of mass, so 1

2mA¯u2

A+1

2mB¯u2

B=1

2mC¯u2

C+1

2mD¯u2

D.qed

Problem 12.3

(a) vG=vAB +vBC;vE=vAB +vBC

1+vAB vBC/c2⇡vG1vAB vBC

c2;)vGvE

vG=vAB vBC

c2.

In mi/h, c= (186,000 mi/s) ⇥(3600 sec/hr) = 6.7⇥108mi/hr.

)vGvE

vG=(5)(60)

(6.7⇥108)2=6.7⇥1016.)6.7⇥1014% error (pretty small!)

(b) 1

2c+3

4c/1+1

2·3

4=5

4c/11

8=10

11c(still less than c)

1+12, or:

2=2

1+212+212+2

(1 + 212+2

12

2)=1+212+2

12

(1 + 212+2

12

2)(1 + 2

12

22

12

(1 + 212+2

12

2)=1(1 2

1)(1 2

(1 + 12)2=1,

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 263

where ⌘(1 2

1)(1 2

2)/(1 + 12)2is clearly a positive number. So 2<1, and hence |vAC |<c. qed.

Problem 12.4

(a) Velocity of bullet relative to ground is 1

2c+1

3c=5

6c=10

12 c.

Velocity of getaway car is 3

4c=9

12 c. Since vb>v

g, bullet does reach target .

(b) Velocity of bullet relative to ground is 1

2c+1

1+ 1

2·1

7c=20

28 c.

Velocity of getaway car is 3

4c=21

28 c. Since vg>v

b, bullet does not reach target .

Problem 12.5

(a) Light from 90th clock took 90⇥109m

3⇥108m= 300 sec = 5 min to reach me, so the time I see on the clock is

11:55 am .

(b) I observe 12 noon .

Problem 12.6

light signal leaves aat time t0

a; arrives at earth at time ta=t0

a+da

light signal leaves bat time t0

b; arrives at earth at time tb=t0

b+db

)t=tbta=t0

bt0

a+(dbda)

c=t0+(vt0cos ✓)

c=t01v

ccos ✓

(Here dais the distance from ato earth, and dbis the distance from bto earth.)

s=vt0sin ✓=vsin ✓t

(1t(1 v

ccos ✓))u=vsin ✓

(1 v

ccos ✓)is the the apparent velocity.

d✓=v(1 v

ccos ✓)(cos ✓)sin ✓(v

csin ✓)

(1 v

ccos ✓)2=0)(1 v

ccos ✓) cos ✓=v

csin2✓

)cos ✓=v

c(sin2✓+ cos2✓)=v

✓max = cos1(v/c) At this maximal angle, u=vp1v2/c2

1v2/c2=v

p1v2/c2

As v!c,u!1, because the denominator !0 — even though v<c.

Problem 12.7

The student has not taken into account time dilation of the muon’s “internal clock”. In the laboratory, the

muon lasts ⌧ =⌧

p1v2/c2, where ⌧is the “proper” lifetime, 2 ⇥106sec. Thus

v=d

t/1v2/c2=d

⌧1v2

c2,where d= 800 meters.

⌧

d2v2=1v2

c2;v2⌧

d2+1

c2= 1; v2=1

(⌧/d)2+ (1/c)2.

c2=1

1+(⌧c/d)2;⌧c

d=(2 ⇥106)(3 ⇥108)

800 =6

8=3

4;v2

c2=1

1+9/16 =16

25;v=4

5c.

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264 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Problem 12.8

(a) Rocket clock runs slow; so earth clock reads t=1

p1v2/c2·1 hr. Here =1

p1v2/c2=1

p19/25 =5

)According to earth clocks signal was sent 1 hr, 15 min after take-o↵.

(b) By earth observer, rocket is now a distance 3

5c5

4(1 hr) = 3

4chr (three-quarters of a light hour) away.

Light signal will therefore take 3

4hr to return to earth. Since it left 1 hr and 15 min after departure, light

signal reaches earth 2 hrs after takeo↵

4·(2 hrs) = 2.5 hrs

Problem 12.9

Lc=2Lv;Lc

c=Lv

v;so 2

c=1

v=11

22=3

4;1

2

c=1v2

c2=v2

c2=3

16 .v2

c2=13

16 =13

16 ;v=p13

Problem 12.10

Say length of mast (at rest) is ¯

l. To an observer on the boat, height of mast is ¯

lsin ¯

✓, horizontal projection

is ¯

lcos ¯

✓. To observer on dock, the former is una↵ected, but the latter is Lorentz contracted to 1

¯

lcos ¯

✓.

Therefore:

tan ✓=¯

lsin ¯

✓

¯

lcos ¯

✓=tan ¯

✓,or tan ✓=tan ¯

✓

1v2/c2

Problem 12.11

Naively, circumference/diameter = 1

(2⇡R)/(2R)=⇡/=⇡1(!R/c)2— but this is nonsense. Point

is: an accelerating object cannot remain rigid, in relativity. To decide what actually happens here, you need a

speciﬁc model for the internal forces holding the disc together.

Problem 12.12

(iv) )t=¯

+vx

c2. Put this into (i), and solve for x:

¯x=xxv¯

+vx

c2=x1v2

c2v¯

t=x1

2v¯

t=x

v¯

t;x=(¯x+v¯

t)X

Similarly, (i) )x=¯x

+vt. Put this into (iv) and solve for t:

t=tv

c2¯x

+vt=t1v2

c2v

c2¯xt

v

c2¯x;t=¯

t+v

c2¯xX

Problem 12.13

Let brother’s accident occur at origin, time zero, in both frames. In system S(Sophie’s), the coordinates

of Sophie’s cry are x=5⇥105m, t= 0. In system ¯

S(scientist’s), ¯

t=(tv

c2x)=vx/c2. Since

this is negative, Sophie’s cry occurred before the accident, in ¯

S.=1

p1(12/13)2=13

p169144 =13

5. So

t=13

512

13 c(5 ⇥105)/c2=12 ⇥105/3⇥108=10

3. 4 ⇥103seconds earlier

Problem 12.14

(a) In Sit moves a distance dy in time dt. In ¯

S, meanwhile, it moves a distance d¯y=dy in time d¯

(dt v

c2dx).

)d¯y

d¯

t=dy

(dt v

c2dx)=(dy/dt)

1v

dt ; or ¯uy=uy

1vux

c2;¯uz=uz

1vux

c2

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 265

(b) S= dock frame; S0= boat frame; we need reverse transformations (v!v):

tan ✓=uy

=¯uy/1+v¯ux

c2

(¯ux+v)/1+ v¯ux

c2=1



¯uy

(¯ux+v). In this case ¯ux=ccos ¯

✓;¯uy=csin ¯

✓, so

tan ✓=1



csin ¯

✓

(ccos ¯

✓+v)=1

sin ¯

✓

cos ¯

✓v/c

[Contrast tan ✓=sin ¯

✓

cos ¯

✓in Prob. 12.10. The point is that velocities are sensitive not only to the transfor-

mation of distances, but also of times. That’s why there is no universal rule for translating angles — you have

to know whether it’s an angle made by a velocity vector or a position vector.]

That’s how the velocity vector of an individual photon transforms. But the beam as a whole is a snapshot

of many di↵erent photons at one instant of time, and it transforms the same way the mast does.

Problem 12.15

Bullet relative to ground: 5

7c. Outlaws relative to police:

4c1

13

4·1

5c.

Bullet relative to outlaws:

7c3

15

7·3

=1

28 c

=1

13c. [Velocity of Arelative to Bis minus the velocity of

Brelative to A, so all entries below the diagonal are trivial. Note that in every case vbullet <v

outlaws, so no

matter how you look at it, the bad guys get away.]

speed of !

relative to #

Ground Police Outlaws Bullet Do they escape?

Ground 0 1

2c3

4c5

7cYes

Police 1

202

5c1

3cYes

Outlaws 3

4c2

5c01

13 cYes

Bullet 5

7c1

3c1

13 c0 Yes

Problem 12.16

(a) Moving clock runs slow, by a factor =1

p1(4/5)2=5

3. Since 18 years elapsed on the moving clock,

3⇥18 = 30 years elapsed on the stationary clock. 51 years old

(b) By earth clock, it took 15 years to get there, at 4

5c,sod=4

5c⇥15 years = 12cyears (12 light years)

(d) ¯

t= 9 yrs, ¯x= 0. [She got on at the origin in ¯

S, and rode along on ¯

S, so she’s still at the origin. If you

doubt these values, use the Lorentz Transformations, with xand tin (c).]

(e) Lorentz Transformations: ˜x=(x+vt)

t=(t+v

c2x)[note that vis negative, since ˜

Sus going to the left]

)˜x=5

3(12cyrs + 4

5c·15 yrs) = 5

3·24cyrs = 40cyears.

t=5

3(15 yrs + 4

c2·12cyrs) = 5

315 + 48

5yrs = (25 + 16)yrs = 41 years.

(f) Set her clock ahead 32 years, from 9 to 41 (¯

t!˜

t). Return trip takes 9 years (moving time), so her clock

will now read 50 years at her arrival. Note that this is 5

3·30 years—precisely what she would calculate if the

stay-at-home had been the traveler, for 30 years of his own time.

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266 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

(g) (i) ¯

t= 9 yrs, x= 0. What is t?t=v

c2x+¯

=3

5·9 yrs = 27

5=5.4 years, and he started at age 21, so he’s

26.4 years old (Younger than traveler (!) because to the traveller it’s the stay-at-home who’s moving.)

(ii) ˜

t= 41 yrs, x= 0. What is t? t=˜

=3

5·41 yrs, or 123

5yrs, or 24.6 yrs, and he started at 21, so he’s

45.6 years old.

(h) It will take another 5.4 years of earth time for the return, so when she gets back, she will say her twin’s

age is 45.6 + 5.4 = 51 years—which is what we found in (a). But note that to make it work from traveler’s

point of view you must take into account the jump in perceived age of the stay-at-home when she changes

coordinates from ¯

Sto ˜

S.)

Problem 12.17

¯a0¯

b0+¯a1¯

b1+¯a2¯

b2+¯a3¯

b3=2(a0a1)(b0b1)+2(a1a0)(b1a0)+a2b2+a3b3

=2(a0b0a0b1

a1b0

+2a1b1a1b1+a1b0

+a0b1

2a0b0)+a2b2+a3b3

=2a0b0(1 2)+2a1b1(1 2)+a2b2+a3b3

=a0b0+a1b1+a2b2+a3b3.qed [Note: 2(1 2) = 1.]

Problem 12.18

(a) 





c¯

¯x

¯y

¯z





=





1000

100

0010

0001















(using the notation of Eq. 12.24, for best comparison)

(b) ⇤=





0 0

0100

 00

0001











¯0¯¯

0

0100

¯¯

0¯0

0001











 00

  00

0010

0001





=





¯¯ ¯¯

0

  00

¯ ¯

¯ ¯

¯0

0001







Yes, the order does matter. In the other order “bars” and “no-bars” would be switched, and this would yield

adi↵erent matrix.

Problem 12.19

(a) Since tanh ✓=sinh ✓

cosh ✓, and cosh2✓sinh2✓= 1, we have:

=1

1v2/c2=1

1tanh2✓=cosh ✓

cosh2✓sinh2✓= cosh ✓; = cosh ✓tanh ✓= sinh ✓.

