Jackson - Electrodynamics Solutions

Prévia do material em texto

Por BRUNO F. AMORIM
 BRUNO JESUS OLIVEIRA
 FABIO NASCIMENTO
 UFPI
1.1
2
In general we take the charge density to be of the form ρ = fr⃗δ, where fr⃗ is determined by
physical constraints, such as ∫ρd3x = Q.
a) variables: r,θ,φ. d3x = dφd cosθr2dr
ρ = fr⃗δr − R = frδr − R = fRδr − R
∫ ρd3x = fR ∫ r2drdΩdr − R = 4πfRR2 = Q → fR = Q
4πR2
ρr⃗ = Q
4πR2
δr − R
b) variables: r,φ, z. d3x = dφdzrdr
ρr⃗ = fr⃗δr − b = fbδr − b
∫ ρd3x = fb ∫ dzdφrdrδr − b = 2πfbbL = λL → fb = λ2πb
ρr⃗ = λ
2πb
δr − b
c) variables: r,φ, z. d3x = dφdzrdr Choose the center of the disk at the origin, and the z-axis
perpendicular to the plane of the disk
ρr⃗ = fr⃗δzθR − r = fδzθR − r
where θR − r is a step function.
∫ ρd3x = f ∫ δzθR − rdφdzrdr = 2π R22 f = Q → f =
Q
πR2
ρr⃗ = Q
πR2
δzθR − r
d) variables: r,θ,φ. d3x = dφdcosθr2dr
ρr⃗ = fr⃗δcosθθR − r = frδcosθθR − r
∫ ρd3x = ∫ frδcosθθR − rdφdcosθr2dr = ∫
0
R
frrrdrdφ = 2πN ∫
0
R
rdr = πR2N = Q
where I’ve used the fact that rdrdφ is an element of area and that the charge density is uniformly
distributed over area.
ρr⃗ = Q
πR2r
δcosθθR − r
1.3
3
Gauss’s Law: ∫ E⃗ ⋅ da⃗ = Qenclosed0
a) Conducting sphere: all of the charge is on the surface σ = Q
4πa2
E4πr2 = 0, r < a E4πr2 = Q0 , r ∑ a
E = 0, r < a E⃗ = Q
4π 0r2
r̂, r ∑ a
b) Uniform charge density: ρ = Q4
3 πa
3 , r < a, ρ = 0, r ∑ a.
E4πr2 = Qr
3
0a3
→ E = Qr
4π 0a3
, r < a
E4πr2 = Q0 → E =
Q
4π 0r2
, r ∑ a
c) ρ = A rn
Q = 4π ∫ r2drArn = 4πAan+3/n + 3 → ρ = n + 3Q
4πan+3
rn, r < a
ρ = 0, r ∑ a
E4πr2 = n + 3Q
4π 0an+3
4πrn+3/n + 3 → E = Q
4π 0r2
rn+3
an+3
, r < a
E = Q
4π 0r2
, r ∑ a
1. n = −2.
E = Q
4π 0ra
, r < a
1.4
4
E = Q
4π 0r2
, r ∑ a
2. n = 2.
E = Qr
3
4π 0a5
, r < a
E = Q
4π 0r2
, r ∑ a
5
φr⃗ =
qe−αr 1 + αr2
4π 0r
∇2φr⃗ = − ρ0 , ∇
2 = 1
r2
∂
∂r
r2 ∂
∂r
∇2φ = q
4π 0
1
r2
∂
∂r
r2 ∂
∂r
e−αr
r +
α
2
e−αr
Using ∇2 1r = −4πδr⃗
∇2φ = q
4π 0
1
r2
∂
∂r
−αre−αr + e−αrr2 ∂
∂r
1
r −
α2
2
r2e−αr
=
q
4π 0
− α
r2
e−αr + α
2e−αr
r +
αe−αr
r2
− 4πδr⃗ − α
2e−αr
r +
α3e−αr
2
= − 10 qδr⃗ −
qα3
8π
e−αr
ρr⃗ = qδr⃗ − qα
3
8π
e−αr
That is, the charge distribution consists of a positive point charge at the origin, plus an
exponentially decreasing negatively charged cloud.
1.5
6
1.6
7
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1.7
10
We will be using Gauss’s law ∫E⃗ ⋅ da⃗ = Qenc 0
a) 1) Parallel plate capacitor
From Gauss’s law E = σ0 =
Q
A0
=
φ12
d
→ Q = A0φ12
d
W = 0
2 ∫ E
2d3x = 0E
2Ad
2
=
0
Q
A0
2
Ad
2
= 1
20
Q2
A
d = 1
20
A0φ12
d
2
A
d = 1
2
0A
φ122
d
2) Spherical capacitor
From Gauss’s law, E = 14π0
Q
r2
, a < r < b.
φ12 = ∫
a
b
Edr = Q
4π 0
∫
a
b
r−2dr = 1
4
Q
π 0
b − a
ba
→ Q =
4π 0baφ12
b − a
W = 0
2 ∫ E
2d3x = 0
2
Q
4π 0
2
∫
a
b
4π r
2dr
r4
=
0
2
Q
4π 0
2
−4π −b + a
ba
= 1
80
Q2
π
b − a
ba
W = 1
80
4π0baφ12
b−a
2
π
b − a
ba
= 2π 0ba
φ122
b − a
3) Cylindrical conductor
1.8
11
From Gauss’ s law, E2πrL = λL
0
=
Q
0
φ12 = ∫
a
b
Edr = Q
2π 0L
∫
a
b dr
r =
Q
2π 0L
ln ba
W = 0
2 ∫ E
2d3x = 0
2
Q
2π 0L
2
2πL ∫
a
b rdr
r2
= 1
4π 0
Q2
L
ln ba
W = 1
4π 0
2π0Lφ12
ln ba
2
L
ln ba = π 0L
φ122
ln ba
b) w = 02 E
2
1) wr = 02
Q
A0
2
= 120
Q2
A2
0 < r < d, = 0 otherwise.
2) wr = 02
1
4π0
Q
r2
2
= 1
320π2
Q2
r4
, a < r < b, = 0, otherwise
3) wr = 02
Q
2π0Lr
2
= 180
Q2
π2L2r2
, = 0 otherwise.
12
I will be using the principle of virtual work. In the figure below, Fδl is the work done by an
external force. If F is along δl (ie. is positive), then the force between the plates is attractive. This
work goes into increasing the electrostatic energy carried by the electric field and into forcing charge
into the battery holding the plates at constant potential φ12.
Conservation of energy gives
Fδl = δW + δQφ12
or
F = ∂W
∂l
+
∂Q
∂l
φ12
From problem 1.8,
a) Charge fixed.
1) Parallel plate capacitor
W = 1
2
0A
φ122
d
, φ12 =
dQ
A 0
→ W = 1
2
0A
dQ
A0
2
d
= d
20A
Q2
∂Q
∂l
= 0, F = ∂W
∂l
=
Q2
20A
(attractive)
2) Parallel cylinder capacitor
φ12 = λ0 ln
d
a , a = a1a2
W = 1
2
Qφ12 → F = ∂W∂l
= 1
2
Q λ0
∂
∂d
ln da  =
1
2
λQ
0d
(attractive)
b) Potential fixed
1) Parallel plate capacitor
Using Gauss’s law, Q =
φ12A 0
d
, ∂∂l Q =
φ12A 0
d2
1.9
13
→ F = − 1
2
0A
φ122
d2
+
φ122 A 0
d2
= 1
2
0A
φ122
d2
= 1
2
0A
Qd
0A
2
d2
= 1
20A
Q2
2) Parallel cylinder capacitor
W = 1
2
Qφ12, and Q =
0Lφ12
ln da 
so
W = 1
2
0Lφ122
ln da 
, ∂W
∂l
= − 1
2
0L
φ122
ln2 da d
∂Q
∂l
= 0L
φ12
ln2 da d
F = − 1
2
0L
φ122
ln2 da d
+ 0L
φ122
ln2 da d
= 1
2
0L
φ122
ln2 da d
14
1.10
15
1.11
R
x
R
y
dx
dy’
dy
dz
x
y
z
E-Field
A
C
B
B’
A’
C’
dx’
16
1.11,Method1 (enlightning but a little cumbersome):
The point of interest is located at the origin of a cartesian coordinate system the xy-plane of which defines
the tangent plane to the curved surface (at the point of interest). Further, the directions of the x- and
y-axes are such that locally the curved surface can be described as the set of points with coordinates
S(x, y) = xx̂ + yŷ − ( 12Rx x2 + 12Ry y2)ẑ. It has been discussed in class that it is always possible to find
orthogonal coordinate axes on the tangent plane that allow such a parametrization of the curved surface.
For the situation depicted in the figure, it would be Rx > 0 and Ry > 0. Due to the orthogonality of x̂ and
ŷ, a surface element with side lengths dx and dy on the curved surface has the area da = dx dy.
