Solutions to Problems in Jackson,
Classical Electrodynamics, Third Edition
Homer Reid
June 15, 2000
Chapter 3: Problems 1-10
Problem 3.1
Two concentric spheres have radii a, b(b > a) and each is divided into two hemispheres by the same horizontal plane. The upper hemisphere of the inner sphere
and the lower hemisphere of the outer sphere are maintained at potential V . The
other hemispheres are at zero potential.
Detemine the potential in the region a ≤ r ≤ b as a series in Legendre polynomials.
Include terms at least up to l = 4. Check your solution against known results in
the limiting cases b → ∞ and a → 0.
The expansion of the electrostatic potential in spherical coordinates for problems with azimuthal symmetry is
Φ(r, θ) =
∞ h
X
l=0
i
Al rl + Bl r−(l+1) Pl (cos θ).
(1)
We find the coefficients Al and Bl by applying the boundary conditions. Multiplying both sides by Pl0 (cos θ) and integrating from -1 to 1 gives
Z 1
i
2 h
Al rl + Bl r−(l+1) .
Φ(r, θ)Pl (cos θ)d(cos θ) =
2l + 1
−1
At r = a this yields
V
Z
1
Pl (x)dx =
0
i
2 h
Al al + Bl a−(l+1) ,
2l + 1
1
2
Homer Reid’s Solutions to Jackson Problems: Chapter 3
and at r = b,
V
Z
0
Pl (x)dx =
−1
i
2 h l
Al b + Bl b−(l+1) .
2l + 1
The integral from 0 to 1 vanishes for l even, and is given in the text for l odd:
Z 1
(l − 2)!!
1
Pl (x)dx = (− )(l−1)/2 l+1 .
2
2 2 !
0
The integral from -1 to 0 also vanishes for l even, and is just the above result
inverted for l odd. This gives
i
2 h
1
(l − 2)!!
Al al + Bl a−(l+1)
V (− )(l−1)/2 l+1 =
2
2l + 1
2 2 !
i
(l − 2)!!
2 h l
1
−V (− )(l−1)/2 l+1 =
Al b + Bl b−(l+1) .
2
2l + 1
2 2 !
or
αl
−αl
with
1
(2l + 1)(l − 2)!!
αl = V (− )a(l−1)/2
.
2
4 l+1
2 !
The solution is
Al = α l
= Al al + Bl a−(l+1)
= Al bl + Bl b−(l+1)
bl+1 + al+1
a2l+1 − b2l+1
Bl = −αl
al+1 bl+1 (bl + al )
a2l+1 − b2l+1
The first few terms of (1) are
2
(a + b2 )r a2 b2 (a + b)
a4 b4 (a3 + b3 )
3
7 (a4 + b4 )r3
− 2 3
− 4 7
Φ(r, θ) = V
P1 (cos θ)−
P3 (cos θ)+· · ·
4
a3 − b 3
r (a − b3 )
16
a7 − b 7
r (a − b7 )
In the limit as b → ∞, the problem reduces to the exterior problem treated
in Section 2.7 of the text. In that limit, the above expression becomes
7 a 4
3 a 2
P1 (cos θ) − V
P3 (cos θ) + · · ·
Φ(r, θ) → V
4
r
16
r
in agreement with (2.27) with half the potential spacing. When a → 0, the
problem goes over to the interior version of the same problem, as treated in
section 3.3 of the text. In that limit the above expression goes to
3 r 7 r 3
Φ(r, θ) → − V
P1 (cos θ) + V
P3 (cos θ) + · · ·
4
b
16
b
This agrees with equation (3.36) in the text, with the sign of V flipped, because
here the more positive potential is on the lower hemisphere.
Homer Reid’s Solutions to Jackson Problems: Chapter 3
3
Problem 3.2
A spherical surface of radius R has charge uniformly distributed over its surface
with a density Q/4πR2 , except for a spherical cap at the north pole, defined by the
cone θ = α.
(a) Show that the potential inside the spherical surface can be expressed as
Φ=
∞
rl
Q X 1
[Pl+1 (cos α) − Pl−1 (cos α)] l+1 Pl (cos θ)
8π0
2l + 1
R
l=0
where, for l = 0, Pl−1 (cos α) = −1. What is the potential outside?
(b) Find the magnitude and direction of the electric field at the origin.
(c) Discuss the limiting forms of the potential (part a) and electric field (part b) as
the spherical cap becomes (1)very small, and (2) so large that the area with
charge on it becomes a very small cap at the south pole.