)⇤=





cosh ✓sinh ✓00

sinh ✓cosh ✓00

0010

0001





Compare: R=



cos sin 0

sin cos 0

001





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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 267

(b) ¯u=uv

1uv

c2)¯u

c=u

cv

c

1u

cv

c)tanh ¯

=tanh tanh ✓

1tanh tanh ✓, where tanh =u/c, tanh ✓=v/c;

tanh ¯

=¯u/c. But a “trig” formula for hyperbolic functions (CRC Handbook, 18th Ed., p. 204) says:

tanh tanh ✓

1tanh tanh ✓= tanh(✓).)tanh ¯

= tanh(✓),or: ¯

=✓

Problem 12.20

(a) (i) I=c2t2+x2+y2+z2=(5 15)2+ (10 5)2+ (8 3)2+ (0 0)2=100 + 25 + 25 = 50

(ii) No. (In such a system ¯

t= 0, so Iwould have to be positive, which it isn’t.)

(iii) Yes.

✲x

✻

246810

❂

!"# $

✻

✲x

world line

of player 1 ✮

world line of

player 2

✲✛

10 ft

✠

world line of

the ball

✒

■

✒

■

✒

✲x

✻

✶

✐

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reproduced, in any form or by any means, without permission inwritingfromthepublisher.

Stravels in the direction from Btoward A,

making the trip in time 10/c.

)v=5ˆx 5ˆy

10/c =c

2ˆx c

2ˆy

Note that v2

c2=1

4+1

4=1

2, so v=1

p2c, safely

less than c.

(b) (i) I=(3 1)2+ (5 2)2+0+0=4+9= 5

(ii) Yes. By Lorentz Transformation: (c¯

t)=(ct)(x). We want ¯

t= 0, so (ct)=(x); or

c=(ct)

(x)=(3 1)

(5 2) =2

3. So v=2

3cin the +xdirection.

(iii) No. (In such a system x=y=z= 0 so Iwould be negative, which it isn’t.)

Problem 12.21

Using Eq. 12.18 (iv): ¯

t=(tv

c2x)=0)t=v

c2x, or v=t

xc2=tBtA

xBxA

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268 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Problem 12.22

(a)

✲x

✻

246810

❂

!"# $

✻

✲x

world line

of player 1 ✮

world line of

player 2

✲✛

10 ft

✠

world line of

the ball

✒

■

✒

■

✒

✲x

✻

✶

✐

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reproduced, in any form or by any means, without permission in writing from the publisher.

Truth is, you never do communicate with

the other person right now—you communicate

with the person he/she will be when the mes-

sage gets there; and the response comes back

to an older and wiser you.

(b) No way It is true that a moving observer

might say she arrived at Bbefore she left A,

but for the round trip everyone must agree

that she arrives back after she set out.

Problem 12.24

(a) 1u2

c2⌘2=u2;u21+⌘2

c2=⌘2;u=1

1+⌘2/c2⌘.

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 269

(b) 1

p1u2/c2=1

p1tanh2✓=cosh ✓

pcosh2✓sinh2✓= cosh ✓)⌘=1

p1u2/c2u= cosh ✓ctanh ✓=csinh ✓.

Problem 12.25

(a) ux=uy=ucos 45=1

p5c=2

5c.

(b) 1

p1u2/c2=1

p14/5=p5

p54=p5. )⌘=u

p1u2/c2)⌘x=⌘y=p2

(d) Eq. 12.45)









¯ux=uxv

1uxv

c2=p2

5cp2

12

= 0.

¯uy=1

uy

1uxv

c2=12

12

=p2/5

p3/5c=2

3c.

(e) ¯⌘x=(⌘x⌘0)=12

5(p2c2

5p5c)= 0.¯⌘y=⌘y=p2c.

(f) 1

p1¯u2/c2=1

p12/3=p3; )¯⌘=p3¯u )¯⌘x=p3¯ux=0X

¯⌘y=p3¯uy=p2cX

Problem 12.26

⌘µ⌘µ=(⌘0)2+⌘2=1

(1u2/c2)(c2+u2)=c2(1u2/c2)

(1u2/c2)=c2.Timelike.

Problem 12.27

Use the result of Problem 12.24(a): u=1

1+⌘2/c2⌘.Here ⌘

c=4

3,so 1

1 + 16/9=3

5, and hence

u=3

5(4 ⇥108)=2.4⇥108m/s. Innocent.

Problem 12.28

(a) From Prob. 11.34 we have =1

bpb2+c2t2.)⌧=1

dt =bdt

pb2+c2t2=b

cln(ct +pb2+c2t2)+k; at

t= 0 we want ⌧=0: 0= b

cln b+k,sok=b

cln b;⌧=b

cln 1

b(ct +b2+c2t2)

(b) px2b2+x=bec⌧/b;px2b2=bec⌧/b x;x2b2=b2e2c⌧/b 2xbec⌧/b +x2;2xbec⌧/b =b2(1 + e2c⌧/b);

x=bec⌧/b+ec⌧/b

2=bcosh(c⌧/b) . Also from Prob. 11.34: v=c2t/pb2+c2t2.

v=c

xpx2b2=c

bcosh(c⌧/b)b2cosh2(c⌧/b)b2=cpcosh2(c⌧/b)1

cosh(c⌧/b)=csinh(c⌧/b)

cosh(c⌧/b)=ctanh c⌧

b.

b= cosh c⌧

b,so⌘µ= cosh c⌧

bc, c tanh c⌧

b,0,0=ccosh c⌧

b,sinh c⌧

b,0,0.

Problem 12.29

(a) mAuA+mBuB=mCuC+mDuD;ui=¯ui+v

1+(¯uiv/c2).

¯uA+v

1+(¯uAv/c2)+mB

¯uB+v

1+(¯uBv/c2)=mC

¯uC+v

1+(¯uCv/c2)+mD

¯uD+v

1+(¯uDv/c2).

This time, because the denominators are all di↵erent, we cannot conclude that

mA¯uA+mB¯uB=mC¯uC+mD¯uD.

As an explicit counterexample, suppose all the masses are equal, and uA=uB=v, uC=uD=0.This is

a symmetric “completely inelastic” collision in S, and momentum is clearly conserved (0=0). But the Einstein

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270 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

velocity addition rule gives ¯uA=0,¯uB=2v/(1 + v2/c2),¯uC=¯uD=v, so in ¯

Sthe (incorrectly deﬁned)

momentum is not conserved:

m2v

1+v2/c26=2mv.

(b) mA⌘A+mB⌘B=mC⌘C+mD⌘D;⌘i=(¯⌘i+¯⌘0

i).(The inverse Lorentz transformation.)

mA(¯⌘A+¯⌘0

A)+mB(¯⌘B+¯⌘0

B)=mC(¯⌘C+¯⌘0

C)+mD(¯⌘D+¯⌘0

D).The gamma’s cancel:

mA¯⌘A+mB¯⌘B+(mA¯⌘0

A+mB¯⌘0

B)=mC¯⌘C+mD¯⌘D+(mC¯⌘0

C+mD¯⌘0

D).

But mi⌘0

i=p0

i=Ei/c, so if energy is conserved in ¯

S(¯

EA+¯

EB=¯

EC+¯

ED), then so too is the momentum

(correctly deﬁned):

mA¯⌘A+mB¯⌘B=mC¯⌘C+mD¯⌘D.qed

Problem 12.30

mc2mc2=nmc2)=n+1= 1

1pu2/c2)1u2

c2=1

(n+1)2

)u2

c2=11

(n+1)2=n2+2n+11

(n+1)2=n(n+2)

(n+1)2;u=n(n+ 2)

n+1 c

Problem 12.31

ET=E1+E2+···;pT=p1+p2+···;¯pT=(pTET/c)=0)=v/c =pTc/ET

v=c2pT/ET=c2(p1+p2+···)/(E1+E2+···)

Problem 12.32

Eµ=(m2

⇡+m2

µ)

2m⇡

c2=mµc2)=(m2

⇡+m2

µ)

2m⇡mµ

1v2/c2;1v2

c2=1

2;

c2=11

2=14m2

⇡m2

(m2

⇡+m2

µ)2=m4

⇡+2m2

⇡m2

µ+m4

µ4m2

⇡m2

(m2

⇡+m2

µ)2=(m2

⇡m2

µ)2

(m2

⇡+m2

µ)2;v=(m2

⇡m2

µ)

(m2

⇡+m2

µ)c.

Problem 12.33

Initial momentum: E2p2c2=m2c4)p2c2= (2mc2)2m2c4=3m2c4)p=p3mc.

Initial energy: 2mc2+mc2=3mc2.

Each is conserved, so ﬁnal energy is 3mc2, ﬁnal momentum is p3mc.

E2p2c2=(3mc2)2(p3mc)2c2=6m2c4=M2c4.)M=p6m⇡2.5m

(In this process some kinetic energy was converted into rest energy, so M>2m.)

v=pc2

E=p3mc c2

3mc2=c

p3=v.

Problem 12.34

First calculate pion’s energy: E2=p2c2+m2c4=9

16 m2c4+m2c4=25

16 m2c4)E=5

4mc2.

Conservation of energy: 5

4mc2=EA+EB

Conservation of momentum: 3

4mc =pA+pB=EA

cEB

c)3

4mc2=EAEB2EA=2mc2

)EA=mc2;EB=1

4mc2.

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 271

Problem 12.35

Classically, E=1

2mv2. In a colliding beam experiment, the relative velocity (classically) is twice the

velocity of either one, so the relative energy is 4E.

✲x

✻

¯x=−3

¯x=−2

¯x=−1

x=0

x=1

¯x=2

x=3

c¯

t=−3

c¯

t=−2

c¯

t=−1

c¯

t=0

c¯

t=1

c¯

t=2

c¯

t=3

✒

✛ ✲

8.7

✻

❄

9.2

✲S

✻

1⃝✲2⃝

✛

✲¯

✻

1⃝2⃝

✛

=⇒

✲

(before)

✒

⑦

60◦

(after)

✲ct

✻

A✒

✒

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Let ¯

Sbe the system in which 1is at rest. Its

speed v, relative to S, is just the speed of 1

in S.

¯p0=(p0p1))¯

c=E

cp, where pis the momentum of 2in S.

E=Mc2,so=E

Mc2;p=MV =Mc.)¯

E=E

c+Mcc=(E+Mc22)

2=1

12)12=1

2)2=11

2=21

2.)¯

E=E

Mc2E+E

Mc221Mc2

E=E2

Mc2+E2

Mc2Mc2;¯

E=2E2

Mc2Mc2.

For E= 30 GeV and Mc2= 1 GeV, we have ¯

E=(2)(900)

11 = 1800 1= 1799 GeV = 60E.

Problem 12.36

✲x

✻

¯x=−3

¯x=−2

¯x=−1

x=0

x=1

¯x=2

x=3

c¯

t=−3

c¯

t=−2

c¯

t=−1

c¯

t=0

c¯

t=1

c¯

t=2

c¯

t=3

✒

✛ ✲

8.7

✻

❄

9.2

✲S

✻

1⃝✲2⃝

✛

✲¯

✻

1⃝2⃝

✛

=⇒

✲

(before)

✒

⑦

60◦

(after)

✲ct

✻

A✒

✒

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One photon is impossible, because in the “center of mo-

mentum” frame (Prob. 12.31) we’d be left with a photon

at rest, whereas photons have to travel at speed c.