At a location S(x, y) on the curved surface, a normal vector to the plane is ∂S∂x × ∂S∂y = xRx x̂+
y
Ry
ŷ+ ẑ. Using
this fact, it can be seen that an area element da′ = dx′ dy′ located at a height dz above da, constructed as
shown in the figure, will have the property that the “sidewalls” of a pillbox with da′ and da as upper and
lower surfaces are orthogonal to the curved surface. Since the electric field, depicted in red in the figure,
also is orthogonal to the surface, the electric flux through the sidewalls is zero. Due to Gauss’s law, the
fluxes through da and da′ must be equal and opposite. Considering that the field is (anti)parallel to both
vectors da and da′, it follows that E′da′ = Eda, with E and E′ being the field magnitudes on da and da′,
respectively.
Simple geometry apparent in the figure yields dx′ = dx(1+ dz
Rx
) and dy′ = dy(1+ dzRy ). Thus, for infinitesimally
small dx, dy and dz it is da′ = (1 + dzRx +
dz
Ry
)da, and
E′ = E
da
da′
= E(1 +
dz
Rx
+
dz
Ry
)−1 = E(1− dz
Rx
− dz
Ry
) (9)
Thus,
1
E
E′ − E
dz
=
1
E
∂E
∂n
= −
(
1
Rx
+
1
Ry
)
q.e.d. (10)
Note: for the case depicted in the figure, both principal radii are > 0, and consequently ∂E
∂n < 0, as expected.
17
1.11,Method2 : (“cleanest” proof):
The electric field is (anti)parallel to both vectors da and da′ of the lower and upper planes in the figure,
respectively. It follows from Gauss’s law thatE′da′ = Eda, with E and E′ being the field magnitudes on da
and da′.
To find da, refer to the points defined in the figure and set A = S(0, 0) = 0, B = S(dx, 0) = dxx̂− 12Rx dx2ẑ,
and C = S(0, dy) = dyŷ− 12Ry dy2ẑ. Then, in lowest order of dx and dy it is da = |(B−A)×(C−A)| = dx dy.
To find da′, we need to determine the points A′, B′, C′ that lie at a distance dz above the points A,
B, C. Thereby, “above” means that the lines A′ − A etc. are parallel to the position-dependent normal
vectors n̂ of the curved surface. The normal vectors are n̂(x, y) = ∂S∂x × ∂S∂y = xRx x̂ +
y
Ry
ŷ + ẑ. Note that
|n̂(x, y)| = 1, up to second- and higher-order corrections. It is thus found A′ = A′ + dzn̂(0, 0) = dzẑ,
B′ = (dx + dx dzRx )x̂ + (dz − dx
2
2Rx
)ẑ, and C′ = (dy + dy dzRy )ŷ + (dz −
dy2
2Ry
)ẑ. Then, in first order of dx and dy
it is found that da′ = |(B′ −A′)× (C′ −A′)| = dx dy
(
1 + dzRx +
dz
Ry
)
. It follows
E′ = E
da
da′
= E(1 +
dz
Rx
+
dz
Ry
)−1 = E(1− dz
Rx
− dz
Ry
) (11)
and
1
E
E′ − E
dz
=
1
E
∂E
∂n
= −
(
1
Rx
+
1
Ry
)
q.e.d. (12)
18
1.11,Method3 : (fast, less enlightning):
Using the coordinate system in the figure, at location S(x, y) on the curved surface the normal vector to the
plane is ∂S∂x × ∂S∂y = xRx x̂ +
y
Ry
ŷ + ẑ. Since we are only interested in the local behavior in the vicinity of the
origin, x ¿ Rx and y ¿ Ry, and the length of the normal vector is 1, with corrections of order (x/Rx)2
and (y/Ry)2. Thus, the electric field on the surface has, locally, a form E(x, y) = E
(
x
Rx
x̂ + yRy ŷ + ẑ
)
with
locally constant magnitude E. At the origin, we therefore have
∂Ex
∂x
=
E
Rx
and
∂Ey
∂y
=
E
Ry
. (13)
Noting that the field is normal to the curved surface, it also is
∂Ez
∂z
=
∂E
∂z
=
∂E
∂n
(14)
From ∇ ·E = 0 it then follows
∂Ez
∂z
=
∂E
∂n
= −
(
∂Ex
∂x
+
∂Ey
∂y
)
= −E
(
1
Rx
+
1
Ry
)
(15)
and
1
E
∂E
∂n
= −
(
1
Rx
+
1
Ry
)
q.e.d. (16)
19
1.12
20
1.14 Consider the electrostatic Green functions of Section 1.10 for Dirichlet and Neumann
boundary conditions on the surface S bounding the volume V . Apply Green’s theorem
(1.35) with integration variable ~y and φ = G(~x, ~y ), ψ = G(~x ′, ~y), with ∇2yG(~z, ~y ) =
−4πδ(~y − ~z ). Find an expression for the difference [G(~x, ~x ′) − G(~x ′, ~x)] in terms of
an integral over the boundary surface S.
Using φ and ψ as indicated in Green’s theorem, we have∫
V
(
G(~x, ~y )∇2yG(~x ′, ~y )−G(~x ′, ~y )∇2yG(~x, ~y )
)
d3y
=
∮
S
(
G(~x, ~y )
∂G(~x ′, ~y )
∂ny
−G(~x ′, ~y )∂G(~x, ~y )
∂ny
)
day
Since ∇2yG(~x, ~y ) = −4πδ(~x− ~y ), the left hand side integrates to −4π[G(~x, ~x ′)−
G(~x ′, ~x )]. Dividing both sides by −4π finally gives
G(~x, ~x ′)−G(~x ′, ~x ) = 1
4π
∮
S
(
G(~x ′, ~y )
∂G(~x, ~y )
∂ny
−G(~x, ~y )∂G(~x
′, ~y )
∂ny
)
day (2)
a) For Dirichlet boundary conditions on the potential and the associated boundary
condition on the Green function, show that GD(~x, ~x ′) must be symmetric in ~x
and ~x ′.
For the Dirichlet Green’s function, GD(~x, ~y ) = 0 for ~y on the boundary S. This
means that the right hand side of (2) vanishes. Then we automatically find
GD(~x, ~x ′) = GD(~x ′, ~x )
b) For Neumann boundary conditions, use the boundary condition (1.45) forGN (~x, ~x ′)
to show that GN (~x, ~x ′) is not symmetric in general, but that GN (~x, ~x ′)− F (~x )
is symmetric in ~x and ~x ′, where
F (~x ) =
1
S
∮
S
GN (~x, ~y ) day
We use the Neumann boundary condition
∂GN (~x, ~y )
∂ny
= −4π
S
for ~y on the boundary S. This means the right hand side of (2) becomes
RHS =
1
S
∮
S
(
GN (~x, ~y )−GN (~x ′, ~y )
)
day = F (~x )− F (~x ′)
where we used the definition of F (~x ) given in the problem. This yields
GnewN (~x, ~x
′) ≡ GN (~x, ~x ′)− F (~x ) = GN (~x ′, ~x )− F (~x ′) (3)
which demonstrates that GnewN (~x, ~x
′) is symmetric in ~x and ~x′.
1.14
21
c) Show that the addition of F (~x ) to the Green function does not affect the potential
Φ(~x ). See problem 3.26 for an example of the Neumann Green function.
What we need to do is to show that the Neumann Green’s function solution
Φ(~x ) = 〈Φ〉S +
1
4π�0
∫
V
ρ(~x ′)GN (~x, ~x ′) d3x′ +
1
4π
∮
S
∂Φ(~x ′)
∂n′
GN (~x, ~x ′) da′ (4)
is unchanged when we replaceGN byGnewN . If we let Φ
new denote the computation
using GnewN , then
Φnew(~x ) = 〈Φ〉S +
1
4π�0
∫
V
ρ(~x ′)GnewN (~x, ~x
′) d3x′ +
1
4π
∮
S
∂Φ(~x ′)
∂n′
GnewN (~x, ~x
′) da′
= 〈Φ〉S +
1
4π�0
∫
V
ρ(~x ′)GN (~x, ~x ′) d3x′ +
1
4π
∮
S
∂Φ(~x ′)
∂n′
GN (~x, ~x ′) da′
− 1
4π�0
∫
V
ρ(~x ′)F (~x ) d3x′ − 1
4π
∮
S
∂Φ(~x ′)
∂n′
F (~x ) da′
= Φ(~x )− F (~x )
4π�0
(∫
V
ρ(~x ′) d3x′ + �0
∮
S
∂Φ(~x ′)
∂n′
da′
)
= Φ(~x )− F (~x )
4π�0
(
qenc − �0
∮
S
~E(~x ′) · n̂′ da′
)
where we used the fact that n̂ · ~E = −n̂ · ~∇Φ = −∂Φ/∂n. A simple application
of Gauss’ law then demonstrates that Φnew = Φ. Hence we have shown that the
addition of F (~x ) leaves the solution unchanged. This demonstrates that we can
always make GN symmetric by appropriate modification with F .
Note that from (3) we could instead have defined the symmetric combination
GbadN (~x, ~x
′) = GN (~x, ~x ′) +F (~x ′). However this is a bad thing to do, as substitu-
tion of GbadN into (4) will generate an incorrect solution for Φ.
