Chapter 1 1-1 R - Springerextras.springer.com/2004/978-1-4020-2699-7/EM_solutions.pdf2 Classical...

Chapter 1

1-1 The expression in (Ex 1.3.4) may be approximated when R� z as

Qk

2πε0R2

(1 − z√

R2 + z2

)≈ Qk

2πε0R2

[1 −

(1 − R2

2z2

)]=

Qk

4πε0z2

1-2 In cylindrical coordinates, the electric field integral becomes:

�E(z) =1

4πε0

∫λ0(1 + sinϕ′)(zk − �r ′)δ(r′ − a)δ(z′)r′dr′dϕ′dz′

|zk − �r ′|3

=λ0

4πε0

∫ 2π

0

(1 + sinϕ′)(zk − a cosϕ′ ı− a sinϕ′j)adϕ′

(z2 + a2)3/2

The integrals over ϕ′ are easily performed to give:

�E(z) =λ0

2πε0

[2πazk

(z2 + a2)3/2− πa2j

(z2 + a2)3/2

]

=λ0a

2ε0(a2 + z2)3/2

(zk − 1

2aj)

1-3 After performing the z′ integration, the electric field integral reduces to:

�E(z) =1

4πε0

∫br′2(zk − r′r)dϕ′r′dr′

(z2 + r′2)3/2

=2π

4πε0

∫br′3zk dr′

(z2 + r′2)3/2

where we have used the fact that∮rdϕ′ = 0. The remaining integral yields:

�E(z) =bzk

2ε0

∫ a

0

r3dr′

(z2 + r′2)3/2=bzk

2ε0

(√z2 + r′2 +

z2

√z2 + r′2

)∣∣∣∣a

0

=bzk

2ε0

(2z2 + a2

√z2 + a2

− 2|z|)

It is easily verified that at large z (compared to a) the field has a 1/z 2 form.

1-4 The field of the cylinders is most easily computed using Gauss’ law. Insidethe inner cylinder the enclosed charge is zero implying that the electric fieldmust vanish. Between the two cylinders, the charge enclosed by a gaussiancylinder of radius r and length L is 2πaLσa leading to a field

Er(a < r < b) =2πaLσa

ε02πrL=aσa

ε0r

—1—

2 Classical Electromagnetic Theory

Outside the outer cylinder the charge enclosed by a concentric gaussian cylin-der is 2πL(aσa + bσb) leading to a field

Er(r > b) =aσa + bσb

ε0r

1-5 The charge contained in a gaussian sphere of radius r is

Qr =∫ r

0

ρ0e−kr′

4πr′2dr′

= −4πe−kr′(r′2

k+

2r′

k2+

2k3

)∣∣∣∣r

0

= 4π[

2k3

− e−kr

(r2

k+

2rk2

+2k3

)]The field we deduce is then

Er(r) =1

ε0r2k3

[2 − e−kr(k2r2 + 2kr + 2)

]

1-6 The charge contained in a gaussian sphere of radius r < a is

Q(r < a) =∫ r

0

ρ0

(1 − 2r′

a+r′2

a2

)4πr′2dr′

= 4πρ0

(r3

3− 2r4

4a+

r5

5a2

)while for radii larger than a, the enclosed charge is 4πρ0a

3/30. The field ineach case may then be deduced as

Er =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

ρ0r

ε0

(13− r

2a+

15r2

a2

)for r < a

ρ0a3

30ε0r2for r > a

1-7 The electric field at either plate is σε0. It would be tempting to conclude thatthe force per unit area on the second plate is therefore σ �E = σ2ε0. This iswrong however! It is worth noting that the electric field between the plates isproduced by both plates; it is then reasonable to assume that only half of theelectric field is effective at producing a force on either plate. It is probablymore convincing to calculate the force from elementary considerations. Con-sider an element of charge dq = σdA on one plate and add up all the forcesarising from all the elements of charge dq′ = σdA′ on the other plate. Theforce is then

d�Fq =σdA

4πε0

∫σ′(�r − �r ′)dA′

|�r − �r ′|3 =−σ2dA

4πε0

∫(zk − x′ ı− yj)dx′dy′

(z2 + x′2 + y′2)3/2

Chapter One Solutions 3

=−σ2dA

2ε0

∫ ∞

0

zk r′dr′

(z2 + r′2)3/2=

−σ2kdA

2ε0

1-8 The calculations to this apparently simple problem are surprisingly cumber-some, giving strong motivation to try the dipole approximation of the nextchapter. The net charge on the line is clearly zero while the total electric fieldis given by:

�E =b

4πε0

∫ a

−a

[xı+ yj+ (z − z′)k

]z′dz′

[x2 + y2 + (z − z′)2]3/2

=b

4πε0

∫ a

−a

(xı+ yj)(z′ − z)dz′

[x2 + y2 + (z′ − z)2]3/2+

bz

4πε0

∫ a

−a

(xı+ yj)dz′

[x2 + y2 + (z′ − z)2]3/2

− bk

4πε0

∫ a

−a

(z′ − z)2dz′

[x2 + y2 + (z′ − z)2]3/2− bzk

4πε0

∫ a

−a

(z′ − z)dz′

[x2 + y2 + (z′ − z)2]3/2

=b

4πε0

{ −(xı+ yj)√x2 + y2 + (z′ − z)2

∣∣∣∣a

−a

+z(z′ − z)(xı+ j)

(x2 + y2)√x2 + y2 + (z′ − z)2

∣∣∣∣a

−a

+k(

z′√x2 + y2 + (z′ − z)2

− ln[(z′ − z) +

√x2 + y2 + (z′ − z)2

])∣∣∣∣a

−a

}

1-9 Any remaining field tangential to the surface would cause further movementof the free charges in the conductor.

1-10 It would clearly have been more efficient to do this before calculating the fieldand then to obtain the field by differentiating the potential.

V (z) =∫

ρ d3r′

4πε0|�r − �r ′| =∫ a

−a

bz′dz′

4πε0√x2 + y2 + (z − z′)2

=b

4πε0

√x2 + y2 + (z − z′)2

∣∣∣∣a

−a

+bz

4πε0ln[(z − z′) +

√x2 + y2 + (z − z′)2

]∣∣∣∣a

−a

1-11 The potential due to the ring along the center line is

V (z) =1

4πε0

∫ 2π

0

λ0(1 + sinϕ)adϕ√z2 + a2

=λ0a

4πε02π√z2 + a2

=λ0a

2ε0√z2 + a2


1-12 The potential above the center of the plate is given by

V (z) =1

4πε0

∫br′2r′dϕ′dr′√z2 + r′2

=2πb4πε0

∫r′3dr′√z2 + r′2

=b

2ε0

[13(z2 + r′2)3/2 − z2

√z2 + r′2

]∣∣∣∣a

0

=b

6ε0

[(a2 − 2z2)

√z2 + a2 + 2|z|3

]

1-13 Integrating the continuity equation, �∇· �J = −∂ρ/∂t, over a sphere surround-ing the point source but excluding the wire, we have 4πR2 �J = −I, where Iis the current leading from the point. We conclude then that

J =I

4πR2

with I the current delivered by the wire.

1-14 The magnetic induction field may be found using the Biot-Savart law as fol-lows:

�B =μ0

4π

∫ �J × (�r − �r ′)d3r′

|�r − �r ′|3

The current density �J = ρ�v = σδ(z′)r′ωϕ for r′ < a. Inserting this into theintegral we have

�B(z) =μ0

4π

∫ρr′ωϕ× (zk − r′r)

(z2 + r′2)3/2r′dr′dϕ′

=μ0ρω

4π

∫r′2zr + r′3k(z2 + r′2)3/2

dϕ′dr′

The integrations over ϕ′ eliminates the radial component leaving

�B(z) =μ0σωk

2

∫r′3

(z2 + r′2)3/2dr′

=μ0σωk

2

(√z2 + r′2 +

z2

√z2 + r′2

)∣∣∣∣a

0

=μ0σωk

2

(2z2 + a2

√z2 + a2

− 2 |z|)

Alternatively we could have integrated the magnetic scalar potential of a ringof radius r′ carrying current dI = σωr′dr′ over the whole disk to obtain:

Vm = −∫

zdI

2√z2 + r′2

= −σωz2

∫ a

0

r′dr′√z2 + r′2


= −σωz2

(√z2 + a2 − |z|

)The z component of the magnetic induction field may then be found by dif-ferentiating

Bz = −μ0∂Vm∂z

=μ0σω

2

(√z2 + a2 +

z2

√z2 + a2

− 2 |z|)

=μ0σω

2

(2z2 + a2

√z2 + a2

− 2 |z|)

While there appears to be no clear advantage to using the scalar potential inthis problem, the next two problems use it more advantageously.

1-15 The magnetic scalar potential of the hollow sphere will be calculated as asum of current loops. A loop of width dz′ subtending angle dθ at height z′

has radius a =√R2 − z′2 = R sin θ and carries current dI = aσωdz′/ sin θ

= Rσωdz′. The contribution to the magnetic scalar potential from one suchloop is then

dVm =−(z − z′)dI

2√

(z − z′)2 + a2= −σRω

2(z − z′) dz′√

(z − z′)2 + (R2 − z′2)

= −Rωσ2

(z − z′) dz′√z2 +R2 − 2zz′

Summing loops from −R to R we find

Vm(z) = −σRωz2

∫ R

−R

dz′√z2 +R2 − 2zz′

+σRω

2

∫ R

−R

z′dz′√z2 +R2 − 2zz′

=σRω

2

√z2 +R2 − 2zz′

∣∣∣∣R

−R

−σRω(2z2 + 2R2 + 2zz′)3(2z)2

√z2 +R2 − 2zz′

∣∣∣∣R

−R

= −σR2ω +σR4ω

3z2

It is now simple to obtain the z -component of the magnetic induction fieldBz:

Bz = −μ0∂Vm∂z

=23μ0ωσR

4

z3

The more complete problem of finding the field anywhere is solved as example(5–10) of the text.

1-16 Although we could integrate the field (or potential) of disks such as problem1-13, it is in fact far simpler to sum spherical shells to fill the sphere. Using


the scalar potential from the problem above and dropping the inconsequentialconstant term, we replace R by r′ and σ by ρdr′ and integrate.

Vm(z) =∫ R

0

ρr′4ω3z2

dr′ =ρR5ω

15z2

Differentiating to find the magnetic induction field we find

Bz(z) =2μ0ρR

5ω

15z3

1-17 The field at z from a circular loop of radius a at ±a is

Bz(z) =μ0I

2a2[

(z ∓ 12a)

2 + a2]3/2

At z = 0 both coils give the same contribution to yield

Bz(0) = μ0Ia2[

( 12a)

2 + a2]3/2

=8μ0I

53/2a

1-18 The magnetic field along the axis of a radius a single turn loop located at z′

is�B(z) =

Iμ0

2a2k

[(z − z′)2 + a2]3/2

The number of turns per length dz′ of solenoid is given by (ndz′)/L, so thatthe total field B is

B(z) =nIμ0a

2k

2L

∫ L/2

−L/2

dz′

[(z − z′)2 + a2]3/2

= −μ0nIa2k

2Lz − z′

a2√

(z − z′)2 + a2

∣∣∣∣L/2

−L/2

=μ0nIk

2L

⎡⎣ 1

2L− z√(z − 1

2L)2 + a2+

12L+ z√

(z + 12L)2 + a2

⎤⎦

1-19 Using Faraday’s law,

for r < a 2πrBϕ = μ0

∫JdS =

μ0Iπr2

πa2⇒ Bϕ =

μ0Ir

2πa2

for a < r < b 2πrBϕ = μ0I ⇒ Bϕ =μ0I

2πr

for r > b 2πrBϕ = 0 ⇒ Bϕ = 0


where the thickness of the outer conductor has been neglected.

1-20 The vector potential of a filamentary current may be deduced from (1–52) as

�A(�r ) =μ0

4π

∫ �J(�r ′)|�r − �r ′|d

3r′ =μ0

4π

∮d�′

|�r − �r ′|Alternatively, we recall the magnetic induction field of a filamentary currentloop (1–35)

�B(�r ) =μ0

4π

∮Id�� ′ × (�r − �r ′)

|�r − �r ′|3 = −μ0

4π

∮Id�� ′ × �∇ 1

|�r − �r ′|

=μ0I

4π

∮�∇× d�� ′

|�r − �r ′| = �∇× μ0I

4π

∮d�� ′

|�r − �r ′|which allows us to identify the vector potential with the integral. We proceedto find the divergence.

�∇ · �A = �∇ · μ0I

4π

∮d�� ′

|�r − �r ′| =μ0I

4π

∮�∇ ·(

d�� ′

|�r − �r ′|)

= −μ0I

4π

∮d�� ′ · �∇′ 1

|�r − �r ′| = −μ0I

4π

∫S′

(�∇′ × �∇′ 1

|�r − �r ′|)dS′ = 0

where we used Stokes’ theorem (18) to effect the last step and the fact that agradient has zero curl. The point of this exercise is to show that the simpleexpression derived from (1–52) expresses �A in the Coulomb gauge.

1-21 The Laplacian of the vector potential may be written using (1–52)

∇2 �A(�r ) =μ0

4π∇2

∫ �J(�r ′)|�r − �r ′|d

3r′ =μ0

4π

∫�J(�r ′)∇2 1

|�r − �r ′|d3r′

∇2(1/|�r− �r ′|) may be rewritten with the aid of (26) as −4πδ(�r− �r ′) so thatthe above becomes

∇2 �A(�r ) = −μ0

∫�J(�r ′)δ(�r − �r ′)d3r′ = −μ0

�J(�r )

1-22 For simplicity we place the z axis along one of the wires. From this wire alone,the nonzero component of the field is given by

Bϕ(r < a) =μ0Ir

2πa2= −∂Az

∂r

from which we conclude that

Az(r < a) = −μ0Ir2

4πa2+ C


and similar reasoning gives the vector potential due to this wire outside thewire as

Bϕ(r > a) =μ0I

2πr⇒ Az(r > a) =

μ0I

2πln r +D

At a point �r outside the two wires, the vector potential is then given by

Az(r > a) = −μ0I

2πln(r

r2

)+ F

where r2 is the distance of the point from the center of the second wire’scenter and F is an arbitrary constant. At a point inside the wire containingthe z-axis, the vector potential is

Az(r < a) = −μ0Ir2

4πa2− μ0I

2πln r2 + E

where E is an arbitrary constant. In either case the distance r2 of the fieldpoint from the second wire may be expressed in terms of the point’s polarangle using

r2 =√h2 + r2 − 2hr cos θ

1-23 In line with the considerations of section 1.2.2, The transverse force on par-ticles must obey F ′ = γ−1F . The inverse Lorentz factor γ−1 =

√1 − β2 =√

1 − .992 = 0.141 The mutual repulsive force is therefore decreased to 14.1%of its electrostatic (rest frame) value.

1-24 The magnetic scalar potential along the axis is given by (p. 29)

Vm(z) = −NI2L

(√(z + 1

2L)2 + a2 −√

(z − 12L)2 + a2

)Differentiating to obtain the magnetic induction field,

�B = −μ0�∇Vm =

μ0Ik

2L

(z + 1

2L√(z + 1

2L)2 + a2+

12L− z√

(z − 12L)2 + a2

)

Figure 1.1: Geometry of solenoid in question 1.24.

With reference to the diagram, it is evident that the two fractions are re-spectively the cosines of the half angles subtended by the ends of the solenoid:

Bz =μ0I

2L(cos θ1 + cos θ2)


1-25 In general a charged particle in an electromagnetic field experiences a Lorentzforce

�F = q( �E + �v × �B)

We must chose �v so that no further acceleration or deceleration occurs, inother words

�E = −�v × �B

To isolate �v, we cross multiply both sides with �B to obtain

�E × �B = −(�v × �B) × �B = (�v · �B) �B −B2�v

Since the component of �v along �B cannot give rise to forces, we find

�v = −�E × �B

B2

Despite my best efforts a few typos have crept into the text, they will bereported at the end of each chapter.

The denominator of the first term of (1–36) is missing an r2 to become

=μ0I2(k × r)

4πz′

r2√r2 + z′2

∣∣∣∣∞

−∞

Chapter 2

2-1 The dipole moment of the ring is

�p =∫�r λδ(r − a)δ(z) cosϕ rdrdϕdz

=∫a(ı cosϕ+ j sinϕ)a cosϕdϕ

= λπa2 ı

2-2 The dipole moment of the rod is

�p =∫ a

−a

zkλzdz =2λa3k

3

2-3 We use (10) to expand the curl of �A of a dipole as given in (2–23):

�∇× �A(�r ) = �∇× μ0

4πr3(�m× �r ) =

μ0

4π

[�m

(�∇ · �r

r3

)− (�m · �∇)

(�r

r3

)]

The �∇ · (�r/r3) occurring in the first term is readily evaluated using (7)

�∇ · �rr3

= �∇(

1r3

)· �r +

1r3�∇ · �r = 0

and the second term may be evaluated from

(�m · �∇)(�r

r3

)= mx

∂

∂x

�r

r3+my

∂

∂y

�r

r3+my

∂

∂y

�r

r3

focussing on the first of these three terms,

mx∂

∂x

�r

r3= mx

(ı

r3− 3x�r

r5

)

and similar expressions obtain for the other two terms. Adding the threeterms gives

�∇× �A(�r ) = − �m

r3+

3(�m · �r )�rr5

2-4 Rather than computing the dipole moment about the origin let us calculate amodified dipole moment about an arbitrary point �a.

�pa =∫

(�r − �a )ρ(r)d3r

=∫�rρ(r)d3r − �a

∫ρ(r)d3r

= �p0 − �aQ

— 10—

Chapter Two Solutions 11

We conclude that whenever the total charge Q vanishes, the dipole momentabout �a is identical to that about the origin.

2-5 The zz component is easily found:

Qzz =∫ρ(3z2 − r2)d3r′ = q(b2 − a2)

and Qxx = q(3〈x2a〉−a2)−q(3〈x2

b〉−b2) = 12q(a

2−b2) = Qyy. The off-diagonalelements vanish.

2-6 The representation of this quadrupole will depend on the relative orientationof the square and the coordinate system. Let us place the corners at ± 1

2awith a positive charges along the y ± 1

2a edges while negative charges resideon the x = ± 1

2a edges. Then

Qzz =∫

(3z2 − r2)ρ(�r )d3r = 0

Qyy =∫

(3y2 − r2)ρ(�r )d3r

=2qa

∫ 12 a

− 12 a

( 34a

2 − x2 − 14a

2)dx− 2qa

∫ 12 a

− 12 a

(3y2 − 14a

2 − y2)dy

=2qa

(a2x

2− x3

3

)∣∣∣∣12 a

− 12 a

− 2qa

(2y3

3− a2y

4

)∣∣∣∣12 a

− 12 a

= qa2

and Qxx = −qa2. The remaining components such as Qxz and Qxy all vanish.

2-7 For an object spinning about the z-axis, let us denote by r the distance of amass or volume element from the axis and let ρm denote the mass density.Then the angular momentum �L is given by

�L = I�ω = kω

∫r2dm = kω

∫r2ρmd

3r

The magnetic moment is given by

�m = 12

∫�r × �Jd3r = 1

2

∫�r × (�ω × �r )ρd3r

= 12

∫r2ωkρd3r

Comparing the two expressions, when the functional form of ρ and ρm is thesame, the ratio of (�m/�L) = 1

2ρ/ρm = 12q/m.

2-8 The moment of inertia of the solid sphere is 25mR

2 while the magnetic momentof the hollow shell may be found as a sum of plane loops:

�m = k

∫πr2dI = k

∫ R

−R

π(R2 − z2)R sin θωσdz

sin θ

= πσωk

(2R4 − 2R4

3

)= 1

34πωσR4k = 13QR

2ωk


The gyromagnetic ratio is then 56Q/m.

2-9 The zz component of the quadrupole moment is given by

Qzz =∫ 1

2 L

− 12 L

η(z2 − L2

12)(3z2 − z2)dz

=L5η

90

while the xx and yy components are each −L5η/180. The off-diagonal ele-ments vanish as would be expected.

2-10 The potential due to the quadrupole when Qxx = Qyy = − 12Qzz is

V (�r ) =1

4πε0

∑ xixjQij

2r5=

18πε0r5

(z2Qzz + x2Qxx + y2Qyy

)=

Qzz

8πε0r5(r2 cos2 θ − 1

2r2 sin2 θ cos2 ϕ− 1

2r2 sin2 θ sin2 ϕ

)=

Qzz

16πε0r3(2 cos2 θ − sin2 θ

)=

Qzz

16πε0r3(3 cos2 θ − 1)

2-11 The charge on a radius R sphere with charge density ρ = ρ0z2 = ρ0r

2 cos2 θis

Q =∫ρd3r = ρ0

∫r2 cos2 θr2dr sin θdθdϕ

=2πρ0R

5

5

∫ π

0

sin θ cos2 θdθ =4πρ0R

5

15

2-12 The zz component of the quadrupole moment of the sphere in problem 2-11is

Qzz =∫

(3z2 − r2)ρd3r

= ρ0

∫3z4d3r − ρ0

∫r2z2d3r

= 2πρ0

∫3r6 cos4 θ sin θdθdr − 2πρ0

∫r6 cos2 θ sin θdrdθ

=12πρ0a

7

5 · 7 − 4πρ0a7

3 · 7 =16πρ0a

7

105

2-13 Expanding the given expression,

V2 = − 14πε0

∑qi�r

(i) · �∇(

1r

)

=1

4πε0

∑qi�r

(i) · �rr3

≡ �p · �r4πε0r3

and we recover (2–4).


2-14 We expand the form given:

V3 =1

8πε0

∑i,j

qi�r(i) · �∇

(−x(i)j xj

r3

)

=−1

8πε0

∑i,j,k

qix(i)k x

(i)j ∂k

(xj

r3

)

=1

8πε0

∑i,j,k

qix(i)k x

(i)j

(3xjxk

r5− δjk

r3

)

=1

8πε0

∑i,j,k

qix(i)k x

(i)j

(3xjxk

r5− δjkxjxk

r5

)=∑j,k

xjxkQjk

8πε0r5

and we recover the point charge form of (2–16).

2-15 Substituting the charge density ρ = ρ0z for �r ′ on the z axis z′ ∈ (−a, a) andzero elsewhere into (2–32), we note that γ = θ, the field point polar angle sothat we may remove it from the integral to obtain form (2–33). Then

V (�r ) =1

4πε0

∞∑n=0

Pn(cos θ)rn+1

∫ a

−a

ρ0z′ z′ndz′

=ρ0

4πε0

∑ Pn(cos θ)rn+1

z′n+2

n+ 2

∣∣∣∣a

−a

=ρ0

2πε0

∑ P2n+1(cos θ)a2n+3

(2n+ 3)r2n+2

2-16 When r is less than a, we must split the integral into two portions, the firstrunning from −r to r where r′ = r< and the second running from r to awhere r′ = r>. Thus

V (r < a, θ) =ρ0

4πε0

∑Pn(cos θ)

{1

rn+1

∫ r

−r

z′n z′dz′

+ rn

[ ∫ a

r

z′dz′

z′n+1+∫ −r

−a

z′dz′

z′n+1

]}

=ρ0

4πε0

∑Pn(cos θ)

{z′n+2

(n+ 2)rn+1

∣∣∣∣r

−r

− rn

(n− 1)

[1

z′n−1

∣∣∣∣a

r

+1

z′n−1

∣∣∣∣−r

−a

]}

=ρ0

4πε0

∑n odd

Pn(cos θ){

2rn+2

(n+ 2)rn+1− rn

(n− 1)

[2

an−1− 2rn−1

]}

=ρ0

2πε0

∑n odd

Pn(cos θ){

r

(n+ 2)+

r

(n− 1)

[1 − rn−1

an−1

]}

2-17 We take the electrons to be at x = ± a at time t = 0 and to rotate in the x -yplane. In (a), the electrons have coordinates:

x1 = a cosωt y1 = a sinωt

x2 = −a cosωt y2 = −a sinωt


The dipole moment is then easily seen to be zero while the quadrupole momenthas nonzero components:

Qyx = Qxy = −6ea2 cosωt sinωt

Qxx = −2ea2(3 cos2 ωt− 1

)Qyy = −2ea2

(3 sin2 ωt− 1

)and Qzz = 2ea2. The counter-rotating electrons of (b) have coordinates

x1 = a cosωt y1 = a sinωt

x2 = −a cosωt y2 = a sinωt

The dipole moment is then �p = −2eaj sinωt while the quadrupole momenthas components (about the nucleus since the result is no longer unique) Qxy =Qzx = Qyz = 0 while the diagonal components are the same as those for (a).

2-18 For simplicity we place one dipole at the origin so that the second is locatedat �r. The force on the dipole may be found from

�F = (�p2 · �∇) �E

where �E is the field at �r due to the first dipole �p1. With the dipole field foundin Example 2.1, and abbreviating ∂/∂x ≡ ∂x, etc.

�F =−1

4πε0(p2x∂x + p2y∂y + p2z∂z)

1r3

[�p1 − 3(�p1 · �r )�r

r2

]

=1

4πε0

{3[(�p2 · �r )�p1 + (�p1 · �r )�p2 + (�p1 · �p2)�r

]r5

− 5(�p1 · �r )(�p2 · �r )�rr7

}

The more general result of arbitrarily located dipoles is obtained by replacing�r by (�r2 − �r1).

2-19 The potential at �r due to a quadrupole at the origin is (assuming summationover repeated indices)

V (�r ) =1

4πε0xixjQij

2r5

The potential energy of a dipole in this potential is

W = −�p · �E = − 14πε0

p�∂�

(xixjQij

2r5

)

The force on the dipole may be found as minus the gradient of its potentialenergy. The k component of the force �F is then

F k =1

4πε0p�∂�∂

k

(xixjQij

2r5

)


=1

4πε0p�∂�

(δikxjQij + δjkxiQij

2r5− 5xkxixjQij

2r7

)

=1

4πε0p�∂�

(xjQk

j + xiQki

2r5− 5xkxixjQij

2r7

)

=p�

4πε0

(Qk

� +Qk�

2r5−5

δk� x

ixjQij +xkxjQ�j + xkxiQi�

2r7+

35x�xkxixjQij

2r9

)

=1

4πε0

(p�Qk

�

r5− 5

pk(xixjQij) + 2xk(xjp�Qj�)2r7

+35xk(p�x�)(xixjQij)

2r9

)

Thus

�F =1

4πε0

(�p · ↔Qr5

− 5�p (xixjQij) + 2�r (xjx�Qj�)

2r7+ 35

�r (�p · �r )(xixjQij)2r6

)

2-20 The potential from the hypothetical monopoles would be

Vm =qm4π

(1

|�r − 12�a|

− 1|�r + 1

2�a|)

=qm4πr

[(1 − �a · �r

r2+

a2

4r2

)−1/2

−(

1 +�a · �rr2

+a2

4r2

)−1/2]

=qm

4πr3�a · �r + O [(a/r)3]

Defining the magnetic moment as �m = qm�a, we find in the limit as a → 0that Vm → �m · �r/4πr3 so that we reproduce the magnetic scalar potential ofthe dipole.

2-21 Generally, the equation of motion is �τ = �L/dt. Substituting �τ = �m× �B and�L = a�m with a−1 = 2.79 e/me we write

�m× �B = ad�m

dt

For simplicity we take �B to be directed along the z axis so that the equationof motion, one component at a time, may be written

myBz = admx

dtmxBz = −admy

dt0 =

dmz

dt

We differentiate the the first of these once more with respect to time andsubstitute for dmy/dt using the second

d2mx

dt2=Bz

a

dmy

dt= −B

2z

a2mx


The solution formx is easily found to bemx = m cos(ω0t+δ) with ω0 = B/a =2.79 eB/me. Reverting to the first order equations we have ω0my = dmx/dt,or my = −m sin(ω0t + δ). The dipole moment has a constant z componentand the x and y components rotate about the z axis at angular frequency ω0.