(a) Let’s denote the charge density on the sphere by σ(θ). At a point infinitesimally close to the surface of the sphere, the electric field is
F = −∇Φ = −
so
σ
r̂
0
σ
∂Φ = .
∂r r=R 0
(2)
The expression for the potential within the sphere must be finite at the origin,
so the Bl in (1) are zero. Differentiating that expansion, (2) becomes
∞
X
∂
Φ(r, θ) =
lAl rl−1 Pl (cos θ)
∂r
l=1
Multiplying by Pl0 and integrating at r = R gives
Z
1 1
2l
σ(θ)Pl (cos θ)d(cos θ) =
Al Rl−1
0 −1
2l + 1
so
2l + 1
Al =
·
2lRl−1
Q
4πR2 0
Z
cos α
Pl (x)dx.
−1
To evaluate the integral we use the identity (eq. 3.28 in the text)
Pl (x) =
d
1
[Pl+1 (x) − Pl−1 (x)]
(2l + 1) dx
Homer Reid’s Solutions to Jackson Problems: Chapter 3
so
Z
cos α
Pl (x)dx =
−1
4
1
[Pl+1 (cos α) − Pl−1 (cos α)] .
2l + 1
(We used the fact that Pl+1 (−1) = Pl−1 (−1) for all l.) With this we have
Al =
Q
[Pl+1 (cos α) − Pl−1 (cos α)]
8π0 lRl+1
so the potential expansion is
Φ(r, θ) =
∞
Q X1
rl
[Pl+1 (cos α) − Pl−1 (cos α)] l+1 Pl (cos θ).
8π0
l
R
l=1
Within the body of the sum, I have an l where Jackson has a 2l + 1. Also,
he includes the l = 0 term in the sum, corresponding to a constant term in
the potential. I don’t understand how he can determine that constant from the
information contained in the problem; the information about the charge density
only tells you the derivative of the potential. There’s nothing in this problem
that fixes the value of the potential on the surface beyond an arbitrary constant.
(b) The field at the origin comes from the l = 1 term in the potential:
∂Φ
1 ∂Φ E(r = 0) = −∇Φ|r=0 = − r̂ +
θ̂ ∂r
r ∂θ r=0
d
Q
[P
(cos
α)
−
1]
P
(cos
θ)r̂
+
P
(cos
θ)
θ̂
= −
2
1
1
8π0 R2
dθ
h
i
Q
3
3
= −
cos θr̂ − sin θ θ̂
cos2 α −
2
8π0 R 2
2
=
3Q sin2 α
k̂.
16π0 R2
The field points in the positive z direction. That makes sense, since a positive
test charge at the origin would sooner fly up out through the uncharged cap
than through any of the charged surface.
5
Homer Reid’s Solutions to Jackson Problems: Chapter 3
Problem 3.3
A thin, flat, conducting, circular disk of radius R is located in the x − y plane
with its center at the origin, and is maintained at a fixed potential V . With the
information that the charge density on a disc at fixed potential is proportional to
(R2 − ρ2 )−1/2 , where ρ is the distance out from the center of the disc,
(a) show that for r > R the potential is
∞
2V R X (−1)l
Φ(r, θ, φ) =
π r
2l + 1
l=0
R 2l
r
!
P2l (cos θ)
(b) find the potential for r < R.
(c) What is the capacitance of the disk?
We are told that the surface charge density on the disk goes like
σ(r)
= K(R2 − r2 )−1/2
1 r 2
K
3 · 1 r 4
5 · 3 · 1 r 6
1+
=
+
+
+···
R
2 R
(2!)(2 · 2) R
(3!)(2 · 2 · 2) R
∞
K X (2n − 1)!! r 2n
(3)
=
R n=0 n! · 2n
R
for some constant K. From the way the problem is worded, I take it we’re
not supposed to try to figure out what K is explicitly, but rather to work the
problem knowing only the form of (3).
At a point infinitesimally close to the surface of the disk (i.e., as θ → π/2),
the component of ∇Φ in the direction normal to the surface of the disk must
be proportional to the surface charge. At the surface of the disk, the normal
direction is the negative θ̂ direction. Hence
σ
1 ∂
=± .
Φ(r, θ)
(4)
r ∂θ
0
θ=(π/2)
with the plus (minus) sign valid for Φ above (below) the disc.
For r < R the potential expansion is
Φ(r, θ) =
∞
X
Al rl Pl (cos θ).