22
a): There are n conductors with surfaces Si, potentials Vi and charges Qi (i = 1, ..., n). Assume that there
is only one conductor with non-zero potential. Call the index of that conductor k, its potential V , and its
charge Q.
²0
∫
V
|∇Φ|2d3x = (by Green I)
−²0
∫
V
Φ∇2Φd3x + ²0
∫
∂V
Φ
∂
∂n
Φda =
0 + ²0
n∑
i=1
Vi
∫
Si
∂
∂n
Φda =
n∑
i=1
Vi
∫
Si
σda = Vk
∫
Sk
σda = V Q
Setting V=1, we have QV = CV 2 = C with capacitance C, and therefore C = ²0
∫
V
|∇Φ|2d3x q.e.d. .
b): Write Ψ = Φ+δψ, with Φ being the exact solution for the potential, and δψ being the difference between
the test function Ψ and Φ. Note that δψ = 0 on all surfaces Si.
²0
∫
V
|∇Ψ|2d3x =
²0
∫
V
|∇Φ|2d3x + 2²0
∫
V
∇Φ · ∇δψd3x + ²0
∫
V
|∇δψ|2d3x = | by a) andGreen I
C − 2²0
∫
V
(∆Φ)δψd3x
︸ ︷︷ ︸
+ 2
∫
∂V
δψ
∂
∂n
Φda
︸ ︷︷ ︸
+ ²0
∫
V
|∇δψ|2d3x
︸ ︷︷ ︸
= | since ² ≥ 0
C + 0 + 0 + ² ≥ C
(1)
Thus, for all Ψ satisfying the boundary conditions it is C ≤ ²0
∫
V
|∇Ψ|2d3x q.e.d. .
1.17
23
For a well-behaved function f(x, y), the k-th order Taylor expansion around (x0, y0) is
f(x, y) =
m+n=k∑
m=0,n=0
1
m!n!
∂(m)x ∂
(n)
y f |(x0,y0)(x− x0)m(y − y0)n (2)
a): Cross average. Since either (x− x0) or (y− y0) is zero (while the other one ±h), the only non-vanishing
terms are ones in which either n = 0 or m = 0. Using Fx = ∂∂xF , F2x =
∂2
∂x2 F etc, it is
F (h, 0) = F (0, 0) + hFx +
h2
2
F2x +
h3
6
F3x +
h4
24
F4x +
h5
120
F5x + h6C6
F (−h, 0) = F (0, 0)− hFx + h
2
2
F2x − h
3
6
F3x +
h4
24
F4x − h
5
120
F5x + h6C6
F (0, h) = F (0, 0) + hFy +
h2
2
F2y +
h3
6
F3y +
h4
24
F4y +
h5
120
F5y + h6C6
F (0,−h) = F (0, 0)− hFy + h
2
2
F2y − h
3
6
F3y +
h4
24
F4y − h
5
120
F5y + h6C6
and the cross sum evidently is
Sc = 4F (0, 0)+h2(F2x+F2y)+
h4
12
(F4x+F4y)+O(h6) = 4F (0, 0)+h2∆F +h
4
12
(F4x+F4y)+O(h6) q.e.d. (3)
b): Square sum. It is easily seen that due to cancellations in the square sum only terms in Eq. 2 contribute
with both m and n even. Thus, it is:
F (±h,±h) = F (0, 0) + h
2
2
(F2x + F2y) +
h4
24
(F4x + F4y + 6F2x2y) + C6h6 + ... , (4)
where the ... stand for terms that cancel when performing the square sum, and C6 for a 6-th order coefficient.
Thus,
Ss = 4F (0, 0) + 2h2(F2x + F2y) +
h4
6
(F4x + F4y) + h4F2x2y +O(h6) . (5)
Noting that ∆(∆F ) = F4x +2F2x2y +F4y, we see that F2x2y = 12 (∆(∆F )− F4x − F4y), and the square sum
becomes
Ss = 4F (0, 0) + 2h2∆F +
h4
2
∆(∆F ) +
(
h4
6
− h
4
2
)
(F4x + F4y) +O(h6) (6)
Ss = 4F (0, 0) + 2h2∆F +
h4
2
∆(∆F )− h
4
3
(F4x + F4y) +O(h6) q.e.d. (7)
1.22
24
Note: An improved average (=sum/4) can be defined as follows:
S̄ =
1
5
Ss +
1
20
Sc = F (0, 0) +
3
10
h2∆F +
1
40
h4∆(∆F ) +O(h6) . (8)
For functions F thatare solutions of the Laplace equation all correction terms up to and excluding the O(h6)
vanish. In a charged space with charge density ρ(x), the expressions
∆F = − ρ
²0
and ∆(∆F ) ≈ − 4ρc
²0h2
+
4ρ
²0h2
, (9)
with ρc being the cross average of the charge density, can be inserted into Eq. 8, yielding an equation that
can be resolved for F (0, 0). This procedure and a consideration of the error in the second equation of Eq. 9
lead to Equation 1.82 in the textbook.
25
We will work in cylidrical coordinate, (ρ, z,φ, with the charge q located at the point
d⃗ = dẑ, and the conducting plane is in the z = 0 plane.
Then we know from class the potential is given by
φx⃗ = 1
4π 0
q
x⃗ − d⃗
− q
x⃗ + d⃗
Ez = −
q
4π 0
∂
∂z
1
z − d2 + ρ2 1/2
− 1
z + d2 + ρ2
1/2
Ez =
q
4π 0
z − d
z − d2 + ρ2 3/2
− z + d
z + d2 + ρ2
3/2
a) σ = 0Ezz = 0
σ = 0
q
4π 0
−d
−d2 + ρ2 3/2
− +d
+d2 + ρ2
3/2
= − q
2πd2
1
12 + ρ
d
2
3
2
Plotting −1
12+r2
3
2
gives
-1
-0.8
-0.6
-0.4
-0.2
0
1 2 3 4 5r
b) Force of charge on plane
F⃗ = 1
4π 0
q−q
2d2
−ẑ = 1
4 × 4π 0
q2
d2
ẑ
c)
F
A
= w = 0
2
E2 = σ
2
20
= 1
20
− q
2πd2
1
12 + ρ
d
2
3
2
2
= 1
80
q2
π2d4 1 + ρ
2
d2
3
F = 2π q
2
80π2d4
∫
0
β ρ
1 + ρ
2
d2
3 dρ = 2π
q2
80π2d4
1
4
d2 = 1
4 × 4π 0
q2
d2
2.1
26
d)
W = ∫
d
β
Fdz = q
2
4 × 4π 0
∫
d
β dz
z2
=
q2
4 × 4π 0d
e)
W = 1
2
1
4π 0
>
i,j,i≠j
q iq j
|x⃗ i − x⃗ j |
= − q
2
2 × 4π 0d
Notice parts d) and e) are not equal in magnitude, because in d) the image moves when q moves.
f) 1 Angstrom = 10−10m, q = e = 1. 6 × 10−19C.
W = q
2
4 × 4π 0d
= e e
4 × 4π 0d
= e 1.6 × 10
−19
4 × 10−10
9 × 109 V = 3.6 eV
27
The system is described by
a) Using the method of images
φx⃗ = 1
4π 0
q
|x⃗ − y⃗|
+ q
′
|x⃗ − y⃗ ′ |
with y ′ = a2y , and q ′ = −q ay
b) σ = −0 ∂∂n φ|x=a = +0
∂
∂x φ|x=a
σ = 0 14π 0
∂
∂x
q
x2 + y2 − 2xycosγ1/2
+ q
′
x2 + y ′2 − 2xy ′ cosγ1/2
σ = −q 1
4π
a 1 − y
2
a2
y2 + a2 − 2aycosγ3/2
Note
q induced = a2 ∫ σdΩ = −q 14π a
22πa 1 − y
2
a2
∫
−1
1 dx
y2 + a2 − 2ayx3/2
, where x = cosγ
q induced = −
q
2
aa2 − y2  2
aa2 − y2 
= −q
c)
|F| = qq
′
4π 0y ′ − y2
= 1
4π 0
q2ay
a2 − y2
, the force is attractive, to the right.
d) If the conductor were fixed at a different potential, or equivalently if extra charge were put on
the conductor, then the potential would be
φx⃗ = 1
4π 0
q
|x⃗ − y⃗|
+ q
′
|x⃗ − y⃗ ′ |
+ V
and obviously the electric field in the sphere and induced charge on the inside of the sphere would
remain unchanged.