2-22 The potential can immediately be written as a summation of the individualdipole fields:

V (�r ) =1

4πε0

∫S

�D · (�r − �r ′)|�r − �r ′| d2r′

2-23 The force on the dipole takes the form �F = −�∇(−�m· �B) = �∇(mxBx+myBy +mzBz) . Assuming that �B has only a variation with z we have

�F =(mx

∂Bx

∂zı+my

∂By

∂zj+mz

∂Bz

∂zk

)

= αm cos θk

2-24 The residence time of the quadrupole in the field is 10−3m/(100 m/s) = 10−5s.The force required to impart the required impulse is then 10−21 N. In generalthe force may be obtained from the potential energy of the quadrupole:

�F = −�∇W = − 16�∇(Q�m

∂2V

∂x�∂xm

)

= 16�∇Qzz

∂

∂z

(− ∂V

∂z

)= 1

6�∇(Qzz

∂Ez

∂z

)

The k component of the force is then

Fk = 16Qzz

∂2Ez

∂z∂xk

leading to a requirement of (∂2Ez)/(∂z∂xk) of order 6×1018V/m3. Assumingthis field is to be established at the tip of a charged needle, it is interestingto consider the size of tip required. Let us assume the tip is charged toV0 = 1000 V; the third derivative of the potential around a spherical tip justbecomes 6V0/R

3. Equating this to the required inhomogeneity above leadsto R = 10−5 m. While a 10 μm radius tip is not unreasonable, it is clearlyimpossible to maintain the required field gradient over distance the order ofa millimeter.

Chapter 3

3-1 From the definition of the magnetic flux and with the aid of Stokes’ theorem(18),

Φ =∫

�B · d�S =∫

(�∇× �A) · d�S =∮

�A · d��

3-2 The current I circulating in the loop may be found from the induced EMF(disregarding signs),

I =ER

=πa2

R

dB

dt

The resulting field at the center of the loop is

Bind =μ0

2ππa2I

a3=μ0πa

2RdB

dt

The direction of the induced field must be chosen, according to Lenz’ law tooppose the increase or decrease of the background field. There is a difficultywith this result: the induced field becomes arbitrarily large as the resistanceis decreased. The solution is of course that the induced field’s dB/dt mustbe included in the flux change, when the resistance vanishes the the inducedcurrent’s flux keeps the total flux in the loop constant.

3-3 To have a stable orbit the magnetic force on the electron must provide thecentripetal acceleration so that, disregarding signs,

�Fcent =mev

2

r= qvB ⇒ |v| =

qrB(�r )me

The tangential electric field that accelerates the electron may be obtained(again disregarding signs) from

∫�E(�r ) · d�� = −

∫d �B(�r )dt

· d�S

2πrEϕ = πr2dB

dt⇒ Eϕ(�r ) =

r

2dB

dt

(where B is the magnetic field averaged over the area included by the elec-tron’s orbit) and substituted into Newton’s second law to obtain the acceler-ation:

med|v|dt

= qE =qr

2dB

dt

in other words,d|v|dt

=qr

2me

dB

dt

Integrating both sides, we find |v| =qrB

2me.

— 17—


Comparing the two expressions we find that we require B = 12B(r). The

space average field inside the orbit must be half the field at the orbit.

3-4 The magnetic induction field in the torus with center line in the x-y planeat distance x from the center, [ x ∈ (a − b, a + b)] is easily seen to be B =μ0NI/2πx. The flux Φ may be found by integrating this field over a cross-section in the x -z plane.

Φ =∫

�B · d�S =∫ a+b

a−b

∫ √b2−(x−a)2

−√

b2−(x−a)2

μ0NI

2πxdzdx

=μ0NI

2π

∫ a+b

a−b

2√b2 − (x− a)2

xdx

=μ0NI

ππ(a−√

a2 − b2)

= μ0NI(a−√

a2 − b2)

The integral is far from trivial, however, it may be verified that as a b, theflux reduces to that of a long solenoid.

3-5 The electric field between the plates is given in terms of the surface chargedensity by �E = σkε0, so that

�∇× �B = μ0ε0∂ �E

∂t= μ0k

dσ

dt= μ0k

I

A

We draw a circle of radius r between the plates, centered on the center ofsymmetry and integrate both sides over the area of this circle. The left handside may be replaced by an integral along the boundary so that, assuming auniform electric field, we obtain∮

Bϕ(r) · d� = μ0k ·∫

I

Ad�S

giving 2πrBϕ(r) =μ0πr

2I

A⇒ Bϕ =

μ0 rI

2A

3-6 At large distances r the electric field included within the loop is that cor-responding to the total charge on the plates. Repeating the reasoning ofproblem 3-5, ∮

�B · d�� = μ0I

giving Bϕ =μ0I

2πr

3-7 The power dissipated by the pendulum written in mechanical terms may beequated to that written in electrical terms to give

ωτ = EI =E2

R

Chapter Three Solutions 19

and the EMF generated is

E = AdB⊥dt

= AdB⊥dθ

dθ

dt=dB⊥dθ

Aω

so that the dissipative torque on the loop due to power generation is

τ =E2

Rω=A2

(dB⊥dθ

)2

ω

R

The moment of inertia of the loop about the point of suspension is m(�2+a2),so the equation of motion becomes

m(�2 + a2)d2θ

dt2+

(πa2

)2R

(dB⊥dθ

)2dθ

dt+mg� sin θ = 0

3-8 The charge delivered by the coil may be written as the time integral of thecurrent in the coil which itself may be calculated from the EMF induced inthe coil as the coil flips.

I =ER

=1R

d(BA)dt

=BA

R

d

dt(cos θ)

Identifying the time before the coil is flipped as t0 and the time after as tπ,we find the charge may be found as

Q =∫ tπ

t0

Idt =∫ tπ

t0

BA

R

d

dt(cos θ)dt

=BA

Rcos θ

∣∣∣∣π

0

=2BAR

It is evident that the charge generated is independent of the speed of flippingthe coil.

3-9 Writing out the Lagrangian in terms of the cartesian components of positionand velocity we have

L = 12m(x2 + y2 + z2) − qV + q(xAx + yAy + zAz)

Then∂L∂x

= mx+ qAx,∂L∂y

= my+ qAy and∂L∂z

= mz+ qAz. The canonical

momentum is then given by �p+ q �A in agreement with (3–38)

3-10 We take the electric field to have the form �E = �E0ei(�k·�r−ωt), with �E0 inde-

pendent of the coordinates. Then �∇ · �E is given by


�∇ · �E = ∂jEj = Ej

0∂j

[ei(�k·�r−ωt)

]= iEj

0ei(�k·�r−ωt)∂j

(kix

i − ωt)

= iEj0δ

ijki = ikjE

j = i�k · �E

In similar fashion, the j component of �∇× �E is given by

(�∇× �E)j = εjk�∂kE� = iεjk�E�∂k (kmrm − ωt)

= iεjk�δmk kmE�

= iεjk�kkE� = i(�k × �E)j

3-11 To obtain a wave equation for �B we take the curl of

�∇× �B = μ0�J + μ0ε0

∂ �E

∂t

and use (13), �∇× (�∇× �B) = �∇(�∇ · �B) −∇2B to get

�∇(�∇ · �B) −∇2 �B = μ0�∇× �J + μ0ε0

∂

∂t(�∇× �E)

The curl of �E is replaced by −∂ �B/∂t, and �∇ · �B and �J both vanish to leave

−∇2 �B = −μ0ε0∂2 �B

∂t2

3-12 To obtain a reasonable ionization rate, we need eV = eE/d. The electric fieldmust therefore be at least

E =10 V

10−9m= 1010V/m

The corresponding irradiance I is found from

I =∣∣∣∣ �E × �B

μ0

∣∣∣∣ = |E|2μ0c

= 2.65 × 1017W/m2

A (modest) 10 MW pulse, focussed to a (10 μm)2 spot would provide I =1017 W/m2 so that we see that air sparks should be easily produced.

3-13 We wish to show that �ξ = �∇ψ and �ξ ′ = �r× �∇ψ solve the vector wave equation

∇2�ξ =1c2∂2�ξ

∂t2


whenever ψ solves the scalar wave equation. We recall the definition of thevector Laplacian (13),

∇2�ξ = �∇(�∇ · �ξ) − �∇× (�∇× �ξ)

and substitute each of the forms �ξ and �ξ ′ in turn. Thus for �ξ = �∇ψ

∇2(�∇ψ) = −�∇× [�∇ × (�∇ψ)] + �∇(�∇ · �∇ψ) = �∇(∇2ψ)

since the gradient of any function is curl free. Substituting for ∇2ψ from thescalar wave equation, we find

∇2(�∇ψ) = �∇(

1c2∂2ψ

∂t2

)=

1c2∂2

∂t2(�∇ψ)

which proves the required result for �∇ψ. The equivalent result for ψ′ is alittle more laborious.

∇2(�r × �∇ψ) = �∇[�∇ · (�r × �∇ψ)] − �∇× [�∇× (�r × �∇ψ)]

= �∇{�∇ · [−�∇× (�rψ)]} − �∇× [�∇× (�r × �∇ψ)]

The first term on the right side vanishes because the curl of any functionhas no divergence. In the remaining term, we expand the expression withinsquare brackets

�∇× (�r × �∇ψ) = �r (�∇ · �∇ψ) − (�∇ψ)(�∇ · �r ) + (�∇ψ · �∇)�r − (�r · �∇)�∇ψ= �r∇2ψ − 3�∇ψ + �∇ψ − (�r · �∇)�∇ψ

The last term is most easily evaluated by adding the null term (�∇ψ · �∇)�r− �∇ψto it. Thus

(�r · �∇)�∇ψ+ (�∇ψ · �∇)�r − �∇ψ= �∇(�r · �∇ψ) − �r × (�∇× �∇ψ) − �∇× (�∇× �r) − �∇ψ

Because both �r and �∇ψ have zero curl, the two middle terms vanish. Further-more, taking the curl once more (as we must) eliminates �∇ψ and �∇(�r · �∇ψ)as well so that we find

�∇× [�∇× (�r × �∇ψ)] = �∇× (�r∇2ψ) = �∇×(�r

1c2∂2ψ

∂t2

)

=1c2

∂2

∂t2

(�∇× �rψ

)= − 1

c2∂2

∂t2

(�r × �∇ψ

)

Finally, ∇2(�r × �∇ψ) =1c2∂2

∂t2(�r × �∇ψ)


The relative difficulty of the foregoing is largely circumvented if tensornotation and the Levi-Cevita symbol are used instead, as will be demonstratedbelow.

[∇2(�r × �∇ψ)]k = ∂i∂i(εk�mx�∂mψ

)= εk�m∂i∂

i(x�∂mψ)

= εk�m(∂iδi�∂mψ + x�∂i∂

i∂mψ)

= εk�m∂�(∂mψ) + εk�mx�∂m(∂i∂iψ)

= [�∇× (�∇ψ)]k + εk�mx�∂m(∇2ψ)

= [�r × �∇(∇2ψ)]k =[�r × �∇

(1c2∂2ψ

∂t2

)]k

=1c2

∂2

∂t2(�r × �∇ψ)k

where the first term in the sum was eliminated because a gradient has no curl.

3-14 Generally, the function �A(�r + δ�r, t) may be expanded to first order as

�A(�r + δ�r, t) = �A(�r, t) +∂ �A(�r, t)∂x

· δx+∂ �A(�r, t)∂y

· δy +∂ �A(�r, t)∂z

· δz

Substituting δx = vxdt, δy = vydt, and δz = vzdt, we find

�A(�r + �vdt, t) = �A(�r, t) + (�v · �∇) �Adt

3-15 The Coulomb gauge poses �∇ · �A = 0. Replacing �A by �A′ = �A+ �∇Λ, we find

�∇ · �A′ = �∇ · �A+ ∇2Λ

If ∇2Λ = 0, the Coulomb gauge is clearly preserved.

3-16 In the Lorenz gauge, �A and V satisfy

�∇ · �A+1c2∂V

∂t= 0

Replacing �A by �A′ = �A+ �∇Λ and V by V ′ = V −∂Λ/dt, the gauge conditionbecomes

�∇ · �A′ +1c2∂V ′

∂t= �∇ · �A+ ∇2Λ +

1c2∂V

∂t− 1c2∂2Λ∂t2

It is evident that in order to preserve the Lorentz gauge, the gauge functionΛ must satisfy

∇2Λ − 1c2∂2Λ∂t2

= 0


3-17 Electrons in the spinning disk experience a force e�v× �B. Neglecting the vectorcharacter and writing only the magnitude this is rωB. Integrating the forceper unit charge from r = 0 to r = a on the rim, we obtain the EMF

E =∫ a

0

rωBdr =r2ωB

2

3-18 The “paradox” arises because the current in the loop has been neglected indetermining the potential measured by the voltmeter. Suppose, for simplicitythat the loop has the same resistance R in the bottom half and in the tophalf of the loop. The current of magnitude

I =1

2RdΦdt

flows in a clockwise direction around the loop in response to the changingmagnetic field indicated. The potential measured in circuit b is then IR =12dΦ/dt and the potential measured in c is dΦ/dt − IR = 1

2dΦ/dt. Thisargument works equally well if we define unequal resistances for the top andbottom half of the loop.

Chapter 4

A number of readers have complained that the integral in (Ex 4.4.12) is notobvious. Admittedly I used tables to to find∫ π

0

ln(a2 − 2ab cos θ + b2)dθ ={

2π ln a a ≥ b ≥ 02π ln b b ≥ a ≥ 0

Indeed, there seems to be no elementary method for obtaining this result.However, computing the potential of a charged cylinder of radius a via twodifferent methods closely reproduces the required result. We begin by Us-ing Gauss’ law to find the electric field and consequent potential around thecylinder. Assuming charge density λ per unit length,

�E =λr

2πε0r⇒ V (r > a) =

−λ2πε0

ln r +A =−λ2πε0

ln(r

b

)

where we have imposed that V vanish at a distant point b. Alternatively,we consider each the cylindrical shell to be composed of filaments runningparallel to the axis, each subtending angle dϕ with respect to the axis andcarrying line charge density (λ/2π)dϕ. Summing the potentials from each ofthese filaments

dV (�r ) =−λdϕ′

2π1

2πε0ln

|�r − �r ′|b

=−λdϕ′

2π1

2πε0ln

√a2 + r2 − 2ar cosϕ′

b2

which may be integrated to give

V (�r ) =−λ2πε0

12π

∫ 2π

0

ln


b2dϕ′

Comparing the two expressions for V (r > a),

∫ 2π

0

ln


b2dϕ′ = 2π ln

(r

b

)

For a not so distant point �b, the correct form to integrate would be

dV (�r ) =−λdϕ′

(2π)2ε0ln

|�r − �r ′||�b− �r ′|

leading to

V (�r ) =−λ2πε0

12π

(∫ 2π

0

ln√a2 + r2 − 2ar cosϕ′dϕ′

−∫ 2π

0

ln√a2 + b2 − 2ab cosϕ′dϕ′

)

— 24—

Chapter Four Solutions 25

Denoting the first integral as F (r), the second is F (b) and we have ln(r/b) =12π

[F (r)−F (b)

]allowing us to conclude that except for a constant multiplier,

F (r) = 2π ln r. Coupled with the foregoing discussion that multiplier shouldbe 1.

4-1 The potential energy of all the charges is given by

W =1

8πε0

8∑i=1

8∑j=1j �=i

qiqjri,j

The inner sum is easily evaluated with the aid of a diagram. Any charge hasthree neighbors at distance a, three at distance

√2a, and one at

√3a. The

inner sum is therefore

8∑j=1j �=i

qiqjri,j

=q2

a

(3 +

3√2

+1√3

)

Every other charge contributes exactly the same to the potential energy sum,therefore

W =8q2

8πε0a

(3 +

3√2

+1√3

)=

5.7 q2

πε0a

4-2 The inductance of the solenoid may be found by equating the energy of theenclosed field to 1

2LI2 or by differentiating the flux at any turn with re-

spect to I and summing over the turns. Since the magnetic induction field isnearly constant through the volume of the solenoid and nearly zero outsidethe solenoid, either method ought to work. For either method we need thefield of a solenoid of length � and N turns: B = μ0NI�.

Using energy:

W = 12LI

2 =∫B2

2μ0d3r

=(μ0NI

�

)2πR2�

2μ0=πμ0I

2N2R2

2�

from which we deduce

L =πμ0N

2R2

�

From the flux:

L = N∂Φ∂I

= N∂

∂I

∫μ0NI

�dS =

μ0πR2N2

�

4-3 For the centerline of the torus in the x-y plane, the magnetic induction fieldover a cross-section in the x-z plane is given by (see also problem 3-4, note


however, that a and b are reversed) By = μ0NI/(2πx). A volume element inthe neighborhood of the x-z plane is d3r = xϕdS so that we may write

W =∫

B2

2μ0d3r =

∫μ2

0N2I2

8μ0π2x2xdϕdS

=μ0N

2I2

4π

∫ b+a

b−a

∫ √a2−(x−b)2

−√

a2−(x−b)2

1xdzdx

Exactly the same integral was evaluated in problem 3-4 to give

W =μ0N

2I2

2

(b−

√b2 − a2

)leading us to deduce that the inductance of the torus is

L = μ0N2(b−

√b2 − a2

)We could of course have differentiated the flux Φ found in problem 3-4 withrespect to the current to obtain the same result.

4-4 In order to have the same cross-section, the outer conductor must have anouter radius c that satisfies π(c2 − b2) = πa2 implying c2 = a2 + b2. Thefield in the various regions is easily obtained; in particular, because a loopencircling both inner and outer conductor has zero net current included, thefield outside the wire must vanish. This fact makes it practical to calculatethe total energy content of the fields established by the currents.

The field inside the inner wire is obtained from∮�B · d� = μ0

∫�J · d�S =

μ0r2

a2I ⇒ �B(r < a) =

μ0Ir

2πa2

Between the wires, 2πr �Bϕ = μ0I ⇒ B = μ0I2πr . Finally, in the outer conductor

2πrBϕ = μ0

(I −

∫ r

b

�J · d�S)

= μ0I

(1 − 1

πa2

∫ r

b

2πr′dr′)

= μ0I

(1 − r2 − b2

a2

)

The energy for a unit length of the field is then

W =μ0I

2

4π

{∫ a

0

r3

a4dr +

∫ b

a

1rdr

+∫ √

a2+b2

b

1r

[(1 +

b2

a2

)2

− 2r2

a2

(1 +

b2

a2

)+r4

a2

]dr

}

=μ0I

2

4π

[14

+ lnb

a+(

1 +b2

a2

)2

ln

√1 +

a2

b2

− (a2 + b2) − b2

a2

(1 +

b2

a2

)+

(a2 + b2

)2 − b4

4a4

]


=μ0I

2

4π

[12

(− 1 − b2

a2

)+ ln

b

a+

12

(1 +

b2

a2

)2

ln(

1 +a2

b2

)]The inductance is now easily deduced to be

L =μ0

4π

[lnb2

a2+(

1 +b2

a2

)2

ln(

1 +a2

b2

)−(

1 +b2

a2

)]

4-5 We place one of the wires along the z axis and the other parallel to the firstin the x-z pane at distance h from the first. The magnetic induction field inthe x-z plane between the wires of the first wire is then,

�B(�r ) =μ0Ij

2πx

and that from the second wire is

�B(�r ) =μ0Ij

2π(h− x)

The flux through a loop of width (h− 2a) between the wires is

μ0I

2π

∫ ∫ (1x

+1

h− x

)dx dz =

μ0I�

2π

∫ h−a

a

(1x

+1

h− x

)dx

=μ0I�

πln(h− a

a

)

The inductance may be found as dΦ/dI

dΦdI

=μ0�

πln(h

a− 1)

When h a, this result is similar to (Ex 4.4.13).

4-6 The electric field inside the uniformly charged sphere is obtained from Gauss’law ∮

�E · d�S =∫

ρ

ε0d3r

or 4πr2E =Qr3

ε0a3⇒ E =

Qr

4πε0a3

Outside the sphere, the field is E =Q

4πε0r2. The energy is then

W =ε02

∫E2d3r

=ε02

∫ a

0

Q2r2

(4πε0)2a64πr2dr +

ε02

∫ ∞

a

Q2

(4πε0)2r44πr2dr

=Q2

8πε0

(∫ a

0

r4

a6dr +

∫ ∞

a

1r2dr

)=

3Q2

20πε0a


4-7 Using Gauss’ law to find the field inside and outside the charge

E(r < a) =1

4πε0r2

∫ r

0

ρ0

(1 − r′ 2

a2

)4πr′ 2dr′ =

ρ0

ε0

(r

3− r3

5a2

)

and E(r ≥ a) =Q

4πε0r2with Q = ρ0

(8πa3

15

)

The energy is then

W =ε02

[∫ ∞

a

Q24πr2dr(4πε0)2r4

+∫ a

0

ρ20r

2

ε20

(13− r2

5a2

)2

4πr2dr]

=Q2

8πε0a+

2πρ20

ε0

(a5

45− 2a7

105a2+

a9

225a4

)

=16πρ2

0a5

315ε0=

5Q2

28πε0a

4-8 The magnitude of the Poynting vector S is

S =E2

2μ0c=

E2

2 × 4π × 10−7 × 3 × 108 m/s

If we set S = 5W/.5 × 10−6m2 = 107 W/m2, we find E = 8.68 × 104 V/m.

4-9 Since �A is parallel to z and �B at sufficiently large distance is azimuthal, �A× �Bis radial and the surface integral over the end faces of our cylinder make nocontribution. The vector potential �A at large r is the difference between theexterior field terms of the two wires, ie. Az(r h) = μ0I/(2π)(ln r2/a −ln r1/a). We rewrite this as

Az(r > h) =μ0I

4πln(r21 + h2 − 2hr1 cosϕ

r21

)=μ0I

4πln(

1 +h2

r2− 2h cosϕ

r

)

When r h we may approximate the logarithm using ln(1 + ε) ≈ ε to writethe vector potential as

Az(r h) ≈ μ0I

4π

(h2

r2− 2h

rcosϕ

)

The magnetic induction field of two opposing currents may be written as

Bϕ =μ0I

2π

(1r− 1√

r2 + h2 − 2rh cosϕ

)

=μ0I

2πr

(1 − 1√

1 + (h2/r2) − 2(h/r) cosϕ

)≈ μ0I

2πr

(h2

2r2− h

rcosϕ

)


in other words it vanishes at least as fast as r−2. The surface integral of(4–43) then becomes

12μ0

∣∣∣∣∮

( �A× �B) · d�S∣∣∣∣ < lim

r→∞μ0I

2

(4π)2

∫C

r32πrdrdz → 0

4-10 As a first approximation, we might be tempted to use the central field B =μ0 I/ 2a of the coil as an estimate of the average field through the second coil.The flux would then be

Φ =πaμ0I

2and M12 = dΦ2/dI1 = 1

2μ0πa. Unfortunately there is no good reason toassume that the magnetic induction field is uniform across the second loop.In particular it seems likely that that the field near the edges, close to theprimary loop might well be much larger. Indeed this is so. We now pursue amore rigorous approach to an approximate solution. We begin by writing theflux threading loop 2 as Φ2 =

∮�A(�r ) · d��2, where the integration is carried

out along loop 2. Replacing �A by its integral form, we find

Φ2 =μ0I

4π

∮loop 2

d��2 ·∮

loop 1

d��1|�r − �r ′|

{�r ′ on loop 1�r on loop 2

The mutual inductance M12 is then

M12 =μ0

4π

∮d��1 ·

∮d��2√

b2 + (2a sin 12ϕ)2

where ϕ is the projection of the angle between �r ′ and �r on the plane of oneof the coils as shown in figure 4.1. Replacing the scalar product of the unittangent vectors d��1 · d��2 by cosϕd�1d�2 we get

M12 =μ0

4π

∮d�1

∮cosϕd�2√

b2 + (2a sin 12ϕ)2

Figure 1.2: The current loops of problem 4-8


For any fixed point r′, we compute the second integral as follows.Except when �r and �r ′ are very close to each other, say within 2a sin 1

2ϕ0 =±Δ b, we make no great mistake neglecting b. For the region inside Δ, weignore the curvature of the loops and set cosϕ = 1. Thus

M12 =μ0

4π

∮loop 2

d�2

(∫ 2π−ϕ0

ϕ0

cosϕd�1|2a sin 1

2ϕ|+∫ ϕ0

−ϕ0

d�1√b2 + �21

)

=μ02πa

4π

(2∫ π

ϕ0

cosϕd�1|2a sin 1

2ϕ|+∫ Δ

−Δ

d�1√b2 + �21

)

where Δ ≈ aϕ0.We will now recast the second integral in parentheses in a form that will

allow us to rewrite it as an integral with the same form as the first, buteliminating the arbitrary Δ (or ϕ0). We integrate the second integral to get

∫ Δ

−Δ

d�1√b2 + �21

≈ 2 sinh−1 Δb

≈ 2 ln2Δb

= 2∫ Δ

12 b

dx

x≈ 2

∫ ϕ0

12 b/a

adϕ

2a sin 12ϕ

The second integral may now be included within the first by merely changingthe lower limit of the first

M12 = μ0a

∫ π

12 b/a

a cosϕdϕ2a sin 1

2ϕ

= μ0a{− ln tan[ 1

4 (b/2a)] − 2 cos[ 12 (b/2a)]

}

= μ0a

[− ln

(b

8a

)− 2]

= μ0a

(ln

8ab

− 2)

The self-inductance of a single loop is found in very similar manner with theradius of the wire replacing b. There will, however, be another term (+1

4μ0a),the contribution of the local current as opposed to that of the remainder ofthe loop.

4-11 Using the magnetic field portion of the Maxwell stress tensor (in Cartesiancoordinates) we find the x and y components of force on an element dSr ofthe curved wall

dFx =B2

2μ0dSx and dFy =

B2

2μ0dSy

leading to

dFr = dFx cosϕ+ dFy sinϕ

=B2

2μ0(dSx cosϕ+ dSy sinϕ) =

B2

2μ0dSr


The pressure ℘ on the coil is the found as

℘ =dFr

dSr=

B2

2μ0=μ0N

2I2

2L2

This pressure must be balanced by the inward resultant of the tension on thewires. There are N/L wires in a unit width sharing the outward directed forcefrom the pressure on those wires. A simple diagram shows that the inwardforce of a wire under tension T is T/R where R is the radius of curvature.Therefore N/L such wires produce an inward force NT/LR per unit area.Equating this to the pressure, we find

T =μ0RNI

2

2L

Alternatively, the average field at the wire is μ0NI2/2L, giving an outward

force μ0NI2/2L on the wire. The tensile force required to sustain this force

is given by equating it to T/R.

T

R=μ0NI

2

2L⇒ T =

μ0NI2R

2LAlthough this seems an easier approach, it is by no means self evident thatone should use half the field in the solenoid as the field at the wire.

4-12 The magnetic force between two hypothetical magnetic monopoles qm, is

Fm =μ0

4πq2mr2

=μ0

4πr2

(2πε0c2h

e

)2

=πh2

μ0e2r2

The force between two electric charges e separated by the same distance is

Fe =e2

4πε0r2

The ratio of the two forces isFe

Fm=

μ0e4

4π2ε0h2 = 2.13 × 10−4.