Cons. of energy: p0c2+m2c4+mc2=EA+EB

Cons. of mom.: horizontal: p0=EA

ccos 60+EB

ccos ✓)EBcos ✓=p0c1

2EA

vertical: 0 = EA

csin 60EB

csin ✓)EBsin ✓=p3

2EAsquare and add:

B(cos2✓+ sin2✓)=p0c2p0cEA+1

4E2

A+3

4E2

)E2

B=p0c2p0cEA+E2

A=p2

0c2+m2c4+mc2EA2

=p0c2+m2c4+2

p2

0c2+m2c4(mc2EA)+m2c42EAmc2+E2

A.Or:

p0cEA=2m2c4+2mc2p2

0c2+m2c42EAp2

0c2+m2c42EAmc2;

)EA(mc2+p2

0c2+m2c4p0c/2) = m2c4+mc2p0c2+m2c4;

EA=mc2(mc2+p2

0c2+m2c4)

(mc2+p2

0c2+m2c4p0c/2) ·(mc2p2

0c2+m2c4p0c/2)

(mc2p2

0c2+m2c4p0c/2)

=mc2(m2c4

p2

0c2m2c4

1

2p0mc3p0c

2p2

0c2+m2c4)

(m2c4

p0mc3+p0c2

4p0c2m2c4

)

=mc2

(mc +2p0+p2

0+m2c2)

(mc +3

4p0).

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272 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Problem 12.37

F=dp

dt =d

1u2/c2=mdu

1u2/c2+u1

21

c22u·du

(1 u2/c2)3/2

1u2/c2a+u(u·a)

(c2u2).qed

Problem 12.38

At constant force you go in “hyperbolic” mo-

tion. Photon A, which left the origin at t<0,

catches up with you, but photon B, which

passes the origin at t>0, never does.

✲x

✻

¯x=−3

¯x=−2

¯x=−1

x=0

x=1

¯x=2

x=3

c¯

t=−3

c¯

t=−2

c¯

t=−1

c¯

t=0

c¯

t=1

c¯

t=2

c¯

t=3

✒

✛ ✲

8.7

✻

❄

9.2

✲S

✻

1⃝✲2⃝

✛

✲¯

✻

1⃝2⃝

✛

=⇒

✲

(before)

✒

⑦

60◦

(after)

✲ct

✻

A✒

✒

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Problem 12.39

(a) ↵0=d⌘0

d⌧=d⌘0

d⌧=d

dt c

1u2/c2 1

1u2/c2

1u2/c21

21

c22u·a

(1 u2/c2)3/2=1

u·a

(1 u2/c2)2.

↵↵↵=d⌘

d⌧=dt

d⌧

d⌘

dt =1

1u2/c2

dt u

1u2/c2=1

1u2/c2a

1u2/c2+u(t)1

c22u·a

(1 u2/c2)3/2

(1 u2/c2)a+u(u·a)

(c2u2).

(b) ↵µ↵µ=(↵0)2+↵↵↵·↵↵↵=1

(u·a)2

(1 u2/c2)4+1

(1 u2/c2)4a(1 u2

c2)+ 1

c2u(u·a)2

(1 u2/c2)41

c2(u·a)2+a21u2

c22

c21u2

c2(u·a)2+1

c4u2(u·a)2

(1 u2/c2)4a21u2

c22

+(u·a)2

c21+22u2

c2+u2

  

(1 u2

c2)

(1 u2/c2)2a2+(u·a)2

(c2u2).

d⌧(⌘µ⌘µ)=↵µ⌘µ+⌘µ↵µ=2↵µ⌘µ= 0, so ↵µ⌘µ=0.

(d) Kµ=d⇢µ

d⌧=d

d⌧(m⌘µ)= m↵µ. Kµ⌘µ=m↵µ⌘µ=0.

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 273

Problem 12.40

KµKµ=(K0)2+K·K. From Eq. 12.69, K·K=F2

(1u2/c2). From Eq. 12.70:

K0=1

d⌧=1

c1u2/c2

dt mc2

1u2/c2=mc

1u2/c21

(1/c2)

(1 u2/c2)3/22u·a=m

(u·a)

(1 u2/c2)2

But Eq. 12.74: u·F=uF cos ✓=m

1u2/c2(u·a)+ u2(u·a)

c2(1 u2/c2)=m(u·a)

(1 u2/c2)3/2, so:

K0=uF cos ✓

c1u2/c2.)KµKµ=F2

(1 u2/c2)u2F2cos2✓

c2(1 u2/c2)=1(u2/c2) cos2✓

(1 u2/c2)F2.qed

Problem 12.41

F=m

1u2/c2a+u(u·a)

c2u2=q(E+u⇥B))a+u(u·a)

(c2u2)=q

m1u2/c2(E+u⇥B).

Dot in u:(u·a)+ u2(u·a)

c2(1 u2/c2)=u·a

(1 u2/c2)=q

m1u2/c2u·E+u·(u⇥B)

  

=0 ;

)u(u·a)

(c2u2)=q

m1u2

u(u·E)

c2.So a=q

m1u2

c2E+u⇥B1

c2u(u·E).qed

Problem 12.42

One way to see it is to look back at the general formula for E(Eq. 10.36). For a uniform inﬁnite plane of

charge, moving at constant velocity in the plane, ˙

J= 0 and ˙⇢= 0, while ⇢(or rather, ) is independent of t

(so retardation does nothing). Therefore the ﬁeld is exactly the same as it would be for a plane at rest (except

that itself is altered by Lorentz contraction).

A more elegant argument exploits the fact that Eis a vector (whereas Bis a pseudovector). This means that

any given component changes sign if the conﬁguration is reﬂected in a plane perpendicular to that direction.

But in Fig. 12.35(b), if we reﬂect in the xy plane the conﬁguration is unaltered, so the zcomponent of Ewould

have to stay the same. Therefore it must in fact be zero. (By contrast, if you reﬂect in a plane perpendicular

to the ydirection the charges trade places, so it is perfectly appropriate that the ycomponent of Eshould

reverse its sign.)

Problem 12.43

(a) Field is 0/✏0, and it points perpendicular to the positive plate, so:

E0=0

✏0

(cos 45ˆx + sin 45ˆy )= 0

p2✏0

(ˆx +ˆy ).

(b) From Eq. 12.109, Ex=Ex0=0

p2✏0;Ey=Ey0=0

p2E0. So E=0

p2✏0

(ˆx +ˆy ).

(d) Let ˆn be a unit vector perpendicular to the plates in S— evidently

ˆn =sin ✓ˆx + cos ✓ˆy ;|E|=0

p2✏01+2.

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274 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

✲x

✻

❨

ˆn

✲x

✻

✲

✲¯x

✻

✛

✲x

✻

✯

❑φ

❑

✯

❑

❄

φ✲¯x

✻

¯y

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So the angle between ˆn and Eis:

E·ˆn

|E|= cos =1

1+2(sin ✓+cos ✓)= cos ✓

1+2(tan ✓+)= 2

1+2cos ✓

But = tan ✓=sin ✓

cos ✓=p1cos2✓

cos ✓=1

cos2✓1)1

cos

2✓=2+1)cos ✓=1

p1+2. So cos =2

1+2.

Evidently the ﬁeld is not perpendicular to the plates in S.

Problem 12.44

(a) E=

2⇡✏0

ˆs

s=

2⇡✏0

x0ˆx +y0ˆy

(x2

0+y2

0).

(b) ¯

Ex=Ex=

2⇡✏0

(x2

0+y2

0),¯

Ey=Ey=

2⇡✏0

(x2

0+y2

0),¯

Ez=Ez=0,¯

E=

2⇡✏0

(x0ˆx +y0ˆy

(x2

0+y2

0).

Using the inverse Lorentz transformations (Eq. 12.19), x0=(x+vt),y

0=y,

E=

2⇡✏0

(x+vt)ˆx +yˆy

[2(x+vt)2+y2]=

2⇡✏0



(x+vt)ˆx +yˆy

[(x+vt)2+y2/2].

Now S=(x+vt)ˆx +yˆy , and y=Ssin ✓, so [(x+vt)2+y2/2] = [(x+vt)2+y2(1v2/c2]=S2(v/c)2S2sin2✓=

S2[1 (v/c)2sin2✓], so

E=

2⇡✏01(v/c)2

1v2sin2✓/c2ˆ

This is reminiscent of Eq. 10.75. Yes, the ﬁeld does point away from the present location of the wire.

Problem 12.45

(a) Fields of Aat B:E=1

4⇡✏0

d2ˆy ;B=0. So force on qBis F=1

4⇡✏0

qAqB

d2ˆy .

✲x

✻

❨

ˆn

✲x

✻

✲

✲¯x

✻

✛

✲x

✻

✯

❑φ

❑

✯

❑

❄

φ✲¯x

✻

¯y

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✲x

✻

❨

ˆn

✲x

✻

✲

✲¯x

✻

✛

✲x

✻

✯

❑φ

❑

✯

❑

❄

φ✲¯x

✻

¯y

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(b) (i) From Eq. 12.67: ¯

F=

4⇡✏0

qAqB

d2ˆy .(Note: here the particle is at rest in ¯

S.)

(ii) From Eq. 12.93, with ✓= 90:¯

E=1

4⇡✏0

qA(1 v2/c2)

(1 v2/c2)3/2

d2ˆy =

4⇡✏0

d2ˆy

(this also follows from Eq. 12.109).

B6= 0, but since vB= 0 in ¯

S, there is no magnetic force anyway, and ¯

F=

4⇡✏0

qAqB

d2ˆy (as before).

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 275

Problem 12.46

System A: Use Eqs. 12.93 and 12.112, with ✓= 90,R=dˆy , and ˆ

=ˆz :

E=q

4⇡✏0



d2ˆy ;B=q

4⇡✏0



d2ˆz ; where =1

1v2/c2.

[Note that (E2B2c2)=q

4⇡✏0d2221v2

c2=q

4⇡✏0d22is invariant, since it doesn’t depend on v(see

Prob. 12.47b for the general proof). We’ll use this as a check.]

F=qE+(vˆx )⇥B=q2

4⇡✏0



d2ˆy v2

c2(ˆx ⇥ˆz )=q2

4⇡✏0



d21+v2

c2ˆy .

System B: The speed of qis vB=v+v

1+v2/c2=2v

(1 + v2/c2)

B=1

14v2/c2

(1+v2/c2)2

=(1 + v2/c2)

12v2

c2+v4

=(1 + v2/c2)

(1 v2/c2)=21+v2

c2;vBB=2v2.

)E=q

4⇡✏0

d221+v2

c2ˆy ;B=q

4⇡✏0

2

d2ˆz .

Check:(E2B2c2)=q

4⇡✏0d2241+ 2v2

c2+v4

c44v2

c2=q

4⇡✏0d2241

4=q

4⇡✏0d22X

F=qE=q2

4⇡✏0

2

d21+v2

c2ˆy (+qat rest )no magnetic force). [Check: Eq. 12.67 )FA=1

FB.X]

System C :vC=0.E=q

4⇡✏0

d2ˆy ;B=0.F=qE=q2

4⇡✏0

d2ˆy .

