Solution Manual for Classical Electromagnetism, 1st Edition

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’s Solutions Manual
to accompany
Classical Electromagnetism
Jerrold Franklin
December 11, 2005

i
Preface
This instructor’s Solutions Manual contains my worked out solutions for all the problems
in the text “Classical Electromagnetism”. I have tried to be complete in each answer, and
to include all the steps leading to a sraightforward solution. As much as possible I have
written down the solutions as I would in doing them on the blackboard, including enough
explanation to make each step reasonable.
When a problem requires a particular equation from the book, I have included a reference
enclosed in square brackets, as, for instance, [Eq. (1.24)]. References to equations in the
solutions manual are given without the square brackets. If I use an equation without a
reference to the text, that is an indication that I have expected the students to know that
equation without going back to the text. References to Sections in the text are also given in
square brackets. For a number of problems, I have included a Note that adds an explanatory
extension of the solution or relates the solution to material in the text. These notes are always
preceded by the word Note in bold type. Most of the problems in the text relate directly
to the material covered. They are meant to extend and deepen the students knowledge of
EM, and to achieve utility with the necessary mathematics. The notes I have added to some
problems are meant to emphasize the connection with the text.
The solutions in this manual represent my teaching philosophy in doing the problems.
You may have a different approach for some problems. In that case, just use my solutions
as one way of doing things. I have tried to keep virtuoso problems, or problems that are
used to learn new things, out of the text. I expect that professors will add some of their own
problems, in line with their specialties or geared to their student’s interests. I don’t have
solutions for those problems.
This is an “Instructors Solutions Manual”. As much as possible, I do not think it would
be helpful to students to make it available to them in toto. My feeling is that struggling
through the problems, at least once, is an effective tool for absorbing the course material,
and also the math techniques needed throughout physics. I have tried to keep the problems
in the text at a level that will challenge, but not discourage, the students. In the text, I try
to lead the students through the material, but the problems are meant to help them see how
to work things out for themselves.
I hope that the text, along with this instructor’s solution manual, will make the teaching
of EM more pleasureable and rewarding for you and your students.
Jerrold Franklin

2 CHAPTER 1. FOUNDATIONS OF ELECTROSTATICS
(b) If the four charges are released from rest, we can write for the acceleration of any of the
charges
F
m = a = dv
dt = dv
dL′
dL′
dt , (1.3)
where L′ is the side length of the square at any time during the motion. The center of
the square remains fixed, and the distance, r, of a charge from the center is related to
L′ by L′ = √2r, so
dL′
dt = √2dr
dt = √2v. Then, a = √2v dv
dL′ , (1.4)
and the squared velocity after a long time is given by
V 2 =
∫ ∞
L
√2a dL′ =
[ q2(1 + 2√2)
√2m
] ∫ ∞
L
dL′
L′2 = q2(1 + 2√2)
√2mL . (1.5)
The velocity is the square root of this.
Problem 1.2:
(a) The four point charges q are located at the corners of a square with sides of length L.
The distance from each charge to a point z above the square, on the perpendicular
axis of the square, is
√
z2 + L2/2. The horizontal fields cancel, and the magnitude of
the vertical field is given by
Ez = E cos θ = 4qz
(z2 + L2/2)3
2
. (1.6)

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3
(b) For small oscillations, z << L, and we can approximate the force on a charge −q as
F = −8√2q2z
L3 = −kz. (1.7)
This is a restoring force proportional to the distance with an effective spring constant
k = 8√2q2
L3 , (1.8)
which leads to simple harmonic motion with period
T = 2π
√m
k − = 2π
[ mL3
8√2q2
] 1
2
. (1.9)
Problem 1.3:
(a) By symmetry, the electric field of a long straight wire is perpendicular to the wire. Its
magnitude a distance r from the wire is given by
Ez =
∫ ∞
−∞
rλdz
(z2 + r2)3/2 = λ
r
∫ π/2
−π/2
cos θdθ = 2λ
r . (1.10)
We made the substitution z = r tan θ in doing the integral. For the configuration of
two wires a distance a apart, the electric field is
E = 2λ(r − a/2)
|r − a/2)|2 + 2λ(r + a/2)
|r + a/2)|2 . (1.11)
(b) The electric field in Cartesian coordinates is
Ex = 2λ(x − a/2)
(x − a/2)2 + y2 + 2λ(x + a/2)
(x + a/2)2 + y2 (1.12)
Ey = 2λy
(x − a/2)2 + y2 + 2λy
(x + a/2)2 + y2 . (1.13)