4-13 We transform each equation in turn:

�∇ · �E′ = �∇ · �E cos θ + c�∇ · �B sin θ

=ρe cos θε0

+ cμ0ρm sin θ

=1ε0

(ρ′m − ρm

csin θ

)+ε0μ0c

ε0ρm sin θ =

ρ′eε0

�∇ · �B′ = �∇ · �B cos θ −�∇ · �Ec

sin θ

= μ0ρm cos θ − ρe

ε0csin θ

= μ0 (ρ′m + cρe sin θ) − ρe

ε0csin θ = μ0ρ

′m


�∇× �E′ = (�∇× �E) cos θ + c(�∇× �B) sin θ

= −∂�B

∂tcos θ +

1c

∂ �E

∂tsin θ − μ0Jm cos θ + μ0cJe sin θ

= −∂�B′

∂t− μ0

�J ′m

�∇× �B′ = �∇×(−�E

csin θ

)+ �∇× �B cos θ

=(

1c

∂ �B

∂t+μ0

c�Jm

)sin θ +

1c2∂ �E

∂tcos θ + μ0

�Je cos θ

=1c2

∂

∂t

(�E cos θ + c �B sin θ

)+μ0

cJm sin θ + μ0Je cos θ

=1c2∂ �E′

∂t+ μ0J

′e

4-14 The capacitance of each sphere is

C =Q

V=

Q

Q/4πε0R= 4πε0R (= 111pF)

The approximate mutual capacitance of the pair of spheres, using the resultsfrom example 4.2 is

C12 � −C1C2

4πε0r= − (4πε0R)2

4πε0r= −4πε0 × 1 m2

10 m= −11.1 pF

4-15 The charge on the concentric cylinders must lie entirely in the region wherethe two cylinders overlap (and the fringing region), because any charge on anexposed long cylinder leads to an infinite (logarithmically divergent) potentialon the cylinder. The electric field between the cylinders is easily found to be

E =V

r ln(b/a)

whence the energy of a length z of concentric cylinders is

W =ε02

V 2

[ln(b/a)]2

∫ z0+z

z0

∫ b

a

1r2

2πrdrdz′ =πε0V

2

[ln(b/a)]z

The energy of the field in the fringing region is irrelevant since it will notchange as one cylinder moves with respect to the other. The force pulling onecylinder into the other is

∂W

∂z=

πε0V2

ln(b/a)


4-16 The electric field between the overlapping plates is E = 2V/d and when theplates don’t overlap the field vanishes. The energy of the 14 gaps between theplates overlapping by angle θ is

W = 14 × ε02

∫4V 2

d2d3r =

14ε0V 2R2θ

d

The torque is given by

∂W

∂θ=

14ε0V 2R2

d

Substitution of reasonable values for the radius and the spacing d, say 10cm and 1 mm, respectively, gives a torque τ = 1.24 × 10−9V 2 Nm/V2 =1.24 V 2 × 10−2 dyne·cm/V2. It would appear to be entirely feasible to buildan electrostatic voltmeter on this principle.

4-17 The flux through the loop is Φ =∫I0e

−iωt/(2μ0r)dA giving rise to an EMF−iωI0e−iωt/(2μ0) ln(b/d)Δz. The potential on the plate (working into an infi-nite impedance) is V0e

−iωtd/ ln(b/a) where d is the distance of the plate fromthe central conductor and a and b are respectively the inner and outer radiiof the coaxial wire. The impedance (1/iωC) of the source (plate) decreaseslinearly with the area of the plate. If the power meter impedance is smallcompared to 1/ωC, we conclude that the inductive and capacitive currenteach increase linearly with ω resulting in a torque proportional to ω2.

4-18 The electrostatic energy of the charged bubble is

W =ε02

∫ ∞

R

(Q

4πε0r2

)24πr2dr =

Q2

8πε01−r∣∣∣∣∞

R

=Q2

8πε0R

The outward pressure from the electrical repulsion is then

℘ =dW

dτ=dW

dR

dR

dτ=

Q2

32πε0R4

Equating this to the pressure 4T/R (there is both an inside and an outsidesurface) required to balance surface tension gives

R =(

Q2

128π2ε0T

)1/3

If the radius were to increase beyond the equilibrium point, the electrical (ex-pansive) pressure decreases as R−4 whereas the (contractile) surface-tensionderived force only decreases as R−1 and if the radius were to decrease, theelectrical force would grow faster. The bubble would tend to return to itsequilibrium radius in either case.


The center term of (4–6) has an erroneous subscript. The expressionshould read:

W4 =1

8πε0

(q1q4 + q4q1

r14+q2q4 + q4q2

r24+q3q4 + q4q3

r34

)(4–6)

Chapter 5

5-1 Although there are elegant ways to prove this we use the brute force approachas it requires little thought. The general expansion for the potential in spher-ical polar coordinates is

V (�r ) =∑(

A�r� +

B�

r�+1

)Ym

� (θ, ϕ)

Each of the constants B� must vanish since otherwise the potential wouldbe infinite at the origin. To evaluate the constants A� we multiply bothsides of the expression above by Y∗m′

�′ (θ, ϕ) and integrate over the surface ofthe enclosing sphere. Using the orthogonality of the spherical harmonics weobtain

A�a� =

∫V (R)Y∗m

� (θ, ϕ)dΩ

At the center of the sphere, r = 0, only the first term does not vanish and weneed evaluate only this term. Thus

A0 =∫

4π

V (R, θ, ϕ)√4π

dΩ =1√4π

∫4π

V dΩ =〈V 〉 4π√

4π

The potential at r = 0 then becomes

V (0) = A0r0Y0

0(θ, ϕ) =〈V 〉 4π

4π= 〈V 〉

5-2 We use plane polar coordinates to describe the geometry. Because the bound-ary conditions do not depend on r we expect V to be independent of r.Laplace’s equation then becomes

1r2∂2V

∂ϕ2= 0

The solution is easily found to have the form V = Aϕ+B. Evaluating A andB from the boundary conditions we obtain

V (ϕ) =V0ϕ

α

5-3 The potential between the cones is most easily solved for in spherical polarcoordinates since the boundary conditions are independent of r and ϕ. Itremains to solve

1r2 sin2 θ

d

dθ

(sin θ

dV

dθ

)= 0

Excluding r = 0, we have

sin θdV

dθ= a

and integrating once more,

— 35—


V = a ln(tan 12θ) + b

Fitting the boundary conditions at θ = 12α1 and θ = 1

2α2 we find

a =V2 − V1

ln[tan(1

4α2)tan(1

4α1)

] and b = V1 − a ln tan(14α1)

Substituting these constants into the solution we find

V = V1 + (V2 − V1)ln[

tan(12θ)

tan(14α1)

]

ln[tan(1

4α2)tan(1

4α1)

]

5-4 The general solution of Laplace’s equation compatible with the boundary con-ditions is

V =∑

λ

Aλ sinλy sinhλx

=∑

n

An sinnπy

bsinh

nπx

b

At x = a,V (a, y) =

∑An sinh

nπa

bsin

nπy

b= V0

then

An sinhnπa

b· b2

=∫ b

0

V0 sinnπy

bdy

= − b

nπ(cosnπ − cos 0)

=

⎧⎨⎩

2V0b

nπfor n odd

0 for n even

We evaluate the first four nonzero coefficients

A1 =4V0

π sinhπa

b

= 0.05019V0

A3 =4V0

3π sinh3πab

= 6.492 × 10−6V0

A5 =4V0

5π sinh5πab

= 1.51 × 10−9V0

Chapter Five Solutions 37

A7 =4V0

7π sinh7πab

= 4.2 × 10−13V0

We therefore write the potential to the required precision as

V = 0.0502V0 sinπy

bsinh

πx

b+ 6.5 × 10−6V0 sin

3πyb

sinh3πxb

+1.51 × 10−9V0 sin5πyb

sinh5πxb

+ 4.2 × 10−13V0 sin7πyb

sinh7πxb

which may be numerically evaluated at the required points to give

V (a/10, b/2) = 0.0502 sinh 18π − 6.5 × 10−6 sinh 3

8π = .0202V0

V (a/2, b/2) = .0502 sinh 58π − 6.5 × 10−6 sinh 15

8 π = 0.1753V0

and

V (9a/10, b/2) = 0.0502 sinh 98π − 6.5 × 10−6 sinh 27

8 π

+ 1.51 × 10−6 sinh 458 π − 4.2 × 10−13 sinh 63

8 π = .758V0

5-5 In cylindrical polars, the potential in the region including r = 0 is

V (r, ϕ) =∑

A�

(r

a

)�

sin �ϕ

At r = a this becomes

V (a) =∑

A� sin �ϕ = ±V0

2

leading to A� =2V0

�π

when � is odd and A� = 0 when � is even. Therefore

V (r < a) =∑

� odd

2V0

�π

(r

a

)�

sin �ϕ

Outside the pipe, the general expansion is

V (r > a) = B�

(a

r

)�

sin �ϕ

The boundary conditions lead to the same coefficients so that

V (r > a) =∑

� odd

2V0

�π

(a

r

)�

sin �ϕ


5-6 The general solution in spherical polar coordinates may be written

V =∑[

A�

(r

a

)�

+B�

(a

r

)�+1]

P�(cos θ)

For the interior solutions all the B coefficients must vanish while the outsidesolutions must have A� = 0. In either case, at the surface of the sphere, thecoefficients must satisfy

∑(A�

B�

)P�(cos θ) = ±V0

leading to

2A�

2�+ 1= 2V0

∫ 12 π

0

P�(cos θ) sin θdθ = 2V0

∫ 1

0

P�(x)dx

for � odd, and 0 when � is even and an identical expression for B�. Theintegral may be evaluated using (F–30), (2�+ 1)P�(x) = P′

�+1(x)−P′�−1(x).∫ 1

0

P�(x)dx =1

2�+ 1

[P�−1(0) − P�+1(0)

]=

(−1)(�+3)/2(�− 2)!!(�+ 1)!!

The coefficients A� for the interior solution and B� for the exterior solutionfollow immediately.

5-7 In Example 1.16 we found the scalar potential along the axis of a solenoid tobe

Vm(z) =NI

4�

(√(z + �)2 +R2 −

√(z − �)2 +R2

]where � = L/2 is the half length of the solenoid. We factor (R2 + �2)1/2 fromthe radicals

√R2 + �2 + z2 ± 2�z =

√R2 + �2

(1 +

z2 ± 2�zR2 + �2

)1/2

and expand the term in parentheses using the binomial theorem(1 +

z2 ± �2

R2 + �2

)1/2

= 1 +12z2 ± 2�zR2 + �2

− 18

(z2 ± 2�z)2

(R2 + �2)2+

116

(z2 ± 2�z)3

(R2 + �2)3− · · ·

yielding

Vm =NI

√R2 + �2

4�

[− 2�zR2 + �2

+�z3

(R2 + �2)2− �3z3

(R2 + �2)3+ · · ·

]

=NI

4√R2 + �2

{−2z +

[+

1R2 + �2

− �2

(R2 + �2)2

]z3 + · · ·

}

The general form of the scalar potential in a region that includes the originis (in spherical polars)

Vm(r, θ) =∑

AlrlPl(cos θ)


which for cos θ = 0 specializes to

Vm(z) =∑

Alzl

along the z axis. Comparing the coefficients of the two series we find

A1 = − NI

2√R2 + �2

, A2 = 0, A3 =NIR2

4(R2 + �2)5/2

so that the general solution in the neighborhood of r = 0 is

Vm = A1r cos θ +A3r3P3(cos θ) + · · ·

= A1r

(z

r

)+A3r

3 12

[5(z

r

)3

− 3(z

r

)]+ · · ·

= A1z + 52A3z

3 − 32A3r

2z + · · ·

We may usefully write this in terms of the cylindrical coordinates ρ and z sothat r2 = ρ2 + z2 and

V (ρ, z) = A1z +A3z3 − 3

2A3ρ2z + · · ·

The resulting second order axial and first order radial magnetic field compo-nents are now easily calculated.

5-8 We express f in terms of its real and imaginary part u and v as

u+ iv = ln(a+ x+ iy

a− x− iy

)= ln

(a2 − x2 − y2 + 2iaya2 − 2ax+ x2 + y2

)

= lnReiα = lnR+ iα

which allows us to identify v as α. But we can also find α from

tanα =Im(f)Re(f)

=2ay

a2 − x2 − y2

On the circle x2 + y2 = a2− we find when y > 0 that

tanα = +∞ ⇒ v = α =π

2

and when y < 0

tanα = −∞ ⇒ v = α = −π2

In other words, the the upper half circle x2 + y2 = a2− is mapped from the

v = π/2 line while the lower half of the interior circle is mapped from the v= −π/2 line. It may be verified using l’Hopital’s rule that as x varies from−a to a, u varies from −∞ to ∞.


The potential V = V0v/π between the v = ±π/2 lines becomes in termsof x and y

V =V0

πtan−1

(2ay

a2 − x2 − y2

)At y = 0.2a and x = 0, this becomes

V =2V0

π× 1

2tan−1

(0.40.96

)=

2V0

π× (0.1973956)

Using the series solution from problem 5-5 to find the potential at this point,we have

V =∑

� odd

2V0

�π

(r

a

)�

sin �ϕ

=2V0

π

[1(.2) − 1

3 (.2)3 + 15 (.2)5 − 1

7 (.2)7 + 19 (.2)9 + · · ·]

=2V0

π× (0.1973956)

5-9 The required mapping must change the polar angle of f from π to α whileleaving a polar angle of zero unaffected. The mapping

z = fα/π =(Reiθ

)α/π= Rα/πeiθα/π

clearly has this property. When v = 0, R = u and

z = x+ iy = Rα/π ⇒ 0 ≤ x <∞ and y = 0

When v = π, R = |u|, then

z = x+ iy = Rα/πeiα

= Rα/π (cosα+ i sinα)

Thus, as R increases, a line with x = Rα/π cosα and y = Rα/π sinα isswept out. Despite the fact that the mapping is obvious it is a good exerciseto use the Schwarz Christoffel transformations to derive it. Thus

dz

df= A(f − 0)α/π−1

which is readily integrated to yield

z =Afα/π

α/π+ k

The denominator may be absorbed into the constant A and the choice ofk = 0 fixes the inflection point to the origin.


5-10 As found on page 118, the potential in the vicinity of the right-angled platehas the form V = V0−2axy. (It is trivial to verify that this satisfies ∇2V = 0and satisfies the boundary conditions.) The electric field is then

�E = −�∇V = 2ayı+ 2axj

Along the x axis (y = 0) this becomes �E = 2axj. We equate the normalcomponent of the electric field to σ/ε0 to obtain σ(x, 0) = 2aε0x. Similarly,along the y axis σ(0, y) = 2aε0y.

5-11 When v = 0, (f = u+ iv), cosh f = 12 (eu + e−u). Therefore, as u varies from

−∞ to +∞, cosh f varies from ∞ to 1 to ∞, meaning that z ∝ ln(cosh f)ranges from ∞ to 0 to ∞ and is pure real. In other words, the image of theu axis is the positive x axis.

When v = − 12π,

cosh f =eu+iπ/2 + e−u−iπ/2

2

=ieu − ie−u

2= i sinhu = (sinhu)eiπ/2

Then x + iy = (2a/π)[ln(sinhu) + 12 iπ]. We conclude then that when v =

π/2, y = a and as u varies from 0 to ∞, sinhu varies from 0 to ∞ and x =(2a/π) ln(sinhu) varies from −∞ to +∞. In the same fashion, the line with v= −π/2 maps to a line at y = −a running from −∞ to ∞. The capacitanceof a unit length and width w (large compared to a) including the end of theplate) capacitor made by the central plate of one polarity and the outsideplate of opposite polarity is easily obtained.

We begin by finding the values of u that correspond to the ends of thesegment x = -∞ to x = w a on the upper plate. We have

πx

2a= ln(sinhu) ⇒ eπx/2a = sinhu =

eu − e−u

2

The value of u corresponding to x large and negative is clearly u = 0. Whenx a, u 1 implying that to an excellent approximation

sinhu = 12e

u

We find then

eu = 2eπw/2a or u = ln 2 +πw

2a

Using the results of p 126,

C = −ε0u2 − u1

v2 − v1= ε0

(ln 2 + πw/2a

π/2

)

= ε0

(2 ln 2π

+w

a

)


The term ε0w/a is just what we would have expected in the absence of fringingfields at the x = 0 edge. The bottom plate makes an identical contributionto the capacitance so that we have for such a capacitor

C = 2ε0

(2 ln 2π

+w

a

)

5-12 Expression (5–74) gives the capacitance per unit length of a single sided stripof parallel plate capacitor. Specializing this result to the circular capacitor wenote that the capacitance should become the sum of the area term ε0πr

2/aand a contribution of the perimeter due to a strip of width r and length 2πrso that

C = ε0

(πr2

a+ r ln

2πra

)W have arbitrarily taken r as the effective width of the strip. However, writingthe result as

C =ε0πr

2

a

(1 +

a

πrln

2πra

)

we note that fractional contribution diminishes as r/a increases.

5-13 The mapping

z =a

2

(f

b+b

f

)=a

2

(u+ iv

b+

b

u+ iv

)

may be rationalized as

x+ iy =a

2

(u+ iv

b+b(u− iv)u2 + v2

)

=a

2

(u

b+

bu

u2 + v2

)+ai

2

(v

b− bv

u2 + v2

)

so that

x =a

2

(u

b+

bu

u2 + v2

)and y =

a

2

(v

b− bv

u2 + v2

)

For v = 1, x =a

2

(u

b+

bu

u2 + 1

)and y =

a

2

(1b− b

u2 + 1

)

and for v = 2, x =au

2

(1b

+b

u2 + 4

)and y = a

(1b− b

u2 + 4

)

We plot these curves parametrically for b = 1 in figure 5.1.

5-14 The mapping indicated may be expanded for f = u+ iv as

z =i− f

i+ f=i− u− iv

i+ u+ iv= −u+ (v − 1)i

u+ (v + 1)i


Figure 5.1: Images of the v = 1 and v = 2 lines in the x-y plane

Along the real axis of the f plane, v = 0, reducing z to

z = −u− i

u+ i= −u

2 − 1 − 2uiu2 + 1

The magnitude of z when v = 0 is

|z| =∣∣∣∣u− i

u+ i

∣∣∣∣ ≡ 1

Moreover, u = 0 → z = 1, u = 1 → z = i, u = −1 → z = −i, u = ∞ → z =−1. More generally, when v = 0 we can write z = e−2iα with α = tan−1(1/u).Evidently the real axis of f maps into the unit circle in the z plane.

When v �= 0, we can again find |z|2,

|z|2 =∣∣∣∣u+ (v − 1)iu+ (v + 1)i

∣∣∣∣2

=[u+ (v − 1)i][u− (v − 1)i][u+ (v + 1)i][u− (v + 1)i]

=u2 + (v − 1)2

u2 + (v + 1)2

=u2 + (v + 1)2 − 4vu2 + (v + 1)2

= 1 − 4vu2 + (v + 1)2

We see then that for v < 0, |z| increases to give points outside the unit circle.In similar fashion the second mapping may be investigated.

z =i+ f

i− f=i+ u+ iv

i− u− iv= −u+ (v + 1)i

u+ (v − 1)i

For f on the real axis,

|z| =∣∣∣∣u+ i

u− i

∣∣∣∣ ≡ 1

Moreover, u = 0 → z = −1, u = 1 → z = i, u = −1 → z = −i, u = ∞ →z = −1. Again, the real axis is mapped to a unit circle. When v �= 0,

|z|2 =∣∣∣∣u+ (v + 1)iu+ (v − 1)i

∣∣∣∣2

=[u+ (v + 1)i][u− (v + 1)i][u+ (v − 1)i][u− (v − 1)i]

=u2 + (v + 1)2

u2 + (v − 1)2

=u2 + (v − 1)2 + 4vu2 + (v − 1)2

= 1 +4v

u2 + (v − 1)2

5-15 We investigate first the behavior of the mapping for f along the real axis.When u is real,

√f is real. Thus we readily find that u = 0 → z = −1, u =


1 → z = 0, u = ∞ → z = +1. In other words, the positive real axis maps tox ∈ (−1, 1). When u is negative, the square root is imaginary and we writeit as ri. Then

|z| =∣∣∣∣ri− 1ri+ 1

∣∣∣∣ ≡ 1

In particular u = −1 maps to i and points ∈ (0,−1) map to the left quadrantquarter circle whereas points more negative than −1 map to the right quartercircle. The whole axis then is mapped into a closed half circle in the upperhalf plane.

The second mapping is easily deduced from the first. Taking the n’throot of any point in the complex plane reduces its magnitude to the n’th rootand divides the argument by n. Thus the points at |z| = 1 remain at thatdistance but have their polar angles reduced by a factor n. In particular, theimage of 0 at x = −1 gets moved to polar angle π/n. The result is that thesemicircle of the previous map now gets compressed like a closing hand-heldfan to become a wedge with interior angle π/n.

5-16 To find the image of the real axis set f = u and note that when u is positive,the mapping is straightforward, u = 0 → x = −∞, u = 1 → x = 0, u = ∞ →x = ∞. In other word, the positive u axis maps to the entire (−∞,∞) x axis.When u < 0, we write the square root as

√u = ir. Then

z = d(ir − 1

ir

)= id

(r +

1r

)Thus the negative u axis maps to a vertical line along the y axis that termi-nates at height 2d above the x axis (the image point of u = −1).

5-17 This mapping presents several points where the image of the real line willabruptly change directions. We note that if u is large and positive, the ln fand −2 ln[(f + 1)1/2 + 1] essentially cancel leaving only the first term whichclearly increases to ∞ as f → ∞. As u→ 0, the ln f will dominate and image0 to −∞ The first singularity occurs when f = 0 at that point the ln f termhas to be replaced by ln(|u|eiπ) = ln |u| + iπ. To sum up to this point then,u ∈ (0,∞) → x ∈ (−∞,∞). At u = 0, the image is displaced upwards adistance π while still at x = −∞. As u becomes more negative, v remains atπ until u reaches −1 when the argument of the two square root terms bothbecome negative. Just before the arguments turn negative, we abbreviate1 + u = ε. The expression for z then becomes

z = 2ε1/2 − 2 ln(1 + ε1/2) + iπ + ln |u| ≈ 2ε1/2 − 2ε1/2 + iπ + ln(1) = 0 + iπ

In other words, u = (0,−1) maps to x = (−∞, 0), y = π After u passes −1,set (1 + u)1/2 = |1 + u|1/2eiπ/2 = riπ/2. We again express z in this domain.

z = 2ir − 2 ln(1 + reiπ/2) + iπ + ln(1 + r2)

= 2ir − 2 ln[eiπ/2(e−iπ/2 + r] + iπ + ln(1 + r2)


= 2ir − iπ − 2 ln(r − i) + iπ + ln(1 + r2)

= 2ir − ln[(1 + r2)e−2iα] + ln(1 + r2) = 2ir + 2iα

with α = tan−1(1/r).We see that z becomes pure imaginary with y = 1 at r = 0 increasing to

infinity as r → ∞. The entire image is illustrated in Figure 5.2.

Figure 5.2: Image of the f plane real axis in the z plane. The correspondingvalues of u are given at the vertices.

5-18 The mapping z = a√f2 − 1 is readily investigated. The real axis has three

regions of interest. When u > 1, z is also real and increases with u. When|u| < 1 we write z = ai

√1 − u2 and finally when u < −1 we revert to the

original. we find the, u ∈ (1,∞) → x ∈ (0,∞), y = 0; u ∈ (−1, 0, 1) → x =0, y ∈ (0, a, 0) and u ∈ (−1,−∞) → x ∈ (0,∞), y = 0. If f is in the firstquadrant then f2 is in the positive half plane as is f2 − 1. Taking the squareroot maps all such points into the first quadrant. When f is in the secondquadrant, f2 will be in the third or fourth, and f2 − 1 will lie in the thirdor the fourth quadrant. The square root therefore maps the second quadrantf ’s into the second quadrant.

5-19 For a cylinder of length L and when the potential has no ϕ dependence,

V (r, z) =∑

λ

Aλ sinhλzJ0(λr)

From V (r = a) = 0 we deduce that λa is a root of J0 say ρ0i ⇒ λ = ρ0i/a.Thus

V (r, z) =∑

i

Ai sinh(ρ0iz

a

)J0

(ρ0ir

a

)

At z = L this specializes to

V (r, L) =∑

i

Ai sinh(ρ0iL

a

)J0

(ρ0ir

a

)


We abbreviate the constant terms by Ci and write

V (r, L) =∑

i

CiJ0

(ρ0ir

a

)= V0

(1 − r2

a2

)

The coefficients Ci may be evaluated by multiplying both sides of the equationby J0(ρ0jr/a) and integrating over the surface.

12Cia

2J21(ρ0i) =

∫ a

0

V0

(1 − r2

a2

)J0

(ρ0ir

a

)rdr

Changing variables to x = ρ0ir/a, the integral on the right becomes

a2

ρ20i

∫ ρ0i

0

xJ0(x)dx− a2

ρ40i

∫ ρ0i

0

x3J0(x)dx

We evaluate each term independently.∫ ρ0i

0

xJ0(x)dx = ρ0iJ1(ρ0i)

∫ ρ0i

0

x3J0(x)dx =∫x2 (xJ0) dx =

∫x2 d

dx(xJ1)dx

= x3J1

∣∣∣∣ρ0i

0

−∫

2x (xJ1)dx

= ρ30iJ1(ρ0i) − 2x2J2

∣∣∣∣ρ0i

0

= ρ30iJ1(ρ0i) − 2ρ2

0iJ2(ρ0i)

Gathering the terms, we may solve for Ci to get

Ci =2V0

ρ0iJ21(ρ0i)

[J1(ρ0i) − J1(ρ0i) +

2ρ0i

J2(ρ0i)]

=4V0J2(ρ0i)ρ20iJ

21(ρ0i)

5-20 The boundary conditions are obtained from Maxwell’s equations:

�∇ · �E =ρ

ε0⇒ Eext

r − Eintr =

σ

ε0=σ0 cos θε0

�∇× �E = 0 ⇒ Eextθ − Eint

θ = 0

Using these boundary conditions for the general solution of ∇2V = 0 inspherical polar coordinates

V (r, θ) =∑(

A�r� +

B�

r�+1

)P�(cos θ)

we have from the radial equation

∂V

∂r

∣∣∣∣a−

− ∂V

∂r

∣∣∣∣a+

=σ0 cos θε0


or∞∑

�=1

�A�a�−1P�(cos θ) −

∞∑�=0

−(�+ 1)B�

a�+2P�(cos θ) =

σ0 cos θε0

which means for � = 12B1

a3+A1 =

σ0

ε0

and for � �= 1(�+ 1)B�

a�+2+ �A�a

�−1 = 0

Similarly, for the θ component of the electric field we find

1a

∂V

∂θ

∣∣∣∣a+

=1a

∂V

∂θ

∣∣∣∣a−

givingB�

a�+2= A�a

�−1

The θ and r equations may be solved simultaneously to give

B1 =σ0a

3

3ε0, A1 =

σ0

3ε0, and A� = B� = 0 for � �= 1

The potentials inside and outside the sphere and the associated fields are then

V (r < a) =σ0r cos θ

3ε0⇒ �E = −σ0k

3ε0=

σ0

3ε0

(−r cos θ + θ sin θ

)V (r > a) =

σ0a3 cos θ

3ε0r2⇒ �E =

σ0a3

3ε0r3(2r cos θ + θ sin θ

)The continuity of Eθ across the surface is evident, as is the discontinuity ofEr by σ/ε0.

5-21 As there is both a ϕ and θ dependence in this problem we use the generalspherical solution to Laplace’s equation.

V (r < a) =∑�,m

A�m

(r

a

)�

Ym� (θ, ϕ)

and

V (r > a) =∑�,m

B�m

(a

r

)�+1

Ym� (θ, ϕ)

The boundary condition V (a) = sin 2θ cosϕ may be written in terms of spher-ical harmonics using (F–44) as

sin 2θ sinϕ =

√8π15

[Y−1

2 (θ, ϕ) − Y12(θ, ϕ)

]


Equating the inside and outside solutions at to the potential at a, we findA2,−1 = A2,1 =

√8π/15 to get

V (r < a) =r2

a2

√8π15

(Y−1

2 − Y12

)=r2

a2sin 2θ cosϕ

Similarly when r > a, we get

V (r > a) =a3

r3

√8π15

(Y−1

2 − Y12

)=a3

r3sin 2θ cosϕ

5-22 We again take the general solution for the potential as and denote the radiusof the sphere by R.