2.2
28
The system is described by
a) Given the potenial for a line charge in the problem, we write down the solution from the figure,
φT = λ4π 0
ln R
2
x⃗ − x⃗o 2
− ln R
2
x⃗ − x⃗o1 2
− ln R
2
x⃗ − x⃗o2 2
+ ln R
2
x⃗ − x⃗o3 2
Looking at the figure when y = 0, x⃗ − x⃗o 2 = x⃗ − x⃗o1 2, x⃗ − x⃗o2 2 = x⃗ − x⃗o3 2, so φT |y=0 = 0
Similarly, when x = 0, x⃗ − x⃗o 2 = x⃗ − x⃗o2 2, x⃗ − x⃗o1 2 = x⃗ − x⃗o3 2, so φT |x=0 = 0
On the surface φT = 0, so δφT = 0, however,
δφT =
∂φT
∂x t
δx t = 0 →
∂φT
∂x t
= 0 → E t = 0
b) We remember
σ = −0
∂φT
∂y
= −λπ
y0
x − x0 2 + y02
− y0
x + x0 2 + y0
2
where I’ve applied the symmetries derived in a). Let
σ/λ = −1π
y0
x − x0 2 + y02
− y0
x + x0 2 + y0
2
This is an easy function to plot for various combinations of the position of the original line charge
(x0,y0.
c) If we integrate over a strip of width Δz, we find, where we use the integral
∫
0
∞ 1
x ∓ x0 2 + y02
dx = 1
2
π ± 2arctan x0y0
y0
ΔQ = ∫
0
∞
σdxΔz → ΔQ
Δz = ∫0
∞
σdx = −2λπ tan
−1 x0
y0
and the total charge induced on the plane is −∞, as expected.
d)
Expanding
2.3
29
ln R
2
x − x0 2 + y − y0 2
− ln R
2
x − x0 2 + y + y0 2
− ln R
2
x + x0 2 + y − y0 2
+ ln R
2
x + x0 2 + y + y
to lowest non-vanishing order in x0, y0 gives
16 xy
x2 + y2 2
y0x0
so
φ → φasym = 4λπ 0
xy
x2 + y2 2
y0x0
This is the quadrupole contribution.
30
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2.4
31
a)
W = ∫
r
∞
|F|dy = q
2a
4π 0
∫
r
∞ dy
y3 1 − a
2
y2
2 =
q2a
8π 0r2 − a2 
Let us compare this to disassemble the charges
− W′ = − 1
8π 0
∑
i≠j
q iq j
|x⃗i−x⃗j |
= 1
4π 0
aq2
r
1
r 1 − a
2
r2
= q
2a
4π 0r2 − a2 
> W
The reason for this difference is that in the first expression W, the image charge is moving and
changing size, whereas in the second, whereas in the second, they don’t.
b) In this case
W = ∫
r
∞
|F|dy = q
4π 0
∫
r
∞ Qdy
y2
− qa3 ∫
r
∞ 2y2 − a2 
yy2 − a2 2
dy
Using standard integrals, this gives
W = 1
4π 0
q2a
2r2 − a2 
−
q2a
2r2
−
qQ
r
On the other hand
− W′ = 1
4π 0
aq2
r2 − a2 
−
qQ + ar q
r =
1
4π 0
aq2
r2 − a2 
−
q2a
r2
−
qQ
r
The first two terms are larger than those found in W for the same reason as found in a), whereas the
last term is the same, because Q is fixed on the sphere.
2.5
32
We are considering two conducting spheres of radii ra and rb respectively. The charges on
the spheres are Qa and Qb.
a) The process is that you start with qa1 and qb1 at the centers of the spheres, and sphere a
then is an equipotential from charge qa1 but not from qb1 and vice versa. To correct this we use
the method of images for spheres as discussed in class. This gives the iterative equations given in the
text.
b) qa1 and qb1 are determined from the two requirements
>
j=1
β
qaj = Qa and >
j=1
β
qbj = Qb
As a program equation, we use a do-loop of the form
>
j=2
n
qaj =
−raqbj − 1
dbj − 1

and similar equations for qbj, xaj, xbj, daj,dbj. The potential outside the spheres is given
by
φx⃗ = 1
4π 0 >j=1
n
qaj
x⃗ − xajk̂
+>
j=1
n
qbj
x⃗ − dbjk̂
This potential is constant on the surface of the spheres by construction.
And the force between the spheres is
F = 1
4π 0
>
j,k
qajqbk
d − xaj − xbk
2
c) Now we take the special case Qa = Qb, ra = rb = R, d = 2R. Then we find, using the iteration
equations
xaj = xbj = xj
x1 = 0, x2 = R/2, x3 = 2R/3, or xj = 
j − 1
j
R
qaj = qbj = qj
qj = q, q2 = −q/2, q3 = q/3, or qj =
−1 j+1
j
q
So, as n → β
>
j=1
β
qj = q>
j=1
β
−1 j+1
j
= q ln2 = Q → q = Q
ln2
The force between the spheres is
F = 1
4π 0
q2
R2
>
j,k
−1 j+k
jk 2 − j−1
j
− 
k−1
k
2 =
1
4π 0
q2
R2
>
j,k
−1 j+kjk
j + k2
Evaluating the sum numerically
2.6
33
F = 1
4π 0
q2
R2
0.0739 = 1
4π 0
Q2
R2
1
ln22
0.0739
Comparing this to the force between the charges located at the centers of the spheres
Fp = 14π 0
Q2
R24
Comparing the two results, we see
F = 41
ln22
0.0739Fp = 0.615Fp
On the surface of the sphere
φ = 1
4π 0
>
j=1
β
qj
R − xj
=
q
4π 0R
>
j=1
β
−1 j+1
Notice 1
1 + 1
= >
j=1
β
−1 j+1
So φ = 1
4π 0
q
2R
= 1
4π 0
Q
2 ln2R
=
Q
C
→ C
4π 0R
= 2 ln2 = 1. 386
34
The system is described by
a) The Green’s function, which vanishes on the surface is obviously
Gx⃗, x⃗ ′ = 1
|x⃗ − x⃗ ′ |
− 1
|x⃗ − x⃗I′ |
where
x⃗ ′ = x ′î + y ′ ̂ + z ′ k̂, x⃗I′ = x ′î + y ′ ̂ − z ′ k̂
b) There is no free charge distribution, so the potential everywhere is determined by the potential
on the surface. From Eq. (1.44)
φx⃗ = − 1
4π ∫S φx⃗
′ ∂Gx⃗, x⃗
′
∂n ′
da ′
Note that n̂ ′ is in the −z direction, so
∂
∂n ′
Gx⃗, x⃗ ′|z′=0 = −
∂
∂z
Gx⃗, x⃗ ′|z′=0 = −
2z
x − x ′ 2 + y − y ′2 + z2
3/2
So
φx⃗ = z
2π
V ∫
0
a
∫
0
2π ρ ′dρ ′dφ ′
x − x ′ 2 + y − y ′2 + z2
3/2
where x ′=ρ ′ cosφ ′,y ′ = ρ ′ sinφ ′.
c) If ρ = 0, or equivalently x = y = 0,
φz = z
2π
V ∫
0
a
∫
0
2π ρ ′dρ ′dφ ′
ρ ′2 + z2 3/2
= zV ∫
0
a ρdρ
ρ2 + z2 3/2
φz = zV −
z − a2 + z2 
z a2 + z2 
= V 1 − z
a2 + z2 
d)
φx⃗ = z
2π
V ∫
0
a
∫
0
2π ρ ′dρ ′dφ ′
ρ⃗ − ρ⃗ ′ 2 + z2
3/2
2.7
35
In the integration choose the x −axis parallel to ρ⃗, then ρ⃗ ⋅ ρ⃗ ′ = ρρ ′ cosφ ′
φx⃗ = z
2π
V ∫
0
a
∫
0
2π ρ ′dρ ′dφ ′
ρ2 + ρ ′2 − 2ρ⃗ ⋅ ρ⃗ ′ + z2 3/2
Let r2 = ρ2 + z2, so
φx⃗ = z
2π
V
r3
∫
0
a
∫
0
2π ρ ′dρ ′dφ ′
1 + ρ
′2−2ρ⃗⋅ρ⃗ ′
r2
3/2
We expand the denominator up to factors of O1/r4, ( and change notation
φ ′ → θ,ρ ′ → α, r2 → 1
β2

φx⃗ = z
2π
V
r3
∫
0
a
∫
0
2π αdαdθ
1 + α
2−2ρ⃗⋅α⃗
r2
3/2
where the denominator in this notation is written
1
1 + β2α2 − 2ραcosθ3/2
or, after expanding,
φx⃗ = z
2π
V
r3
∫
0
a
αdα ∫
0
2π
1 − 3
2
β2α2 + 3β2ραcosθ + 15
8
β4α4 − 15
2
β4α3ρcosθ + 15
2
β4ρ2α2 cos2θ dθ
Integrating over θ gives
φx⃗ = z
2π
V
r3
∫
0
a
α 2π + 15
4
β4α4π − 3β2α2π + 15
2
β4ρ2α2π dα
Integrating over α yields
φx⃗ = z
2π
V
r3
5
8
β4πa6 − 3
4
a4β2π + 15
8
a4β4ρ2π + πa2
or
φx⃗ = Va
2
2ρ2 + z2 3/2
1 − 3
4
a2
ρ2 + z2 
+ 5
8
a4 + 3ρ2a2
ρ2 + z2 2
36
The system is pictured below
a) Using the known potential for a line charge, the two line charges above give the potential
φr⃗ = 1
2π 0
λ ln r
′
r = V, a constant. Let us define V
′ = 4π 0V
Then the above equation can be written
r ′
r
2
= e V
′
λ or r ′2 = r2e V
′
λ
Writing r ′2 = r⃗ − R⃗
2
, the above can be written
r⃗ + ẑ R
e
V′
λ − 1
2
= R
2e
V′
λ
e
V′
λ − 12
The equation is that of a circle whose center is at −ẑ R
e
V′
λ −1
, and whose radius is a = Re
V′
2λ
e
V′
λ −1
b) The geometry of the system is shown in the fugure.