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4 CHAPTER 1. FOUNDATIONS OF ELECTROSTATICS
Problem 1.4:
(a) The field on the axis of the uniformly charged wire, a distance z from the center of the
wire, is given for z > L/2 by
Ez = Q
L
∫ +L/2
−L/2
dz′
(z − z′)2
= Q
L
[ 1
z − L/2 − 1
z + L/2
]
= Q
z2 − L2/4 . (1.14)
(b) For −L/2 < z < L/2, the point z is a distance (L/2 − z) from the end of the wire. The
wire can be thought of as two parts.
The part of the wire from z′ = z − (L/2 − z) = 2z − L/2 to z′ = 2L is symmetric about
the point z. This means that the field due to that portion of the wire will cancel. The
remaining part of the wire has a length L′ = L − 2(L/2 − z) = 2z and a charge Q′ =
2zQ/L. The midpoint of this part of the wire is at z0 = (2z − L/2 − L/2)/2 = z − L/2.
Thus the electric field from this part of the wire is
Ez = Q′
[(z − z0)2 − L′2/4]
= 2Qz
L[(z − z + L/2)2 − z2]
= 2Qz
L[L2/4 − z2]. (1.15)

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5
Problem 1.5:
The parallel component of the electric field a distance d from a uniformly charged straight
wire of length L is given by the integral
Ex = Q
L
∫ x2
x1
dx′ cos θ
r2 , (1.16)
where x1 and x2 are the two endpoints of the wire. Let
x′ = −d cot θ
dx′ = d csc2 θdθ
r = d csc θ. (1.17)
Then
Ex = Q
Ld
∫ π−θ2
θ1
cos θ dθ = Q
Ld(sin θ2 − sin θ1), (1.18)
where θ1 and θ2 are the angles shown.
For the perpendicular component of E, the same substitution for z, leads to
Ey = Q
Ld
∫ π−θ2
θ1
sin θ dθ = Q
Ld(cos θ2 + cos θ1). (1.19)

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6 CHAPTER 1. FOUNDATIONS OF ELECTROSTATICS
Problem 1.6:
(a) Every point on the uniformly charged ring is the same distance from a point a distance
z along the axis of the ring, and a line from any point on the ring makes the same
angle θ with the z-axis. Thus the electric field at z is
Ez = Q cos θ
r2 = Qz
(z2 + R2)3
2
. (1.20)
(b) The disk has a surface charge density σ = Q/πR2. It can be considered as a collection
of rings, each of radius r′ with a charge
dq = 2πr′σdr′ = 2Qr′dr′
R2 . (1.21)
The electric field a distance z along the axis of the disk is given as (using part a)
Ez =
∫ R
0
2Qr′
R2
zdr′
(z2 + r′2)3
2
= 2Q
R2
[
1 − z
√z2 + R2
]
. (1.22)
(c) (i) For z = 0+ (just above the disk),
Ez = 2Q
R2 = 2πσ. (1.23)
(ii) For z >> R, we write E as
Ez = 2Q
R2

1 −
(
1 + R2
z2
)− 1
2

 . (1.24)
Using the binomial theorem, we get
Ez = 2Q
R2
[
1 −
(
1 − R2
2z2 + . . .
)]
.
' Q
z2 . (1.25)
This limit, equal to the field of a point charge Q, can be used as a check on the original
result.

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7
Problem 1.7:
(a) Each charge q, at the corner of a square with sides of length L is a distance
√
z2 + L2/2
from a point z along the perpendicular axis of the square. (See the figure in Prob. 1.2.)
Thus the potential at the point z is φ = 4q
√
z2 + L2/2 . (1.26)
(b) The electric field is
Ez = −∂zφ = 4qz
(z2 + L2/2)3
2
, as in Prob. 1.2. (1.27)
Problem 1.8:
(a) The potential on the axis of the uniformly charged wire, a distance z from the center of
the wire, is given for z > L/2 by
φ = Q
L
∫ +L/2
−L/2
dz′
z − z′ = Q
L ln
( z + L/2
z − L/2
)
. (1.28)
(b) The electric field is
Ez = −∂z φ = Q
L
[ 1
z − L/2 − 1
z + L/2
]
= Q
z2 − L2/4 , as in Prob. 1.4(a). (1.29)
Problem 1.9:
(a) Every point on the uniformly charged ring is the same distance √z2 + R2 from a point
a distance z along the axis of the ring, so the potential at z is
φ = Q
√z2 + R2 . (1.30)
(b) The uniformly charged disk of radius R has a surface charge density σ = Q/πR2. It can
be considered as a collection of rings, each of radius r′ with a charge
dq = 2πr′σdr′ = 2Qr′dr′
R2 . (1.31)
The potential a distance z along the axis of the disk is given as (using part a)
φ = 2Q
R2
∫ R
0
r′dr′
√z2 + r′2 = 2Q
R2
[√z2 + R2 − z]
. (1.32)
(c) The electric field of the ring is
Ez = −∂z
[ Q
√z2 + R2
]
= Qz
(z2 + R2)3
2
, as in Prob. 1.6(a). (1.33)
The electric field of the disk is
Ez = −2Q
R2 ∂z
[√z2 + R2 − z]
= 2Q
R2
[
1 − z
√z2 + R2
]
, as in Prob. 1.6(b). (1.34)