V�m(r, θ, ϕ) =∞∑

�=0

�∑m=−�

[A�m

(r

R

)�

+B�m

(R

r

)�+1]

Ym� (θ, ϕ)

Inside the spherical shell, the solution takes the form

V (r < R) =∑�,m

A�m

(r

R

)�

Ym� (θ, ϕ)

whereas outside the shell V is given by

V (r > R) =∑�,m

B�m

(R

r

)�+1

Ym� (θ, ϕ)

V is continuous across the boundary since otherwise its gradient would divergeleading to infinite electric field. The linear independence of the sphericalharmonics require A�m = B�m The boundary condition on Er is

∂V

∂r

∣∣∣∣R−

− ∂V

∂r

∣∣∣∣R+

=σ

ε0=σ0 sin θ sinϕ

ε0

The trigonometric terms may be written in terms of spherical harmonics using(F–44)

Y11(θ, ϕ) + Y−1

1 (θ, ϕ) = −√

38π

sin θ(eiϕ − e−iϕ

)= −i

√32π

sin θ sinϕ

so that the condition above becomes

1R

∑(�A�m + (�+ 1)B�m

)Ym

� = i

√2π3σ0

ε0

(Y1

1 + Y−11

)and the continuity of V at R gives A�m = B�m. Thus for (�,m) �= (1,±1),A�m = B�m = 0 and

A1,1 = B1,1 = A1,−1 = B1,−1 =R

3i

√2π3σ0

ε0


Inserting these values into the general solutions above, we obtain

V (r < R) = A1,1

(r

R

)Y1

1 +A1,−1

(r

R

)Y−1

1

=r

3i

√2π3σ0

ε0

(Y1

1 + Y−11

)=

σ0

3ε0r sin θ sinϕ

and

V (r > R) = B1,1

(R2

r2

)Y1

1 +B1,−1

(R2

r2

)Y−1

1

=R3

3r2i

√2π3σ0

ε0

(Y1

1 + Y−11

)=σ0R

3

3ε0r2sin θ sinϕ

5-23 Using the methods of Apendix B, the nonzero elements of the metric tensorfor oblate ellipsoidal coordinates are

gρρ =a2(ρ2 − cos2 α)

ρ2 − 1gαα = a2(ρ2 − cos2 α) gϕϕ = a2(ρ2 − 1) sin2 α

Since on a conducting prolate ellipsoid (the coordinate surface for ρ heldconstant), the potential is independent of α and ϕ we anticipate the potentialeverywhere will be independent of α and ϕ. Therefore eliminating all termsin ∂V/∂α and ∂V/∂ϕ from the Laplacian we obtain as the only remainingterm

∂

∂ρ

[(ρ2 − 1

) ∂V∂ρ

]= 0

which is easily integrated to give

∂V

∂ρ=

k

ρ2 − 1

with k an arbitrary constant of integration. This result may be integratedonce more to give

V = 12k ln

(ρ− 1ρ+ 1

)+ C

As ρ goes to infinity we anticipate that V would tend to zero, implying thatC = 0. For the ellipsoid defined by ρ = ρ0,

k =2V0

ln(ρ0 − 1ρ0 + 1

)

Unfortunately, this result is not usable for the infinitely thin needle (ρ0 = 1)as k would would be indeterminate. Instead, we relate k to the charge on


the surface. The electric field is the negative gradient of V so that in thesecoordinates,

�E = − ρ√gρρ

∂V

∂ρ= − ρ

√ρ2 − 1

a√ρ2 − cos2 α

k

ρ2 − 1

We equate this field to the surface charge density ε0 and integrate over thesurface to obtainQ

ε0=∮

�E · d�S =∮Eρ

√gααgϕϕdαdϕ

= −k∮

a2√ρ2 − 1

a(ρ2 − 1)√ρ2 − cos2 α

√(ρ2 − cos2 α)(ρ2 − 1) sin2 αdαdϕ

= −ka∫ 2π

0

∫ π

0

sinαdαdϕ = −4πka

We conclude then that k = −Q/(4πε0a).Instead of integrating over α, we can change variables to integrate over z.

From the definition, dz = −aρ0 sinαdα, leading to

Q

ε0=

2πkρ0

∫ −a

a

dz = −4πkaρ0

The charge per unit length λ is evidently −2πkε0ρ0. Substituting the valueof k we get the constant linear charge density λ = Q/(2aρ0). For the needleof length 2a with ρ0 = 1 this becomes λ = 1

2Q/a.

5-24 The solution to this problem is rather lengthy, closely following the discussionof page 364 to 367. Completing exercise (B-12) should be a prerequisite.Taking the form of the Laplace equation from (B-12),

∇2V =(cosh ρ− cosα)3

a3 sinα

{sinα

∂

∂ρ

(1

cosh ρ− cosα∂V

∂ρ

)

+∂

∂α

(sinα


∂α

)}+

(cosh ρ− cosα)2

a2 sin2 α

∂2V

∂ϕ2

We place the origin midway between the two spheres so that their centers lieat z = a coth ρ = ±5 m and the radius a/ sinh ρ = 1 m. Thus the surfaces lieat cosh ρ = ±5 which implies a = sinh ρ =

√24 m. It should be clear that

the solution will be independent of ϕ and we expect the solution to dependonly on ρ. Unfortunately, setting ∂V/∂α = 0 does not make the remainingequation α independent. The equation remaining from ∇2V = 0 is

∂

∂ρ

(1


∂ρ

)+

1sinα

∂

∂α

(sinα


∂α

)= 0

Substituting V =√

cosh ρ− cosα U as suggested, changes the equation to

1√cosh ρ− cosα

(∂2U

∂ρ2+∂2U

∂2α+ cotα

∂U

∂α− 1

4U

)= 0


which may be separated. Trying U = R(ρ)A(α) we obtain

d2Rdρ2

R+

d2Adα2

+ cotαdAdα

A− 1

4= 0

resulting in

d2Rdρ2

− ( 14 + λ)R = 0 and

d2Adα2

+cosαsinα

dAdα

+ λA = 0

The equation for A is just the Legendre equation, and insisting on reasonablybehaved solutions we require λ = �(� + 1) meaning A(α) = P�(cosα). Theremaining equation for R is now easily solved.

d2Rdρ2

− [ 14 + �(�+ 1)]R =d2Rdρ2

− (�+ 12 )2R = 0 ⇒ R(ρ) = ±e±(�+ 1

2 )ρ

We can now write the general solution as

V (ρ, α) =√

cosh ρ− cosα∑

�

[C�e

(�+ 12 )ρ + C−�e

−(�+ 12 )ρ]P�(cosα)

on the upper sphere, ρ = cosh−1(5) = ρ0 and v = Va is independent of α. Wesubstitute these values into the general solution to get

Va =√

5 − cosα∑

�

B�P�(cosα)

where we have abbreviated B� = C�e(�+ 1

2 )ρ0 + C−�e−(�+ 1

2 )ρ0 . Evidently, theB� are just the expansion coefficients in the Legendre polynomial expansion ofVa/

√5 − cosα (see example D.3). At the lower sphere ρ = −ρ0 and defining

D� = C�e−(�+ 1

2 )ρ0 + C−�e(�+ 1

2 )ρ0 we similarly get

Vb =√

5 − cosα∑

�

D�P�(cosα)

After solving for B� and D� we may solve their definitions simultaneouslyto obtain C� =

(B�e

(�+ 12 )ρ0 − D�e

−(�+ 12 )ρ0

)/[sinh(2� + 1)ρ0] and C−� =(

D�e(�+ 1

2 )ρ0 −B�e−(�+ 1

2 )ρ0)/[sinh(2�+ 1)ρ0].

For the special case of Va = const. a Legendre polynomial expansion ofVa/

√cosh ρ− cosα may be constructed from the generating function for Leg-

endre polynomials:

1√1 + t2 − 2t cosα

=∑

�

t�P�(cosα)

with |t| < 1. If we substitute t = e−ρ the radical becomes:

1√1 + e−2ρ − 2e−ρ cosα

=1

√2e−ρ/2

√12 (eρ + e−ρ) − cosα


=1√

2e−ρ/2√

cosh ρ− cosα=∑

�

e−�ρP�(cosα)

or1√

cosh ρ− cosα=

√2∑

�

e−(�+ 12 )|ρ|P�(cosα)

The boundary condition at ρ = ρa = cosh−1 5 may then be written

Va√cosh ρa − cosα

= Va

√2∑

�

e−(�+ 12 )|ρa|P�(cosα)

=∑

�

[C�e

(�+ 12 )ρa + C−�e

−(�+ 12 )ρa

]P�(cosα)

Equating coefficients gives

Va

√2e−(�+ 1

2 )|ρa| = C�e(�+ 1

2 )ρa + C−�e−(�+ 1

2 )ρa

similarly at ρb = −ρa we have

Vb

√2e−(�+ 1

2 )|ρa| = C�e−(�+ 1

2 )ρa + C−�e(�+ 1

2 )ρa

with solutions

C� =√

2e−(�+ 12 )|ρa|

[Vae

(�+ 12 )ρa − Vbe

−(�+ 12 )ρa

2 sinh((2�+ 1)ρa)

]

and

C−� =√

2e−(�+ 12 )|ρa|

[Vbe

(�+ 12 )ρa − Vae

−(�+ 12 )ρa

2 sinh((2�+ 1)ρa)

]We could of course accommodate spheres of any size simply by moving thex-y plane up or down the z axis. If we pick our zero of potential so thatVb = −Va we get

C� = −C−� =√

2Vae−(�+ 1

2 )|ρa| cosh(�+ 12 )ρa

sinh(2�+ 1)ρa=

√2Vae

−(�+ 12 )|ρa|

2 sinh(�+ 12 )ρa

Finally the compete solution is

V (ρ, α) =√

2Va

√cosh ρ− sinhα

∞∑�=0

e−(�+ 12 )|ρa| sinh(�+ 1

2ρ)sinh(�+ 1

2 )ρa

P�(cosα)

which, incidentally, also constitutes the solution for a charged sphere in theneighborhood of a conducting, neutral plane.

Errata: Ex 5.4.3 and 5.4.4 should each have s a summation symbol preced-ing the left hand side. Just above Ex 5.4.4, the phrase should have read:Multiplying by sinmϕ and integrating, we have

Chapter 6

6-1 The charge q located at distance h from the center of the isolated conductingsphere has an image charge q′ = −(R/h)q located at h − R2/h from thecharge and neutrality requires us to place a neutralizing charge (+R/h)q atthe center of the sphere. The force on q is therefore

F =q2R

4πε0h

[ −1(h− b)2

+1h2

]

with b = R2/h.

6-2 If we differentiate the given potential

−∂V∂z

= − d

dz

q

8πε0z=

q

8πε0z2

Comparing this to the field we expect at z from the charge at −z,E =

q

4πε0(2z)2=

q

16πε0z2

The reason for the discrepancy is that when we take the gradient of V in theneighborhood of z, we are determining how it varies when the source is heldconstant. By expressing the source coordinate in terms of the field coordinatewe are allowing it to vary as well in the differentiation.

6-3 We assume the sphere of radius R centered on the origin. For a charge qlocated at �r, its image q′ = −q(R/r) is located at �r ′ = (R2/r2)�r. Thepotential at r due to its image is

V (�r ) =q′

4πε0|�r − �r ′| =−q(R/r)

4πε0|�r − (R2/r2)�r | =−qR

4πε0r2(1 −R2/r2)

Then−(�∇V )r =

d

dr

qR

4πε0(r2 −R2)=

−2qrR4πε0(r2 −R2)2

The r component of the electric field we compute directly is

Er(�r ) =1

4πε0q′

(r − r′)2=

14πε0

−qR(r2 −R2)2

We see again that the gradient overcalculates the field by precisely a factorof two for the same reason as in the previous problem.

6-4 The dipole, which we take to lie along the z axis at distance z from the plane,has an image �p ′ = −�p + 2pz k located at z′ = −z. The electric field atdisplacement �r from the dipole �p ′ is in general given by

�E = −�∇V = −�∇ 14πε0

(�p ′ · �rr3

)

=1

4πε0

[3�r(�r ′ · �p ′)

r5− �p ′

r3

]

— 53—


In this case r = z + z′. The potential energy of dipole �p in this field is

W = −�p · �E =−1

4πε0

[3(�r · �p )(�r · �p ′)

r5− �p · �p ′

r3

]

=−1

4πε0

[3pzp

′z

(z + z′)3− �p · �p ′

(z + z′)3

]=

−(p2 + p2z)

4πε0(z + z′)3

The force on the dipole �p is then given by (note that had we written z + z′

as 2z, then dW = −2Fdz)

Fz = −∂W∂z

=−3(p2 + p2

z)4πε0(z + z′)4

=−3(p2 + p2

z)64πε0z4

If we express the potential energy of the dipole in terms of the angle it makeswith the z axis we easily obtain the torque (we have combined the z and z′

coordinates in the energy expression and recognize that changing θ changesθ′ whereas the virtual work requires that only θ change):

τθ = −12∂W

∂θ=

12∂

∂θ

p2(1 + cos2 θ)32πε0z3

=−p2 sin(2θ)

64πε0z3

where the θ axis is normal to the plane containing the dipole moment and thez axis.

6-5 We place the right plate along the x-y plane meaning that the charge is locatedat z = −a. The various images formed are shown in figure 6.1. It is evidentthat the forces from all the positive images cancel. Writing the remainingterms, we have

Fz =q2

4πε0

(1

(2a)2+

1(2D + 2a)2

+1

(4D + 2a)2+

1(6D + 2a)2

+ · · ·)

− q2

4πε0

(1

(2D − 2a)2+

1(4D − 2a)2

+1

(6D − 2a)2+ · · ·

)

=q2

16πε0

(1a2

− 4aD(D2 − a2)2

− 8aD(4D2 − a2)2

− 16aD(9D2 − a2)2

− · · ·)

Figure 6.1: Images of the charge located at z = −a.

Chapter Six Solutions 55

6-6 Since the sphere is conducting, the potential on it must be constant, andoutside it must be spherically symmetric. The total charge enclosed is zero sothat we conclude the potential everywhere outside the sphere is zero. To findthe potential and hence the field inside the sphere, we pick the z axis alongthe dipole axis and construct an extended dipole �p at the center of the sphereby placing a positive charge p/a at z = 1

2a and a negative charge at − 12a.

The point dipole will be recovered by letting a tend to 0. The charges at ± 12a

have images of magnitude ∓2R/a located at h = ±2R/a. The potential alongthe z axis is then

V (z) =p

a

14πε0

(1

z − 12a

− 1z + 1

2a− 2R/a

2R2/a− z+

2R/a2R2/a+ z

)

=p

4πε0a

[1z

(1

1 − 12a/z

− 11 + 1

2a/z

)− 1R

(1

1 − 12az/R

2− 1

1 + 12az/R

2

)]

=p

4πε0a

(a

z2− az

R3+ O(a2)

)taking the limit as a tends to 0, we find

V (z) =p

4πε0

(1z2

− z

R3

)Comparing this to the general form of the potential in spherical polar coor-dinates, we conclude immediately that V inside the sphere is

V (�r) =p

4πε0

(1r2

− r

R3

)cos θ

and the electric field inside the sphere is

�E = −�∇V =p

4πε0

(2r3

+1R3

)r cos θ − p

4πε0

(1r3

− 1R3

)θ sin θ

Outside the sphere the electric field vanishes.

6-7 The ring bearing charge Q has an image ring with charge Q′ = (−b/a)Qandradius a′ = −b2a. The potential along the z axis is then

V =Q

4πε0

(1√

z2 + a2− b/a√

z2 + b4/a2

)

To generalize this result, we expand each of the terms using the binomialtheorem. The first term may be expanded for z ∈ (a, b) as

1√z2 + a2

=1z

(1 +

a2

z2

)−1/2

=1z

[1 − 1

2a2

z2+

12

32

12!

(a2

z2

)2

− 12

32

52

13!

(a2

z2

)3

+ · · ·]

=1z

∞∑n=0

(−1)n(2n− 1)!!2nn!

a2n

z2n


The second term in V,

b/a√z2 + b4/a2

=1√

z2a2/b2 + b2=

1b

(1 +

z2a2

b4

)−1/2

is expanded as

1b

(1 +

z2a2

b4

)−1/2

=1b

∞∑n=0

(−1)n(2n− 1)!!2nn!

z2na2n

b4n

The potential between a and b along the z axis may therefore be expressedas

V (z) =Q

4πε0

∑ (−1)n(2n− 1)!!2nn!

a2n

(1

z2n + 1− z2n

b4n+1

)Comparing this to the general spherical polar solution

V =∑(

Anrn +

Bn

rn+1

)Pn(cos θ)

we find the general solution in the region (a, b) to be

V (a < r < b) =Q

4πε0

∑ (−1)n(2n− 1)!!2nn!

a2n

(1

r2n+1− r2n

b4n+1

)P2n(cos θ)

6-8 According to example 6.3 on page 148, a dipole �p at �r with respect to thecenter of a grounded conducting sphere has an image

�p ′ =R3

r3

[−�p+

3(�p · �r)�rr2

]

located at �r ′ = (R2/r2)�r. We can use the expression for the potential energyof two dipoles from problem 6-4, noting that the vector running from �r ′ to �ris �r ′′ = (1 −R2/r2)�r.

W =−1

4πε0r′′3

[3(�p · �r ′′)(�p ′ · �r ′′)

r′′2− (�p · �p ′)

]

=−R3

4πε0r6(1 −R2/r2)3

[p2 +

3(�r · �p )2

r2

]

For r R, this gives a 1/r6 potential for the dipole–induced dipole interac-tion.

6-9 The force and the torque on the dipole of the preceding problem may be foundby differentiating the potential energy. If we take the dipole to be inclined atangle θ with respect to its position vector, we may write (�r · �p )/r = p cos θand

W =−R3p2(1 + 3 cos2 θ)4πε0r6 (1 −R2/r2)3

=−R3p2(1 + 3 cos2 θ)

4πε0 (r2 −R2)3


so that

Fr = −12∂W

∂r=

−3R3p2r(1 + 3 cos2 θ)4πε0 (r2 −R2)4

and

τθ = −12∂W

∂θ= −3R3p2 sin θ cos θ

4πε0 (r2 −R2)3

6-10 Let the distance of closest approach of the wire to the center of the spherebe b and choose the x axis parallel to the wire lying in the x y plane, with x= 0 at the point of closest approach. A point x on the line charge is imagedat r′ = R2/r where r2 = x2 + b2. Calling the angle between r and the y axisθ, we note that the image falls at the same angle θ. In terms of θ, we mayfind r′ as

r′2 =R4

b2 + x2=

R4

b2(1 + tan2 θ)=R4 cos2 θ

b2

the equation of a circle with radius R2/2b centered at y = R2/2b as is shownbelow.

y′ = r′ cos θ =R2 cos2 θ

b=R2

2b(1 + cos 2θ)

x′ = r′ sin θ =R2 cos θ sin θ

b=R2

2bsin 2θ

leading to

x′2 +(y′ − R2

2b

)2

=(R2

2b

)2

The charge dq′ = λ′dθ on a segment subtending dθ about θ of the ring is theimage of the charge dq = λbdθ/ cos2 θ. The segment on the ring lies a distancer′ = R2 cos θ/b from the origin so that

dq′ = − R

b/ cos θdq =

−R cos θb

dq = −R cos θb

λdx = − λR

cos θdθ

so that, with the help of d� = R2/b dθ

dq′

dθ= − λR

cos θ, and

dq′

d�=dq′

dθ

dθ

d�= − λR

cos θb

R2= −λb

Rcos θ

6-11 In order to establish an electric field E0 at the origin, we place a charge Q =−2πε0L2E0 at +L and Q = +2πε0L2E0 at −L. These charges would haveimages Q′ = −(R/L)Q at +R2/L and −R2/L, respectively. The potentialalong the axis arising from the four charges is

V (z) = − E0

4πε0

(2πε0L2

L− z− 2πε0L2

L− z+

2πε0RLR2/L+ z

− 2πε0RLz −R2/L

)

=−E0

2

[L

(1

1 − z/L− 1

1 + z/L

)+RL

z

(1

1 +R2/Lz− 1

1 −R2/Lz

)]


We expand this expression to first order to obtain

V (z) =−E0

2

[L

(1 +

z

L− 1 +

z

L

)+LR

z

(1 − R2

Lz− 1 − R2

Lz

)+ O

(1L

)]

=E0

2

(2R3

z2− 2z

)

Comparing this to the general solution in spherical polar coordinates we gen-eralize immediately to obtain

V (�r > R) = E0

(2R3

r2− r

)cos θ

6-12 The images produced by a 60◦ plate vertex are shown in figure 6.2. It is readilyascertained that the y component of force vanishes whereas the x componentmay be calculated with the aid of figure 6.2 as

Figure 6.2: The location of the images when the vertex angle is α = 60◦

Fx =q

4πε0

∑ qi(x− xi)|�r − �ri|3

=q2

4πε0

(−2(1

2 )bb3

+2( 3

2 )b

(√

3b)3− 2b

(2b)3

)

=q2

4πε0

(− 1b2

+1√3b2

− 1(2b)2

)= −0.6726

q2

4πε0b2

6-13 Although the diagram above suggests a 60◦ angle between the plates, thegeneralization to α = π/n is immediate. When the angle α of the bend is notcommensurate with π, a solution may be interpolated from the π/n solutions.We take the x axis to contain the vertex and the charge. Consider the forcedue to the first images located at distance b along a line at ±α to the x axis.


The distance between the charge and each of these images is 2b sin 12α so that

we deduce a force

F =q2

4πε0(2b)2 sin2 12α

from each of these. The y components exactly cancel, and the x componentsare given by

Fx = F sin 12α =

−q24πε0(2b)2 sin 1

2α

The second pair of images lie at 2α and each contribute

Fx =+q2

4πε0(2b)2 sinα

to the x component of the force, while the each charge of the third paircontributes

Fx =−q2

4πε0(2b)2 sin 32α

and so forth until the successive images reach the angle π at which point asingle charge of sign (−1)π/α contributes

Fx =(−1)π/α+1q2

4πε0(2b)2

The net force from all these images is

Fx =−q2

4πε0(2b)2

π/α−1∑i=1

2(−1)(π/α+1)

sin(12 iα)

+ (−1)π/α

It is not immediately obvious how to extend this result to non- integralvalues of π/α.

Because we anticipate that the the force willvary smoothly with the angle of the vertex itis useful to examine how the sum varies as afunction of integral π/α. Calculating the firstfew sums, we find for n = π/α = 1, 2, 3, . . .andso forth.

One is struck by the near constancy of theincrease as we go to successive n. In fact,to excellent precision, the value of the sumis given by (try it for other values)

Sn = 0.8825425n+0.13n

This result must surely work for the non-integral values of n as well. In all cases theforce is one of attraction toward the vertex ofthe bend.

n α Sum1 180◦ 12 90◦ 1.828427063 60◦ 2.690598864 45◦ 3.562609035 36◦ 4.438744116 30◦ 5.316984877 25.71◦ 6.196442358 22.5◦ 7.076664969 20◦ 7.95739990

10 18◦ 8.8384949820 9◦ 17.6573902

100 1.8◦ 88.25554491000 0.18◦ 8882.54251


6-14 The images must produce a zero potential line along the plane interface, whichcan be accomplished by any symmetric arrangement of charges. Therefore,we place one image an equal distance below the plane. The original andthis image produce images in the sphere extrapolated from the hemisphere.These three image charges complete the set required to satisfy the boundarycondition.

6-15 We pick the axes so that the line charge lies at x = b and y = 0 and runsparallel to the z axis. Then an image line charge −λ located at h = R2/b willensure a constant potential on the cylinder. We write the potential as

V (r, θ) =−λ4πε0

(ln r1 − ln r2) + C

=−λ4πε0

ln(r21r22

)+ C =

−λ4πε0

ln(r2 + b2 − 2rb cos θr2 + h2 − 2rh cos θ

)+ C

At r = R the ratio r21/r22 = b/h, so that in order to obtain zero potential

on the cylinder we must set

C =λ

4πε0ln(b

h

)

The potential inside the cylinder is therefore

V (r, θ) =−λ4πε0

ln[h(r2 + b2 − 2rb cos θ)b(r2 + h2 − 2rh cos θ)

]

with h = R2/b.

6-16 We know that two parallel line charges ±λ produce a set of nonconcentriccylindrical equipotentials. Our task will be to pick the locations of these linecharges so that their equipotentials coincide with the two cylinders. Let hdenote displacement of the nearer line charge from axis of cylinder a whileh′ denotes the distance of the farther line charge from the axis of cylinder a.The distances of these line charges from the axis of cylinder b are then h+Dand h′ +D. The geometry is illustrated in Figure 6.3. In order that the twoline charges produce the equipotential desired, they must satisfy

hh′ = a2

(D + h)(D + h′) = b2

Expanding the second of these and substituting hh′ = a2 we obtain

D2 + (h+ h′)D + a2 = b2

Substituting a2/h for h′ we obtain the quadratic

Dh2 + (D2 + a2 − b2)h+ a2D = 0


Figure 6.3: The nested conducting cylinders of problem 6-16.

which may be solved to yield

h =(b2 − a2 −D2) ±√(b2 − a2 −D2)2 − 4a2D2

2D

The negative sign gives h while the + sign gives h′.The potentials on the two cylinders are given by

Va =−λ2πε0

ln(a

h

)Vb =

−λ2πε0

ln(

b

h+D

)

so that the capacitance per unit length, λ/ΔV , becomes

C

�=

2πε0

ln(a

h· h+D

b

)

with h given above.

6-17 The zero potential plane lies halfway between the wires of example 6.4. Thepotential difference between ground and one of the wires is just half thatbetween the wires so that we deduce the capacitance per unit length is twicethat of the pair of wires (D = 2d).