Note that
d = R + d1 + d2
with
d1 = R
e
Va′
λ −1
, d2 = R
e
−Vb
′
λ −1
and
a = Re
Va′
2λ
e
Va′
λ − 1
, b = Re
−Vb
′
2λ
e
−Vb
′
λ − 1
Forming
2.8
37
d2 − a2 − b2 = R + R
e
Va′
λ −1
+ R
e
−Vb
′
λ −1
2
− Re
Va′
2λ
e
Va′
λ − 1
2
− Re
−Vb
′
2λ
e
−Vb
′
λ − 1
2
or
d2 − a2 − b2 =
R2 e
Va′ −Vb
′
λ + 1
e
Va′
λ − 1e
−Vb
′
λ − 1
Thus we can write
d2 − a2 − b2
2ab
=
e
Va′ −Vb
′
λ + 1
2e
Va′
2λ e
−Vb
′
2λ
= e
Va′ −Vb
′
2λ + e
− Va′ −Vb
′
2λ
2
= cosh Va
′ − Vb′
2λ
or
Va − Vb
λ =
1
2π 0
cosh−1 d
2 − a2 − b2
2ab
Capacitance/unit length = C
L
= Q/L
Va − Vb
= λ
Va − Vb
= 2π 0
cosh−1 d
2−a2−b2
2ab
c) Suppose a2 << d2, and b2 << d2, and a ′ = ab , then
C
L
= 2π 0
cosh−1 d
2−a2−b2
2a ′2
= 2π 0
cosh−1
d2 1−a2+b2/d2
2a ′2
cosh−1
d21 − a2 + b2/d2 
2a ′2
= 2π 0L
C
d21 − a2 + b2/d2 
2a ′2
= e
2π0L
C
2
+ negligible terms if 2π 0L
C
>> 1
or
ln
d21 − a2 + b2/d2 
a ′2
= 2π 0L
C
or
C
L
= 2π 0
ln
d2 1−a2+b2/d2
a ′2
Let us defind α2 = a2 + b2/d2, then
C
L
= 2π 0
ln
d2 1−α2
a ′2
= 2π 0
ln d
2
a ′2
+ 2π 0
ln2 d
2
a ′2
α2 + Oα4 
The first term of this result agree with problem 1.7, and the second term gives the appropriate
correction asked for.
d) In this case, we must take the opposite sign for d2 − a2 − b2, since a2 + b2 > d2. Thus
C
L
= 2π 0
cosh−1 a
2+b2−d2
2a ′2
If we use the identiy, lnx + x2 − 1  = cosh−1x, G.&R., p. 50., then for d = 0
38
C
L
= 2π 0
ln a
2+b2
2ab
+ a2−b2
2ab
= 2π 0
ln a
b
in agreement with problem 1.6.
39
The system is pictured below
a) We have treated this probem in class. We found the charge density induced was
σ = 30E0 cosθ
We also note the radial force/unit area outward from the surface is σ2/20. Thus the force on the
right hand hemisphere is, using x = cosθ
Fz = 120
∫ σ2 ẑ ⋅ da⃗ = 120 30E0 
22πR2 ∫
0
1
x3dx = 1
20
30E0 22πR2/4 = 94
π 0E02R2
An equal force acting in the opposite direction would be required to keep the hemispheres from
sparating.
b) Now the charge density is
σ = 30E0 cosθ +
Q
4πR2
= 30E0x +
Q
4πR2
= 30E0 x +
Q
12π 0E0R2
Fz = 120
∫ σ2 ẑ ⋅ da⃗ = 120 30E0 
22πR2 ∫
0
1
x x + Q
12π 0E0R2
2
dx
Thus
Fz = 94
π 0E02R2 +
1
2
QE0 + 1
320πR2
Q2
An equal force acting in the opposite direction would be required to keep the himispheres from
separating.
2.9
40
As done in class we simulate the electric field E0 by two charges at ∞
σ = 30E0 cosθ
a) This charge distribution simulates the given system for cosθ > 0.We have treated this probem
in class. The potential is given by
φx⃗ = −E0 1 − a
3
r3
r cosθ
Using
σ = −0 ∂∂n
φ|surface
We have the charge density on the plate to be
σplate = −0 ∂∂z
φ|z=0 = 0E0 1 − a
3
ρ3
For purposes of plotting, consider
σplate
0E0
= 1 − 1
x3
0
0.2
0.4
0.6
0.8
1
2 4 6 8 10x
σboss = 30E0 cosθ =
For plotting, we use σboss30E0 = cosθ
2.10
41
0
0.2
0.4
0.6
0.8
1
0.2 0.4 0.6 0.8 1 1.2 1.4
b)
q = 30E02πa2 ∫
0
1
xdx = 3π 0E0a2
c) Now we have
φr⃗ = 1
4π 0
q
r⃗ − d⃗
+ q
′
r⃗ − d⃗ ′
− q
r⃗ + d⃗
− q
′
r⃗ + d⃗ ′
where q ′ = −q a
d
, d ′ = a2
d
.
σ = −0 ∂∂r
φ|r=a =
−q
4π
d2 − a2 
a a⃗ − d⃗
3 −
d2 − a2 
a a⃗ + d⃗
3
q ind = 2πa2
−q
4π ∫0
1 d2 − a2 
a a⃗ − d⃗
3 −
d2 − a2 
a a⃗ + d⃗
3 dx
q ind =
−qa2d2 − a2 
2a
1
da
1
d − a
− 1
a2 + d2
+ 1
d + a −
1
a2 + d2
q ind = −12 q
d2 − a2
d
2d
d2 − a2
− 2
a2 + d2
= −q 1 −
d2 − a2 
d a2 + d2
42
The system is pictured in the following figure:
a) The potential for a line charge is (see problem 2.3)
φr⃗ = λ
2π 0
ln r∞r
Thus for this system
φr⃗ = τ
2π 0
ln r∞
r⃗ − R⃗
+ τ
′
2π 0
ln r∞
r⃗ − R⃗ ′
To determine τ ′ and R ′, we need two conditions:
I) As r → ∞, we want φ → 0, so τ ′ = −τ.
II) φr⃗ = b�  = φr⃗ = −b�  or
ln b − R
′
R − b = ln
b + R ′
b + R
or
b − R ′
R − b =
b + R ′
b + R
This is an equation for R ′ with the solution
R ′ = b
2
R
The same condition we found for a sphere.
b)
2.11
43
φr⃗ = τ
4π 0
ln
r2 + b4
R2
− 2r b2R cosφ
r2 + R2 − 2rRcosφ
as r → ∞
φr⃗ = τ
4π 0
ln
1 − 2b2 cosφ/rR
1 − 2Rcosφ/r
= τ
4π 0
ln 1 − 2
rR 
b2 − R2 cosφ
Using
ln1 + x = x − 1
2
x2 + 1
3
x3 + Ox4 
φr⃗ = − τ
2π 0
1
rR 
b2 − R2 cosφ
c)
σ = −0 ∂∂r
φ|r=b = − τ4π
∂
∂r
ln
r2 + b4
R2
− 2r b2R cosφ
r2 + R2 − 2rRcosφ
r=b
σ = τ
2πb
1 − y2
y2 + 1 − 2ycosφ
where y = R/b. Plotting σ/ τ
2πb =
1−y2
y2+1−2ycosφ
, for y = 2, 4, gives
gy = 1 − y
2
y2 + 1 − 2ycosφ
g2,g4 = − 35−4cosφ , −
15
17−8cosφ
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5 1 1.5 2 2.5 3
d) If the line charges are a distance d apart, then the electric field at τ from τ ′ is, using Gauss’s law
E = τ
′
2π 0d
The force on τ is τLE, ie,
F = ττ
′L
2π 0d
= − τ
2L
2π 0d
, and the force is attractive.
44
2.12 Starting with the series solution (2.71) for the two-dimensional potential problem with
the potential specified on the surface of a cylinder of radius b, evaluate the coefficients
formally, substitute them into the series, and sum it to obtain the potential inside the
cylinder in the form of Poisson’s integral:
Φ(ρ, φ) =
1
2π
∫ 2π
0
Φ(b, φ′)
b2 − ρ2
b2 + ρ2 − 2bρ cos(φ′ − φ)
dφ′
What modification is necessary if the potential is desired in the regionof space
bounded by the cylinder and infinity?