[The relative velocity of Band Cis 2v/(1 + v2/c2), and corresponding is 2(1 + v2/c2). So Eq. 12.67

)FC=1

2(1+v2/c2)FB.X]

Summary:

q

4⇡✏0d2ˆy q

4⇡✏0d221+v2/c2ˆy q

4⇡✏0d2ˆy

q

4⇡✏0d2v

c2ˆz q

4⇡✏0d22v

c22ˆz 0

q2

4⇡✏0d21+v2/c2ˆy q2

4⇡✏0d221+v2/c2ˆy q2

4⇡✏0d2ˆy

Problem 12.47

(a) From Eq. 12.109:

E·¯

B=¯

Ex¯

Bx+¯

Ey¯

By+¯

Ez¯

Bz=ExBx+2(EyvBz)(By+v

c2Ez)+(Ez+vBy)(Bzv

c2Ey)

=ExBx+2{EyBy+v

c2EyEz

vByBz

v2

c2EzBz+EzBzv

c2EyEz

+vByBz

v2

c2EyBy}

=ExBx+2EyBy1v2

c2+EzBz1v2

c2=ExBx+EyBy+EzBz=E·B.qed

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276 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

(b) ¯

E2c2¯

B2=E2

x+2(EyvBz)2+2(Ez+vBy)2c2B2

x+2By+v

c2Ez2+2Bzv

c2Ey

=E2

x+2E2

y2EyvBz

+v2B2

z+E2

z+2EzvBy

+v2B2

yc2B2

yc22v

c2ByEz



c2v2

c4E2

zc2B2

z+c22v

c2BzEy

c2v2

c4E2

yc2B2

=E2

xc2B2

x+2E2

y1v2

c2+E2

z1v2

c2c2(B2

y)1v2

c2c2B2

z1v2

c2

=(E2

x+E2

y+E2

z)c2(B2

x+B2

y+B2

z)=E2B2c2.qed

any system. )E6=0in all systems.

Problem 12.48

(a) Making the appropriate modiﬁcations in Eq. 9.48 (and picking = 0 for convenience),

E(x, y, z, t)=E0cos(kx !t)ˆy ,B(x, y, z, t)=E0

ccos(kx !t)ˆz ,where k⌘!

(b) Using Eq. 12.109 to transform the ﬁelds:

Ex=¯

Ez=0,¯

Ey=(EyvBz)=E0cos(kx !t)v

ccos(kx !t)=↵E0cos(kx !t),

Bx=¯

By=0,¯

Bz=(Bzv

c2Ey)=E01

ccos(kx !t)v

c2cos(kx !t)=↵E0

ccos(kx !t),

where ↵⌘1v

c=1v/c

1+v/c.

Now the inverse Lorentz transformations (Eq. 12.19) )x=(¯x+v¯

t) and t=¯

t+v

c2¯x, so

kx !t=k(¯x+v¯

t)!¯

t+v

c2¯x=k!v

c2¯x(!kv)¯

t=¯

k¯x¯!¯

where, recalling that k=!/c): ¯

k⌘k!v

c2=k(1 v/c)=↵kand ¯!⌘!(1 v/c)=↵!.

Conclusion:

E(¯x, ¯y, ¯z, ¯

t)= ¯

E0cos(¯

k¯x¯!¯

t)ˆy ,¯

B(¯x, ¯y, ¯z, ¯

t)= ¯

ccos(¯

k¯x¯!¯

t)ˆz ,

where ¯

E0=↵E0,¯

k=↵k, ¯!=↵!,and ↵⌘1v/c

1+v/c.

1+v/c.This is the Doppler shift for light. ¯

=2⇡

k=2⇡

↵k=

↵.The velocity of the wave

in ¯

Sis ¯v=¯!

2⇡¯

=!

2⇡=c. Yup, this is exactly what I expected (the velocity of a light wave is the same

in any inertial system).

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 277

(d) Since the intensity goes like E2, the ratio is ¯

I=¯

=↵2=1v/c

1+v/c.

Dear Al,

The amplitude, frequency, and intensity of the light will all decrease to zero as you run faster

and faster. It’ll get so faint you won’t be able to see it, and so red-shifted even your night-vision

goggles won’t help. But it’ll still be going 3 ⇥108m/s relative to you. Sorry about that.

Sincerely,

David

Problem 12.49

t02 =⇤0

⇤2

t =⇤0

0⇤2

2t02 +⇤0

1⇤2

2t12 =t02 +()t12 =(t02 t12).

t03 =⇤0

⇤3

t =⇤0

0⇤3

3t03 +⇤0

1⇤3

3t13 =t03 +()t13 =(t03 t13)=(t03 +t31).

t23 =⇤2

⇤3

t =⇤2

2⇤3

3t23 =t23.

t31 =⇤3

⇤1

t =⇤3

3⇤1

0t30 +⇤3

3⇤1

1t31 =()t30 +t31 =(t31 t03).

t12 =⇤1

⇤2

t =⇤1

0⇤2

2t02 +⇤1

1⇤2

2t12 =()t02 +t12 =(t12 t02).

Problem 12.50

Suppose t⌫µ=±tµ⌫(+ for symmetric, for antisymmetric).

t =⇤

µ⇤

⌫tµ⌫

t =⇤

µ⇤

⌫tµ⌫=⇤

⌫⇤

µt⌫µ[Because µand ⌫are both summed from 0 !3,

it doesn’t matter which we call µand and which call ⌫.]

=⇤

µ⇤

µ(±tµ⌫) [Using symmetry of tµ⌫, and writing the ⇤’s in the other order.]

=±¯

t.qed

Problem 12.51

Fµ⌫Fµ⌫=F00F00 F01F01 F02F02 F03F03 F10F10 F20F20 F30F30

+F11F11 +F12F12 +F13F13 +F21F21 +F22F22 +F23F23 +F31F31 +F32F32 +F33F33

=(Ex/c)2(Ey/c)2(Ez/c)2(Ex/c)2(Ey/c)2(Ez/c)2+B2

z+B2

y+B2

z+B2

x+B2

y+B2

=2B22E2/c2= 2B2E2

c2,

which, apart from the constant factor 2/c2, is the invariant we found in Prob. 12.47(b).

Gµ⌫Gµ⌫= 2(E2/c2B2) (the same invariant).

Fµ⌫Gµ⌫=2(F01G01 +F02G02 +F03G03) + 2(F12G12 +F13G13 +F23G23)

=21

cExBx+1

cEyBy+1

cEzBz+ 2[Bz(Ez/c)+(By)(Ey/c)+Bx(Ex/c)]

=2

c(E·B)2

c(E·B)= 4

c(E·B),

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278 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

which, apart from the factor 4/c, is the invariant we found of Prob. 12.47(a). [These are, incidentally, the

only fundamental invariants you can construct from Eand B.]

Problem 12.52

E=1

4⇡✏0

2

xˆx =µ0

2⇡

c2

xˆx

B=µ0

4⇡

2v

xˆy =µ0

2⇡

v

xˆy 



Fµ⌫=µ0

2⇡x





0c00

c00v

0000

0v00





Gµ⌫=µ0

2⇡x





00v0

0000

v00c

00c0







Problem 12.53

@⌫Fµ⌫=µ0Jµ. Di↵erentiate: @µ@⌫Fµ⌫=µ0@µJµ.

But @µ@⌫=@⌫@µ(the combination is symmetric) while F⌫µ=Fµ⌫(antisymmetric).

)@µ@⌫Fµ⌫= 0. [Why? Well, these indices are both summed from 0 !3, so it doesn’t matter which we

call µ, which ⌫:@µ@⌫Fµ⌫=@⌫@µF⌫µ=@µ@⌫(Fµ⌫)=@µ@⌫Fµ⌫. But if a quantity is equal to minus itself,

it must be zero.] Conclusion: @µJµ= 0. qed

Problem 12.54

We know that @⌫Gµ⌫= 0 is equivalent to the two homogeneous Maxwell equations, r·B= 0 and r⇥E=

@B

@t. All we have to show, then, is that @Fµ⌫+@µF⌫ +@⌫Fµ= 0 is also equivalent to them. Now this

equation stands for 64 separate equations (µ=0!3, ⌫=0!3, =0!3, and 4 ⇥4⇥4 = 64). But many

of them are redundant, or trivial.

Suppose two indices are the same (say, µ=⌫). Then @Fµµ +@µFµ=@µFµ= 0. But Fµµ = 0 and

Fµ=Fµ, so this is trivial: 0 = 0. To get anything signiﬁcant, then, µ,⌫,must all be di↵erent. They

could beall spatial (µ, ⌫,=1,2,3=x, y, z — or some permutation thereof), or one temporal and two spatial

(µ= 0, ⌫,=1,2 or 2,3, or 1,3 — or some permutation). Let’s examine these two cases separately.

All spatial:say,µ= 1, ⌫=2,= 3 (other permutations yield the same equation, or minus it.)

@3F12 +@1F23 +@2F31 =0)@

@z(Bz)+ @

@x(Bx)+ @

@y(By)=0)r·B=0.

One temporal:say,µ=0,⌫= 1, = 2 (other permutations of these indices yield same result, or minus it).

@2F01 +@0F12 +@1F31 =0)@

@yEx

c=@

@(ct)(Bz)+ @

@x+Ey

c=0.

or: @Bz

@t+@Ex

@y@Ey

@x= 0, which is the z-component of @B

@t=r⇥E.(Ifµ=0,⌫=1,= 2, we get the

ycomponent; for ⌫=2,= 3 we get the xcomponent.)

Conclusion:@Fµ⌫+@µF⌫ +@⌫Fµ= 0 is equivalent to r·B= 0 and @B

@t=r⇥E, and hence to

@⌫Gµ⌫= 0. qed

Problem 12.55

K0=q⌘⌫F0⌫q(⌘1F01 +⌘2F02 +⌘3F03)=q(⌘·E)/c =q

cu·E.Now from Eq. 12.70 we know that

K0=1

d⌧, where Wis the energy of the particle. Since d⌧=1

dt, we have:

cdW

dt =q

c(u·E))dW

dt =q(u·E)

This says the power delivered to the particle is force (qE)times velocity (u) — which is as it should be.

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 279

Problem 12.56

@0=@

@¯x0

=1

@¯

t=1

c@

@¯

t+@

@¯

t+@

@¯

t+@

@¯

t

From Eq. 12.19, we have: @t

@¯

t=,@x

@¯

t=v,@y

@¯

t=@z

@¯

t= 0.

So @0=1

c@

@t+v@

@xor (since ct =x0=x0): @0=@

@x0v

@

@x1=(@0)(@1).

@1=@

@¯x=@

@x1+@

@¯x+@

@¯x=v

@

@t+@

@x=@

@x1v

@

@x0=(@1)(@0).

@2=@

@¯y=@

@¯y+@

@¯y=@

@y=@2.

@3=@

@¯z=@

@¯z+@

@¯z=@

@z=@3.

Conclusion: @µtransforms in the same way as aµ(Eq. 12.27)—and hence is a contravariant 4-vector. qed

Problem 12.57

According to Prob. 12.54, @Gµ⌫

@x⌫= 0 is equivalent to Eq. 12.130. Using Eq. 12.133, we ﬁnd (in the notation

of Prob. 12.56):

@Fµ⌫

@x+@F⌫

@xµ+@Fµ

@x⌫=@Fµ⌫+@µF⌫ +@⌫Fµ

=@(@µA⌫@⌫Aµ)+@µ(@⌫A@A⌫)+@⌫(@Aµ@µA)

=(@@µA⌫@µ@A⌫)+(@µ@⌫A@⌫@µA)+(@⌫@Aµ@@⌫Aµ)=0.qed

[Note that @@µA⌫=@2A⌫

@x@x⌫=@2A⌫

@x⌫@x=@⌫@A⌫, by equality of cross-derivatives.]