C

�=

2πε0cosh−1 (d/R)

6-18 We again use the equipotentials around two line charges to do this problem.Denoting the distance of the hypothetical line charge λ from the center ofcylinder a by ha and that of −λ from the axis of b as hb we write, referringto Figure 6.4


Figure 6.4: The equipotential cylinders of problem 6-18.

ha(D − hb) = a2

hb(D − ha) = b2

Using the second equation to eliminate hb from the first equation, we find

Dh2a − (D2 − b2 + a2)ha + a2D = 0

which we solve to obtain

ha =D2 − b2 + a2 ±√(D2 − b2 + a2)2 − 4a2D2

2D

The negative root must be chosen to give ha as illustrated. hb is found in thesame manner to give

hb =D2 − a2 + b2 ±√(D2 − a2 + b2)2 − 4b2D2

2D

Again the positive root must be chosen. The potentials on cylinder the cylin-ders are

Va =−λ2πε0

ln(a

ha

)Vb =

λ

2πε0ln(b

hb

)from which we conclude the capacitance per unit length is

C

�=

2πε0

ln(a

ha· bhb

)

6-19 When there is no inner sphere, and the charge distribution is independent ofϕ, the Green’s function (6-48) reduces to

G(�r, �r ′) =∞∑

�=0

P�(cos θ′)P�(cos θ)r�<

(1r�>

− r�> + 1b2�+1

)


With the charge density given,

ρ =Q

2b2πr′2[δ(cos θ′ + 1) + δ(cos θ′ − 1)]

the potential becomes

V (�r ) =Q

2b1

4πε0

∞∑�=1

[P�(1) + P�(−1)] P�(cos θ)∫ b

0

r�<

(1

r�+1>

− r�>

b2�+1

)dr′

The integral must be split into a region where r′ is r< and a second regionwhere r′ is r>.∫ b

0

(. . .)dr′ =∫ r

0

(r′ = r<)dr′ +∫ b

r

(r′ = r>)dr′

=(

1r�+1

− r�

b2�+1

)∫ r

0

r′�dr′ + r�

∫ b

r

(1

r′�+1− r′�

b2�+1

)dr′

=1

�+ 1

(1 − r2�+1

b2�+1

)+ r�

( −1�r′�

− r′�

(�+ 1)b2�+1

)∣∣∣∣b

r

=(2�+ 1)�(�+ 1)

(1 − r�

b�

)The � = 0 term is best obtained by direct integration(

1r− 1b

)∫ r

0

dr′ +∫ b

r

(1r′

− 1b

)dr′ = r

(1r− 1b

)+ ln r′

∣∣∣∣b

r

− 1b(b− r)

= 1 − r

b+r

b− 1 +

ln br

=ln br

The Legendre polynomial P�(−1) = (−1)� so that P�(1) + P�(−1) = 2 foreven � and vanishes for odd �. Then

V (�r ) =Q

4πε0b

{lnb

r+

∞∑�=1

P2�(cos θ)4�+ 1

2�(2�+ 1)

[1 −

(r

b

)2� ]}

6-20 The charge density may be written ρ(�r ′) = qδ[�r ′− 12 (a+ b)k] = qδ[r ′− 1

2 (a+b)]δ(cos θ − 1)/2πr′2. Both surfaces are grounded; hence the surface integralof (6–34) vanishes. We conclude, therefore, that

V (�r ) =1ε0

∫τ

ρ(�r ′)G(�r, �r ′)d3r′

with G given by (6–74). As the solution should have no ϕ dependence wereplace

∑m Ym

� (θ, ϕ)Y∗m� (θ′, ϕ′) by [(2�+ 1)/4π]P�(cos θ), leading to

V (�r ) =1

4πε0

∑�

P�(cos θ)1 − (a/b)2�+1

×∫τ

(r�< − a2�+1

r�+1>

)(1

r�+1>

− r�>

b2�+1

)δ[r′ − 1

2 (a+ b)]dr′


When r < 12 (a + b) it is r< so that performing the trivial integration, the

potential reduces to

V (r < r′) =q

4πε0

∑�

P�(cos θ)1 − (a/b)2�+1

(r − a2�+1

r�+1

)(2�+1

(a+ b)�+1− (a+ b)�

2�b2�+1

)

When r > 12 (a+ b), r = r> so that we may write

V (r > r′) =q

4πε0

∑�

P�(cos θ)1 − (a/b)2�+1

(1

r�+1− r�

b2�+1

)((a+ b)�

2�− 2�+1a2�+1

(a+ b)�+1

)

6-21 We develop the Dirichlet Green’s function for the interior of a cylindrical boxof radius a and length L. The equation we need to solve is

1r

∂

∂r

(r∂G(�r, �r ′)

∂r

)+

1r2∂2G(�r, �r ′)

∂ϕ2+∂2G(�r, �r ′)

∂z2= −δ(�r − �r ′)

We consider G(�r, �r ′) as a function of r and expand it in terms products of ofBessel functions that vanish at r = a and complex exponentials as

G(�r, �r ′) =∑�,m

Amj(θ′, ϕ′)F(z, z′)Jm

(ρmjr

a

)eimϕ

Inspection of Bessel’s equation, (E–1) gives

1r

d

dr

(rdJm

dr

)=(m2

r2− 1)Jm

so that substituting the expansion of G into ∇2G = δ(�r − �r ′) we simplify

∑�,m

(d2F(z, z′)

dz2− ρ2

mj

a2F(z, z′)

)Amj(θ′, ϕ′)Jm

(ρmjr

a

)eimϕ = δ(�r − �r ′)

Next we must expand the δ function in terms of Bessel functions andcomplex exponentials. Using Example D.4 we can write

δ(ϕ− ϕ′) =12π

∞∑m=−∞

eim(ϕ−ϕ′)

and using (D–15) we write

δ(r − r′) =∑m,j

r

νmjJm

(ρmjr

a

)Jm

(ρmjr′

a

)

so that δ(�r − �r ′) =δ(z − z′)δ(r − r′)δ(ϕ− ϕ′)

rmay be written

δ(�r − �r ′) = δ(z − z′)∑m,j

1νmj

Jm

(ρmjr

a

)Jm

(ρmjr′

a

)eim(ϕ−ϕ′)

2π


with

νmj =∫ a

0

J(ρmjr

a

)J(ρmjr

a

)r′dr =

a2

2[Jm+1(ρmj)]2

Substituting the δ function expansion into the equation for F(z, z′) above, weimmediately identify the expansion coefficients as

Ajm(r′, ϕ′) =Jmj

(ρmjr′

a

)e−imϕ′

a2π[Jm+1(ρmj)]2

leaving only the equation in (z, z′) to be solved.

d2F(z, z′)dz2

− ρ2mj

a2F(z, z′) = −δ(z − z′)

Recognizing the discontinuity provided by the δ function we solve thehomogeneous equation when z < z′ and when z > z′

F(z < z′) = Cmj(r′) sinh(ρmjz

a

)

F(z > z′) = Dmj(r′) sinh(ρmj(L− z)

a

)We impose the symmetry between z and z′ and the requirement of continuityto write the solution as

F(z < z′) = C ′mj sinh

(ρmjz

a

)sinh

(ρmj(L− z′)a

)and

F(z > z′) = C ′mj sinh

(ρmjz′

a

)sinh

(ρmj(L− z)a

)Finally, we determine the constant C ′ by integrating the differential equa-

tion over a small region containing z′.∫ z′+ε

z′−ε

d2Fdz2

dz −∫ z′+ε

z′−ε

ρ2mj

a2Fdz = −

∫ z′+ε

z′−ε

δ(z − z′)dz = −1

ordFdz

∣∣∣∣∣z′+ε

− dFdz

∣∣∣∣∣z′−ε

= −1

sinh(ρmjz

′

a

)ρmj

acosh

(ρmj(L− z′)a

)

+ρmj

acosh

(ρmjz′

a

)sinh

(ρmj(L− z′)a

)=ρmj

asinh

(ρmjL

a

)=

1C ′

mj

We conclude F(z, z′) may be written

F(z, z′) =a

ρmj sinh(ρmjL/a)sinh

(ρmjz<

a

)sinh

(ρmj(L− z>)a

)


We can now reassemble the Green’s function to get

G(�r, �r ′) =1aπ

∞∑m=−∞

∞∑j=1

sinh(ρmjz<

a

)sinh

(ρmj(L− z>)a

)ρmj sinh(ρmjL/a)J2

m+1(ρmj)

× eim(ϕ−ϕ′)Jm

(ρmjr

a

)Jm

(ρmjr′

a

)The Green’s function for locations surrounding a cylinder or for regions

between two cylinders may be constructed in similar (laborious) fashion.

Errata: The line above Ex 6.4.5 should read: The capacitance per unit lengthis now obtained as λ/ΔV .

The discussion following (6–33) has the wrong sign for the surface charge andthe dipole layer.

Chapter 7

7-1 This problem is essentially solved in the example 7.2. It suffices to replace the0.9 by t/d and 0.1 by 1 − t/d so that

Ed =V

κ(d− t) + tand Eair =

κV

κ(d− t) + t

7-1 Since the needle like cavity is parallel to the polarization, the bound surfacecharge has little effect on the field inside the cavity; instead, we must usethe continuity of the parallel component of �E to determine the field in thecavity. Taking the dielectric to be linear and isotropic, we have �Ediel. =�P/ε0χ. Therefore, in the cavity

�Ecav =�P

ε0χand �Dcav = ε0 �E =

�P

χ

7-3 This time the exposed ends of the dipoles contribute an field P/ε0 to the fieldin the cavity. We use the continuity of the component of �D perpendicularto the interface to determine the fields in the cavity. In the dielectric, thedisplacement field is

�D = ε0 �E + �P =�P

χ+ �P

Since �D⊥ is continuous, we find

Dcav = P

(1 +

1χ

)and Ecav =

Dcav

ε0=P

ε0+

P

χε0=P

ε0+ Ediel.

7-4 The electric field outside the dielectric cylinder may be found as the fieldcreated by the exposed ends of the dipoles on the two end faces. Thus,

�E(z) =1

4πε0

∫z=0

(�P · n)(zk − �r ′)|�r − �r ′|3 dS′ +

∫z=L

(�P · n)(zk − �r ′)|�r − �r ′|3 dS′

=1

4πε0

∫ a

0

∫ 2π

0

−P (zk − r′r)r′dr′dϕ′

(z2 + r′2)3/2

+1

4πε0

∫ a

0

∫ 2π

0

P [(z − L)k − r′r]r′dr′dϕ′

[(z − L)2 + r2]3/2

Inspection shows that these integrals are indistinguishable from those of(Ex 7.1.1). The electric field when z > L is then:

�E(z) =P k

2ε0

{z

(z2 + a2)1/2− z − L

[(z − L)2 + a2]1/2

}

— 67—


7-5 When the slot is parallel, the internal field strength �H is the same as that ofthe magnetized material. In other words, for Hmat = M/χ, Hslot = M/χ andB = μ0Hslot = μ0M/χ. When the slot is perpendicular to the magnetization,B⊥ is continuous. Therefore Bdisk = μ0(H + M) = μ0M(1/χ + 1) andHdisk = M(1/χ+ 1).

7-6 To find the force on charge 2, we use �∇ · �D = ρ and Gauss’ law to conclude

D =q1

4πr212

The force on charge 2 is

q2E =q2D

ε1=

q2

4πε1r212

7-7 If q2 lies in the material with permittivity ε2, the field at this point is thatdue to the screened charge q′′1 and that due to the image q′2

q′′1 =2ε2

ε1 + ε2q1 and q′2 = −ε1 − ε2

ε1 + ε2q2

Hence the apparent charge at position q2 is

3ε2 − ε1ε1 + ε2

q

and the forces on the charges neglecting the signs are

F2 =(3ε2 − ε1)q2

4πε2(ε1 + ε2)d2and F1 =

(3ε1 − ε2)q2

4πε1(ε1 + ε2)d2

where d is the distance between the charges. The difference reflects the forcethat the discontinuity of polarization at the interface exerts on the charges.

7-8 The potential V (R) at the surface of the sphere is Q/(4πε0R) so that thepotential at the center of the sphere may be found from

V (0) − V (R) = −∫ 0

R

�E(r) · d�r =1

4πε1

∫ R

0

Qr

R3dr

=Q

8πε1R

where we have used Gauss’ law to find �E(r) inside the sphere. The potentialat the center is therefore

V (0) =Q

4πR

(1ε0

+1

2ε1

)

Chapter Seven Solutions 69

7-9 Using expression (7–114), we find for a uniform magnetization

�B = −μ0

4π

∮�∇( �M · d�S′

|�r − �r ′|)

=μ0

4π

∮( �M · n)(�r − �r ′)dS′

|�r − �r ′|3

When r is 0, the integral over the inside and outside hemisphere preciselycancel leaving only the integral over the disk. On this surface, ( �M · n)�rdS′ =Mrr′dr′dϕ′ . Clearly this also vanishes since

∮2πrdϕ′ ≡ 0.

7-10 Noting that the image dipole �p ′ is given by

�p ′ =ε0 − ε

ε0 + ε(�p− 2pz k)

we adopt the expression for the energy of the dipoles found in problem 6-4:

W =1

32πε0z3(3pzp

′z − �p · �p ′)

=1

32πε0z3

ε1 − ε0ε0 + ε1

p2(1 + cos2 θ)

where the z axis was chosen to lie along the line joining the dipole and itsimage. The force on the dipole and the torque are now easily obtained.

Fz = −12∂W

∂z=ε0 − ε

ε0 + ε

3p2(1 + cos2 θ)64πε0z4

and

τθ = −12∂W

∂θ=ε0 − ε

ε0 + ε

p2 sin(2θ)64πε0z3

7-11 In the region of the capacitor where the dielectric slab lies between the plates,we may use the results of problem 7-1 to write the electric field

Eair =2εV

(ε0 + ε)dand Ediel =

2ε0V(ε0 + ε)d

and the field in the remaining area is E = V/d. The potential energy of thecapacitor when charged to voltage V is then

W =∫

12εE

2dr3

=ε0(ab− ΔxΔy)V 2

2d+ε0ΔxΔy 1

2d

2(2εV )2

(ε0 + ε)2d2+εΔxΔy 1

2d

2(2ε0V )2

(ε0 + ε)2d2

The force in the x direction (when V is held constant) is

Fx =∂W

∂x=[−ε0

2d+

ε0ε2

(ε0 + ε)2d+

εε20(ε0 + ε)2d

]ΔyV 2 =

ε0(ε− ε0)2(ε0 + ε)d

ΔyV 2


and the y component is

Fy =∂W

∂y=ε0(ε− ε0)2(ε0 + ε)d

ΔxV 2

7-12 The electric field between the cylinders takes the form

E =Va − Vb

r ln(a/b)≡ ΔVr ln(b/a)

so that, assuming that oil fills the capacitor of length � to height z, the po-tential energy may be written

W =π(ΔV )2

ln(b/a)[εz + ε0(�− z)]

The vertical electrical force on the oil may be equated to the gravitationalforce

Fz =(ε− ε0)π(ΔV )2

ln(b/a)= ρgzπ(b2 − a2)

to obtain the height to which the oil rises:

z =(ε− ε0)(ΔV )2

ρg(b2 − a2) ln(b/a)

7-13 We start this problem by recognizing that any line charge λ at distance d fromthe plane dielectric interface has an image λ′ = (ε0 − ε1)/(ε0 + ε1)λ ≡ Kλa distance d from the interface inside the dielectric. Assuming |K| is smalla rapidly diminishing series of image line charges will establish a constantpotential on the cylinder surface. Let us assume the cylinder center is atdistance D from the interface. As a first approximation we place a line chargeλ (not the total charge per length on the cylinder which will be the sum of allthe line charges in the cylinder) at the center which we denote by x = 0. Animage charge λ′ = Kλ at x′1 = 2D is required. This image must be balancedby an image −Kλ placed at x1 = a2/(2D). This charge in turn produces animage −K2λ in the plane at x′2 = 2D−2a2/2D. This image in turn yields animage K2λ inside the cylinder located at x2 = 2a2D/(4D2 − a2). We iterateonce more to get an image K3λ in the dielectric at x′3 = 2D − 2a2D/(4D2 −a2) whose image −K3λ inside the cylinder lies at x3 = a2/x′3 = a2(4D2 −a2)/[2D(4D2 − 3a2)]. This process could be continued ad infinitum althoughthe location of images hardly changes after the first few iterations. The chargedensity on he cylinder is λ(1−K+K2−K3 + · · ·) = λ/(1+K) = λ0 meaningthe total image charge is λ0K. We would make little error by approximatingthe charges inside the cylinder by a central line charge density λ and theremaining −Kλ0 at x1 = a2/2D. The image line charge in the dielectric may


be approximated at Kλ0 located at x′1 = 2D. The potential just outside thesurface of the cylinder is then

V (�r ) =−λ0

2πε0

(ln r −K ln

√r2 + x2

1 − 2rx1 cosϕ

+K ln√r2 + (2D)2 − 4rD cosϕ

whence

Er(a) =−λ0

2πε0

(1a− K(a− x1 cosϕ)√

a2 + x21 − 2ax1 cosϕ

+K(a− 2D cosϕ)√

a2 + (2D)2 − 4Da cosϕ

)

=σ

ε0

If higher accuracy is desired, it suffices to carry more terms of the series asoutlined. The term for the central charge in relation to the total charge iscorrect to all orders

7-14 The field strength between the faces of the magnetron magnet is

H = − 1L

∫PM

HPMd�

where L is the length of the gap. Assuming that inserting the screwdriverinto the gap does not significantly perturb the field internal to the magnet,we find the energy of the field increased by

W =(μ− μ0)

( ∫PM

HPMd�)2Ar

L

when the screwdriver is inserted to a distance r. The radial force is then

Fr = −∂W∂r

=A(μ− μ0)

( ∫PM

HPMd�)2

L

In reality , of course, the field in the magnet will be affected by the change ofreluctance as the screwdriver is inserted.

7-15 Several different methods may be used to solve this problem. We will explorethree taking the z axis along the magnetization. For the first method we useequation (7–110) and proceed much like the example 7.10.

Vm = − 14π

�∇ ·∫

τ

M0kd3r′

|�r − �r ′|To effect the integration, we expand |�r − �r ′|−1 as a Legendre series:

1|�r − �r ′| =

1r>

∑�

(r<r>

)�

P�(cos θ)


All the � �= 0 terms vanish when integrated over the solid angle, leaving only

Vm = −M0

4π∂

∂z

∫r′2 dr′dΩ′

r>

= −M0∂

∂z

∫r′2dr′

r>

r′ must always be less than a, the radius of the sphere. Likewise, when r liesoutside the sphere it is always r>, leading to

Vm(r > a) = −M0∂

∂z

∫ a

0

r′2dr′

r= −M0

∂

∂z

a3

3r

=M0a

3z

3r3

Differentiating, we obtain

�B(r > a) = −μ0�∇Vm = μ0M0a

3

(z�r

r5− k

3r3

)

When r lies inside the sphere, some care is required. r′ is r< from 0 untilit reaches r whereupon it becomes r> for the rest of the integration. Thus

Vm(r < a) = −M0∂

∂z

(∫ r

0

r′2dr′

r+∫ a

r

r′2dr′

r′

)

= −M0∂

∂z

(r2

3− r2

2+a2

2

)= 1

3M0z

The induction field this time is given by

�B = μ0( �H + �M) = μ0(−�∇Vm + �M)

= μ0(− 13M0k +M0k) = 2

3μ0M0k

An alternative approach is to replace the magnetization by a body currentJm = �∇ × �M and a surface current �j = �M × n. The body current vanishesleaving only the surface current

�M × n = Mk × r = M sin θ ϕ

Comparing this current to that on the rotating charged sphere (Example 5.10)we identify �M × n with σaω and adapt the solution immediately.

We could also have treated this problem as a boundary condition problem.The scalar potential takes the form

Vm(r < a) =∑

A�r�P�(cos θ)

Vm(r > a) =∑ B�

r�+1P�(cos θ)


From this we find

�H = −�∇Vm ⇒ Hr =

⎧⎨⎩∑

− �r�−1A�P�(cos θ) (r < a)∑ (�+ 1)B�

r�+2P�(cos θ) (r > a)

and

Hθ =

⎧⎪⎨⎪⎩∑

−A�r�−1 ∂P�(cos θ)

∂θ(r < a)∑

− B�

r�+2

∂P�(cos θ)∂θ

(r > a)

The magnetic induction field may generally be found as �B = μ0( �H + �M),whence we obtain the radial (perpendicular) component of �B

Br =

⎧⎨⎩

μ0

∑−�A�r

�−1P�(cos θ) +M cos θ (r < a)

μ0

∑ B�(�+ 1)r�+2

P�(cos θ) (r > a)

We equate the interior and exterior forms of Br at the boundary

−A1 +M =2B1

a3and �A�a

�−1 =(�+ 1)B�

a�+2when � �= 1

The requirement of continuity of Hθ at the boundary gives

A1 =B1

a3and A�a

�−1 =B�

a�+2when � �= 1

Solving the � = 1 equation gives us A1 = 13M while the � �= 1 equations give

A� =B�

(�+ 1�

)a2�+1

=B�

a2�+1

having no nonzero solutions. The scalar magnetic potential may now bewritten

Vm(r < a) = 13Mr cos θ = 1

3Mz

Vm(r > a) =a3

3r2M cos θ

The evaluation of �B is performed as in the first solution.

7-16 We write the general solution for the scalar potential in spherical polar coor-dinates, including explicitly the term to give �H0 at large distances.

Vm(r ≤ R) =∑

A�r�P�(cos θ)

Vm(r > R) = −H0r cos θ +∑ B�

r�+1P�(cos θ)


The boundary conditions to be applied to this solution are

Br(R+) = Br(R−) and Hθ(R+) = Hθ(R−)

The first of these yields∑�A�r

�−1P�(cos θ) =∑ −(�+ 1)B�

r�+2P�(cos θ) −H0 cos θ −M cos θ

while the second yields∑A�r

�−1P′�(cos θ) = −H0P′

1(cos θ) +∑ B�

r�+2P′

�(cos θ)

The � = 1 terms yield

A1 = −H0 +B1

R3andA1 = −H0 −M − 2B1

R3

which we solve to obtain B1 = − 13MR3 and A1 = −H0 − 1

3M . The � �= 1terms must vanish to give the following interior and exterior potentials

Vm(r ≤ R) = −(H0 + 13M)r cos θ = −(H0 + 1

3M)z

Vm(r > R) = −(H0 +

MR3

3r3

)z

The magnetic induction field is now easily obtained:

�B(r ≤ R) = μ0(H0 + 13M)k

�B(r > R) = μ0

(H0 +M +

MR3

3r3

)k − μ0MR3z�r

r5

7-17 Adapting the results for the magnetized cylinder to this problem by lettinga→ ∞, we find Vm = Mz inside the slab where we have taken M to lie alongthe z axis. We deduce that �H = −�∇Vm = − �M . Hence �B = μ0( �H + �M) = 0inside the slab. That �B = 0 outside the slab is easily demonstrated bycalculating the field of an infinite sheet of dipoles as follows:

�B = −μ0�∇∫ �M · (�r − �r ′)

4π|�r − �r ′|3 dS′ = −μ0

�∇∫ ∞

0

�M · (zk − ρr)4π(z2 + ρ′2)3/2

2πρ′dρ′ = 0

7-18 The contribution from the face centered at 12b to the scalar potential at z

assuming that M is z directed is

Vm(z) =14π

∫ �M · d�S′

|�r − �r ′| = −M4π

∫ a

0

2πr′dr′√r′2 + (z − 1

2b)2

= −M2

√r′2 + (z − 1

2b)2

∣∣∣∣a

0

= −M2

[√a2 + (z − 1

2b)2 −

√(z − 1

2b)2]

Similarly the face at − 12b contributes


Vm(z) =M

2

[√a2 + (z + 1

2b)2 − (z + 1

2b)]

to the scalar potential. Adding the two terms, we obtain

Vm(z) =M

2

[√a2 + (z + 1

2b)2 −

√a2 + (z − 1

2b)2 − 2z

]

7-19 At large distances the magnetic field intensity must tend to �H0, implying thatVm → −H0r cosϕ. Thus

Vm(r > a) = −H0r cosϕ+∑ Bn

rncosnϕ

andVm(r < a) =

∑Anr

n cos nϕ

The boundary conditions require the continuity of Br = −μ∂Vm/∂r andHϕ = −(1/r)∂Vm/∂ϕ at r = a. Applying the first of these,

μ0

(H0 cosϕ+

∑ nBn

rn+1cosnϕ

)∣∣∣∣r=a

= μ∑

nAnrn−1 cosnϕ

∣∣∣∣r=a

which gives

−μ0

(H0 +

B1

a2

)= μA1 and − Bn

a2n=

μ

μ0An (n �= 1)

The continuity of Hϕ gives

−H0 +B1

a2= A1 and

Bn

a2n= An (n �= 1)

The n �= 1 equations admit only the trivial solution when μ �= μ0 while the n= 1 equations give

(μ− μ0)A1 = −2μB1

a2and (μ− μ0)H0 = (μ+ μ0)

B1

a2

or

B1 =(μ− μ0

μ+ μ0

)H0a

2 and A1 =−2μ0

μ+ μ0H0

Thus, for an x -directed field �B0

V (r < a) = − 2μ0

μ+ μ0H0r cosϕ = − 2B0x

μ+ μ0

and

V (r > a) = −B0

μ0

(r cosϕ− μ− μ0

μ+ μ0

a2

rcosϕ

)= −B0

μ0

(x− μ− μ0

μ+ μ0

a2

r2x

)


The fields are now easily computed. Interior to the cylinder,

�B(r < a) = −μ�∇Vm =2μ

μ+ μ0

�B0

and

�B(r > a) = �B0 − μ− μ0

μ+ μ0

(a2 �B0

r2− 2a2( �B0 · �r)�r

r4

)

7-20 The magnetic dipole moment∫�Md3r must lie along the symmetry axis which

we take to be the z axis. The magnetic dipole moment is therefore simplyMτ , where τ is the volume of the magnet.

7-21 The magnet at perpendicular distance z from the interface induces an imagedipole −(1 − 2 × 10−4)V ( �M − 2Mz k) inside the iron. Following the methodof problem 6-4, we find the force on the dipole to be

Fz =−3μ0(m2 +m2

z)64πz4

We specialize this result to the dipole being parallel to the plate, mz = 0, �Fz =−3μ0(MV )2/(64πz4), or perpendicular, mz = m, �Fz = −3μ0(MV )2/(32πz2).The force is directed towards the iron surface.

7-22 We consider the magnetic circuit constituted by the yoke, the airspace andthe magnet. For the entire circuit∮

�H · d�� = 0 ⇒ HPM�PM = −∫

rest

�H · d��

Further,∫

rest

�H · d�� = Φ�, with

� =∫

d�

μA=

20 cm25, 000 μ0 cm2

+1 cmμ0 cm2

=25, 020 cm−1

25, 000 μ0

so that the flux may be found in terms of the PM field,

Φ =−HPM × 10cm

(25, 020/25, 000)cm= −9.992μ0cm2HPM

Φ is assumed constant around the circuit so that we may relate it the mag-netic induction field in the magnet to obtain a relation between BPM andHPM : BPM = ΦA = −9.992μ0HPM . Solving this simultaneously with thehysteresis curve relation between B and H, we obtain BPM ∼ 0.54T andHPM ∼ −0.054T/4π×10−7 = −4.3×104A/m. Because the cross section doesnot vary, B is constant around the loop, so that Hair = B/μ0 = 4.3×105A/mand Hyoke = B/(25, 000μ0) = 17.2A/m.


7-23 The displacement field is generally given �D = ε0 �E + �P . Adding the polariza-tion to ε0 �E as given in (Ex 7.7.1) gives

�D(z) =�P

2

{z

(z2 + a2)1/2+

L− z

[(L− z)2 + a2]1/2

}

We translate the origin to z = 12L so that z, the distance to the bottom face

is replaced by z + 12L and (L− z) is replaced by 1

2L− z to obtain

�D(z) =�P

2

{ 12L+ z

[(12L+ z)2 + a2]3/2

+12L− z

[(12L− z)2 + a2]3/2

}

Comparison to the result of Exercise 7.4 shows that replacing �P by μ0�M

makes the results identical.

7-24 This problem is very similar to Example 7.10 (or 7.12) and we proceed in thesame way. Thus,

Vm(�r ) = − 14π

�∇ ·∫

τ

M0x′ ı

|�r − �r ′|d3r′

= −M0

4π∂

∂x

∫τ

x′

|�r − �r ′|d3r′

where the volume τ includes the boundary. We again resort to sphericalharmonics to effect the integration. x′ = r′ sin θ′ cosϕ′ may be written as

x′ = 2r′√

8π3

[Y−1

1 (θ′, ϕ′)−Y11(θ

′, ϕ′)]

and we use (F–47) to expand 1/|�r−�r ′|resulting in

Vm(�r ) = −2M0

4π∂

∂x

∑�,m

Ym� (θ, ϕ)

√8π3

4π2�+ 1

×∫

r�<

r�+1>

[Y−1

1 (Ω′) − Y11(Ω

′)]Y∗m

� (Ω′)r′r′2dr′dΩ′

The integration over Ω′ eliminates all but the � = 1,m = ±1 terms each ofwhich integrates to unity.

Vm(�r ) = −M0

3∂

∂x

x

r

∫ a

0

r<r2>r′3dr′

When r > a, the scalar magnetic potential is

Vm(�r ) = −M0

3∂

∂x

x

r

∫ a

0

r′

r2r′3dr′ = −M0a

5

15∂

∂x

x

r3= −M0a

5

15

( 1r3

− 3x2

r5

)When r < a we need to break the integral into two intervals.

Vm(�r ) = −M0

3∂

∂x

x

r

[ ∫ r

0

r′

r2r′3dr′ +

∫ a

r

r

r′2r′3dr′

]


= −M0

3∂

∂x

(xr25

+xa2

2− xr2

2

)= −M0

30

(5a2 − 3(3x2 + y2 + z2)

)The magnetic induction field may be found as �B = −μ0

�∇Vm

�B(r > a) =μ0M0a

5

15

(− �r

r5− 3xı

r5+

15x2�r

r7

)Similarly, the interior induction field is

�B(r < a) = −μ0M0

5

(3xı+ yj+ zk

)Alternatively, we might use the methods of Example 7.12. The divergence

of the magnetization �∇·M0xı = M0 which is easily integrated over the volumeto give the first of the two integrals in (7–114).