(5)
l=0
Combining (3), (4), and (5) we have
∞
∞
X
K X (2n − 1)!! r 2n
d
=±
Al rl−1
Pl (cos θ)
.
dθ
R0 n=0 n! · 2n
R
cos θ=0
l=0
(6)
Homer Reid’s Solutions to Jackson Problems: Chapter 3
6
For l even, dPl /dx vanishes at x = 0. For l odd, I used some of the Legendre
polynomial identities to derive the formula
(2l − 1)!!
d
P2l+1 (x)
.
= (−1)l (2l + 1)
dx
l! · 2l
x=0
This formula reminds one strongly of expansion (3). Plugging into (6) and
equating coefficents of powers of r, we find
A2l+1 = ±
so
Φ(r, θ) = A0 ±
(−1)l K
(2l + 1)R2l+1 0
∞
K X (−1)l r 2l+1
P2l+1 (cos θ).
0
2l + 1 R
l=1
I wrote A0 explicitly because we haven’t evaluated it yet–the derivative condition
we used earlier gave no information about it. To find A0 , observe that, on the
surface of the disk (cos θ = 0), all the terms in the above sum vanish ( because
Pl (0) is 0 for odd l) so Φ = A0 on the disk. But Φ = V on the disk. Therefore,
A0 = V . We have
Φ(r, θ) = V ±
∞
K X (−1)l r 2l+1
P2l+1 (cos θ)
0
2l + 1 R
(7)
l=1
where the plus (minus) sign is good for θ less than (greater than)π/2. Note that
the presence of that ± sign preserves symmetry under reflection through the z
axis, a symmetry that is clearly present in the physical problem.
(a) For r > R, there is no charge. Thus the potential and its derivative must be
continuous everywhere–we can’t have anything like the derivative discontinuity
that exists at θ = π/2 for r < R. Since the physical problem is symmetric under
a sign flip in cos θ, the potential expansion can only contain Pl terms for l even.
The expansion is
Φ(r, θ) =
∞
X
B2l r−(2l+1) P2l (cos θ).
l=0
At r = R, this must match up with (7):
V ±
∞
∞
X
K X (−1)l
P2l+1 (cos θ) =
B2l R−(2l+1) P2l (cos θ).
0
2l + 1
l=1
l=0
Multiplying both sides by P2l (cos θ) sin(θ) and integrating gives
Z 0
Z 1
Z 1
∞
K X (−1)l
2R−(2l+1)
P2l+1 (x)Pl (x)dx
Pl (x)dx +
P2l+1 (x)P2l (x)dx +
B2l
= V
−
4l + 1
0
2l + 1
0
−1
−1
l=1
Z
∞
2K X (−1)l 1
= 2V δl,0 +
P2l+1 (x)P2l (x)dx.
0
2l + 1 0
l=1
Homer Reid’s Solutions to Jackson Problems: Chapter 3
7
but I can’t do this last integral.
Problem 3.4
The surface of a hollow conducting sphere of inner radius a is divided into an even
number of equal segments by a set of planes; their common line of intersection is
the z axis and they are distributed uniformly in the angle φ. (The segments are like
the skin on wedges of an apple, or the earth’s surface between successive meridians
of longitude.) The segments are kept at fixed potentials ±V , alternately.
(a) Set up a series representation for the potential inside the sphere for the general
case of 2n segments, and carry the calculation of the coefficients in the series
far enough to determine exactly which coefficients are different from zero. For
the nonvanishing terms exhibit the coefficients as an integral over cos θ.
(b) For the special case of n = 1 (two hemispheres) determine explicitly the potential up to and including all terms with l = 3. By a coordinate transformation
verify that this reduces to result (3.36) of Section 3.3.
(a) The general potential expansion is
Φ(r, θ, φ) =
∞ X
l
h
X
l=0 m=−l
i
Alm rl + Blm r−(l+1) Ylm (θ, φ).
(8)
For the solution within the sphere, finiteness at the origin requires Blm = 0.
∗
Multiplying by Yl0m0
and integrating over the surface of the sphere we find
Z
1
∗
Alm =
Φ(a, θ, φ) Ylm
(θ, φ) dΩ
al
Z
Z
n
π
2kπ/n
V X
∗
k
Ylm
(θ, φ) sin θ dφ dθ
=
(−1)
al
0
2(k−1)π/n
k=1
(Z
)
1/2 Z 1
X
n
2kπ/n
V 2l + 1 (l − m)!
k
−imφ
m
=
Pl (x) dx
(−1)
e
dφ .
al
4π (l + m)!