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8 CHAPTER 1. FOUNDATIONS OF ELECTROSTATICS
Problem 1.10:
(a) A uniformly charged spherical shell has a surface charge density
σ = Q
4πR2 . (1.35)
The spherical surface can be thought of as composed of rings of radius r = R sin θ,
where θ is the angle from the z axis. Each ring has a charge
dq == 2πrσR dθ = 1
2 Q sin θ dθ. (1.36)
The distance from the plane of a ring to the point d is z = d − R cos θ. Thus the
potential at the point d from the center of the sphere is
φ =
∫ dq
√z2 + r2 = Q
2
∫ π
0
sin θ dθ
√
(d − R cos θ)2 + R2 sin2 θ
= Q
2
∫ π
0
sin θ dθ
√d2 + R2 − 2Rd cos θ = Q
2
[ √d2 + R2 − 2Rd cos θ
−Rd
]π
0
= Q
2Rd [(d − R) − (d + R)] = Q
d , d ≥ R. (1.37)
Note that, for d ≤ R,
φ = Q
2Rd [(R − d) − (d + R)] = Q
R, d ≤ R. (1.38)
For d outside the sphere, the potential is the same as that of a point charge. For d
inside the sphere, the potential is constant, and hence the electric field is zero.
(b) The potential at a distance d ≥ R from the center of a uniformly charged spherical shell
is ∆q/d, where ∆q is the charge on the shell. The potential does not depend on the
radius R of the shell, as long as d ≥ R. A uniformly charged sphere can be considered
a collection of the uniformly charged shells, so that the potential due to the uniformly
charged sphere will be φ = Q/d, where Q is the net charge on the sphere. (Q is the
sum of all the ∆q on each spherical shell.)
Note that the charged sphere need not be uniformly charged as long as its charge
distribution is spherically symmetric.
(c) Since the potentials in (a) and (b) are the same as the potential of a point charge, the
electric fields are the same as the electric field of a point charge:
E = −∇
( Q
r
)
= Qˆr
r2 , r ≥ R. (1.39)

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9
Problem 1.11:
(a) The long straight wire has radius R and uniform charge density ρ. By symmetry, the
electric field is perpendicular to the axis of the wire. We consider a Gaussian cylinder
of length L and radius r, concentric with the wire. Then, for r ≤ R, Gauss’s law
becomes
∮
E·dA = 4πQ
2πrLE = 4π2r2ρL
E = 2πρr, r ≤ R. (1.40)
For r ≥ R, Gauss’s law is
∮
E·dA = 4πQ
2πrLE = 4π2R2ρL
E = 2πρR2
r , r ≥ R. (1.41)
(b) We have to pick some radius for which φ = 0. We choose φ(R) = 0, so that the surface
of the wire is at zero potential. Then, for r ≤ R,
φ(r) = −
∫ r
R
E·dr = −
∫ r
R
2πρrdr = πρ(R2 − r2). (1.42)
For r ≥ R,
φ(r) = −
∫ r
R
E·dr = −
∫ r
R
2πR2ρdr
r = 2πρR2 ln(R/r). (1.43)
Problem 1.12:
(a) By symmetry, the electric field of the uniformly charged hollow sphere is in the radial
direction. We consider a Gaussian sphere of radius r, concentric with the hollow sphere.
Then, for r ≤ R, Gauss’s law becomes
∮
E·dA = 4πQ
πr2E = 0
E = 0. (1.44)
For r ≥ R, Gauss’s law is
∮
E·dA = 4πQ
πr2E = 4πQ
E = Q
r2 . (1.45)