The series solution (2.71) is given as
Φ(ρ, φ) = a0 + b0 ln ρ +
∞∑
n=1
[anρn sin(nφ + αn) + bnρ−n sin(nφ + βn)]
Since we want an inside solution, we take bn = 0 so the potential is well behaved
at ρ = 0. With some rewriting of the series, we can then turn it into the equivalent
form
Φ(ρ, φ) = 12c0 +
∞∑
n=1
(ρ
b
)n
[cn cos(nφ) + dn sin(nφ)] (1)
Breaking up the sin(nφ + αn) terms into sines and cosines is convenient because
we now end up with a standard Fourier series. The Fourier coefficients are
cn =
1
π
∫ 2π
0
Φ(b, φ′) cos(nφ′) dφ′
dn =
1
π
∫ 2π
0
Φ(b, φ′) sin(nφ′) dφ′
Substituting this back into (1) yields
Φ(b, φ) =
1
2π
∫ 2π
0
Φ(b, φ′)
×
[
1 + 2
∞∑
n=1
(ρ
b
)n (
cos(nφ) cos(nφ′) + sin(nφ) sin(nφ′)
)]
dφ′
=
1
2π
∫ 2π
0
Φ(b, φ′)
[
1 + 2
∞∑
n=1
(ρ
b
)n
cos n(φ′ − φ)
]
dφ′
(2)
2.12
45
The series in the square brackets can be summed as a geometric series
1 + 2
∞∑
n=1
(ρ
b
)n
cos n(φ′ − φ)
= <
[
1 + 2
∞∑
n=1
(ρ
b
)n
ein(φ
′−φ)
]
= <
[
−1 + 2
∞∑
n=0
(ρ
b
ei(φ
′−φ)
)n]
= <
[
−1 + 2
1− ρb ei(φ
′−φ)
]
= <
1 + ρb e
i(φ′−φ)
1− ρb ei(φ
′−φ)
= <
(
1 + ρb e
i(φ′−φ)
) (
1− ρb e
−i(φ′−φ)
)
(
1− ρb e
i(φ′−φ)
) (
1− ρb e
−i(φ′−φ)
)
= <1− (ρ/b)
2 + 2i(ρ/b) sin(φ′ − φ)
1 + (ρ/b)2 − 2(ρ/b) cos(φ′ − φ)
=
b2 − ρ2
b2 + ρ2 − 2ρb cos(φ′ − φ)
Inserting this into (2) finally yields the result
Φ(ρ, φ) =
1
2π
∫ 2π
0
Φ(b, φ′)
b2 − ρ2
b2 + ρ2 − 2ρb cos(φ′ − φ)
dφ′ (3)
Note that this can be obtained more directly from the Cauchy integral formula
of complex analysis (Jackson problem 2.21).
For the exterior solution, we may simply take ρ/b→ b/ρ in (1). This corresponds
to making the replacement ρ ↔ b in the fraction in the integrand of (3). Since
this only affects the numerator, the simple result is that we change the sign of
Φ(ρ, φ).
46
The system is pictured in the following figure:
a) Notice from the figure, Φρ,−φ = Φρ,φ; thus from Eq. (2.71) in the text,
Φρ,φ = a0 +∑
n=1
∞
anρn cosnφ
∫
−π/2
3π/2
Φb,φ = 2πa0 = πV1 + πV2 → a0 =
V1 + V2
2
Using
∫
−π/2
3π/2
cosmφ cosnφdφ = δnmπ
Applying this to Φ, only odd terms m contribute in the sum and
am =
2V1 − V2 
πmbm
−1
m−1
2
Thus
Φρ,φ = V1 + V2
2
+
2V1 − V2 
π Im ∑
m odd
imρme imφ
mbm
Using
2 ∑
m odd
xm
m = ln
1 + x
1 − x
and
Im lnA + iB = tan−1B/A
we get
Φρ,φ = V1 + V2
2
+
V1 − V2 
π tan
−1 2
ρ
b
cosφ
1 − ρ
2
b2
as desired.
b)
σ = −0 ∂∂ρ
Φρ,φ|ρ=b
2.13
47
σ = −0
V1 − V2 
π
∂
∂ρ
tan−1
2 ρ
b
cosφ
1 − ρ
2
b2 ρ=b
σ = −20
V1 − V2
π bcosφ
b2 + ρ2
b4 − 2b2ρ2 + ρ4 + 4ρ2b2 cos2φ
|ρ=b = −0
V1 − V2
πbcosφ
48
2.14 A variant of the preceeding two-dimensional problem is a long hollow conducting
cylinder of radius b that is divided into equal quarters, alternate segments being held
at potential +V and −V .
a) Solve by means of the series solution (2.71) and show that the potential inside
the cylinder is
Φ(ρ, φ) =
4V
π
∞∑
n=0
(ρ
b
)4n+2 sin[(4n+ 2)φ]
2n+ 1
The general series solution for the two-dimensional problem in polar coordinates
is given by (2.71)
Φ(ρ, φ) = a0 + b0 log ρ+
∞∑
n=1
anρ
n sin(nφ+ αn) + bnρ−n sin(nφ+ βn)
Since we are interested in the interior solution, we demand that the potential
remains finite at ρ = 0. This indicates that the bn coefficients must all vanish.
We are thus left with
Φ(ρ, φ) = a0 +
∞∑
n=1
anρ
n sin(nφ+ αn)
which we may choose to rewrite as
Φ(ρ, φ) =
A0
2
+
∞∑
k=1
[Akρk cos(kφ) +Bkρk sin(kφ)] (1)
This form of the series is supposed to be reminiscent of a Fourier series.
The boundary condition for this problem is that the potential at ρ = b is either
+V or −V , depending on which quadrant we are in
VV−
V −V
b
49
This can be plotted as a function of φ
Φ
−π π
−V
V
ϕ
), ϕb(
It should be obvious that Φ(b, φ) is an odd function of φ. As a result, we im-
mediately deduce that the Ak Fourier coefficients in (1) must vanish, leaving us
with
Φ(ρ, φ) =
∞∑
k=1
Bkρ
k sin(kφ) (2)
On the interior surface of the conducting cylinder, this reads
Φ(b, φ) =
∞∑
k=1
Bkb
k sin(kφ)
where Φ(b, φ) is given by the figure above. In particular, we see that the quantities
Bkb
k are explicitly the Fourier expansion coefficients of a square wave with period
π (which is half the usual 2π period). As a result, we may simply look up the
standard Fourier expansion of the square wave and map it to this present problem.
Alternatively, it is straightforward to calculate the coefficients directly
Bkb
k =
1
π
∫ π
−π
Φ(b, φ) sin(kφ)dφ
=
V
π
(∫ −π/2
−π
−
∫ 0
−π/2
+
∫ π/2
0
−
∫ π
π/2
)
sin(kφ)dφ
=
V
kπ
(
− cos(kφ)
∣∣∣−π/2
−π
+ cos(kφ)
∣∣∣0
−π/2
− cos(kφ)
∣∣∣π/2
0
− cos(kφ)
∣∣∣π
π/2
)
=
2V
kπ
(
1− 2 cos
(kπ
2
)
+ cos(kπ)
)
=
8V
kπ
k = 2, 6, 10, 14, . . . (ie k = 4n+ 2)
Substituting Bk = 8V/kπbk into (2) and usiing k = 4n+ 2 then gives
Φ(ρ, φ) =
4V
π
∞∑
n=0
(ρ
b
)4n+2 sin[(4n+ 2)φ]
2n+ 1
(3)
50
b) Sum the series and show that
Φ(ρ, φ) =
2V
π
tan−1
(
2ρ2b2 sin 2φ
b4 − ρ4
)
This series is easy to sum if we work with complex variables. Since sin θ is the
imaginary part of eiθ, we write (3) as
Φ(ρ, φ) =
4V
π
=
∞∑
n=0
(ρ/b)4n+2e(4n+2)iφ
2n+ 1
=
4V
π
=
∞∑
n=0
z2n+1
2n+ 1
=
4V
π
=
(
z + 13z
3 + 15z
5 + · · ·
) (4)
where
z ≡ ρ
2
b2
e2iφ (5)
Now recall that the Taylor series expansion for log(1 + z) is given by
log(1 + z) =
∞∑
k=1
(−1)k+1
k
zk = z − 12z
2 + 13z
3 − 14z
4 + 15z
5 − · · ·
We may eliminate the even powers of z by taking the difference between log(1+z)
and log(1− z). This allows us to derive the series expression
z + 13z
3 + 15z
5 + · · · = 1
2
log
1 + z
1− z
Substituting this into (4) gives
Φ(ρ, φ) =
2V
π
= log 1 + z
1− z
=
2V
π
arg
1 + z
1− z
Since arg(x+ iy) = tan−1(y/x), a bit of algebra gives
Φ(ρ, φ) =
2V
π
tan−1
(
2=z
1− |z|2
)
Using the expression for z given in (5), we finally obtain
Φ(ρ, φ) =
2V
π
tan−1
(
2(ρ2/b2) sin 2φ
1− (ρ2/b2)2
)
=
2V
π
tan−1
(
2ρ2b2 sin 2φ
b4 − ρ4
)
(6)
51
c) Sketch the field lines and equipotentials.