−1
2(k−1)π/n
k=1
(9)
The φ integral is easy:
Z 2kπ/n
i
1 h −2imkπ/n
e
− e−2im(k−1)π/n .
e−imφ dφ = −
im
2(k−1)π/n
This is to be summed from k = 1 to n with a factor of (−1)k thrown in:
i
X
1 h −2mπi(1/n)
= −
(e
− 1) − (e−2mπi(2/n) − e−2mπi(1/n) ) + · · · − (1 − e−2mπi((n−1)/n) )
im
o
2 n
=
1 − e−2mπi/n + e2(−2mπi/n) − e3(−2mπi/n) + · · · + e(n−1)(−2mπi/n) .
(10)
im
Homer Reid’s Solutions to Jackson Problems: Chapter 3
8
Putting x = − exp(−2mπi/n), the thing in braces is
1 + x + x2 + x3 + · · · + xn−1 =
1 − xn
1 − e−2mπi
=
,
1−x
1 + e−2imπ/n
Note that the numerator vanishes. Thus the only way this thing can be nonzero
is if the denominator also vanishes, which only happens if the exponent in the
denominator equates to -1. This only happens if m/n = 1/2, 3/2, 5/2, · · · . In
that case, the 2mπi/n term in the exponent of the terms in (10) equates to πi,
so all the terms with a plus sign in (10) come out to +1, while all the terms
with a minus sign come out to -1, so all n terms add constructively, and (10)
equates to
(
2n
X
,
m = n/2, 3n/2, 5n/2, · · ·
= im
0,
otherwise.
Then the expression (9) for the coefficients becomes
Alm
1/2 Z 1
2nV 2l + 1 (l − m)!
=
Plm (x)dx,
imal
4π (l + m)!
−1
m=
n 3n
,
, · · · = 0, otherwise.
2 2
(b) As shown above, the only terms that contribute are those with m = n/2,
m = 3n/2, et cetera. Of course there is also the constraint that m < l. Then,
with n = 2, up to l = 3 the only nonzero terms in the series (9) are those with
l = 1, m = ±1, and l = 3, m = ±1 or ±3. We need to evaluate the θ integral
for these terms. We have
Z 1
Z 1
1
P1 (x) dx = −
(1 − x2 )1/2 dx = −π
Z
Z
−1
1
−1
1
−1
P31 (x) dx
P33 (x) dx
= −
Z
−1
1
(1 − x2 )1/2
−1
= −15
Z
1
3π
15 2 3
dx = −
x −
2
2
8
(1 − x2 )3/2 dx = −
−1
15π
.
4
Using these results in (??), we have
A1±1
A3±1
A3±3
1/2
3
4πV i
a
4π · 2
1/2
3πV i 7 · 2
= ±
2a3 4π · 4!
1/2
5πV i
7
= ± 3
a
4π · 6!
= ±
Now we can plug these coefficients into (8) to piece together the solution.
This involves some arithmetic in combining all the numerical factors in each
9
Homer Reid’s Solutions to Jackson Problems: Chapter 3
coefficient, which I have skipped here.
h r
7 r 3
sin θ(5 cos2 θ − 1) sin φ
sin θ sin φ +
Φ(r, θ, φ) = V 3
a
16 a
7 r 3 3
+
sin θ sin 3φ + · · ·
144 a
Problem 3.6
Two point charges q and −q are located on the z azis at z = +a and z = −a,
respectively.
(a) Find the electrostatic potential as an expansion in spherical harmonics and
powers of r for both r > a and r < a.
(b) Keeping the product qa = p/2 constant, take the limit of a → 0 and find the
potential for r 6= 0. This is by definition a dipole along the z azis and its
potential.
(c) Suppose now that the dipole of part b is surrounded by a grounded spherical
shell of radius b concentric with the origin. By linear superposition find the
potential everwhere inside the shell.
(a) First of all, for a point on the z axis the potential is
1
q
1
Φ(z) =
−
4π0 |z − a| z + a
a a 2
a a 2
q
+··· − 1−
···
=
1+
+
+
4π0 z
z
z
z
z
q
a
a 3
=
+
+···
2π0 z
z
z
P
for z > a. Comparing this with the general expansion Φ =
Bl r−(l+1) Pl (cos θ)
at θ = 0 we can identify the Bl s and write
a 3
a
q
P3 (cos θ) + · · ·
P1 (cos θ) +
Φ(r, θ) =
2π0 r
r
r
for r > a. For r < a we can just swap a and r in this equation.