Problem 12.58

From Eqs. 12.40 and 12.42, ⌘µ=(c, v), while

µ=(ct ctr,rw(tr)) = (

), so ⌘⌫

⌫=

c

+v·

=(

c

·v).

q

4⇡✏0c

⌘0

(⌘⌫

⌫)=q

4⇡✏0c

c

(

c

·v)=1

4⇡✏0c

(

c

·v)=1

(Eq. 10.46),

q

4⇡✏0c

⌘

(⌘⌫

⌫)=q

4⇡✏0c

v

(

c

·v)=1

4⇡✏0c

(

c

·v)=A

(Eq. 10.47), so (Eq. 12.132)

q

4⇡✏0c

⌘µ

(⌘⌫

⌫)=Aµ.X

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280 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Problem 12.59

✲x

✻

❨

ˆn

✲x

✻

✲

✲¯x

✻

✛

✲x

✻

✯

❑φ

❑

✯

❑

❄

φ✲¯x

✻

¯y

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Step 1 : rotate from xy to XY , using Eq. 1.29:

X= cos x+ sin y

Y=sin x+ cos y

Step 2 : Lorentz-transform from XY to ¯

X¯

Y, using

Eq. 12.18:

X=(Xvt)=[cos x+ sin yct]

Y=Y=sin x+ cos y

Z=Z=z

c¯

t=(ct X)=ct (cos x+ sin y)

Step 3 : Rotate from ¯

X¯

Yto ¯x¯y, using Eq. 1.29 with negative :

¯x= cos ¯

Xsin ¯

Y=cos [cos x+ sin yct]sin [sin x+ cos y]

=(cos2+ sin2)x+(1) sin cos y cos (ct)

¯y= sin ¯

X+ cos ¯

Y=sin (cos x+ sin yct) + cos (sin c+ cos y)

=(1) sin cos x+(sin2+ cos2)y sin (ct)

In matrix form: 





c¯

¯x

¯y

¯z





=





 cos  sin 0

 cos (cos2+ sin2)(1) sin cos 0

 sin (1) sin cos (sin2+ cos2)0

00 01

















Problem 12.60

✲

π✛pbefore (CM)

KΣafter (CM)

✲

πp

Before

✲

KΣ

After

In CM :✲ ✛

Before

✒

✠

rµ

❘

sµ

■

After

✲x

✻

(p= magnitude of 3-momentum

in CM, φ= CM scattering angle)

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In center-of-momentum system, threshold occurs when incident en-

ergy is just suﬃcient to cover the rest energy of the resulting particles,

with none “wasted” as kinetic energy. Thus, in lab system, we want

the outgoing Kand ⌃to have the same velocity, at threshold:

✲

✛

pbefore (CM)

KΣafter (CM)

✲

πp

Before

✲

KΣ

After

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Initial momentum: p⇡; Initial energy of ⇡:E2p2c2=m2c4)E2

⇡=m2

⇡c4+p2

⇡c2.

Total initial energy: mpc2=m2

⇡c4+p2

⇡c2. These are also the ﬁnal energy and momentum: E2p2c2=

(mK+m⌃)2c4.

mpc2+m2

⇡c4+p2

⇡c22p2

⇡c2=(mK+m⌃)2c4

pc4

+2mpc2

c4m2

⇡c2+p2

⇡c+m2

⇡c4

+p2

⇡c2

p2

⇡c2

=(mK+m⌃)2c4



2mp

cm2

⇡c2+p2

⇡=(mK+m⌃)2m2

pm2

⇡

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 281

(m2

⇡c2+p2

⇡)4m2

c2=(mK+m⌃)42(m2

p+m2

⇡)(mK+m⌃)2+m4

p+m4

⇡+2m2

pm2

⇡

4m2

c2p2

⇡=(mK+m⌃)42(m2

p+m2

⇡)(mK+m⌃)2+(m2

pm2

⇡)2

p⇡=c

2mp(mK+m⌃)42(m2

p+m2

⇡)(mK+m⌃)2+(m2

pm2

⇡)2

(2mpc2)c(mKc2+m⌃c2)42(mpc2)2+(m⇡c2)2(mKc2+m⌃c2)2+(mpc2)2(m⇡c2)22

2c(900) (1700)42(900)2+ (150)2(1700)2+(900)2(150)22

1800c(8.35 ⇥1012)(4.81 ⇥1012) + (0.62 ⇥1012)= 1

1800c(2.04 ⇥106)= 1133 MeV/c

Problem 12.61

✲

π✛pbefore (CM)

KΣafter (CM)

✲

πp

Before

✲

KΣ

After

In CM :✲ ✛

Before

✒

✠

rµ

❘

sµ

■

After

✲x

✻

(p= magnitude of 3-momentum

in CM, φ= CM scattering angle)

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Outgoing 4-momentua: rµ=E

c,pcos ,psin ,0;sµ=E

c,pcos ,psin ,0.

In Lab:✲

Before

✒

❘

¯rµ

¯sµ

θProblem: calculate θ, in terms of p,φ.

τ/sin φ

√1+(τ/sin φ)2

✲x

✻

d/2

−d/2

−q

!"# $

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Lorentz transformation: ¯rx=(rxr0); ¯ry=ry;¯sx=(sxs0); ¯sy=sy.

Now E=mc2;p=mv (vhere is to the left;E2p2c2=m2c4,so=pc

)¯rx=pcos +pc

c=p(1 + cos ); ¯ry=psin ;¯sx=p(1 cos ); ¯sy=psin .

cos ✓=¯r ·¯s

¯r¯s=2p2(1 cos2)p2sin2

2p2(1 + cos )2+p2sin22p2(1 cos )2+p2sin2

=(21) sin2

2(1 + cos )2+ sin22(1 cos )2+ sin2

=(21)

21+cos 

sin 2+1

21cos 

sin 2+1

=(21)

2cot2

2+1

tan2

2+1



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282 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

cos ✓=!

1 + cot2

2+!cot2

21 + tan2

2+!tan2

2(where !⌘21)

csc2

2+!cot2

2sec2

2+!tan2

2=!sin 

2cos 

1+!cos2

21+!sin2

2

2!sin 

1+!1

2(1 + cos )1+!1

2(1 cos )=sin 

2

!+1

+ cos 2

!+1

cos 

=sin 

2

!+1

2cos2

=sin 

4

!2+4

!+ sin2

1+⌧2

sin2,where ⌧2=4

!2+4

sin ✓=⌧

sin .⌧2=4

!2(1 + !)= 4

(21)22, so tan ✓=2

(21) sin .

Or, since (21) = 211

2=2v2

c2, tan ✓=2c2

v2sin 

In Lab:✲

Before

✒

❘

¯rµ

¯sµ

θProblem: calculate θ, in terms of p,φ.

τ/sin φ

√1+(τ/sin φ)2

✲x

✻

d/2

−d/2

−q

!"# $

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Problem 12.62

d⌧=K(a constant) )dp

d⌧=K.Butdt

d⌧=1

p1u2/c2;p=mu

p1u2/c2.

dt u

p1u2/c2=K

m1u2/c2. Multiply by dt

dx =1

dtu

1u2/c2=d

dxu

1u2/c2=K

m1u2/c2

u.Let w=u

1u2/c2.

dx =K

w;wdw

dx =1

dxw2=k

m;d(w2)

dx =2K

m)d(w2)=2K

m(dx).

)w2=2K

mx+ constant. But at t= 0, x= 0 and u= 0 (so w= 0), and hence the constant is 0.

w2=2K

mx=u2

1u2/c2;u2=2Kx

m2Kx

mc2u2;u21+2Kx

mc2=2Kx

u2=2Kx/m

1+2Kx

mc2

=c2

1+mc2

2Kx;dx

dt =c

1+mc2

2Kx;ct =1+mc2

2Kxdx

Let mc2

2K=a2;ct =px+a2

pxdx. Let x=y2;dx =2y dy;px=y.

ct =y2+a2

y2y dy =2y2+a2dy =yy2+a2+a2ln(y+y2+a2)+ constant.

At t= 0, x=0)y= 0. )0=a2ln a+ constant, so constant = a2ln a.

)ct =yy2+a2=a2lny/a +(y/a)2+1

=a2y

ay

a2+1+lny

a+y

a2+1



Let: z=y

a=px2K

mc2=2Kx

mc2=z. Then 2Kt

mc =z1+z2+ ln(z+1+z2).

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 283

Problem 12.63

In Lab:✲

Before

✒

❘

¯rµ

¯sµ

θProblem: calculate θ, in terms of p,φ.

τ/sin φ

√1+(τ/sin φ)2

✲x

✻

d/2

−d/2

−q

!"# $

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(a) x(t)= c

↵1+(↵t)21, where ↵=F

mc . The force of +qon

qwill be the mirror image of the force of qon +q(in the x-axis),

so the net force is in the xdirection (the net magnetic force is zero).

So all we need is the x-component of E.

The ﬁeld at +qdue to qis: (Eq. 10.72)

E=q

4⇡✏0

(

·u)3u(c2v2)+u(

·a)a(

·u).

u=c

v)ux=cl

v=1

(cl v

);

·u=c



·v=(c

lv);

·a=la. So:

Ex=q

4⇡✏0

vl)31

(cl v

)(c2v2)= 1

(cl v

)la a(c

lv)

  

ca(l2

2)=cad2/



=q

4⇡✏0

vl)3(cl v

)(c2v2)cad2.

The force on +qis qEx, and there is an equal force on q, so the net force on the dipole is:

F=2q2

4⇡✏0

lv)3(cl v

)(c2v2)cad2ˆx It remains to determine

,l,

v, and a, and plug these in.

v(t)=dx

dt =c

↵

1+(↵t)22↵2t=c↵t

1(↵t)2;v=v(tr)=c↵tr

T,where T=1+(↵tr)2.

a(tr)= dv

dtr

=c↵

T+c↵tr1

22↵2tr

T3=c↵

T31+(↵tr)2(↵tr)2=c↵

Now calculate tr:c2(ttr)2=

2=l2+d2;l=x(t)x(tr)= c

↵1+(↵t)21+(↵tr)2,so

2ttr+t2

=1

↵21+(↵t)2

+1+(↵tr)2

21+(↵t)21+(↵tr)2+(d/c)2

(F)1+(↵t)21+(↵tr)2=1+↵2ttr+1

2↵d

c2. Square both sides:

1+(↵t)2+(↵tr)2+↵4t2t2

=1+↵4t2t2

+1

4↵d

c4+2↵2ttr+↵d

c2+↵2ttr↵d

c2

t2+t2

r2ttrttr↵d

c2d

c2↵2

4d

c4=0

At this point we could solve for tr(in terms of t), but since vand aare already expressed in terms of tr, it is

simpler to solve for t(in terms of tr), and express everything in terms of tr:

t2ttr2+↵d

c2+t2

rd

c2↵2

4d

c4=0=)

t=1

2tr2+↵d

c2±t2

r4+4

↵d

c2+↵d

c44t2

+4

d

c2+↵2d

c4

=tr1+1

2↵d

c)2±1+(↵tr)2d

c21+↵d

2c

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284 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

Which sign? For small ↵we want t⇡tr+d/c, so we need the + sign:

t=tr1+1

2↵d

c2+d

cTD, where D=1+↵d

2c2

=c(ttr))

=ctr

2↵d

c2+dT D. Now go back to Eq. (F) and solve for 1+(↵t)2:

1+(↵t)2=1

T1+1

2↵d

c2+↵2trtr1+1

2↵d

c2+d

cTD

T1+(↵tr)2

  

T21+1

2↵d

c2+↵2trd

cTD=1+ 1

2↵d

c2T+↵2trd

l=c

↵1+(↵t2)1+(↵tr)2=c

↵1+1

2↵d

c2T+↵2trd

cDT=↵dd

2cT+trD

Putting all this in, the numerator in square brackets in Fbecomes:

[]=

c↵dd

2cT+trDc↵tr

Tctr

2↵d

c2+dT Dc2c2↵2t2

T2cc↵

=c↵dd

2cT+trDd(atr)2

2cT trDc2

T21+(↵tr)2

(↵tr)2

c2↵d2

=c2↵d2

T31

2T21

2(↵tr)21=c2↵d2

2T31+(↵tr)2(↵tr)22=c2↵d2

2T3

)F=q2

4⇡✏0

c2↵d2

(c

lv)T3ˆx . It remains to compute the denominator:

lv)T=cctr

2↵d

c2+dT D↵dd

2cT+trDc↵tr

TT

=1

2↵2trd2

+cdT D 1

2↵2trd2

cd(↵tr)2

TDT=cdDT2(↵tr)2

  

1+(↵tr)2

(↵tr)2

=dcD

)F=q2

4⇡✏0

c2d2↵

c3d3D3ˆx =q2

4⇡✏0

↵

cd1+↵d

2c23/2ˆx ↵=F

mc

Energy must come from the “reservoir” of energy stored in the electromagnetic ﬁelds.