− 14π

∫τ

M0

|�r − �r ′|r′2dr′dΩ′ = −M0

∫ a

0

1r>r′2dr′

where the volume τ does not include the boundary.

Vm(r > a) = −M0a3

3rVm(r < a) = −M0

(a2

2− r2

6

)

The remaining surface integral may be evaluated using ı·d�S = sin θ′ cosϕ′a2dΩ′

and on the surface x′ = a sin θ′ cosϕ′ so that

14π

∮ �M · d�S|�r − �r ′| =

M0

4π

∮a sin2 θ′ cos2 ϕ′a2dΩ′

|�r − �r ′|

= M0a3∑�,m

Ym� (θ, ϕ)2�+ 1

∮Y∗m

� (θ′, ϕ′) sin2 θ′ cos2 ϕ′r�<dΩ

′

r�+1>

sin2 θ cos2 ϕ may be written as 14 sin2 θ(eiϕ+e−iϕ)2 = 1

4 sin2 θ(e2iϕ+e−2iϕ+2)

=√

π30

[Y−2

2 − Y22

]+ 1

2 sin2 θ =√

π30

[Y−2

2 − Y22

] −√ 4π45 Y0

2 +√

4π9 Y0

0 whichallows us to eliminate the terms other than � = 2,m = 0,±2 and � = 0,m = 0from the sum. the sum above, integrated over Ω then becomes

M0a3

√π

90

[√3Y−2

2 (θ, ϕ) −√3Y2

2(θ, ϕ) −√8Y0

2

5r2<r3>

+√

40Y00(θ, ϕ)r>

]

=M0a

3

5r2<r3>

sin2 θ cos2 ϕ+M0a

3

3

( 1r>

− r2<5r3>

)When r > a, meaning r< = a, the scalar magnetic potential, using both thevolume and surface integral becomes

Vm(r > a) =M0a

5

15r5(3x2 − r2)


The interior (r> = a) magnetic scalar potential becomes

Vm(r < a) = −M0

(a2

2− r2

6

)+M0x

2

5+M0

3

(a2 − r2

5

)

=M0

30(9x2 + 3y2 + 3z2 − 5a2)

7-25 Following the hint, we compute

〈 �D〉 =⟨(〈ε〉 + δε)

( 〈 �E〉 + δ �E)⟩

= 〈ε〉〈 �E〉 + 〈δεδ �E〉

as 〈δε〉 and 〈δ �E〉 vanish. To a zeroth approximation, the permittivity of themixture is just the average of the permittivities. To obtain the first ordercorrection we must evaluate 〈δεδ �E〉. Consider �∇ · �D = 0,

�∇ ·[(〈ε〉 + δε)(〈 �E〉 + δ �E)

]= �∇ · (〈ε〉δ �E + δε〈 �E〉)= 〈ε〉(�∇ · δ �E) +

(〈 �E〉 · �∇)δε = 0

In other words,〈ε〉∂kδE

k = −〈Ek〉∂kδε

We differentiate once more with respect to xj to get

〈ε〉∂j∂kδEk = −〈Ek〉∂j∂kδε

If the medium is isotropic, ∂j∂kε = 13δ

jk∇2ε while ∂j(�∇ · �E) = ∇2Ej so that

we rewrite the equation as

∇2δEj = −〈Ej〉3〈ε〉 ∇

2δε

⇒ δEj = −〈Ej〉3〈ε〉 δε

Multiplying both sides by δε and averaging, we get the required correction to〈 �D〉. Thus

〈δEjδε〉 = −〈Ej〉〈(δε)2〉3〈ε〉

Substitute this into the expression for 〈 �D〉 we have

〈 �D〉 =[〈ε〉 − (δε)2

3〈ε〉]〈 �E 〉

From this expression we read immediately

εmix = 〈ε〉 − (δε)2

3〈ε〉To show that this is the desired result, we expand


(〈ε〉 + δε)1/3 = 〈ε〉1/3 + 1

3 〈ε〉−2/3(δε) − 19 〈ε〉−5/3(δε)2 + · · ·

We take the average of both sides to get⟨(〈ε〉 + δε)1/3

⟩= 〈ε〉1/3 − 1

9 〈ε〉−5/3(δε)2 + · · ·= 〈ε〉1/3

[1 − (δε)2

9〈ε〉2 + · · ·]

Finally we cube both sides to obtain⟨(〈ε〉 + δε)1/3

⟩3

= 〈ε〉[1 − (δε)2

3〈ε〉2 + · · ·]

completing the demonstration.

7-26 Equation (7–54) readily reduces to the diffusion equation

∇2 �B = gμ∂ �B

∂t

when the conductor is stationary. This problem is better described as aninitial value problem than a boundary condition problem. Let us look at theform of the solutions to this equation. If �B were a scalar (it is not of course) theseparable solutions of the form R(r)f(t) would have spherical Bessel functionsR(r) = j0(kr) = (sin kr)/(kr). B would then satisfy ∇2B = −k2B so thatthe temporal function f(t) satisfies

df

dt= − k2

gμf ⇒ f = e−(k2/gμ)t

The characteristic decay time for a disturbance of length k−1 is gμ/k2. Beforeit is objected that B is a vector field and all this is irrelevant, we recall that thegradient (as well as �r× �∇) of the scalar solution solves the vector equation sothat conclusions about the temporal behavior are valid. Continuing thereforewith the scalar solution, we can synthesize an arbitrary B in the sphere bysuperimposing solutions with increasing k. The lowest k (slowest decay) isk = π/R. Thus, τ = gμR2/π2 is the decay time for global fields. For acopper sphere of 1 m radius this implies a decay time of

τ =5.9 × 107 × 4π × 10−7

π2= 7.5s

For the hypothetical earth (R = 3 × 106 m for the core),

τ =106 × 4π × 10−7 × 9 × 1012

π2≈ 1.15 × 1012s ≈ 3.6 × 104y

These considerations would suggest that it would be very difficult to producelarge-scale changes in the earth’s magnetic field in times less than 104 years.

Errata: Ex 7.8.3 on page 187 has a misplaced superscript (�+2) which shouldbe in the denominator R of the last term.

On page 188, following Ex 7.8.6 the phrase should read: which, when solvedfor A1, give A1 = −E0 − 1

3P/ε0.

Chapter 8

8-1 Generally, in a conductor, the wave vector �K of a plane wave satisfies (8–123)

K2 = μεω2

(1 +

ig

εω

)

A good conductor is characterized by g εω, so that to a good approximation

K2 = μω2

(ig

ω

)⇒ K =

√μgωeiπ/4

Writing �H in terms of �E using (8–122), �H0 = ( �K× �E)/(ωμ) it is evident thatH and E oscillate 45◦ out of phase.

8-2 The skin depth δ is given by

δ =2g

√βε

2μwith β = 1 +

√1 +

(g

εω

)2

Putting the explicit values in we find

g

εω=

4.380 × 8.85 × 10−12 × 2π × 100

= 9.7 × 106 1

so that

δ =√

2gωμ0

=

√2

4.3 × 4π × 10−7 × 2π × 102= 24.3 m

Such ELF waves would be useful only if the submarine was within a fewmeters of the water surface.

8-3 A spherical membrane with surface tension σ and radius of curvature R exertsinward pressure ℘ = σ/2R per surface. The outward pressure from blackbodyradiation is 1

3U = 43σS-BT 4/c.

Putting in numeric values as far as possible, the interior pressure supportedby two surfaces is

σ

R=

4 × 5.67 × 10−8 × (300)4

3 × 3 × 108

Nm2

= 0.51 × 10−6Pascal

⇒ R =σ

0.51 × 10−6= 1.96 × 106σ

8-4 According to (8–39) the momentum density is given by �D × �B, whereas thePoynting vector �S = �E × �H according to (8–36). In the dielectric,

�D × �B = ε �E ×�B

μ=ε

μ�S

— 81—


8-5 Forces arise from two different sources—from the change of momentum of thereflected light and from the change of momentum of the transmitted light.

The reflected power is just [(1−n)/(1+n)]2dW/dt, with attendant changein momentum flux

Fr =2c

(1 − n

1 + n

)2dW

dt

The portion of the beam that will be transmitted has energy flux (4n)/(1 +n)2dW/dt and carries (in air) momentum flux

4n(dW/dt)c(1 + n)2

toward the interface while it carries momentum at a rate

4n(dW/dt)v(1 + n)2

away from the interface. The net change in momentum of the light beam perunit time is therefore

Fr =

2c(1 − n)2 +

4nc

− 4nv

(1 + n)2dW

dt

=2 − 4n+ 2n2 + 4n− 4n2

(1 + n)2cdW

dt

=2(1 − n2)(1 + n)2c

dW

dt

The corresponding force on the dielectric is just minus this result.It is instructive to obtain this result from the Maxwell stress tensor using

dFz = −TzzdSz with

Tzz = −EzDz −BzHz + 12 ( �E · �D + �B · �H) = 1

2 ( �E · �D + �B · �H)

for a z directed transverse wave. In accordance with the discussion on page84, the force from a parallel field is directed toward the wall.

On the incident side,

Etot = Ei +Er =(

1 +1 − n

1 + n

)Ei =

2n+ 1

Ei

and

Htot =√ε0μ0

(Ei − Er) =√ε0μ0

(1 − 1 − n

1 + n

)=√ε0μ0

2n1 + n

Ei

leading to

Tzz = 〈 12 ( �E · �D + �B · �H)〉 = 1

2

[ε0

(2

n+ 1

)2

+ μ0ε0μ0

(2nn+ 1

)2]〈E2

i 〉

Chapter Eight Solutions 83

=2(1 + n2)c(n+ 1)2

ε0c〈E2i 〉 =

2(1 + n2)c(n,+1)2

d2W

dAdt

In the same fashion, on the dielectric side of the interface

Et =2

n+ 1Ei Ht =

√ε

μ0

2n+ 1

Ei

whence

12 〈 �E · �D + �B · �H〉 = 1

2

[4ε

(n+ 1)2+

4ε(n+ 1)2

]〈E2

i 〉

=4n2

c(n+ 1)2ε0c〈E2

i 〉 =4n2

c(n+ 1)2d2W

dAdt

The force on the dielectric is then

F =2

c(n+ 1)2[2n2 − (1 + n2)]

dW

dt=

2(n2 − 1)c(n+ 1)2

dW

dt

8-6 The decay constant is 1/α, with

α = kt

√sin2 θi

n2− 1 = ki

√sin2 θi − n2 =

2πλ

√sin2 60◦ − (1/1.7)2

=3.993λ

Thus the decay distance is nearly 14λ.

8-7 Using the formulas for the half angles, the phase angles are easily computed.

tan 12ϕs =

√sin2 θi − n2

cos θi= 1.27 ⇒ ϕs = 103.6◦

tan 12ϕp = (1.7)2 × 1.27 = 3.67 ⇒ ϕp = 149.52◦

8-8 An EM plane wave propagating through a conducting medium satisfies

−k2 + μεω2

(1 +

ig

ωε

)= 0

Taking the view that the plasma is a conducting rather than a polarizablemedium, we set ε = ε0 and using

m�v = q �E ⇒ �v =iq �E

mω⇒ �J = nq�v =

inq2

mω�E

we get the conductivity g = inq2/mω. We substitute this into the dispersionrelation above to find

k2 = με0ω2

(1 − nq2

mω2ε0

)


With the identification ω2p = nq2/(mε0) this becomes

k2 =ω2

c2

(1 − ω2

p

ω2

)

8-9 Newton’s equation of motion for a charge e in an axial field B and an electricfield �E (in the x-y plane) is

mr = e �Ee−iωt + eyBı− exBj

or

x =eB

my +

eEx

me−iωt ≡ ωcy +

eEx

me−iωt

y = −eBmx+

eEy

me−iωt ≡ −ωcx+

eEy

me−iωt

We untangle these equations by taking (complex) linear combinations of xand y.

x+ iy = ωc(y − ix) +e(Ex + iEy)

me−iωt

x− iy = ωc(y + ix) +e(Ex − iEy)

me−iωt

Now assuming a harmonic time dependence with angular frequency ω for eachof the components, we write

− ω2(x+ iy) = ωc(−iω)(y − ix) +e(Ex + iEy)

m= −ωωc(x+ iy) +

eE+

m

−ω2(x− iy) = ωc(−iω)(y + ix) +e(Ex − iEy)

m= ωωc(x− iy) +

eE−m

Grouping terms we find

−(ω2 ± ωωc)(x± iy) =eE±m

⇒ x± iy = − eE±m(ω2 ± ωωc)

The corresponding components of the polarization of the medium are

P± ≡ Px ± iPy = ne〈x± iy〉 =−ne2E±

m(ω2 ± ωωc)=

−ω2pε0E±

(ω2 ± ωωc)

from which we deduce that

ε± = ε0

[1 − ω2

p

(ω2 ± ωωc)

]

The dispersion relation k2 = ω2μ0ε becomes

k2± =

ω2

c2

(1 − ω2

p

ω2 ± ωωc

)

Chapter Eight Solutions 85

A right hand circularly polarized wave propagating in the +z direction haselectric field components E− = E0 and E+ = 0 and therefore propagates withk− while a left hand circularly polarized wave propagates with k+.

8-10 The total current consists of both the real current and the bound current aris-ing from the changing polarization of the medium. Thus the current traversinga cross sectional area A is

I =(�J +

∂ �P

∂t

)· �A = �E0(g cosωt− ωχε0 sinωt) · �A

=VA

d

(g cosωt− (ε− ε0)ω sinωt

)The amplitude of the current is

|I| =VA

d

[g2 + (ε− ε0)2

]1/2

=VA

10−2m[4.32 + (3.14 × 109 × 79 × 8.85 × 10−12)2

]1/2

= 482.8 × VA (A/V m)

8-11 The permittivities may be computed from the principal refractive indices,εxx = 1.69ε0, εyy = 2.25ε0 and εzz = 2.89ε0. The electric displacement vector�D associated with �E is then �D = ε0(0.976ı, 1.26j, 1.674k). The Poyntingvector is now easily computed, except for a numerical constant

�S = �E × �H ∝ (ı, j,−2k)

The direction of the wave vector �k may similarly be found

�k = �D × �B ∝ (ı, j,−1.36k)

The non-colinearity of the Poynting and wave vector is evident.

8-12 The total power incident on the interface where it has footprint S is

Pi = ( �Ei × �Hi) · �S = S( �Ei × �Hi) cos θi ∝ niS cos θiE2i

On the other side of the interface, the footprint is the same, but the crosssection of the beam is different,

Pt = ( �Et × �Ht) · �S ∝ ntS cos θtE2t

The ratio of transmitted to incident power is therefore

Pt

Pi=nt cos θt

ni cos θit2

It is readily verified that r2 + Pt/Pi = 1.


8-13 If the expansion occurred adiabatically the expanded volume must have thesame energy contained within it as did the unexpanded. The radiation doeswork as it is expanding and we use the adiabatic expansion law PV γ = con-stant with γ = 4

3 and noting that V ∝ L3 and P ∝ T 4, we have T 4L4 =constant. Thus

Lf

Li=Ti

Tf= 366.3

8-14 The ratio of reflected to incident power is given (for a good conductor)

∣∣∣∣E0,r

E0,i

∣∣∣∣2

= 1 − 2√

2ωε0g

= 0.9898

The transmitted and reflected field “amplitudes” may be found from

E0,t =2

1 + ηE0,i and E0,r =

1 − η

1 + ηE0,i

with η given by

η =√

g

2ωε0(1 + i) = 196 + 196i

Substituting these values we find

E0,t = 0.0072e−0.7828iE0,i and E0,r = −(0.9949 + .005i)E0,i

yielding phase shifts of 44.85◦ for the transmitted wave and 179.72◦ for thereflected wave. The reflection coefficient found earlier is easily verified fromthe amplitude of the reflected wave.

Chapter 9

9-1 TM modes are characterized by Bz = 0. It remains, therefore, to solve for Ez

from∇2

tEz + (μεω2 − k2)Ez = 0

The solutions are of the form

Ez ={

cosαxsinαx

}{cosβysinβy

}

Applying the boundary conditions: Ez = 0 at x = 0, x = a, y = 0 and y =b the solution reduces to

Ez = A sinnπx

asin

mπy

b

If this is substituted into the wave equation (above) we find immediately therelevant dispersion relation(

nπ

a

)2

+(mπ

b

)2

+ k2 − μεω2 = 0

The transverse components of the magnetic field intensity Ht are found from(9–23)

�Ht =iεωk × �∇tEz

μεω2 − k2

=iεωA

μεω2 − k2k ×

(nπ

aı cos

nπx

asin

mπy

b+mπ

bj sin

nπx

acos

mπy

b

)

=iεωA

μεω2 − k2

(nπ

aj cos

nπx

asin

mπy

b− mπ

bı sin

nπx

acos

mπy

b

)

9-2 TE modes satisfy(∇2 + γ2)H0,z = 0

with boundary conditions H0,z(0) = H0,z(c) = 0 and ∂H0,z/∂n = 0 at theside walls, leading to solutions of the form

H0,z = H0 sin�πz

ccos

nπx

acos

mπy

b

whence we find the transverse fields (page 254)

�H0,t = −�πH0

cγ2cos

�πz

c

[ınπ

asin

nπx

acos

mπy

b+ j

mπ

bcos

nπx

asin

mπy

b

]and

�E0,t =iμωH0

γ2sin

�πz

c

[jnπ

a sinnπx

acos

mπy

b− ı

mπ

bcos

nπx

asin

mπy

b

]

— 87—


where

γ2 = μεω2 −(�π

c

)2

=(nπ

a

)2

+(mπ

b

)2

The second equality may be rewritten to give

μεω2 =(�π

c

)2

+(nπ

a

)2

+(mπ

b

)2

or

f =ω

2π=

12√με

√(na

)2+(mb

)2+(�

c

)2

The � = 0 term yields only the trivial solution so that we must have � ≥ 1.Some of the lower modes have then the mode numbers (n,m, �) = (0, 0, 1),(1, 0, 1), (0, 1, 1), (0, 0, 2), (2, 0, 1), (0, 2, 1) and (1, 1, 1). It is not possible,without further information about the dimensions a, b, and c to judge whichof the second to sixth term will be the lowest.

9-3 The fundamental modes of the air-spaced parallel plates are TEM modes.(Note that the argument forbidding these fails when the enclosure is notcomplete.) If we ignore the fringing field, (more accurately we should use thefield distribution illustrated in figure 5.12 to account for the fringing field)the TEM electric field may be written

�E(�r, t) = E0 ıei(kz−ωt)

where we have taken the x axis perpendicular to the plates. The correspondingmagnetic field intensity is

�H =�∇× �E

iμ0ω=E0j

μ0cei(kz−ωt)

As pointed out these waves are non-dispersive.The parallel plates also support are also TE and TM modes. The TM

modes satisfy[∇2

t + (μεω2 − k2)]E0z = 0

with solutions

E0,z = A

{cosαxsinαx

}{sinβycosβy

}with α2 + β2 = μεω2 − k2

The boundary condition at the plates, x = 0 and x = a is that Ez mustvanish, leading to

E0,z = A sinnπx

a

{cosβysinβy

}

Chapter Nine Solutions 89

At the open sides, y = 0 and y = b, the most obvious boundary condition isthat the surface current must vanish. Writing the surface current in terms ofH we have

�j = n× �H = ±j× (H0,x ı+H0,y j) = H0,xk

which, in turn, implies that ∂Ez/∂y = 0. The form of the z component ofthe electric field is therefore

ETM0,z = A sin

nπx

acos

mπy

b

TE modes similarly have

H0,z = α cosnπx

a

{sinβycosβy

}

Using Hx ∝ ∂ �Hz/∂x, and the fact that this vanishes at y = 0 and at y = b,we obtain

HTE0,z = A cos

nπx

asin

mπy

b

9-4 We first calculate kz and then find λ as 2π/kz. To this end we evaluate

ω

c=

2π × 12 × 109s−1

3 × 108m/s= 80π m−1

so thatkz = π

√802 − (0.0228)−2 = 66.9π m−1

From this we conclude that λ = 2π/kz = 2.99 cm.

9-5 Using the results of example 9.3, we have

dP/dzP

=2δ

γ2kab

[(a+ b)γ4 + π2k2

(n2

a+m2

b

)]

For the TE1,0 mode, m = 0 and n = 1, meaning γ = π/a and ωc = (cπ/a)which we use to rewrite the expression above as

dP/dzP

=2δkb

((a+ b)π2

a3+ k2

)

where

k2 =ω2

c2− π2

a2

We simplify by assuming that b = 2a, (the usual shape for rectangular wave-guides) and compute

k(2ωc) =√

3πa

and k(1.05ωc) = 0.32π

a


as well asδ(2ωc) = 4.89 × 10−6a1/2 m1/2

and δ(1.05ωc) = 9.314 × 10−6a1/2 m1/2

Substituting these values we get

dP/dzP

= 5.32 × 10−5a−3/2m1/2 ω = 2ωc

dP/dzP

= 2.84 × 10−4a−3/2m1/2 ω = 1.05ωc

For a typical value of a = 1 cm, these attenuation coefficients reduce to0.0532/m and 0.283/m respectively.

9-6 TE modes have Ez = 0. For simplicity we restrict ourselves to TE0,m modesin a rectangle of width 2a so that

H0,z =

{A sinαx+B cosβx −a < x < a

Ce−|γx| |x| > a

We further subdivide the modes into symmetric and antisymmetric modes.Fixing our attention on the antisymmetric modes we have

E0,y =−iωμα cosαx

α2inside the guide

and E0,y = ±C iωμγe−γ|x|

γ2outside the guide

with the + sign pertaining to positive x. (E0,x = 0 in either case.) MatchingE‖ and H‖ at the boundary (x = ±a), we find

A sinαa = Ce−γa A sin(−αa) = Ceγa

−A cosαaα

=e−γa

γ

−A cos(−αa)α

=−eγa

γ

In either case, αa must be a root of

− tanαaαa

=e−|γa|

γa

where α and γ are related by α2 + γ2 = (n2 − 1)ω2/c2. The roots may bedetermined graphically or numerically.

The symmetric modes are found in the same fashion to obey

−cotβaβa

=e−|γa|

γa

9-7 The general solution the the wave equation in spherical polar coordinates isgiven on page 66. For TE modes we have

Br =1c

�(�+ 1)r

j�(kr)Ym� (θ, ϕ)


Br, Eϕ, and Eθ must each vanish at the conducting wall, which leads us toconclude that ka must be a root of j�. The � = 0 solution would have non-zero divergence for B and is therefore eliminated. The first few zeros are r11= 4.4934, r21 = 5.76349, r31 = 6.987932, r12 = 7.72525, .... The radial Bcomponent is therefore

Br =A

rj�(r�iar)Ym

� (θ, ϕ)

For TM modes we find Er =�(�+ 1)

rj�(kr)Ym

� (θ, ϕ) and setting Eθ = Eϕ = 0

d

dr(rj�(kr))

∣∣∣∣r=a

= 0 ⇒ j�(ka) + akj′�(ka) = 0

Again � = 0 does not give valid solution. For � �= 0, solutions must be foundnumerically. In particular, when � = 1, the first two solutions are ka =2.74371, and ka = 6.11676.

9-8 For the TM1,0,1 mode,

E0,z = A sinπx

acos

πz

c

and from (9–50, 51)

{�E0,t

�H0,t

}=

A

εμω2 − k2

⎧⎪⎨⎪⎩

− ıπ2

acsin

πz

ccos

πx

ajiωεπ

acos

πz

ccos

πx

a

⎫⎪⎬⎪⎭

The Q of the cavity may be defined as the ω0× energy stored divided by theenergy lost per second. The resonant frequency is easily found to be

ω0 ≡ ω1,0,1 =1√με

√(π

a

)2

+(π

c

)2

and the energy stored may be computed from

W = 14

∫ (ε|E|2 + μ|H|2) d3r

(as 〈E2〉 = 12 |E2| and 〈H2〉 = 1

2 |H2|) or

W = 14A

2

∫ [ε

(π4

γ4a2c2cos2

πx

asin2 πz

c

)

+ μ

(ω2ε2π2

γ4a2cos2

πz

ccos2

πx

a

)]d3r

The integration is straightforward and gives


W = 14A

2ε

(π4

γ4a2c2+μεω2π2

γ4a2

)abc

4

The rate of energy loss by surface currents may be expressed as

P =μωδ

4

∮|H‖|2dS

=μωδ

4

∫ (H2

y

∣∣∣x=0

dydz +∫H2

y

∣∣∣x=a

dydz +

+∫H2

y

∣∣∣z=0

dxdy

∫H2

y

∣∣∣z=c

dxdy

)

=μωδ

2A2

γ4

ω2ε2π2

a2

(∫cos2

πx

adxdy +

∫cos2

πz

cdydz

)

=μωδ

4A2

γ4

ω2ε2π2

a2

(ab

2+bc

2

)

The Q of the cavity is then

ω0W

P=π2/c2 + μεω2

2μεω2δ

abc

ab+ bc=(

2a2 + c2

a2 + c2

)abc

2δ(a+ c)b

where we have used μεω2 =π2

a2+π2

c2. We see that the Q of the cavity

increases as the volume to surface area.

9-9 A TM mode has Hz = 0 and in a circular waveguide Ez must take the form

E0,z =

⎧⎪⎨⎪⎩

J0(γr) with γ2 = με1ω2 − k2; r ≤ a

AK0(βr) with β2 = k2 − με0ω2; r ≥ a

The corresponding transverse fields are

E0,r =ik

γJ′0(γr)

H0,ϕ =iωε1γ

J′0(γr)

⎫⎪⎬⎪⎭ r ≤ a

E0,r = − ikβAK′

0(βr)

H0,ϕ = − iωε0β

AK′0(βr)

⎫⎪⎬⎪⎭ r ≥ a

At the fiber-air interface, we require that εEr, H ϕ, and Ez be continuous. Inother words,

J0(γa) = AK0(βa)iε1k

γJ′0(γa) = − iε0k

βAK′

0(βa)

iωε1γ

J′0(γa) = − iωε0β

AK′0(βa)

The second and third of these equations are equivalent. Dividing the thirdequation by the first, we obtain the characteristic equation (apart from acommon factor a) (9–101)

ε1J′0(γa)γJ0(γa)

+ε0K′

0(βa)βK0(βa)

= 0


9-10 TEM nodes have Ez= 0 and Bz = 0. The wave equation (9–15) then reducesto

(∇2 + μεω2)( �Et

�Ht

)=(∇2

t +∂2

∂z2+ εμω2

)(�Et

�Ht

)= ∇2

t

(�Et

�Ht

)= 0

Just the equation governing the static fields. In order that �Et and �Ht avoidvanishing, we have set εμω2 = k2. �Et and �Ht have the simple solution

�Et = Ar

rand �Ht =

εω

k

A

rϕ

There are also TE and TM modes of propagation with ψ =∑

[AmJm(γr) +BmNm(γr)]e±imϕ.