(b)
Φ(r, θ)
=
=
→
qa
2π0 r2
p
4π0 r2
p
4π0 r2
P1 (cos θ) +
P1 (cos θ) +
cos θ
a 2
r
a 2
r
as a → 0.
P3 (cos θ) + · · ·
P3 (cos θ) + · · ·
Homer Reid’s Solutions to Jackson Problems: Chapter 3
10
(c) When we put the grounded sphere around the two charges, a surface charge
distribution forms on the sphere. Let’s denote by Φs the potential due to this
charge distribution alone (not including the potential of the dipole) and by Φd
the potential due to the dipole. To calculate Φs , we pretend there are no charges
within the sphere, in which case we have the general expansion (1), with Bl = 0
to keep us finite at the origin. The total potential is just the sum Φs + Φd :
∞
Φ(r, θ) =
X
p
cos θ +
Al rl Pl (cos θ).
2
4π0 r
l=0
The condition that this vanish at r = b ensures, by the orthogonality of the Pl ,
that only the l = 1 term in the sum contribute, and that
p
.
A1 = −
4π0 b3
The total potential inside the sphere is then
r
p P1 (cos θ).
1
−
Φ(r, θ) =
4π0 b2
b
Problem 3.7
Three point charges (q, −2q, q) are located in a straight line with separation a and
with the middle charge (−2q) at the origin of a grounded conducting spherical shell
of radius b, as indicated in the figure.
(a) Write down the potential of the three charges in the absence of the grounded
sphere. Find the limiting form of the potential as a → 0, but the product
qa2 = Q remains finite. Write this latter answer in spherical coordinates.
(b) The presence of the grounded sphere of radius b alters the potential for r < b.
The added potential can be viewed as caused by the surface-charge density
induced on the inner surface at r = b or by image charges located at r > b. Use
linear superposition to satisfy the boundary conditions and find the potential
everywhere inside the sphere for r < a and r > a. Show that in the limit
a → 0,
Q
r5
Φ(r, θ, φ) →
1 − 5 P2 (cos θ).
2π0 r3
b
(a) On the z axis, the potential is
q
2
1
1
Φ(z) =
− +
+
4π0
z |z − a| z + a
a a 2
a a 2
q
··· + 1−
+···
=
−2 + 1 +
+
+
4π0 r
z
z
z
z
q
a 2 a 4
=
+
+··· .
2π0 z
z
z
Homer Reid’s Solutions to Jackson Problems: Chapter 3
11
As before, from this result we can immediately infer the expression for the
potential at all points:
a 4
a 2
q
P2 (cos θ) +
P4 (cos θ) + · · ·
Φ(r, θ) =
2π0 r
r
r
2
2
qa
a
=
P
(cos
θ)
+
·
·
·
P
(cos
θ)
+
4
2
2π0 r3
r
Q
→
P2 (cos θ) as a → 0
(11)
2π0 r3
(b) As in the previous problem, the surface charges on the sphere produce an
extra contribution Φs to the potential within the sphere. Again we can express
Φs with the expansion (1) (with Bl = 0), and we add Φs to (11) to get the full
potential within the sphere:
∞
Φ(r, θ) =
X
Q
P2 (cos θ) +
Al rl Pl (cos θ)
3
2π0 r
l=0
From the condition that Φ vanish at r = b, we determine that only the l = 2
term in the sum contributes, and that
A2 = −
Q
.
2π0 b5
Then the potential within the sphere is
r 5 Q
Φ(r, θ) =
1
−
P2 (cos θ).
2π0 r3
b
Problem 3.9
A hollow right circular cylinder of radius b has its axis coincident with the z axis
and its ends at z = 0 and z = L. The potential on the end faces is zero, while the
potential on the cylindrical surface is given as V (φ, z). Using the appropriate separation of variables in cylindrical coordinates, find a series solution for the potential
anywhere inside the cylinder.
The general solution of the Laplace equation for problems in cylindrical coordinates consists of a sum of terms of the form
R(ρ)Q(φ)Z(z).
The φ function is of the form
Q(φ) = A sin νφ + B cos νφ
Homer Reid’s Solutions to Jackson Problems: Chapter 3
12
with ν an integer. The z function is of the form
Z(z) = Cekz + De−kz .
In this case, Z must vanish at z = 0 and z = L, which means we have to take
k imaginary, i.e.