The equipotentials correspond to φ(ρ, φ) = Φ0. To see what this looks like, we
may invert (6) to solve for b as a function of φ at fixed Φ0. The result is
2ρ2b2 sin 2φ
b4 − ρ4
= tan
(
πΦ0
2V
)
⇒ (ρ/b)2 = − sin 2φ
tan(πΦ0/2V )
+
√
1 +
sin2 2φ
tan2(πΦ0/2V )
A plot of the equipotentials is given by
VV
−V
−
V
where we have also shown the electric field lines (the curves with arrows).
52
2.15 a) Show that the Green function G(x, y;x′, y′) appropriate for Dirichlet boundary
conditions for a square two-dimensional region, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, has an
expansion
G(x, y;x′, y′) = 2
∞∑
n=1
gn(y, y′) sin(nπx) sin(nπx′)
where gn(y, y′) satisfies(
∂2
∂y′2
− n2π2
)
gn(y, y′) = −4πδ(y′ − y) and gn(y, 0) = gn(y, 1) = 0
We start by recalling the the Green’s function is defined by
(∂2x′ + ∂
2
y′)G(x, y;x
′, y′) = −4πδ(x′ − x)δ(y′ − y) (7)
Although this is symmetric in x′ and y′, the problem suggests that we begin by
expanding in x′ (and also x). This of course breaks the symmetry in the expanded
form of the Green’s function by treating x′ somewhat differently. Nevertheless
G(x, y;x′, y′) is unique for the given boundary conditions; it just may admit
53
different expansions, and we are free to choose whatever expansion is the most
convenient.
Given the boundary condition that G vanishes for x′ = 0 and x′ = 1, this suggests
an expansion in a Fourier sine series
G(x, y;x′, y′) =
∞∑
n=1
fn(x, y; y′) sin(nπx′)
Substituting this into (7) then gives
∞∑
n=1
(∂2y′ − n2π2)fn(x, y; y′) sin(nπx′) = −4πδ(x′ − x)δ(y′ − y) (8)
However this is not particularly useful (yet), since the δ(x′−x) on the right hand
side does not match with the Fourier sine series on the left. We can get around
this by invoking the completeness relation for the sine series
∞∑
n=1
sin(nπx) sin(nπx′) = 12δ(x− x
′)
By replacing the delta function in (8) by this sum, we end up with
∞∑
n=1
(∂2y′ − n2π2)fn(x, y; y′) sin(nπx′) = −8πδ(y′ − y)
∞∑
n=1
sin(nπx) sin(nπx′)(9)
Matching left and right sides of the Fourier sine series indicates that the x be-
havior of fn(x, y; y′) must be given by sin(nπx). Putting in a factor of two for
convenience
fn(x, y; y′) = 2gn(y, y′) sin(nπx)
finally motivates the expansion
G(x, y;x′, y′) = 2
∞∑
n=1
gn(y, y′) sin(nπx) sin(nπx′)
When this is inserted into (9), we match the x and x′ behavior perfectly, and we
are left with an equation in y′
(∂2y′ − n2π2)gn(y, y′) = −4πδ(y′ − y) (10)
The boundary conditions are that G vanishes at y′ = 0 and y′ = 1. Hence we
must also demand gn(y, 0) = gn(y, 1) = 0.
b) Taking for gn(y, y′) appropriate linear combinations of sinh(nπy′) and cosh(nπy′)
in the two regions, y′ < y and y′ > y, in accord with the boundary conditions
54
and the discontinuity in slope required by the source delta function, show that
the explicit form of G is
G(x, y;x′, y′) = 8
∞∑
n=1
1
n sinh(nπ)
sin(nπx) sin(nπx′) sinh(nπy<) sinh[nπ(1− y>)]
where y<(y>) is the smaller (larger) of y and y′.
To find the Green’s function for (10), we begin with the solution to the homoge-
neous equation (∂2y′ − n2π2)gn(y, y′) = 0. This clearly has exponential solutions
e±nπy
′
, or equivalently sinh(nπy′) and cosh(nπy′). As a result, we can write the
Green’s function as
gn(y, y′) =
{
g< ≡ a< sinh(nπy′) + b< cosh(nπy′) y′ < y
g> ≡ a> sinh(nπy′) + b> cosh(nπy′) y′ > y
(11)
We wish to solve for the four constants a<, b<, a>, b> given the boundary condi-
tions gn(y, 0) = 0, gn(y, 1) = 0 and the continuity and jump conditions
g> = g< ∂y′g> = ∂y′g< − 4π when y′ = y
We start with the boundary conditions. For g< to vanish at y′ = 0 we must take
the sinh solution, while for g> to vanish at y′ = 1 we end up with a> sinh(nπ) +
b> cosh(nπ) = 0 or b> = −a> tanh(nπ). Thus
gn(y, y′) =
{
a< sinh(nπy′) y′ < y
a>[sinh(nπy′)− tanh(nπ) cosh(nπy′)] y′ > y
(12)
The continuity and jump conditions yield the system of equations(
sinh(nπy) − sinh(nπy) + tanh(nπ) cosh(nπy)
cosh(nπy) − cosh(nπy) + tanh(nπ) sinh(nπy)
)(
a<
a>
)
=
(
0
4/n
)
which is solved by(
a<
a>
)
= − 4
n tanh(nπ)
(
sinh(nπy)− tanh(nπ) cosh(nπy)
sinh(nπy)
)
= − 4
n sinh(nπ)
(
cosh(nπ) sinh(nπy)− sinh(nπ) cosh(nπy)
cosh(nπ) sinh(nπy)
)
Inserting this into (12) gives
gn(y, y′) =
4
n sinh(nπ)
×
{
sinh(nπy′)[sinh(nπ) cosh(nπy)− cosh(nπ) sinh(nπy)] y′ < y
sinh(nπy)[sinh(nπ) cosh(nπy′)− cosh(nπ) sinh(nπy′)] y′ > y
55
This is simplified by noting
sinh[nπ(1− y)] = sinh(nπ) cosh(nπy)− cosh(nπ) sinh(nπy)
and by using the definition y< = min(y, y′) and y> = max(y, y′). The result is
gn(y, y′) =
4
n sinh(nπ)
sinh(nπy<) sinh[nπ(1− y>)]
which yields
G(x, y;x′, y′) =
∑
n
8
n sinh(nπ)
sin(nπx) sin(nπx′) sinh(nπy<) sinh[nπ(1− y>)]
Alternatively, instead of using (11), note that we can automatically solve the
boundary conditions gn(y, 0) = gn(y, 1) = 0 by writing
gn(y, y′) =
{
g< ≡ a< sinh(nπy′) y′ < y
g> ≡ a> sinh[nπ(1− y′)] y′ > y
Solving the continuity and jump conditions then gives directly
a< =
4
n
sinh[nπ(1− y)]
sinh(nπ)
, a> =
4
n
sinh(nπy)
sinh(nπ)
so that
gn(y, y′) =
4
n sinh(nπ)
{
sinh[nπ(1− y)] sinh(nπy′) y′ < y
sinh(nπy) sinh[nπ(1− y′)] y′ > y
which is the same result as above. Finally, we note that the one-dimensional
Green’s function gn(y, y′) can also be obtained through Sturm-Liouville theory
as
gn(y, y′) = −
1
A
u(y<)v(y>)
where u(y′) and v(y′) are solutions to the homogeneous equation satisfying bound-
ary conditions at y′ = 0 and y′ = 1, respectively. Here A is a constant given by
W (u, v) = A/p where W is the Wronskian, and the self-adjoint differential oper-
ator is
L =
d
dy′
p(y′)
d
dy′
+ q(y′)
56
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57
2.22 a) For the example of oppositely charged conducting hemispherical shells separated
by a tiny gap, as shown in Figure 2.8, show that the interior potential (r < a) on
the z axis is
Φin(z) = V
a
z
[
1− (a
2 − z2)
a
√
a2 + z2
]
58
Find the first few terms of the expansion in powers of z and show that they agree
with (2.27) with the appropriate substitutions.
As we have seen, the Green’s function for the interior conducting sphere problem is
equivalent to that for the exterior problem. The only difference we need to account
for is that, for the interior problem, the outward pointing normal indeed points
away from the center of the sphere. This indicates that the interior potential may
be expressed as
Φ(r.Ω) =
1
4π
∫
Φ(a,Ω′)
a(a2 − r2)
(r2 + a2 − 2ar cos γ)3/2
dΩ′
where
cos γ = cos θ cos θ′ + sin θ sin θ′ cos(φ− φ′)
In fact, introducing the absolute value |a2−r2|, it is easy to see that the expression
Φ(r,Ω) =
a|a2 − r2|
4π
∫
Φ(a,Ω′)
(r2 + a2 − 2ar cos γ)3/2
dΩ′
is valid for both the interior and the exterior problem.