9-11 The boundary conditions to these waves require that Eϕ and Eθ vanish atr = a and r = b. TE modes, having only these components to the field musthave at least half a wave between a and b leading to ω ∼ c/(a− b). For TMmodes, on the other hand, it is possible to have Eϕ and Eθ vanish while Er

varies only slightly over the interval (a, b). In this case we expect ω ∼ c/a.In terms of the expressions for the tangential electric fields the boundaryconditions are,

1r

∂(rf�)∂r

∣∣∣∣a

=1a

[f�(ka) + ak f ′�(ka)] = 0

and1r

∂(rf�)∂r

∣∣∣∣b

=1b

[f�(kb) + bk f ′�(kb)] = 0

where f� ≡ Aj� +Bn�.Writing the expressions out in full and rearranging the terms, we obtain

A j�(ka) −Aka j′�(ka) = −B n�(ka) +Bkan′�(ka)

A j�(kb) −Akb j′�(kb) = −B n�(kb) +Bkbn′�(kb)

Dividing one equation by the other eliminate the expansion constants A andB to produce the characteristic equations for TM modes

j�(ka) − ak j′�(ka)j�(kb) − bk j′�(kb)

= −n�(ka) − ak n′�(ka)

n�(kb) − bk n′�(kb)

The lowest (non-trivial) frequency mode will correspond to � = 1; numericalmeans must be employed to find the roots of the characteristic equation.Jackson, on page 376 of Classical Electrodynamics, 3rd ed., gives an alternativesolution by solving the wave equation for Bϕ when this is the only non-vanishing component of B obtaining

ω� �√�(�+ 1)

c

a


9-12 The characteristic impedance of the coaxial line carrying a TEM mode is mosteasily determined from Z = V/I as given by (Ex 9.1.8):

I =2πV

vμ ln(b/a)⇒ Z =

V

I=

12π

√μ

εln(b/a)

Inserting numerical values gives Z = 74 Ω.

9-13 In terms of the non-vanishing longitudinal field Hz, the transverse fields aregiven by

�Et =−iωk × �∇tHz

γ2and �Ht =

ik�∇tHz

γ2

Re-expressing �Ht in terms of �Et, we find

| �S| = | �E × �H| =k

μω|E|2

We conclude that the impedance Z is given by

Z =μω

k=μ2πv/λ0

2π/λ= μ

√1με

λ

λ0=√μ

ε

λ

λ0

9-14 We begin with wave equation valid for either Hz or Ez either of which maybe denoted by ψe−iωt

∇2tψ +

∂2ψ

∂z2+ μ0ε0ω

2ψ = 0

we separate variables by setting ψ(r, ϕ, z) = R(r)Φ(ϕ)Z(z). Substituting thisinto the differential equation we obtain after dividing by ψ

∇2t (RΦ)RΦ

+ μ0ε0ω2 = −

d2Zdz2

Z= λ2

where we have chosen a positive separation constant to ensure that either Zor its derivative will vanish at the ends. Moreover as the sine or cosine needsroots at 0 and L we can conclude that λ = nπ/L with n an integer. Theremaining equation in will be separated again this time in polar coordinates.[

1r

∂

∂r

(r∂

∂r

)+

1r2

∂2

∂ϕ2+ γ2

]RΦ = 0

Where we have abbreviated γ2 = μ0ε0ω2−n2π2/L2. Anticipating the solution

we try Φ = e±imϕ which reduces the equation to Bessel’s equation.

r2d2Rdr2

+ rdRdr

+ (γ2r2 −m2)R = 0


with general solution Rm(r) = AmJm(γr) + BmN(γr) we conclude thereforethat ψ has the form

ψ ={

Jm(γr)Nm(γr)

}{cos(nπz/L)sin(nπz/L)

}e±imϕ

It remains to pick out the solutions that fit the boundary conditions.

TE modes: TE modes must have Hz = 0 at the ends, meaning that if theends lie at z = 0 and z = L. The solution must be of the form

Hz ={

Jm(γr)Nm(γr)

}sin

nπz

Le±imϕ

The magnetic field intensity satisfies dHz/dn = 0 at the side walls meaningthat the (n,m) mode must satisfy

AnmJ′m(γR1) +BnmN′m(γR1) = 0

andAnmJ′m(γR2) +BnmN′

m(γR2) = 0

In order that these equation have a solution, we must have

J′m(γR1)N′m(γR2) = J′m(γR2)N′

m(γR1)

Plotting each product as a function of γ we find roots γm,i at the intersections.

The resonant frequency is then given by ω2n,m,i = (με0)−1

√γ2

m,i + (nπ/L)2.

TM modes:For TM modes, Ez = ψ. The normal derivative of Ez mustvanish at the end walls meaning that Ez simplifies to

Ez ={

Jm(γr)Nm(γr)

}cos

nπz

Le±imϕ

At the side walls, Ez = E‖ = 0, meaning

AnmJm(γR1) +BnmNm(γR1) = 0

andAnmJm(γR2) +BnmNm(γR2) = 0

In order that these equation have a solution, we must have

Jm(γR1)Nm(γR2) = Jm(γR2)Nm(γR1)

The roots of this equation which we denote ρm,i give the resonant frequencies

ω2n,m,i = (με0)−1

√ρ2

m,i + (nπ/L)2.

9-15 According to the discussion on page 262, the cut-off angular frequency ωc =2.405/(a

√εcladμclad − εcoreμcore ). We replace με by 1/(n2c2) to get

ωc =2.405c

a√

1/n2clad − 1/n2

core

=2.405 × 3 × 108

12 × 10−5 × .03009

≈ 4.80 × 1015s−1

The lowest frequency radiation that the fiber can transmit in the TE1,0 modeis fc = ωc/2π = 7.63 × 1014 Hz.

Chapter 10

10-1 The first step is to compute the quadrupole moment neglecting the time vari-ation.

Qxx =∑qi(3xi 2 − ri 2)

= q

[3(b

2

)2

− 3(b

2

)2

+ 3(b

2

)2

− 3

(b

2

)2

± 14 (a2 + b2)

]= 0

Similarly Qyy = Qzz = 0. The off-diagonal elements are

Qxy =∑

3qixiyi = 3qab

and Qxz = Qyz = 0. The terms of the angular distribution (10–77) of powerto be calculated are

xβQαβQαγx

γ = (xQyx + yQyx)2 = (x2 + y2 + 2xy)Q2xy

= 9(qab)2(x2 + y2 + 2xy)

= 9(qab)2r2 sin2 θ(1 + 2 cosϕ sinϕ)

and

(xβQαβxα)2 = (xQxyy + yQyxx)2

= 36(qab)2x2y2 = 36(qab)2r4 sin4 θ sin2 ϕ cos2 ϕ

= 9(qab)2r4 sin4 θ sin2 2ϕ

so that

dPdΩ

=μ0ω

6

128π2c3(qab)2 sin2 θ(1 + sin 2ϕ− sin2 θ sin2 2ϕ)

The total power output is given by (10–81) as

P =μ0ω

6

1440πc3(Q2

xy +Q2yx) =

μ0ω6(qab)2

80πc3

10-2 Starting with the expression (10–59) on page 280, the power emitted by anoscillating electric dipole may be written

〈P〉 =μ0ε0ω

4p20

12πε0c=

14πε0

|p|23c3

The expressions (10–67) and (10–68) may be used to obtain for the magneticdipole, ⟨

dPM

dΩ

⟩=

12μ0

k2μ0

4πkωμ0

4π|m sin θ|2 =

μ0

4πω4m2 sin2 θ

8πc3

— 96—

Chapter Ten Solutions 97

We can integrate this over the complete solid angle (or simply compare to theelectric dipole result) to obtain

〈PM 〉 =μ0

4πω4m2

0

3c3=μ0

4π|m|23c3

It is worth noting that this result closely parallels that for the electric dipole;replacing 1/4πε0 by μ0/4π and p by m converts the electric dipole result tothat for the magnetic dipole.

The electric quadrupole result may be written

〈PQ〉 =ω6

4πε0QαβQ

αβ

360c5=

14πε0

¨Qαβ¨Q

αβ

360c5

A reasonable guess for the equivalent gravitational result is obtained replacing1/4πε0 by G and Qαβ by the mass quadrupole moment. For cylindricallysymmetric quadrupoles, Qxx = Qyy = − 1

2Qzz in the principal axis system.In this case the single number Qxx suffices. The term QαβQ

αβ may be writtenas Q2

xx +Q2yy +Q2

zz = 6Q2xx. The formula above then reduces to

P =G ¨Q2

xx

60c5

Landau gives G ¨Q2/45c5 for the equivalent gravitational result.

10-3 The rotating magnet may be thought of as two linear oscillators oscillatingin quadrature. The magnetic dipole moment of the magnet is just �m = �Mτwhere τ is the volume. The power emitted is

〈P〉 = 2 × μ0

4πω4m2

3c3

10-4 The equation of motion for a nucleus with spin angular momentum �I andmagnetic moment �m = γ�I in a z -directed magnetic induction field �B0 is

dIxdt

= myB0 = γIyB0dIydt

= −mxBz = −γIx, BzdIzdt

= 0

with solution

mx = m0 cos(ωt+ ϕ), my = m0 sin(ωt+ ϕ), mz = m0

with ω = γB0. A rotating dipole can always be viewed as two orthogonallinear oscillating dipoles in quadrature, meaning that the energy is radiatedat a rate

〈P〉 = 2μ0

4πω4(γ�I)2

3c3=μ0γ

6B40( 1

2 h)2

6πc3

For a typical field of 2T and γ = 2.68 × 108 (SI units) this amounts to2 × 10−50 W. In a realistic situation, one would have not one spin, but the


appropriate Boltzman fraction of about 1022 spins. In the same 2T field atroom temperature, W/kT = 7 × 10−4 so that the total magnetic momentwould be that of 7× 1018 nuclei. The power output would be about 10−12 W.The total magnetic energy of the spins is 2 × 10−7 J leading us to estimatea relaxation time of order 105 seconds due to radiation. In fact relaxationdue to the induction field would be considerably larger than this, and localphenomena dominate the relaxation.

10-5 (a) The angular frequency ω of the postulated dipole is 2πc/λ = 3.77 ×1015s−1. Taking the dipole moment as e×(10−10m) we find the power emittedby such a rotating dipole to be

P =μ0ω

4p2

6πc= 5.74×, 10−12W ≈ 3 × 107eV/s

The lifetime of such an atom with total energy ∼ 2 eV would be about2eV/(3× 107eV/s) ∼ 10−7s. (b) A quadrupole of the same dimensions wouldtypically have Qαβ ∼ 3e× (10−20m2) to give emitted power of order

P � μ0ω6

1440πc3QαβQ

αβ ∼ 10−7ω6e2 × 10−40

40c3= 6.8 × 10−17W

approximately a factor of 105 smaller than the power emitted by the dipole.(c) The magnetic dipole radiates according to

P =μ0ω

4m2

12πc3

Taking the magnetic moment as m ∼ ( 12e/m)L and taking L = h, we find

P ∼ 2 × 10−17W

about the same as that emitted by the quadrupole.When the radiation has a wavelength of 50 nm, the frequency will be

ten times as large giving 104 times the power output for the dipoles and 106

times the output for the quadrupole. Since the energy of the emitting statemust initially have been 10 times as large as that required to produce 500nm radiation, the lifetimes would be decreased by 103 for the dipole and106 for the quadrupole. It should be clear that for very short wavelengthx-rays, the various multipoles lead to comparable lifetimes. Of course, oncethe wavelength becomes of order 0.1 nm, the multipole approximation willfail.

10-6 The magnetic moment of a single neutron is −1.913 × 5.501 × 10−27J/T =−9.657 × 10−27J/T and the number of neutrons in the star is

N =2 × 1030kg

1.675 × 10−27kg= 1.194 × 1057


The magnetic moment of the star would then be �m = 1.153 × 1031 J/T. Thepower emitted by the rotating dipole would be

P =μ0ω

4|m|26πc3

= 3.28 × 1029ω4J s3

The energy of rotation of the star is W = 12Iω

2 = 15MR2ω2 = 4 × 1037ω2Js2

which we abbreviate as aω2. Writing P = bω4, we have

dW

dt= −bω4 = −bW

2

a2

or ∫ W0/10

W0

dW

W 2=∫ t9/10

0

− b

a2dt

This is easily integrated to give

9W0

=b

a2t9/10 ⇒ t9/10 =

9a2

bW0

Substituting the numerical values of a, b, and W0, we obtain

t9/10 =9 × (4 × 1037)2

3.3 × 1029 × 3.6 × 1042= 1.2 × 104s

This is clearly much faster than we would expect the rotation of the neutronstar to decay. The difficulty lies in the assumption that all the neutrons pointin the same direction. This alignment would impose a huge degeneracy energyto the star and is clearly unrealistic. If we assume a polarization of 0.1%, theradiation rate decreases by a factor of 106 and the lifetime increases by acorresponding factor of 106, still somewhat short but probably in the rightballpark.

10-7 To find the radiation resistance, we equate the power P radiated to 12I

20R.

For a half wave antenna, kd = π, whence

R =2PI2

=μ0c

4π× 2.438 = 10−7 × 3 × 108 × 2.438 = 73Ω

The radiation resistance of the full-wave antenna is similarly found to be200Ω.

10-8 (a) When kd � 1, the supply current I0 = I sin 12kd ∼ 1

2Ikd. The charge oneach side of the wire may be found from

I0 =dQ

dt= −iωQ ⇒ Q =

Ikd

−2iωe−iωt

which leads to a charge per unit length of Q/(2d). The dipole moment of thetwo segments assumed to lie along the z axis is then

�p = k

∫ 0

d/2

−2Qzdzd

+ k

∫ d/2

0

2Qzdzd

=12kQd =

Ikd2k

−4iωe−iωt


This result could have been more easily obtained using (10–44), where thethe dipole moment is related to the current.

(b) The power per solid angle radiated by such an oscillating dipole is

dPdΩ

=μ0ω

4|p|2 sin2 θ

32π2c=μ0ω

2k2I2d4 sin2 θ

29π2c

(c) We approximate the various terms as follows:

cos( 12kd cos θ) � 1 − 1

2!( 12kd cos θ)2 and cos 1

2kd � 1 − 12!

( 12kd)

2

to write

cos( 12kd cos θ) − cos 1

2kd = 12 ( 1

2kd)2(1 − cos2 θ) = 1

8k2d2 sin2 θ

Equation (10–125) then becomes

dPdΩ

=2μ0

(4π)2I2c

( 18k

2d2 sin2 θ

sin θ

)2

=μ0k

4d4I2c sin2 θ

29π2=μ0k

2ω2I2d4 sin2 θ

29π2c

10-9 The large conducting sheet (the z-y plane) may be maintained at zero poten-tial by an image half wave antenna placed λ/4 behind the conducting sheetthat has, at any instant, the opposite polarity of the real antenna. The currentdistribution is therefore given by

�J(�r, t) = kI sin(12kd− k|z|)[δ(x− 1

4λ) − δ(x+ 14λ)]δ(y)e−iωt

Following (10–121), we approximate eikR/R in the radiation zone by

eik|�r−�r ′|

|�r − �r ′| � eikr

r· e−ik(�r·�r ′)/r =

eikr

r· e−i(kz′ cos θ± 1

2 π sin θ cos ϕ)

With this approximation, the vector potential may be written

�A0(�r ) =μ0

4πeikr

rk(e

12 π sin θ cos ϕ − e−

12 π sin θ cos ϕ

)∫ d/2

−d/2

I sin(12kd− k|z|)e−ikz′ cos θdz′

The angular distribution of power becomes

dPdΩ

=μ0I

2c

2π2sin2( 1

2π sin θ cosϕ)cos2( 1

2π cos θ)sin2 θ


To obtain the total power emitted we must integrate this expression over thehalf sphere above the plane. By numerical integration, we find P = 42.8 I2.We deduce that the radiation resistance is 85.7Ω.

10-10 The angular velocity of the electron about the nucleus may be computed bysetting

mω2r =1

4πε0e2

r2⇒ ω2 =

e2

4πε0mr3

and the dipole moment is p = er. The power emitted by the two orthogonaldipoles constituted by the orbiting electron is then

P =2

4πε0ω4

3c3p2 =

2e6

(4πε0)33c3m2r4

If the energy is changed by only a small fraction during each revolution, theorbit will remain roughly circular and we can equate the rate of energy lossto the power emitted.

P = −dWdt

=e2

8πε0d

dt

(1r

)

so that

− 1r2dr

dt=

e4

12π2ε30c3m2r4

or − r2dr

dt=

e4

12π2ε30c3m2

Integrating both sides over the trajectory, from r = r0 to 0, we have∫ 0

r0

−r2dr =e4

12π2ε20c3m2

∫ tf

0

dt

−r33

∣∣∣∣0

r0

=e4tf

12π2ε20c3m2

or tf =4π2ε20c

3m2r30e4

= 1.5 × 10−11 s

where we have set r0 = 0.52 × 10−10 m.

10-11 The two counter-rotating electrons form an oscillating dipole of magnitude2er. The power radiated by such a dipole is

〈P〉 =ω4(2er)2

4πε03c3

10-12 The power radiated by a nonrelativistic accelerated particle is according to(10–149)

P =q2a2

6πε0c3


Applying this to the classical hydrogen atom we have

a =e2

4πε0mer2

which we substitute into (10-149) to obtain

P =e6

96π2ε30c3m2

er4

10-13 Rather than dealing with a varying acceleration for an electron in an ellipticalorbit, we consider the motion as the superposition of orthogonal two dipoles ofamplitude A and B respectively, oscillating in quadrature. The power emittedtwo such by harmonically oscillating dipoles is

P =ω4q2A2

12πε0c3+ω4q2B2

12πε0c3=ω4q2(A2 +B2)

12πε0c3

10-14 At maximum,

d

dθ

(dPdΩ

)= 0 or

d

dθ

[sin2 θ

(1 − β cos θ)5

]= 0

Expanding the derivative, we obtain

2 sin θ cos θ(1 − β cos θ)5

− 5β sin2 θ sin θ(1 − β cos θ)6

= 0

which may be reduced to

3β cos2 θ + 2 cos θ − 5β = 0

⇒ cos θ =

√1 + 15β2 − 1

3β

When β ≈ 1, we set β = 1− ε, leading to cos θ � 1− 14ε. But cos θ � 1− 1

2θ2.

We conclude that θ2 � 12ε. Now, using 1+β � 2, we have 1−β2 = 1/γ2 � 2ε

so that we find θ2 � 1/(4γ2) ⇒ θ � 1/(2γ). At the angle of maximum power,θmax = 1/(2γ) the power per solid angle is readily calculated.

dPdΩ

∣∣∣∣θmax

� q2β2

(4π)2ε0csin2(1/2γ)

[1 − β cos(1/2γ)]5∝ (2γ)−2

[1 − β

(1 − 1

8γ2

)]−5

For β ≈ 1, the square bracketed term becomes

[1 −

(1 − 1

2γ2

)(1 − 1

8γ2

)]−5

=(

58γ2

− 116γ4

)−5


We retain only the leading term to replace the square bracketed term by

[· · · · · ·]−5 =

(5

8γ2

)−5

Gathering terms, we find

dPdΩ

=q2β2

(4π)2ε0c

(85

)5 14γ8

10-15 The angular distribution of power in (10–162) goes to zero when the term insquare brackets vanishes. Taking for simplicity the angle ϕ = 0, the angle θmay be found from

sin2 θ = γ2(1 − β cos θ)2 =(1 − β cos θ)2

1 − β2

we rationalize this and write

(1 − β2)(1 − cos2 θ) = (1 − β cos θ)2

1 − cos2 θ − β2 + β2 cos2 θ = 1 − 2β cos θ + β2 cos θ2

which reduces to

cos2 θ − 2β cos θ + β2 = 0 ⇒ cos θ = β

Substituting β for cos θ in first equation, we have

sin2 θ = 1 − β2 =1γ2

⇒ θ =1γ

A direction parallel to the velocity has θ = 0 and ϕ = 0, so that theangular power density becomes

dPdΩ

∣∣∣∣θ=0

=q2β2

(4π)2ε0c(1 − β)3

Substituting 2γ2 for (1 − β)−1 this becomes

dPdΩ

∣∣∣∣θ=0

=q2β2γ6

2π2ε0c

in agreement with (10–164) except for the numerical factor which would resultfrom the integration over the beam width.

Chapter 11

11-1 The phase of a wave may be written as

�k · �r − ωt = (ω/c,�k ) · (−ct, �r ) = KμXμ

The quotient theorem now guarantees that because the phase is a zero ordertensor and Xμ is a first rank tensor, Kμ must also be a first rank tensor.When �k is parallel to the frame velocity �V, we take K1 = k and when it isperpendicular to �V we take K1 = 0. The transformation law for the frequencyis then obtained from that of the 0-component of Kμ:

ω′

c=(K0)′

= ΓK0 − βΓK1

For �k perpendicular to �V this leads to ω′ = Γω. When �k is parallel to �V wefind

ω′

c= Γ

(ω

c− V

ck

)=

Γωc

(1 − V

c

)=ω

c

√1 − V/c

1 + V/c

11-2 Differentiating the scalar VμVμ = c2 with respect to the proper time τ , we

find

0 =d(VμV

μ)dτ

= AμVμ + VμA

μ = 2AμVμ

11-3 We evaluate the components of f j one-at-a time

f1 = F 10J0 + F 11J1 + F 12J2 + F 13J3

=Ex

cρc+ 0 −BzJy +ByJz

= ρEx + ( �J × �B)x

In the same fashion

f2 = F 20J0 + F 21J1 + F 22J2 + F 23J3

=Ey

cρc+BzJx + 0 −BxJz

= ρEy + ( �J × �B)y

f3 = F 30J0 + F 31J1 + F 32J2 + F 33J3

=Ez

cρc−ByJx +BxJy + 0

= ρEz + ( �J × �B)z

11-4 Expression (11-12), ∂μFμν = μ0J

ν leads to

∂μFμ,0 = ∂0F

00 + ∂1F10 + ∂2F

20 + ∂3F30 = �∇ ·

�E

c= μ0J

0 = μ0ρc =ρ

ε0c

— 104—

Chapter Eleven Solutions 105

μ0J1 = ∂0F

01 + +∂2F21 + ∂3F

31

= − 1c2∂Ex

∂t− ∂Bz

∂y+∂By

∂z

= (�∇× �B)x − 1c2∂Ex

∂t

μ0J2 = ∂0F

0,2 + +∂1F1,2 + ∂3F

3,2

= − 1c2∂Ey

∂t+∂Bz

∂x− ∂Bx

∂z

= (�∇× �B)y − 1c2∂Ey

∂t

The fourth component is found in the same fashion. We similarly interpret(11-14) to obtain

∂0F12 + ∂1F20 + ∂2F01 = −∂Bz

c∂t− ∂Ey

c∂x+∂Ex

c∂y= −1

c

(∂ �B

∂t+ �∇× �E

)z

= 0

∂0F23 + ∂2F30 + ∂3F02 = −∂Bx

c∂t− ∂Ez

c∂y+∂Ey

c∂z= −1

c

(∂ �B

∂t+ �∇× �E

)x

= 0

∂0F13 + ∂1F30 + ∂3F01 =∂By

c∂t− ∂Ez

c∂x+∂Ex

c∂z=

1c

(∂ �B

∂t+ �∇× �E

)y

= 0

∂1F23 + ∂2F31 + ∂3F12 = −∂Bx

∂x− ∂By

∂y− ∂Bz

∂z= −�∇ · �B = 0

Explicitly evaluating the expression implicit in (11–23) and noting thatHμν is just Hμν with rows and columns interchanged, we have

J0 = ∂0H00 + ∂1H

10 + ∂2H20 + ∂3H

30 = ∂xcDx + ∂ycDy + ∂zcDz

ρc = c�∇ · �D

J1 = ∂0H01 + ∂1H

11 + ∂2H21 + ∂3H

31 = −∂tDx + 0 − ∂yHz + ∂zHy

Jx = −∂tDx + (�∇× �H)x

J2 = ∂0H02 + ∂1H

12 + ∂2H22 + ∂3H

32 = −∂tDy + ∂xHz + 0 + ∂zHx

Jy = −∂tDy + (�∇× �H)y

J3 = ∂0H03 + ∂1H

13 + ∂2H23 + ∂3H

33 = −∂tDz − ∂xHy + ∂yHx + 0

Jz = −∂tDx + (�∇× �H)z

In summary, the four equations resulting from (11-23) yield �∇ · �D = ρ and�∇× �H = �J + ∂ �D/∂t.


11-5 Using the definition of the dual tensor Gμν we calculate

G01 = 12

(ε0123F23 + ε0132F32

)= 1

2 (F23 − F32) = −Bx = −G10

G02 = 12

(ε0213F13 + ε0231F31

)= 1

2 (−F13 + F31) = −By = −G20

G03 = 12

(ε0312F12 + ε0321F21

)= 1

2 (F12 − F21) = −Bz = −G30

G12 = 12

(ε1230F30 + ε1203F03

)= 1

2 (−F30 + F03) =Ez

c= −G21

G13 = 12

(ε1320F20 + ε1302F02

)= 1

2 (F20 − F02) = −Ey

c= −G31

G23 = 12

(ε2310F10 + ε2301F01

)= 1

2 (−F10 + F01) =Ex

c= −G32

The entire array may be written in matrix form as

Gμν =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

0 −Bx −By −Bz

Bx 0Ez

c−Ey

c

By −Ez

c0

Ex

c

BzEy

c−Ex

c0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

I will be noted that Gμν may be generated from Fμν by replacing the termsEi/c → Bi and Bi → −Ei/c. The requisite divergence is now easily calcu-lated:

∂μGμ0 = �∇ · �B

∂μGμ1 =

1c

[−∂Bx

dt+(− ∂Ez

∂y+∂Ey

∂z

)]= −1

c

[∂ �B

∂t+ (�∇× �E)

]x

∂μGμ2 =

1c

[−∂By

dt+(∂Ez

∂x− ∂Ex

∂z

)]= −1

c

[∂ �B

∂t+ (�∇× �E)

]y

∂μGμ3 =

1c

[−∂Bz

dt+(− ∂Ey

∂x+∂Ex

∂y

)]= −1

c

[∂ �B

∂t+ (�∇× �E)

]z

Equating each of these terms to zero gives the required result.

11-6 We wish to demonstrate that B′‖ = B‖ and �B′

⊥ = Γ(�B⊥−�v× �E⊥/c2

). Taking,

as usual, the velocity of the frame along the x axis

B′x =

(F 32

)′ = α3μα

2νF

μν = δ3μδ2νF

μν = F 32 = Bx

B′y =

(F 13

)′ = α1μα

3νF

μν = α1μF

μ3 = α10F

03 + α11F

13

= −βΓF 03 + ΓF 13 = Γ(By + β

Ez

c

)


B′z =

(F 21

)′ = α2μα

1νF

μν = α1μF

2ν = α10F

20 + α11F

21

= −βΓF 20 + ΓF 21 = Γ(Bz − β

Ey

c

)Combining the last two expression as

�B′⊥ = B′

y j+B′zk = Γ

(�B⊥ +

vx

c

Ez

cj− vx

c

Ey

ck

)= Γ

(�B⊥ − �v

c2× �E⊥

)

we obtain the requisite relation.

11-7 To obtain the new potential we merely apply the Lorentz transformation tothe four vector Φμ = (V/c, �A) to obtain

Φ0′ = ΓΦ0 − ΓβxΦ1

or V ′ = Γ(V −vx ·Ax). Similarly, A′x = Γ

(Ax−β V

c

), A′

y = Ay, and A′z = Az.

11-8 Consider a charged particle moving with velocity v along the x axis passing theobserver at distance b. For simplicity we take the rest frame of the observerand that of the particle to coincide at t = t′ = 0. In the frame of the particleat time t′, the observer is located at x′ = −vt′ and y′ = b. The non vanishingcomponents of the field at the observer in the particle’s frame are therefore

E′x =

−qvt′4πε0r′3

E′y =

qb

4πε0r′3

Expressing r′ and t′ in terms of r and t we have

t′ = γ′t and r′ =√b2 + γ2v2t2

so that

E′x =

−qγvt4πε0(b2 + γ2v2t2)3/2

and E′y =

qb

4πε0(b2 + γ2v2t2)3/2

We transform these fields to those measured in Σ using (11–19) to obtain

Ex = E′x =

−qγvt4πε0(b2 + γ2v2t2)3/2

and Ey = γE′y =

qγb

4πε0(b2 + γ2v2t2)3/2

We can relate this result to the geometry by noting that r2 = b2 + (vt)2,whence

b2 + γ2v2t2 = γ2r2 + (1 − γ2)b2

= γ2r2(

1 +1 − γ2

γ2

b2

r2

)

= γ2r2(

1 − β2 b2

r2

)

= γ2r2(1 − β2 sin2 θ)


where θ is the angle between �r and �v. Substituting this result, we find

�E =q�r

4πε0γ2r3(1 − β2 sin2 θ)3/2

where �r is the location of the field point with respect to the current (notretarded) position of the particle. As was shown in figure 10.9, �r = (n′− �β′)r.