πn
, n = 1, 2, 3, · · ·
Z(z) = C sin(kn z) with kn =
L
With this form for Z, R must be taken to be of the form
R(ρ) = EIν (kn ρ) + F Kν (kn ρ).
Since we’re looking for the potential on the inside of the cylinder and there is
no charge at the origin, the solution must be finite as ρ → 0, which requires
F = 0. Then the potential expansion becomes
Φ(ρ, φ, z) =
∞ X
∞
X
[Anν sin νφ + Bnν cos νφ] sin(kn z)Iν (kn ρ).
(12)
n=1 ν=0
Multiplying by sin ν 0 φ sin kn0 z and integrating at r = b, we find
Z L Z 2π
πL
Iν (kn b)Anν
V (φ, z) sin νφ sin(kn z) dφ dz =
2
0
0
so
Z L Z 2π
2
Anν =
V (φ, z) sin(νφ) sin(kn z) dφ dz.
πLIν (kn b) 0 0
Similarly,
Z L Z 2π
2
Bnν =
V (φ, z) cos(νφ) sin(kn z) dφ dz.
πLIν (kn b) 0 0
(13)
(14)
Problem 3.10
For the cylinder in Problem 3.9 the cylindrical surface is made of two equal halfcylinders, one at potential V and the other at potential −V , so that
V for −π/2 < φ < π/2
V (φ, z) =
−V for π/2 < φ < 3π/2
(a) Find the potential inside the cylinder.
(b) Assuming L >> b, consider the potential at z = L/2 as a function of ρ and φ
and compare it with two-dimensional Problem 2.13.
The potential expansion is (12) with coefficients given by (13) and (14). The
relevant integrals are
Z L Z 2π
V (φ, z) sin(νφ) sin(kn z) dφ dz
0
0
13
Homer Reid’s Solutions to Jackson Problems: Chapter 3
= V
= 0
(Z
L
sin(kn z) dz
0
Z
= V
=
=
(Z
L
0
Z
) (Z
π/2
sin(νφ) dφ −
−π/2
Z
3π/2
sin(νφ) dφ
π/2
)
2π
V (φ, z) cos(νφ) sin(kn z) dφ dz
0
L
sin(kn z) dz
0
) (Z
π/2
cos(νφ) dφ −
−π/2
Z
3π/2
cos(νφ) dφ
π/2
)
o
2V n
π/2
3π/2
(n odd)
|sin νφ|−π/2 − |sin νφ|π/2
νkn

0 , n or ν even

8V /kn ν , n odd, ν = 1, 5, 9, · · ·

−8V /kn ν , n odd, ν = 3, 7, 11, · · ·
Hence, from (13) and (14),
Anν
Bnν
= 0
= 0,
= (−1)(ν−1)/2 · 16V /(nνπ 2 Iν (kn b)),
n or ν even
n and ν odd
The potential expansion is
Φ(ρ, θ, z) =
16V X (−1)(ν−1)/2
cos(νφ) sin(kn z)Iν (kn ρ)
π 2 n,ν nνIv (kn b)
(15)
where the sum contains only terms with n and ν odd.
(b) At z = L/2 we have
Φ(ρ, θ, L/2) =
Iν (kn ρ)
16V X (−1)(n+ν−2)/2
cos(νφ)
.
2
π n,ν
nν
Iν (kn b)
As L → ∞, the arguments to the I functions become small. Using the limiting
form for Iν quoted in the text as equation (3.102), we have
Φ(ρ, θ) =
ρ ν
16V X (−1)(n+ν−2)/2
cos(νφ)
.
π 2 n,ν
nν
b
The sums over n and ν are now decoupled:
(∞
)( ∞
)
ρ ν
X (−1)ν
16V X (−1)n
Φ(ρ, θ) =
cos(νφ)
π2
2n + 1
2ν + 1
b
n=0
ν=0
(∞
)
ρ ν
16V n π o X (−1)ν
cos(νφ)
=
2
π
4
2ν + 1
b
ν=0
4V
2ρb cos φ
=
tan−1
π
b2 − ρ 2
Homer Reid’s Solutions to Jackson Problems: Chapter 3
14
This agrees with the result of Problem 2.13, with V1 = −V2 = V . The first
series is just the Taylor series for tan−1 (x) at x = 1, so it sums to π/4. The
second series can also be put into the form of the Taylor series for tan−1 (x),
using tricks exactly analogous to what I did in my solution for Problem 2.13.