(b) F=mc↵=1

4⇡✏0

↵

cd1+↵d

2c23/2)1+↵d

2c23/2=q2

8⇡✏0mc2d=µ0q2

8⇡md.

(force on one end only)

)↵=2c

dµ0q2

8⇡md2/31,so F=2mc2

dµ0q2

8⇡md2/31.

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 285

Problem 12.64

(a) Aµ=(V/c, Ax,A

y,A

z) is a 4-vector (like xµ=(ct, x, y, z)), so (using Eq. 12.19): V=(¯

V+v¯

Ax). But

V= 0, and

Ax=µ0

4⇡

(m⇥¯r )x

¯r3

Now (m⇥¯r )x=my¯zmz¯y=myzmzy. So

V=vµ0

4⇡

(myzmzy)

¯r3

Now ¯x=(xvt)=Rx,¯y=y=Ry,¯z=z=Rz,whereRis the vector (in S) from the (instantaneous)

location of the dipole to the point of observation. Thus

¯r2=2R2

x+R2

y+R2

z=2(R2

x+R2

y+R2

z) + (1 2)(R2

y+R2

z)=2R2v2

c2R2sin2✓

(where ✓is the angle between Rand the x-axis, so that R2

y+R2

z=R2sin2✓).

)V=µ0

4⇡

v(myRzmzRy)

3R31v2

c2sin2✓3/2;v·(m⇥R)=v(m⇥R)x=v(myRzmzRy),so

V=µ0

4⇡

v·(m⇥R)1v2

c2

R31v2

c2sin2✓3/2,

or, using µ0=1

✏0c2and v·(m⇥R)=R·(v⇥m): V=1

4⇡✏0

R·(v⇥m)1v2

c2

c2R21v2

c2sin2✓3/2

(b) In the nonrelativistic limit (v2⌧c2):

V=1

4⇡✏0

R·(v⇥m)

c2R2=1

4⇡✏0

R·p

R2,with p=v⇥m

c2,

which is the potential of an electric dipole.

Problem 12.65

(a) B=µ0

2Kˆy (Eq. 5.58); N=m⇥B(Eq. 6.1), so N=µ0

2mK(ˆz ⇥ˆy ).

N=µ0

2mKˆx =µ0

2(vl2)(v)ˆx =µ0

2v2l2ˆx .

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286 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

(b)



Charge density in the front side: 0(=0);

Charge density on the back side: ¯

=¯0, where ¯v=2v

1+v2/c2,

so ¯=1

14v2/c2

(1+v2/c2)2

=(1 + v2/c2)

1+2v2

c2+v4

c44v2

=1+v2/c2

12v2

c2+v4

=(1 + v2/c2)

(1 v2/c2)=21+v2

c2

Length of front and back sides in this frame: l/. So net charge on back side is:

q+=¯

l

=21+v2

c2



=1+v2

c2l

Net charge on front side is:

q=0

=



=1

2l

So dipole moment (note: charges on sides are equal):

p=(q+)l

2ˆy (q)l

2ˆy =1+v2

c2ll

21

2ll

2ˆy =l2

21+v2

c21+v2

c2ˆy =l2v2

c2ˆy .

E=0

2✏0ˆz , where =0,soN=p⇥E=l2v2



2✏0(ˆy ⇥ˆz )= 1



µ0

2l2v2ˆx .

So apart from the relativistic factor of the torque is the same in both systems — but in Sit is the torque

exerted by a magnetic ﬁeld on a magnetic dipole, whereas in ¯

Sit is the torque exerted by an electric ﬁeld on

an electric dipole.

Problem 12.66

Choose axes so that Epoints in the zdirection and Bin the yz plane: E= (0,0,E); B=(0,Bcos ,Bsin ).

Go to a frame moving at speed vin the xdirection:

E=0,vB sin ,(E+vB cos );¯

B0,(Bcos +v

c2E),Bsin .

(I used Eq. 12.109.) Parallel provided vB sin 

(Bcos +v

c2E)=(E+vB cos )

Bsin ,or

vB2sin2=Bcos +v

c2E(E+vB cos )=EB cos +vB2cos2+v

c2E2+v2

c2EB cos 

0=vB2+v

c2E2+EB cos 1+v2

c2;v

1+v2/c2=EB cos 

B2+E2/c2

Now E⇥B=

ˆx ˆy ˆz

00 E

0Bcos Bsin =EB cos . So v

1+v2/c2=E⇥B

B2+E2/c2. qed

No, there can be no frame in which E?B, for (E·B) is invariant, and since it is not zero in Sit can’t be

zero in ¯

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 287

Problem 12.67

✕v

☛

✛✛

¯v

✲x

−q✲

❄

✻

✎❲

✗ ❖

☛❯

✕ ❑

✛

✻

❄

✗❖

✎ ❲

✕❑

☛ ❯

✻

❄

✗

❖

✎❲

✍

▼

✌◆

◆ ❲ ❄

✎✌

✍ ✗ ✻

❖▼

✲x

+−

✶

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Just before:

Field lines emanate

from present position

of particle.

✕v

☛

✛✛

¯v

✲x

−q✲

❄

✻

✎❲

✗ ❖

☛❯

✕ ❑

✛

✻

❄

✗❖

✎ ❲

✕❑

☛ ❯

✻

❄

✗

❖

✎❲

✍

▼

✌◆

◆ ❲ ❄

✎✌

✍ ✗ ✻

❖▼

✲x

+−

✶

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Just after: Field lines outside sphere of radius ct emanate from

position particle would have reached, had it kept going on its

original “ﬂight plan”. Inside the sphere E= 0. On the sur-

face the lines connect up (since they cannot simply terminate

in empty space), as suggested in the ﬁgure.

This produces a dense cluster of tangentially-directed ﬁeld

lines, which expand with the spherical shell. This is a pic-

torial way of understanding the generation of electromagnetic

radiation.

Problem 12.68

Equation 12.67 assumes the particle is (instantaneously) at rest in S. Here the particle is at rest in ¯

S.So

F?=1

¯

F?,F

k=¯

Fk. Using ¯

F=q¯

E, then,

Fx=¯

Fx=q¯

Ex,F

y=1

¯

Fy=1

q¯

Ey,F

z=1

¯

Fz=1

q¯

Ez.

Invoking Eq. 12.109:

Fx=qEx,F

y=1

q(EyvBz)=q(EyvBz)Fz=1

q(Ez+vBy)=q(Ez+vBy).

But v⇥B=vBzˆx +vByˆz , so F=q(E+v⇥B). qed

Problem 12.69

✲y

✻

✰

✻

✰

✲¯y

✻

¯z

✰

¯x

=⇒v

✲¯y

✻

¯z

✰

¯x

✶

ωt

✒

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Rewrite Eq. 12.109 with x!y,y!z,z!x:

Ey=Ey¯

Ez=(EzvBx)¯

Ex=(Ex+vBz)

By=By¯

Bz=Bz+v

c2Ex¯

Bx=Bxv

c2Ez

This gives the ﬁelds in system ¯

Smoving in the ydirection at speed v.

Now E=(0,0,E

0); B=(B0,0,0), so ¯

Ey= 0, ¯

Ez=(E0vB0), ¯

Ex= 0.

If we want ¯

E=0, we must pick vso that E0vB0= 0; i.e. v=E0/B0

(The condition E0/B0<cguarantees that there is no problem getting to such a system.)

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288 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

With this, ¯

By= 0, ¯

Bz= 0, ¯

Bx=(B0v

c2E0)=B01v2

c2=B01

2=1

B0;¯

B=1

B0ˆx .

The trajectory in ¯

S: Since the particle started out at rest at the origin

in S, it started out with velocity vˆy in ¯

S. According to Eq. 12.71

it will move in a circle of radius R, given by

p=qBR, or mv =q1

B0R)R=m2v

qB0

✲y

✻

✰

✻

✰

✲¯y

✻

¯z

✰

¯x

=⇒v

✲¯y

✻

¯z

✰

¯x

✶

ωt

✒

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The actual trajectory is given by ¯x=0; ¯y=Rsin !¯

t;¯z=R(1 cos !¯

t); where !=v

The trajectory in S: The Lorentz transformations Eqs. 12.18 and 12.19, for the case of relative motion in

the y-direction, read:

¯x=xx=¯x

¯y=(yvt)y=(¯y+v¯

¯z=zz=¯z

t=tv

c2yt=¯

t+v

c2¯y

So the trajectory in Sis given by:

x=0;y=(Rsin !¯

t+v¯

t)=Rsin!tv

c2y+vtv

c2y,or

y1+2v2

c2

  

2y(1v2

c2+v2

c2)=2y

=2vt Rsin!tv

c2y(yvt)=Rsin!tv

c2y;

z=R(1 cos2!¯

t)=R1cos !tv

c2y.

So: x=0; y=vt R

sin!tv

c2y;z=RRcos!tv

c2.

We can get rid of the trigonometric terms by the usual trick:

(yvt)=Rsin !(tv

c2y)

zR=Rcos !(tv

c2y))2(yvt)2+(zR)2=R2.

Absent the 2, this would be the cycloid we found back in Ch. 5 (Eq. 5.9). The 2makes it, as it were, an

elliptical cycloid — same picture as Fig. 5.7, but with the horizontal axis stretched out.

Problem 12.70

(a) D=✏0E+Psuggests E!1

✏0D

H=1

µ0BMsuggests B!µ0Hbut it’s a little cleaner if we divide by µ0while we’re at it, so that

E!1

µ0✏0D=c2D,B!H. Then:

Dµ⌫=









0cDxcDycDz

cDx0HzHy

cDyHz0Hx

cDzHyHx0











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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 289

Then (following the derivation in Sect. 12.3.4):

@x⌫D0⌫=cr·D=c⇢f=J0

f;@

@x⌫D1⌫=1

@t(cDx)+(r⇥H)x=(Jf)x; so @Dµ⌫

@x⌫=Jµ

where Jµ

f=(c⇢f,Jf).Meanwhile, the homogeneous Maxwell equations r·B=0,E=@B

@tare unchanged,

and hence @Gµ⌫

@x⌫=0.