11-9 Let the rectangle have length a along the x direction taken to be the directionof motion, and width b in the y direction. The loop will have a magneticmoment �m = Iab in the ±z direction. Taking the current to run in the +xdirection along the top of the loop (so that m points in the −z direction) weuse the Lorentz transformation to find Jμ in the rest frame:

ρc = 0 + βΓJ ′x

Integrating this over the cross section of the wire we obtain a line chargedensity λ

λ = ΓV

c2I ′

The total charge on the top of the loop is found by summing this over thecontracted length a/Γ to give

Q =aV I

c2

Similarly along the bottom, Q = −aIV/c2, while the sides perpendicular to�v remain charge free. The dipole moment of the moving loop becomes

�p = Qb =abIV

c2= − �m× �V

c2

The direction is easily verified with the aid of a diagram.

11-10 In the moving dielectric with �J = 0 and ρ = 0, the absence of magneticmonopoles implies �∇ · �B = 0 and

�∇× �E = −∂�B

∂t

remains unchanged as it involves only the fields, independent of any media.The remaining two Maxwell equations do involve the media and therefore willhave to take account of the movement. Letting �P ′ and �M ′ be the polarizationand magnetization of the medium as measured by a stationary observer, wehave

�∇ · (ε0 �E + �P ′) = 0

and�∇×

( �B

μ0− �M ′

)=∂ �D′

∂t


where, for small V/c,

�P ′ = �P +�V × �M

c2and �M ′ = �M − �V × �P

leading to �D′ = ε0 �E + �P ′ = ε0 �E + �P +�V × �M

c2so that �∇ · �D = −

�V × �M

c2and

�∇×( �B

μ0− �M

)=∂ �P ′

∂t− �∇×

( �V × �P

c2

)or

�∇× �H =∂ �D

∂t+∂

∂t

�V × �M

c2− �∇× (�V × �P )

The movement imparts to the medium a magnetization �P × �V and a polar-ization (�V × �M)/c2.

11-11 The moving magnetization appears in the lab frame to have polarization

�P =�v × �M

c2=ωρϕ× �M

c2=ωM(xı+ yj)

c2

The bound charge density is then −�∇· �P = −2ωM/c2 distributed throughoutthe sphere and bound surface charge �P · n = aωM sin2 θ/c2 on the surface. Itis readily verified that the “net charge” resulting from these bound charges iszero. We find the potential of these distributions as

V (�r ) =1

4πε0

∫τ

−�∇′ · �P (�r ′)|�r − �r ′| r′2dΩ′dr′ +

14πε0

∫S

�P · n|�r − �r ′|a

2dΩ′

=−2ωM4πε0c2

∫τ

r′2dΩ′dr′

|�r − �r ′| +a3ωM

4πε0c2

∫S

sin2 θdΩ′

|�r − �r ′|To perform the integration, we expand

1|�r − �r ′| =

∑�,m

4π2�+ 1

r�<

r�+1>

Ym� (θ, ϕ)Ym∗

� (θ′, ϕ′)

and write 3 sin2 θ = (3−3 cos2 θ) = 2+(1−3 cos2 θ) = 2√

4πY00 −√

16π/5Y02.

Thus in the first integral only the Y00 term and in the second, only the Y0

0

and Y02 terms of the sum survives the integration over Ω′.

When r > a, r< = r′ and r> = r so that

V (r > a) =−2ωM4πε0c2

4πr

∫ a

0

r′2dr′ +a3ωM

4πε0c223

4πr

∫4π

|Y00(θ

′, ϕ′)|2dΩ′

− ωMa5

15ε2c2r3

√16π5

Y02(θ, ϕ)

∫4π

|Y02(θ

′, ϕ′)|2dΩ′


The first two terms cancel precisely leaving only

V (r > a) =ωMa5(1 − 3 cos2 θ)

15ε0c2r3

Evidently, the external potential is that of an electric quadrupole. Theinterior potential is calculated in much the same fashion. For the surfaceintegral r′ = r> and only a single evaluation is required. For the volumeintegral, however, for values of r′ up to r, r′ = r< while for the remainingvolume it is r>.

V (r < a) =−2ωMε0c2

∫ r

0

r′2

rdr′ − 2ωM

ε0c2

∫ a

r

r′dr′

+23a

3ωM

4πε0ac2− r2

a3

ωMa3

15ε0c2

√16π5

Y02(θ, ϕ)

∫4π

|Y02(θ

′, ϕ′)|2dΩ′

= − ωM

15ε0c2[5(a2 − r2) + r2(3 cos2 θ − 1)

]The electric field may be now be found by differentiating the potentials.

It is instructive to solve this problem using an alternative approach. If wewrite V as the integral of �∇ · �P/R over a region of space that includes theboundary, then there is no need to include an integral over the boundaryseparately. However, as we noted, �∇· �P has a singularity at the discontinuity.We can sidestep this discontinuity in the following fashion.

V (�r ) = − 14πε0

∫τ

�∇′ · �P (�r ′)|�r − �r ′| d3r′

= − 14πε0

∫τ

[�∇′ ·

( �P (�r ′)|�r − �r ′|

)+

14πε0

∫τ

�P (�r ′) · ∇′(

1|�r − �r ′|

)]d3r′

The first of the two integrals may be converted to a vanishing surface integralover a surface outside the sphere (where �P vanishes). In the second integral,we replace �∇′(1/|�r − �r ′|) with −�∇(1/|�r − �r ′|) to obtain

V (�r ) = − 14πε0

∫τ

�P (�r ′) · ∇(

1|�r − �r ′|

)]d3r′ = − 1

4πε0�∇ ·∫

τ

�P (�r ′)d3r′

|�r − �r ′|

= − Mω

4πε0c2�∇ ·∫

τ

xı+ yj

|�r − �r ′|r′2dr′dΩ′

To perform the integration, we expand

1|�r − �r ′| =

∑�,m

4π2�+ 1

r�<

r�+1>

Ym� (θ, ϕ)Ym∗

� (θ′, ϕ′)


and write

x′ = −√

2π3r′[Y1

1(θ′, ϕ′) − Y−1

1 (θ′, ϕ′)]

y′ = i

√2π3r′[Y1

1(θ′, ϕ′) + Y−1

1 (θ′, ϕ′)]

It is evident that only the � = 1,m = ±1 terms will survive the integrationover Ω′. Performing the integration over Ω′, we are left with

V (�r ) = − ωM

3ε0c2�∇ ·(xı+ yj

r

∫ a

0

r<r2>r′2dr′

)

When r > a, r = r> and r′ = r<, so that

V (r > a) = − ωM

3ε0c2�∇ ·[(xı+ yj)

a5

5r5

]

= −ωMa5

15ε0c2

[2r3

− 3(x2 + y2)r5

]

= −ωMa5(1 − 3 cos2 θ)15ε0c2r3

The internal potential is obtained as

V (r < a) = − ωM

3ε0c2�∇ · xı+ yj

r

(∫ r

0

r′4dr′

r2+∫ a

r

rr′3dr′

r′2

)

= − ωM

3ε0c2�∇ · (xı+ yj)

(r5

5r3+a2

2− r2

2

)

= − ωM

15ε0c2[(a2 − r2) − r2(3 cos2 −1)

]

11-12 As in section 11.4 we start with the four force density Kμ = Fμν = FμνJν

generalization of the Lorentz force and wish to express Kμ as Kμ = −∂ληλμ

where ημν depends only on the fields. Replacing Jν by the field derivatives,we get

Kμ = FμνJν = Fνμ∂λH

λν = ∂λ(FμνHλν) −Hλν∂λFμν

The first term is in the required form, the second can be expanded to give

Hλν∂λFμν = 12H

λν∂λFμν + 12H

νλ∂νFμλ

= 12H

λν∂λFμν + 12H

λν∂νFλμ

= 12H

λν(∂λFμν + ∂νFλμ)

The source-free field equation ∂λFμν +∂μFνλ +∂νFλμ = 0 allows us to replacethe two terms in parentheses by −∂μFνλ, leading to


Hλν∂λFμν = − 12H

λν∂μFνλ = 12H

λν∂μFλν

If the medium is linear, we can replace the latter term by 14∂μ(HλνFλν) and

writeKμ = ∂λ(FμνH

λν) − 14∂μ(HλνFλν)

= ∂λ

(FμνH

λν − 14δ

λμFλνH

λν)

We can restate the equation in contravariant form by multiplying by gμα toobtain

Kα = gαμKμ = ∂λ

(gαμFμνH

λν − 14g

αμδλμFλνH

λν)

= ∂λ

(gαμFμνH

λν − 14g

αλFλνHλν)

The resulting Stress-Energy-Momentum tensor is

ηλα = −(gαμFμνHλν − 1

4gαλFλνH

λν)

11-13 We start with the form of the Lorentz transformation when the motion isalong one axis, labelling the component of �r along this axis r‖:

(X0)′ = Γ(X0 − �β · �r )

r′‖ = Γ(r‖ − βX0)

�r ′⊥ = �r⊥

In general, the parallel component may be found as r‖ =(�r · �β)β

so that �r ′

may be written

�r ′ = �r ′‖ + �r ′

⊥ = �r⊥ +Γ(�r · �β)�β

β2− Γ�βX0

= �r⊥ + �r‖ +(Γ − 1)(�r · �β)�β

β2− Γ�βX0

This transformation may be written in matrix form as

L(�β) =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

Γ −Γβx −Γβy −Γβz

−Γβx 1 +(Γ − 1)β2

x

β2

(Γ − 1)βxβy

β2

(Γ − 1)βxβz

β2

−Γβy(Γ − 1)βxβy

β21 +

(Γ − 1)β2y

β2

(Γ − 1)βyβz

β2

−Γβz(Γ − 1)βxβz

β2

(Γ − 1)βyβz

β21 +

(Γ − 1)β2z

β2

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

11-14 If the invariant εμνρσFμνF ρσ = − �E · �B/c is 0 in any frame then it must be

zero in all frames, proving the required result.


11-15 Take the direction of the moving frame to be perpendicular to both �E and�B. Then, according to (11–19) when |E| < c |B| it is possible to pick �v sothat �v × �B = − �E , making �E vanish. When |E| > c|B| we can pick �v sothat �B c2 = �v × �E , making �B vanish. When |E| = c|B| the required framevelocity would be c.

11-16 In the moving medium, the polarization is produced by the effective field,�E + �V × �B. Thus,

�P = ε0( �E + �V × �B)

Neglecting the term in �M , and limiting ourselves to terms linear in �V , wehave for the curl of �B,

�∇× �B = μ0

{ε0∂ �E

∂t+ ε0(κ− 1)

∂ �E

∂t+ ε0

(κ− 1)κ

[− �V × ∂ �B

∂t+ �∇× ( �E × �V )

]}

=κ

c2

{∂ �E

∂t+κ− 1κ

[− �V × (�∇× �E) + �∇× ( �E × �V )

]}

=κ

c2

{∂ �E

∂t+κ− 1κ

[�∇(�V · �E) − (�V · �∇) �E − (�V · �∇) �E

]}

We have used the vector identities for �∇× ( �E× �V ) and �∇(�V · �E) and assumed�V constant. Taking the curl once more we get

∇2 �B =κ

c2

[∂2 �B

∂t2+ 2(κ− 1κ

)(�V · �∇)

∂ �B

∂t

]

For a plane wave, travelling with wave vector �k we have

(�V · �∇) �B = (�V · �k) �B = −�V · �kω

∂ �B

∂t=�V · nv0

∂ �B

∂t

so that the wave equation becomes

∇2 �B =κ

c2∂2 �B

∂t2

[1 − 2

(κ− 1κ

) �V · nv0

]

from which we deduce the speed v of the wave through the moving medium:

v = v0

[1 − 2

κ− 1κ

�V · nv0

]−1/2

� v0 +(κ− 1κ

)�V · n

11-17 Using (11–83), we find that the total emitted power is

P =q2c

6πε0(m0c2)2

∣∣∣∣dpdt∣∣∣∣2

=q2c

6πε0(m0c2)2

∣∣∣∣dWdS∣∣∣∣2

=6 × 109 × (1.6 × 10−19)2 × 3 × 108

0.511MeV)2× (1MeV/m)2

= 1.76 × 10−19W ∼ 1.1eV/s


Clearly this is not a major energy loss mechanism.

11-18 Using (11–84), the power may be written

P =q2γ2

6πε0m20c

3

(d�p

dt

)2

=q2γ4a2

6πε0c3=

q2γ4

6πε0c3

(v2

r

)2

When v � c, this becomes

P =q2c

6πε0γ4

r2= 6 × 109 × (1.6 × 10−19)2 × 3 × 108 γ

4

r2

= 4.6 × 10−20 γ4

r2Wm2

A 10-GeV electron has γ = 104/0.511 = 19569 ⇒ γ4 � 1.5 × 1017, so thatfor R = 20 m, P = 1.69× 10−5W � 108 MeV/s. A revolution takes 2πr/c =4.19×10−7 seconds; hence the energy loss per revolution is about 44 MeV. Asit is difficult to increase the electron’s energy by this much in one revolutionthe radiation loss represents a major loss. Much larger radii are needed toget the losses down to a manageable amount.

Chapter 12

12-1 Making the substitution suggested, we have y = (t′ − t)/τ , t′ = t+ yτ and dt′

= τdy. The integral (12–41) then becomes

�v =∫ ∞

0

F (t+ τy)e−ydy

m0

12-2 Using the expression above, the acceleration of the particle may be written

�v =∫ ∞

0

−ω20x(t+ τy)e−ydy

and trying a solution of the form x = x0e−αt we find on substitution

α2x0e−αt =

∫ ∞

0

−ω20x0e

−α(t+τy)e−ydy

orα2 =

∫ ∞

0

−ω20e

−(ατ+1)ydy

In order that the integral converge, it is required that Re(ατ +1) > 0. Underthis condition the integral is easily evaluated to give

α2 =ω2

0e−(ατ+1)y

−(ατ + 1)

∣∣∣∣∞

0

= − ω20

ατ + 1

Upon rationalization and multiplication by τ2 we obtain the characteristicequation

α3τ3 + α2τ2 + ω20τ

2 = 0

which may in principle be solved for α. When ω0τ is very small, an ap-proximate solution may be found. It is probably easiest to use an iterativeapproach. We abbreviate ατ = z and ω0τ = r and note that when r issmall z must also be small. Therefore a zeroth order solution may be foundfrom z2 + r2 = 0, or z0 = ±ir. We now use Newton-Raphson iteration[zi+1 = zi − f(zi)/f ′(zi)] to obtain successively better estimates of the root.Thus, taking z0 = ir

z1 = ir − (ir)3 − r2 + r2

2ir= ir + 1

2r2

where we have neglected terms in r3 in the denominator. We iterate oncemore, carrying terms to order r5 to obtain

z2 = ir + 12r

2 − (ir + 12r

2)3 + (ir + 12r

2)2 + r2

2ir= 1

2r2 + i(r − 5

8r3)

The solution correct through third order in ωτ is then

— 115—


α = 12ω

20τ ± i(ω0 − 5

8ω30τ

2)

The decay constant is 12ω

20τ and the damping decreases the resonant angular

frequency by 58ω

30τ

2.

12-3 The differential equation of motion for the charged particle is

d2x

dt2+ ω2

0x− τd3x

dt3= 0

Again trying a solution of the form x = x0e−αt, we find

α2x+ ω20x+ α3τx = 0

which gives the characteristic equation

α3τ3 + α2τ2 + ω20τ

2 = 0

identical to the characteristic equation found in 12-2.

12-4 (a) Neglecting radiation damping, the relativistic equation of motion is

m0cdβμ

dτ= Fμ

ext

We note that in terms of the three-vectors, βμ = γ(1, �v/c) and �Fμ = γ(�f ·�β, �f).Taking �f to be in the x direction, we rewrite

m0cdβ1

dτ= m0γ

d(γvx)dt

= F 1 ⇒ d(γvx)dt

=fx

m0

This equation is easily integrated to give

γvx =fxt

m0

Equating the expression for γv to the numerical value fxt/m0 = c, we solvefor v to obtain v = .707c.

(b) In terms of the four-momentum, the covariant equation of motion (12–46)may be written

dPμ

dτ− q2

6πε0m0c3

(d2Pμ

dτ2+

Pμ

m20c

2

dP ν

dτ

dPν

dτ

)= Fμ

Writing

dP ν

dτ=

d

dτ

(W

c, �p

)=

d

dτ

(√p2 +m2

0c2 , �p

)

=(

pdp/dt√p2 +m2

0c2,d�p

dt

)

Chapter Twelve Solutions 117

we find

dP ν

dτ

dPν

dτ=(

p2

p2 +m20c

2− 1)(

dp

dτ

)2

=−m2

0c2

p2 +m20c

2

(dp

dτ

)2

Further,

Fμ = γ

( �f · �vc, �f

)=

√1 +

p2

m20c

2

( �f · �vc, �f

)

Taking all motion and forces to be directed along x we extract the μ = 1component of the resulting equation to obtain

dpx

dτ− a

[d2px

dτ2− px

p2 +m20c

2

(dpx

dτ

)2]

=

√1 +

p2

m20c

2fx

where we have relabelled q2/(6πε0m0c3) as a to avoid confusion with the

proper time τ . The equation may be considerably simplified with the sub-stitution px = m0c sinh s. With this substitution, the equation of motionbecomes

cosh sds

dτ− a

[cosh s

d2s

dτ2− sinh s

(ds

dτ

)2

+ sinh s(ds

dτ

)2]

= cosh sfx

m0c

ords

dτ= a

d2s

dτ2=

fx

m0c

The solution to this equation is easily seen to be

ds

dτ= Aeτ/a +

fx

m0c

We might try to set A = 0 in order to avoid the runaway solution and integrateonce more to obtain

s =fxτ

m0c

whence,

p = m0c sinh s = m0c sinh(fxτ

m0c

)

It remains to relate t to τ . We note that

t =∫ τ

0

γ(τ ′)dτ ′ =∫ τ

0

dτ ′√

1 +p2

m20c

2

where

p = sinhfxτ

m0c, and

√1 +

p2

m20c

2= cosh

fxτ

m0c


We abbreviate b = fx/m0c and integrate,

t =∫ τ

0

cosh bτ ′dτ ′ =sinh bτb

=m0c sinh s

fx=

p

fx

which leads to p = fxt or γv = c ⇒ v = .707c, precisely the same resultas obtained in (a). Clearly setting A = 0 eliminates all effects of radiationdamping.

(c) We may convert the differential equation in s and τ to an integral equationusing exactly the same stratagem as in section 12.4 to obtain

ds

dτ=

1a

∫ ∞

τ

fx(τ ′)m0c

e−(τ ′−τ)/adτ ′

If the force continues constantly for all time, then we find as before

ds

dτ=

fx

m0c

and radiation damping appears to have no effect. Only when the force termi-nates at some time τ0 do we get a different effect. Now

ds

dτ=

1a

∫ τ0

τ

fxe−(τ ′−τ)/a

m0cdτ ′

=fxe

τ/a

am0c

∫ τ0

τ

e−τ ′/adτ ′ =fx

m0c

(1 − e−τ0/a

)As we have no information about the continuation of the force, we abandonthe calculation at this point. It should be clear that it would be considerablyeasier to estimate the energy loss due to radiation assuming the accelerationis f/γm0 and calculate the final energy from

dW

dt= �f · �v − af2

γ2m0

Appendix B

B-1 The result is most easily obtained by using raising a lowering operators.

Bj∂iAj = gjkB

k∂igj�A� = gjkg

j�Bk∂iA� = δ�kB

k∂iA� = B�∂iA�

B-2 To demonstrate the equivalence of the second and third line of (Ex B.8.18) itsuffices to show that

sinh2 α− 2 coshα(coshα− cosβ) + sin2 β

(coshα− cosβ)2= −1

To this end we expand the numerator

sinh2 α− 2 cosh2 α+ 2 coshα cosβ + sin2 β

= −1 − cosh2 α+ 2 coshα cosβ + sin2 β

= − cos2 β + 2 coshα cosβ − cosh2 α

by inspection, this is just minus the denominator.

B-3 A particle’s four momentum is

Pμ = m0Vμ = γm0(c,�v)

where m0 is the particle’s proper or rest mass. The components of the 3-momentum have the same transformation law as those of the three-velocity.

p‖ =p‖ −m0βc

1 − βvx/cand p⊥ =

p⊥Γ(1 − βvx/c)

B-4 The standard expansion of a determinant in terms of its cofactors is preciselywhat this expression describes.

B-5 The square interval (dS)2 is generally given by

(dS)2 = dqidqi = gijdq

jdqi

For spherical polar coordinates this becomes

grr(dr)2 + gθθ(dθ)2 + gϕϕ(dϕ)2 = (dr)2 + r2(dθ)2 + r2 sin2 θ(dϕ)2

B-6 Again this should be almost self-evident from the definition of gik = �ei · �ek

We transform to the primed system

(gik)′ = �ei′ · �ek

′ = αji′�ej · α�

k′�e� = αji′α

�k′gj�

Comparing this with (B–30) we find this to be exactly the transformation lawfor a second rank fully covariant tensor.

— 119—


B-7 The simplest method is probably to use the definition Aj = �A · �ej althoughthe alternative Aj = gijA

i could also be used.

A1 = (5ı+ 6j+ 7k) · (2ı+ 3j+ k) = 10 + 18 + 7 = 35

A2 = (5ı+ 6j+ 7k) · (ı− j+ k) = 5 − 6 + 7 = 6

A3 = (5ı+ 6j+ 7k) · (j+ k) = 6 + 7 = 13

We quickly verify that AiAi is independent of which coordinate system is

chosen. Indeed, Calculate in the Cartesian frame, |A|2 = 25 + 36 + 49 = 110and in the oblique frame AiA

i = 35 × 32 + 6 × 2 + 13 × 7

2 = 110.As an illustration of using the alternate method we compute A1 = g1jA

j .We first must find g1j .

g11 = �e1 · �e1 = 14 g12 = �e1 · �e2 = 0 g13 = �e1 · �e3 = 4

leading to

A1 = g11A1 + g12A

2 + g13A3 = 14 × 1.5 + 0 × 2 + 4 × 3.5 = 35

The other components are computed in the same manner.

B-8 We continue to find the remaining components of the metric tensor in additionto the g1j already obtained above.

g21 = 0 g22 = 3 g23 = 0g31 = 4 g32 = 0 g33 = 2

Applying these elements to find |A|2 we have

|A|2 = g11A1A1 + g13A

1A3 + +g22A2A2 + g31A1A3 + g33A

3A3

= 14 × (1.5)2 + 4 × (1.5)(3.5) + 3 × (2)(2) + 4 × (3.5)(1.5) + 2 × (3.5)2

= 110

B-9 The angular momentum of the rigid system is

�L =∑

j

m(j)�r(j) × (�ω × �r(j))

Dropping the summation over particles (we will restore it at the end of theexercise) to avoid the rather cumbersome notation, we write

Li = mεijkxjεk�mω�xm = mεkijεk�mxjω

�xm

= m(δi�δ

jm − δj

�δim)xjω

�xm

= m(ωixmxm − xjω

jxi)

= ωjm(δijxmx

m − xjxi) = ωjIi

j

Appendix B Solutions 121

We can generally exchange the upper and lower indices of an inner product(AjBj = AjB

j) so that we can write the latter expression as

Li = ωjIij with Iij = m(δijxmx

m − xixj)

Restoring the sum over particles we have

Iik =∑

j

m(j)

(δikr2(j) − xi

(j)xk(j)

)

In this sum, r2 is a scalar and we know each of δik and xixk to be a secondrank contravariant tensor, which allows us to conclude that Iik is a secondrank tensor.

B-10 It is simple to construct the mixed tensor

U ji = 1

2 (∂iUj + ∂jU

i)

The fully covariant tensor is obtained by multiplying by gik to get

U jk = gikU ji = 1

2 (∂kU j + gik∂jUi)

Now, ∂jUi = ∂jUi so that the second term in the sum may be written ∂jUk

to complete the problem.

B-11 The hyperbolic coordinates are defined by

x =√ρ+ v y =

√ρ− v z = z

with ρ =√u2 + v2. The basis may be constructed as

�eu =d�r

du=dx

duı+

dy

duj =

u

ρ

( 12 ı√ρ+ v

+12 j√ρ− v

)

�ev =d�r

dv=dx

dvı+

dy

dvj =

( 12 (v/ρ+ 1)ı√

ρ+ v+

12 (v/ρ− 1)j√

ρ− v

)

and, of course �ez = k. The nonzero metric tensor elements are gzz = 1,

guu = �eu · �eu =u2

ρ2

( 14

ρ+ v+

14

ρ− v

)=

12ρ

=1

2√u2 + v2

gvv = �ev · �ev =

(v + ρ

2ρ

)2

v + ρ+

(ρ− v

2ρ

)2

ρ− v=

12ρ

=1

2√u2 + v2

The remaining elements all vanish. The Laplacian may be constructed fol-lowing the prescription of (B–69). The determinant of the metric tensor,G = 1/[4(u2 + v2)] and

√Gguu =

√Ggvv = 1 so that ∇2V becomes

∇2V = 2√u2 + v2

(∂2V

∂u2+∂2V

∂v2

)+d2V

dz2


B-12 We begin by defining r =√x2 + y2 to write

r =a sin θ

cosh η − cos θand z =

a sinh ηcosh η − cos θ

Referring to Example B.8, we note that r and z have been reversed, butotherwise the problems are identical. The toroidal system has a circle in thecentered on and perpendicular to the x-y plane that, rotated about the z axis,sweeps out a torus, whereas the bispherical coordinates have a circle centeredon the z axis which rotated about the z axis sweeps out a sphere. Adaptingthe results of Example B.8 simply exchanging z and r we have from (ExB.8.8)

r2 + (z − coth η)2 =a2

sinh2 η

the equation of a circle of radius a/ sinh η centered at z = coth η. Positiveη lead to spheres centered on positive z whereas negative η yield spheres atnegative z.

To obtain the metric tensor we first construct the local basis.

�eφ =d�r

dφ=

d

dφ(xı+ yj+ zk) =

a sin θcosh η − cos θ

(−ı sinφ+ j cosφ)

�eη =−a sin θ sinh η(ı cosφ+ j sinφ)

(cosh η − cos θ)2+[

a cosh ηcosh η − cos θ

− a sinh2 η

(cosh η − cos θ)2

]k

�eθ =[

a cos θcosh η − cos θ

− a sin2 θ


](ı cosφ+ j sinφ) − a sinh η sin θk


The metric tensor elements may now be computed,

gφφ = �eφ · �eφ =a2 sin2 θ


gθθ =[

a cos θcosh η − cos θ

− a sin2 θ


]2+

a sinh2 η sin2 θ


=a2


{cos2 θ + sin2 θ

×[sin2 θ − 2 cos θ(cosh η − cos θ) + sinh2 η


]}

=a2


Appendix B Solutions 123

and

gηη =a2 sin2 θ sinh2 η

(cosh η − cos θ)4+[

a cosh ηcosh η − cos θ

− a sinh2 η


]2

=a2


{cosh2 η + sinh2 η

×[sin2 θ − 2 cosh η(cosh η − cos θ) + sinh2 η


]}

=a2


The off-diagonal elements vanish.The Laplacian is constructed according to the prescription of (B–69):

∇2V =(cosh η − cos θ)3

a3 sin θ

{sin θ

∂

∂η

(1

cosh η − cos θ∂V

∂η

)

+∂

∂θ

(sin θ

cosh η − cos θ∂V

∂θ

)}+


a2 sin2 θ

∂2V

∂φ2

c©Jack Vanderlinde, Reproduction without express permission of the authoris strictly prohibited.

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