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10 CHAPTER 1. FOUNDATIONS OF ELECTROSTATICS
(b) The electric field for r ≥ R is the same as that for a point charge, so the potential is
the same as the potential of a point charge:
φ(r) = Q
r . (1.46)
For r ≤ R, E = 0, so the interior of a uniformly charged shell is an equipotential with
φ = Q
R . (1.47)
Problem 1.13:
(a) The charge density inside the uniformly charged sphere is
ρ = 3Q
4πR3 . (1.48)
By symmetry, the electric field is in the radial direction. We consider a Gaussian sphere
of radius r, concentric with the uniformly charged sphere. Then, for r ≤ R, Gauss’s
law becomes ∮
E·dA = 4πQ
πr2E =
[ 3Q
4πR3
] [ 4πr3
3
]
E = Qr
R3 . (1.49)
For r ≥ R, Gauss’s law is
∮
E·dA = 4πQ
πr2E = 4πQ
E = Q
r2 . (1.50)
(b) The electric field for r ≥ R is the same as that for a point charge, so the potential is
the same as the potential of a point charge:
φ(r) = Q
r , r ≥ R. (1.51)
For r ≤ R,
φ(r) − φ(R) = −
∫ r
R
E·dr
φ(r) = Q
R − Q
R3
∫ R
r
rdr = Q
R
[
1 − 1
2R3 (r2 − R2)
]
= Q
2R (3R2 − r2). (1.52)

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11
Problem 1.14:
(a) For φ = qe−μr/r, the electric field is
E = −q∇(e−μr/r) = −qe−μr∇
(1
r
)
− q
r ∇(e−μr )
= −qe−μr
(
− ˆr
r2 − μˆr
r
)
= qˆre−μr
r2 (1 + μr) . (1.53)
(b) The charge density is given by
ρ = 1
4π ∇·E = q
4π ∇·
[
e−μr
( r
r3 + μr
r2
)]
, (1.54)
where we have written E in a more convenient form for taking its divergence. We
differentiate each term in turn, leading to
ρ = q
4π
{
e−μr
[
∇·
( r
r3
)
+ ∇·
(μr
r2
)]
+
[ r
r3 + μr
r2
]
·∇(e−μr )
}
= qe−μr
4π
[
4πδ(r) + 3μ
r2 − 2μ
r2 − μ
r2 − μ2
r
]
= qδ(r) − μ2qe−μr
4πr . (1.55)
Note that above we isolated the term ∇· (r/r3), because we knew that it equals
4πδ(r). This correctly accounted for the singular behavior at the origin. The term
qδ(r) corresponds to a point charge q at the origin. The other part of Eq. (1.55)
represents a negative charge distribution surrounding the point charge.
(c) Gauss’s law is ∮
E·dA = 4πQ. (1.56)
We choose a sphere of radius R as our Gaussian surface. The integral of E·dS over
the surface of the sphere is
∮
E·dA = 4πR2Er(R) = 4πqe−μR (1 + μR) , (1.57)
where we have taken Er from Eq. (1.53) The charge within the Gaussian sphere is
given (using Eq. (1.55) by
Q =
∫
r≤R
ρdτ = q − 4π
∫ R
0
μ2qe−μrr2dr
4πr = 4πqe−μR (1 + μR) , (1.58)
which agrees with Eq. (1.57), and Gauss’s law is satisfied. (The last integral above was
done using integration by parts.)
Note the importance of the proper treatment of the delta function at the origin.

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12 CHAPTER 1. FOUNDATIONS OF ELECTROSTATICS
Problem 1.15:
(a) F= (r×p)(r·p), with p a constant vector.
∇×F = ∇×[(r×p)(r·p)]
= (r·p)[∇×(r×p)] − (r×p)×[∇(r·p)]
= (r·p)[(p·∇)r − p(∇·r)] − (r×p)×p
= (r·p)(p − 3p) − p(r·p)+p2r
= p2r − 3p(r·p). (1.59)
∇·F = ∇·[(r×p)(r·p)]
= (r·p)[∇·(r×p)] + (r×p)·[∇(r·p)]
= (r·p)[p·(∇×r)] + (r×p)·p
= 0 + r·(p×p) = 0. (1.60)
(b) F = (r·p)2r, with p a constant vector.
∇×F = ∇×[(r·p)2r]
= (r·p)2(∇×r) − r×[∇(r·p)2]
= 0 − r×[2(r·p)p]
= −2(r·p)(r×p) (1.61)
∇·F = ∇·[(r·p)2r]
= (r·p)2(∇·r) + r·[∇(r·p)2]
= 3(r·p)2 + r·[2(r·p)p]
= 5(r·p)2. (1.62)

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