(b)

Hµ⌫=









0HxHyHz

Hx0cDzcDy

HycDz0cDx

HzcDycDx0











Dµ0⌘0=c2✏Fµ0⌘0)Dµ0=c2✏Fµ0)cD=c2✏E

c)D=✏E(Eq.4.32),X

Hµ0⌘0=1

µGµ0⌘0)Hµ0=1

µGµ0)H=1

µB)H=1

µB(Eq.6.31).X

(d) In general, ⌘⌫=(c, u), so, for µ= 0:

D0⌫⌘⌫=D01⌘1+D02⌘2+D03⌘3=cDx(ux)+cDy(uy)+cDz(uz)=c(D·u),

F0⌫⌘⌫=F01⌘1+F02⌘2+F03⌘3=Ex

c(ux)+Ey

c(uy)+Ez

c(uz)=

c(E·u),so

D0⌫⌘⌫=c2✏F0⌫⌘⌫)c(D·u)=c2✏

c(E·u))D·u=✏(E·u).[1]

H0⌫⌘⌫=H01⌘1+H02⌘2+H03⌘3=Hx(ux)+Hy(uy)+Hz(uz)=(H·u),

G0⌫⌘⌫=G01⌘1+G02⌘2+G03⌘3=Bx(ux)+By(uy)+Bz(uz)=(B·u),so

H0⌫⌘⌫=1

µG0⌫⌘⌫)(H·u)= 1

µ(B·u))H·u=1

µ(B·u).[2]

Similarly, for µ=1:

D1⌫⌘⌫=D10⌘0+D12⌘2+D13⌘3=(cDx)(c)+Hz(uy)+(Hy)(uz)=(c2Dx+uyHzuzHy)

=c2D+(u⇥H)x,

F1⌫⌘⌫=F10⌘0+F12⌘2+F13⌘3=Ex

c(c)+Bz(uy)+(By)(uz)=(Ex+uyBzuzBy)

=[E+(u⇥B)]x,so D1⌫⌘⌫=c2✏F1⌫⌘⌫)

c2D+(u⇥H)x=c2✏ [E+(u⇥B)]x)D+1

c2(u⇥H)=✏[E+(u⇥B)] .[3]

H1⌫⌘⌫=H10⌘0+H12⌘2+H13⌘3=(Hx)(c)+(cDz)(uy)+(cDy)(uz)

=c(HxuyDz+uzDy)=c[H(u⇥D)]x,

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290 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

G1⌫⌘⌫=G10⌘0+G12⌘2+G13⌘3=(Bx)(c)+Ez

c(uy)+Ey

c(uz)

=

c(c2BxuyEz+uzEy)=

cc2B(u⇥E)x,so H1⌫⌘⌫=1

µG1⌫⌘⌫)

c[H(u⇥D)]x=1



cc2B(u⇥E)x)H(u⇥D)= 1

µB1

c2(u⇥E).[4]

Use Eq. [4] as an expression for H, plug this into Eq. [3], and solve for D:

D+1

c2u⇥(u⇥D)+ 1

µB1

c2(u⇥E)=✏[E+(u⇥B)] ;

D+1

c2(u·D)uu2D=✏[E+(u⇥B)] 1

µc2(u⇥B)+ 1

µc4[u⇥(u⇥E)] .

Using Eq. [1] to rewrite (u·D):

D1u2

c2=✏

c2(E·u)u+✏[E+(u⇥B)] 1

µc2(u⇥B)+ 1

µc4(E·u)uu2E

=✏1u2

✏µc4E1

c211

✏µc2(E·u)u+(u⇥B)11

✏µc2.

Let ⌘1

1u2/c2,v⌘1

p✏µ.Then

D=2✏1u2v2

c4E+1v2

c2(u⇥B)1

c2(E·u)u.

Now use Eq. [3] as an expression for D, plug this into Eq. [4], and solve for H:

Hu⇥1

c2(u⇥H)+✏[E+(u⇥B)]=1

µB1

c2(u⇥E);

H+1

c2(u·H)uu2H=1

µB1

c2(u⇥E)+✏(u⇥E)+✏[u⇥(u⇥B)].

Using Eq. [2] to rewrite (u·H):

H1u2

c2=1

µc2(B·u)u+1

µB1

c2(u⇥E)+✏(u⇥E)+✏(B·u)uu2B

µ1µ✏u2B+✏µ1

c2[(u⇥E)+(B·u)u].

H=2

µ1u2

v2B+1

v21

c2[(u⇥E)+(B·u)u].

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CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 291

Problem 12.71

We know that (proper) power transforms as the zeroth component of a 4-vector K0=1

d⌧. The Larmor

formula says that for v= 0, dW

d⌧=1

4⇡✏0

c3a2(Eq. 11.70). Can we think of a 4-vector whose zeroth component

reduces to this when the velocity is zero?

Well, a2smells like (↵⌫↵⌫), but how do we get a 4-vector in here? How about ⌘µ, whose zeroth component

is just c, when v= 0? Try, then:

Kµ=1

4⇡✏0

c5(↵⌫↵⌫)⌘µ

This has the right transformation properties, but we must check that it does reduce to the Larmor formula

when v= 0:

dt =1



d⌧=1

cK0=1

cµ0q2

6⇡c3(↵⌫↵⌫)⌘0, but ⌘0=c, so dW

dt =µ0q2

6⇡c(↵⌫↵⌫) . [Incidentally, this tells

us that the power itself (as opposed to proper power) is a scalar. If this had been obvious from the start, we

could simply have looked for a Lorentz scalar that generalizes the Larmor formula.]

In Prob. 12.39(b) we calculated (↵⌫↵⌫) in terms of ordinary velocity and acceleration:

↵⌫↵⌫=4a2+(v·a)2

(c2v2)=6a22+1

c2(v·a)2

=6a21v2

c2+1

c2(v·a)2=6a21

c2v2a2(v·a)2.

Now v·a=va cos ✓, where ✓is the angle between vand a, so:

v2a2(v·a)2=v2a2(1 cos2✓)=v2a2sin2✓=|v⇥a|2.

↵⌫↵⌫=6a2v⇥a

c

2.

dt =µ0q2

6⇡c6a2v⇥a

c

2,which is Li´enard’s formula (Eq. 11.73).

Problem 12.72

(a) It’s inconsistent with the constraint ⌘µKµ= 0 (Prob. 12.39(d)).

(b) We want to ﬁnd a 4-vector bµwith the property that d↵µ

d⌧+bµ⌘µ= 0. How about bµ=d↵⌫

d⌧⌘⌫⌘µ? Then

d↵⌫

d⌧+bµ⌘µ=d↵µ

d⌧⌘µ+d↵⌫

d⌧⌘⌫(⌘µ⌘µ). But ⌘µ⌘µ=c2, so this becomes d↵µ

d⌧⌘µc2d↵⌫

d⌧⌘⌫, which is zero,

if we pick =1/c2. This suggests Kµ

rad =µ0q2

6⇡cd↵µ

d⌧+1

d↵⌫

d⌧⌘⌫⌘µ.Note that ⌘µ=(c, v), so the spatial

components of bµvanish in the nonrelativistic limit v⌧c, and hence this still reduces to the Abraham-Lorentz

formula. [Incidentally, ↵⌫⌘⌫=0)d

d⌧(↵⌫⌘⌫)=0)d↵⌫

d⌧⌘⌫+↵⌫d⌘⌫

d⌧= 0, so d↵⌫

d⌧⌘⌫=↵⌫↵⌫, and hence bµcan

just as well be written 1

c2(↵⌫↵⌫)⌘µ.]

Problem 12.73

Deﬁne the electric current 4-vector as before: Jµ

e=(c⇢e,Je), and the magnetic current analogously:

Jµ

m=(c⇢m,Jm). The fundamental laws are then

@⌫Fµ⌫=µ0Jµ

e,@⌫Gµ⌫=µ0

cJµ

m,K

µ=qeFµ⌫+qm

cGµ⌫⌘⌫.

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292 CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY

The ﬁrst of these reproduces r·E= (1/✏0)⇢eand r⇥B=µ0Je+µ0✏0@E/@t, just as before; the second

yields r·B=(µ0/c)(c⇢m)=µ0⇢mand (1/c)[@B/@t+r⇥E]=(µ0/c)Jm, or r⇥E=µ0Jm@B/@t

(generalizing Sec. 12.3.4). These are Maxwell’s equations with magnetic charge (Eq. 7.44). The third says

K1=qe

1u2/c2[E+(u⇥B)]x+qm

cc

1u2/c2(Bx)+ uy

1u2/c2Ez

c+uz

1u2/c2Ey

c,

K=1

1u2/c2qe[E+(u⇥B)] + qmB1

c2(u⇥E),or

F=qe[E+(u⇥B]+qmB1

c2(u⇥E),

which is the generalized Lorentz force law (Eq. 7.69).

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CONCORDANCE 293

4th ed Same? 3rd ed

1.1 X1.1

1.2 X1.2

1.3 X1.3

1.4 X1.4

1.5 X1.5

1.6 X1.6

1.7 X1.7

1.8 X1.8

1.9 X1.9

1.10 X1.10

1.11 X1.11

1.12 X1.12

1.13 X1.13

1.14 X1.14

1.15 X1.15

1.16 X1.16

1.17 X1.17

1.18 X1.18

1.19 new –

1.20 X1.19

1.21 X1.20

1.22 X1.21

1.23 X1.22

1.24 X1.23

1.25 X1.24

1.26 X1.25

1.27 X1.26

1.28 X1.27

1.29 X1.28

1.30 X1.29

1.31 X1.30

1.32 X1.31

1.33 X1.32

1.34 X1.33

1.35 X1.34

1.36 X1.35

1.37 X1.36

1.38 X1.37

1.39 X1.38

1.40 X1.39

1.41 X1.40

1.42 X1.41

1.43 X1.42

1.44 X1.43

1.45 X1.44

4th ed Same? 3rd ed

1.46 X1.45

1.47 X1.46

1.48 X1.47

1.49 X1.48

1.50 X1.49

1.51 X1.50

1.52 X1.51

1.53 X1.52

1.54 X1.53

1.55 X1.54

1.56 X1.55

1.57 X1.56

1.58 X1.57

1.59 X1.58

1.60 X1.59

1.61 X1.60

1.62 X1.61

1.63 X1.62

1.64 new –

2.1 X2.1

2.2 mod 2.2

2.3 X2.3

2.4 X2.4

2.5 X2.5

2.6 X2.6

2.7 X2.7

2.8 X2.8

2.9 X2.9

2.10 X2.10

2.11 X2.11

2.12 X2.12

2.13 X2.13

2.14 X2.14

2.15 mod 2.15

2.16 X2.16

2.17 X2.17

2.18 X2.18

2.19 X2.19

2.20 X2.20

2.21 X2.21

2.22 X2.22

2.23 X2.23

2.24 X2.24

2.25 X2.25

2.26 X2.26

4th ed Same? 3rd ed

2.27 X2.27

2.28 X2.28

2.29 X2.29

2.30 X2.30

2.31 X2.31

2.32 new –

2.33 new –

2.34 X2.32

2.35 X2.33

2.36 X2.34

2.37 new –

2.38 X2.35

2.39 X2.36

2.40 new –

2.41 X2.37

2.42 X2.38

2.43 X2.39

2.44 X2.40

2.45 X2.41

2.46 new –

2.47 X2.43

2.48 X2.44

2.49 X2.45

2.50 X2.46

2.51 new –

2.52 X2.47

2.53 X2.48

2.54 mod 2.49

2.55 X2.50

2.56 X2.51

2.57 X2.52

2.58 new –

2.59 new –

2.60 new –

2.61 new –

3.1 X3.1

3.2 X3.2

3.3 X3.3

3.4 new –

3.5 X3.4

3.6 X3.5

3.7 X3.6

3.8 X3.7

3.9 X3.8

3.10 X3.9

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

294 CONCORDANCE

4th ed Same? 3rd ed

3.11 X3.10

3.12 X3.11

3.13 X3.12

3.14 X3.13

3.15 X3.14

3.16 mod 3.15

3.17 X3.16

3.18 X3.17

3.19 X3.18

3.20 X3.19

3.21 X3.20

3.22 X3.21

3.23 X3.22

3.24 X3.23

3.25 X3.24

3.26 X3.25

3.27 X3.26

3.28 new –

3.29 X3.27

3.30 X3.28

3.31 X3.29

3.32 X3.30

3.33 X3.31

3.34 X3.32

3.35 new –

3.36 X3.33

3.37 new –

3.38 new –

3.39 X3.35

3.40 X3.36

3.41 new –

3.42 new –

3.43 X3.37

3.44 X3.38

3.45 X3.39

3.46 X3.40

3.47 X3.41

3.48 X3.42

3.49 new –

3.50 X3.43

3.51 X3.44

3.52 mod 3.45

3.53 X3.46

3.54 X3.47

3.55 X3.48

4th ed Same? 3rd ed

3.56 X3.49

3.57 new –

3.58 new –

4.1 X4.1

4.2 X4.2

4.3 X4.3

4.4 X4.4

4.5 X4.5

4.6 X4.6

4.7 X4.7

4.8 X4.8

4.9 X4.9

4.10 X4.10

4.11 X4.11

4.12 X4.12

4.13 X4.13

4.14 X4.14

4.15 X4.15

4.16 mod 4.16

4.17 X4.17

4.18 X4.18

4.19 X4.19

4.20 X4.20

4.21 X4.21

4.22 X4.22

4.23 X4.23

4.24 X4.24

4.25 X4.25

4.26 X4.26

4.27 X4.27

4.28 X4.28

4.29 X4.29

4.30 X4.30

4.31 new –

4.32 new –

4.33 X4.31

4.34 new –

4.35 X4.32

4.36 X4.33

4.37 X4.34

4.38 X4.35

4.39 X4.36

4.40 X4.37

4.41 X4.38

4.42 X4.39

4th ed Same? 3rd ed

4.43 X4.40

5.1 X5.1

5.2 X5.2

5.3 X5.3

5.4 X5.4

5.5 X5.5

5.6 X5.6

5.7 X5.7

5.8 X5.8

5.9 X5.9

5.10 X5.10

5.11 X5.11

5.12 new –

5.13 X5.12

5.14 X5.13

5.15 X5.14

5.16 X5.15

5.17 X5.16

5.18 X5.17

5.19 X5.18

5.20 X5.19

5.21 X5.20

5.22 X5.21

5.23 X5.22

5.24 X5.23

5.25 X5.24

5.26 X5.25

5.27 X5.26

5.28 X5.27

5.29 X5.28

5.30 X5.29

5.31 X5.30

5.32 X5.31

5.33 X5.32

5.34 X5.33

5.35 X5.34

5.36 X5.37

5.37 mod 5.35, 5.36

5.38 X5.60

5.39 new –

5.40 X5.38

5.41 X5.39

5.42 X5.40

5.43 X5.41

5.44 X5.42

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

CONCORDANCE 295

4th ed Same? 3rd ed

5.45 X5.43

5.46 X5.44

5.47 X5.46

5.48 mod 5.47

5.49 X5.48

5.50 X5.49

5.51 new –

5.52 X5.50

5.53 X5.51

5.54 X5.52

5.55 X5.53

5.56 X5.54

5.57 X5.55

5.58 X5.56

5.59 X5.57

5.60 X5.58

5.61 X5.59

5.62 X5.61

6.1 X6.1

6.2 X6.2

6.3 X6.3

6.4 X6.4

6.5 X6.5

6.6 X6.6

6.7 X6.7

6.8 X6.8

6.9 X6.9

6.10 X6.10

6.11 X6.11

6.12 X6.12

6.13 X6.13

6.14 X6.14

6.15 X6.15

6.16 X6.16

6.17 X6.17

6.18 X6.18

6.19 X6.19

6.20 X6.20

6.21 X6.21

6.22 X6.22

6.23 X6.25

6.24 new –

6.25 X6.23

4th ed Same? 3rd ed

6.26 X6.24

6.27 X6.26

6.28 X6.27

6.29 X6.28

7.1 X7.1

7.2 X7.2

7.3 X7.3

7.4 X7.4

7.5 X7.5

7.6 X7.6

7.7 X7.7

7.8 X7.8

7.9 X7.9

7.10 X7.10

7.11 X7.11

7.12 X7.12

7.13 X7.13

7.14 X7.14

7.15 X7.15

7.16 X7.16

7.17 X7.17

7.18 X7.18

7.19 X7.19

7.20 new –

7.21 new –

7.22 X7.20

7.23 X7.21

7.24 X7.22

7.25 X7.23

7.26 X7.24

7.27 X7.25

7.28 X7.26

7.29 X7.27

7.30 X7.28

7.31 X7.29

7.32 X7.30

7.33 new –

7.34 X7.31

7.35 X7.32

7.36 X7.33

7.37 X7.34

7.38 X7.35

7.39 X7.36

4th ed Same? 3rd ed

7.40 X7.37

7.41 X7.39

7.42 X7.41

7.43 X7.57

7.44 X7.42

7.45 X7.43

7.46 X7.44

7.47 X7.45

7.48 mod 7.46

7.49 mod 7.47

7.50 X7.48

7.51 new –

7.52 X7.49

7.53 X7.50

7.54 new –

7.55 X7.51

7.56 X7.52

7.57 X7.53

7.58 X7.54

7.59 new –

7.60 X7.55

7.61 X7.56

7.62 X7.58

7.63 X7.59

7.64 X7.60

8.1 X8.1

8.2 X8.2

8.3 X8.3

8.4 X8.4

8.5 new –

8.6 mod 8.6

8.7 mod 8.5

8.8 X8.7

8.9 new –

8.10 X8.8

8.11 new –

8.12 new –

8.13 X8.9

8.14 new –

8.15 new –

8.16 X8.10

8.17 X8.11

8.18 new –

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

296 CONCORDANCE

4th ed Same? 3rd ed

8.19 X8.12

8.20 new –

8.21 mod 8.13

8.22 X8.14

8.23 X8.15

8.24 new –

9.1 X9.1

9.2 X9.2

9.3 X9.3

9.4 X9.4

9.5 X9.5

9.6 X9.6

9.7 X9.7

9.8 X9.8

9.9 X9.9

9.10 X9.10

9.11 new –

9.12 X9.11

9.13 X9.12

9.14 X9.13

9.15 X9.14

9.16 X9.15

9.17 X9.16

9.18 X9.17

9.19 X9.18

9.20 X9.19

9.21 X9.20

9.22 X9.21

9.23 X9.22

9.24 X9.23

9.25 X9.24

9.26 new –

9.27 X9.26

9.28 X9.27

9.29 X9.28

9.30 X9.29

9.31 X9.30

9.32 X9.31

9.33 X9.32

9.34 new –

9.35 X9.33

9.36 X9.34

9.37 X9.35

9.38 X9.36

9.39 X9.37

9.40 X9.38

4th ed Same? 3rd ed

10.1 X10.1

10.2 X10.2

10.3 mod 10.3, 10.5

10.4 X10.4

10.5 X10.6

10.6 X10.7

10.7 new –

10.8 new –

10.9 new –

10.10 X10.8

10.11 X10.9

10.12 X10.10

10.13 X10.11

10.14 X10.12

10.15 X10.13

10.16 X10.14

10.17 X10.15

10.18 X10.16

10.19 X10.17

10.20 X10.18

10.21 mod 12.43a

10.22 X10.19

10.23 X10.20

10.24 X10.21

10.25 X10.22

10.26 new –

10.27 X10.23

10.28 X10.24

10.29 new –

10.30 new –

10.31 X10.25

10.32 X10.26

10.33 new –

10.34 new –

11.1 X11.1

11.2 X11.2

11.3 X11.3

11.4 X11.4

11.5 X11.5

11.6 X11.6

11.7 X11.7

11.8 new –

11.9 X11.8

11.10 X11.9

11.11 X11.12

11.12 X11.10

4th ed Same? 3rd ed

11.13 new –

11.14 X11.14

11.15 X11.15

11.16 X11.16

11.17 X11.17

11.18 new –

11.19 X11.19

11.20 mod 11.20

11.21 new –

11.22 X11.21

11.23 X11.22

11.24 X11.11

11.25 X11.23

11.26 new –

11.27 new –

11.28 X11.24

11.29 X11.26

11.30 X11.27

11.31 X11.28

11.32 X11.29

11.33 X11.30

11.34 X11.31

11.35 new –

12.1 X12.1

12.2 X12.2

12.3 X12.3

12.4 X12.4

12.5 X12.5

12.6 X12.6

12.7 X12.7

12.8 X12.8

12.9 X12.9

12.10 X12.10

12.11 X12.11

12.12 X12.12

12.13 X12.13

12.14 mod 12.14

12.15 X12.15

12.16 X12.16

12.17 X12.17

12.18 X12.18

12.19 X11.19

12.20 X12.20

12.21 X12.21

12.22 X12.22

12.23 X12.23

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

CHAPTER 12. ELECTRODYNAMICS AND RELATIVITY 297

4th ed Same? 3rd ed

12.24 X12.24

12.25 X12.25

12.26 mod 12.26

12.27 new –

12.28 X12.27

12.29 X12.28

12.30 X12.29

12.31 X12.30

12.32 X12.31

12.33 X12.32

12.34 X12.33

12.35 X12.34

12.36 X12.35

12.37 X12.36

12.38 X12.37

12.39 X12.38

12.40 X12.39

12.41 X12.40

12.42 X12.41

12.43 X12.42

12.44 new –

12.45 X12.44

12.46 X12.45

12.47 X12.46

12.48 X12.47

4th ed Same? 3rd ed

12.49 X12.48

12.50 X12.49

12.51 X12.50

12.52 X12.51

12.53 X12.52

12.54 X12.53

12.55 X12.54

12.56 X12.55

12.57 X12.56

12.58 new –

12.59 X12.57

12.60 X12.58

12.61 X12.59

12.62 X12.60

12.63 X12.61

12.64 X12.62

12.65 X12.63

12.66 X12.64

12.67 X12.65

12.68 X12.66

12.69 X12.67

12.70 X12.68

12.71 X12.69

12.72 X12.70

12.73 X12.71

protected under all copyright laws as they currently exist. No portion of this material may be

reproduced, in any form or by any means, without permission in writing from the publisher.

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