For the oppositely charged hemisphere problem, this integral takes the form
Φ(r,Ω) =
V a|a2 − r2|
4π
∫ 2π
0
dφ′
∫ 1
0
d(cos θ′)[
(r2 + a2 − 2ar cos γ)−3/2 − (r2 + a2 + 2ar cos γ)−3/2
]
This simplifies on the z axis, where θ = 0 implies cos γ = cos θ′. We find
Φ(z) =
V a|a2 − z2|
4π
∫ 2π
0
dφ′
∫ 1
0
d(cos θ′)[
(z2 + a2 − 2az cos θ′)−3/2 − (z2 + a2 + 2az cos θ′)−3/2
]
=
V |a2 − z2|
2z
[
(z2 + a2 − 2az cos θ′)−1/2 − (z2 + a2 + 2az cos θ′)−1/2
]1
0
=
V |a2 − z2|
2z
(
1
|z − a|
+
1
|z + a|
− 2√
z2 + a2
)
=
V
z
(
max(a, z)− |a
2 − z2|√
z2 + a2
)
(13)
For the interior, z < a, this may be rewritten as
Φin(z) = V
a
z
(
1− a
2 − z2
a
√
a2 + z2
)
(14)
59
while for the exterior, z > a, this becomes
Φout(z) = V
(
1− z
2 − a2
z
√
a2 + z2
)
The interior solution, (14), may be expanded for z ≈ 0. The result is
Φin(z) =
3V
2
(z
a
)[
1− 7
12
(z
a
)2
+
11
24
(z
a
)4
− · · ·
]
(15)
This may be compared with the exterior solution (2.27)
Φout(r, θ) =
3V
2
(a
r
)2 [
P1(cos θ)−
7
12
(a
r
)2
P3(cos θ) + · · ·
]
⇒ Φout(z) =
3V
2
(a
z
)2 [
1− 7
12
(a
z
)2
+ · · ·
]
This demonstrates that the expansion coefficients agree, and that in fact the
interior and exterior expressions are identical up to the substitution
in:
( r
a
)l
↔ out:
(a
r
)l+1
b) From the result of part a and (2.22), show that the radial electric field on the
positive z axis is
Er(z) =
V a2
(z2 + a2)3/2
(
3 +
a2
z2
)
for z > a, and
Er(z) = −
V
a
[
3 + (a/z)2
(1 + (z/a)2)3/2
− a
2
z2
]
for |z| < a. Show that the second form is well behaved at the origin, with the
value, Er(0) = −3V/2a. Show that at z = a (north pole inside) it has the value
−(
√
2− 1)V/a. Show that the radial field at the north pole outside has the value√
2V/a.
The radial electric field on the positive z axis is given by
Er(z) = −
∂
∂z
Φ(z)
Rewriting the potential Φ(z) in (13) as
Φ(z) = V
(
max(a/z, 1)− |a
2 − z2|
z
√
z2 + a2
)
60
we find
Er(z) =

−V
(
− a
z2
+
a2(a2 +3z2)
z2(z2 + a2)3/2
)
z < a
−V
(
− a
2(a2 + 3z2)
z2(z2 + a2)3/2
)
z > a
This may be simplified to read
Er(z) =

−V
a
(
3 + (a/z)2
(1 + (z/a)2)3/2
− a
2
z2
)
z < a
V
a
3 + (a/z)2
(1 + (z/a)2)3/2
z > a
(16)
which is the desired result. As z → 0, we may Taylor expand the interior solution
to obtain
Er(z) = −
3V
2a
[
1− 7
4
(z
a
)2
+
55
24
(z
a
)4
− · · ·
]
Hence Er(0) = −3V/2a. Note that this result could have been obtained directly
by differentiating (15). Finally, the value of the radial electric field at z = a−
(immediately inside) and z = a+ (immediately outside) may be obtained from
(16)
Er(a±) =

−V
a
(
√
2− 1) z = a−
V
a
√
2 z = a+
c) Make a sketch of the electric field lines both inside and outside the conducting
hemispheres, with directions indicated. Make a plot of the radial electric field
along the z axis from z = −2a to z = +2a.
A rough sketch of the electric field lines is as follows
61
Note that the field lines are not necessarily continuous from the inside to the
outside of the hemispheres. The z component of the electric field along the z axis
is given by (16)
-2 -1 1 2
-1.5
-1
-0.5
0.5
1
z /a
/( V /a)zE
Note that Ez is positive (pointed upwards) outside the sphere, and negative
(pointed downwards) inside the sphere. By symmetry, Ez is the only non-
vanishing component of the electric field along the axis. The radial or r com-
ponent of the electric field, Er, is the same as Ez on the +z axis, but has the
opposite sign on the −z axis
-2 -1 1 2
-1.5
-1
-0.5
0.5
1
1.5
z /a
/( V /a)rE
62
2.23 A hollow cube has conducting walls defined by six planes x = 0, y = 0, z = 0, and
x = a, y = a, z = a. The walls z = 0 and z = a are held at a constant potential V .
The other four sides are at zero potential.
a) Find the potential Φ(x, y, z) at any point inside the cube.
The potential may be obtained by superposition
Φ = Φtop + Φbottom
where Φtop (Φbottom) is the solution for a hollow cube with the top (bottom) held
at constant potential V and all other sides at zero potential. As we have seen,
the series solution for Φtop is given by
Φtop =
∑
n,m
An,m sin
(nπx
a
)
sin
(mπx
a
)
sinh
(√n2 + n2 πz
a
)
where
An,m =
4
a2 sinh(
√
n2 +m2 π)
∫ a
0
dx
∫ a
0
dy V sin
(nπx
a
)
sin
(mπx
a
)
63
Noting that∫ a
0
sin
(nπx
a
)
dx = − a
nπ
cos
(nπx
a
)∣∣∣a
0
=
a
nπ
(1− (−1)n) = 2a
nπ
for n odd
we have
An,m =
16V
nmπ2 sinh(
√
n2 +m2 π)
n,m odd
and hence
Φtop =
16V
π2
∑
n,m odd
1
nm sinh(
√
n2 +m2 π)
× sin
(nπx
a
)
sin
(mπx
a
)
sinh
(√n2 + n2 πz
a
)
To obtain Φbottom, it is sufficient to realize that symmetry allows us to take
z → a− z. More precisely
Φbottom(x, y, z) = Φtop(x, y, a− z)
As a result
Φ = Φtop + Φbottom
=
16V
π2
∑
n,m odd
1
nm sinh(
√
n2 +m2 π)
sin
(nπx
a
)
sin
(mπx
a
)
×
[
sinh
(√n2 + n2 πz
a
)
+ sinh
(√n2 +m2 π(a− z)
a
)]
Note that this may be simplified using
sinh ζ + sinh(α− ζ) = 2 sinh(α/2) cosh(ζ − α/2)
to read
Φ =
16V
π2
∑
n,m odd
1
nm cosh(
√
n2 +m2 π/2)
× sin
(nπx
a
)
sin
(mπx
a
)
cosh
(√n2 +m2 π(z − a/2)
a
)
(17)
b) Evaluate the potential at the center of the cube numerically, accurate to three
significant figures. How many terms in the series is it necessary to keep in order
to attain this accuracy? Compare your numerical result with the average value
of the potential on the walls. See Problem 2.28.
64
At the center of the cube, (x, y, z) = (a/2, a/2, a/2), the potential from (17) reads
Φ(center) =
16V
π2
∑
n,m odd
sin(nπ/2) sin(mπ/2)
nm cosh(
√
n2 +m2 π/2)
=
16V
π2
∞∑
i,j=0
(−1)i+j
(2i+ 1)(2j + 1) cosh(
√
(2i+ 1)2 + (2j + 1)2 π/2)
Numerically, the first few terms in this series are given by
n m Φn,m/V running total
1 1 .347546 .347546
1 3 −.007524
3 1 −.007524 .332498
3 3 .000460 .332958
1 5 .000215
5 1 .000215 .333389
3 5 −.000023
5 3 −.000023 .333343
This table indicates that we need to keep at least the first four terms to achieve
accuracy to three significant figures. To this level of accuracy, we have
Φ(center) ≈ .333V
If we went to higher orders, it appears that the potential at the center is precisely
Φ(center) = 13V
which is the average value of the potential on the walls. In fact, we can prove (as
in Problem 2.28) that the potential at the center of a regular polyhedron is equal
to the average of the potential on the walls. Hence this value of V/3 is indeed
exact.
c) Find the surface-charge density on the surface z = a.
For the surface-charge density on the inside top surface (z = a), we use
σ = −�0
∂Φ
∂n
∣∣∣∣
S
= �0
∂Φ
∂z
∣∣∣∣
z=a
where the normal pointing away from the top conductor is n̂ = −ẑ. This is what
accounts for the sign flip in the above. Substituting in (17) gives
σ =
16�0V
πa
∑
n,m odd
√
n2 +m2
nm
tanh(
√
n2 +m2 π/2) sin
(nπx
a
)
sin
(mπy
a
)
65
The functions
{
Um(φ) = Am sin
(
mπφ
β
)}
with integer m and arbitrary non-zero constants Am form a com-
plete orthogonal set on the interval 0 ≤ φ ≤ β with Dirichlet BC. Note the analogy of these functions with
the complete set of eigenfunctions of a quantum particle in an infinitely deep square well.
Consider the expansion of a function f(φ) satisfying Dirichlet BC, f(φ) =
∑∞
m=1 amUm(φ) with coefficients
am, multiply both sides with Um′(φ) and integrate over φ:
∫ β
0
f(φ)Am′ sin
(
m′πφ
β
)
dφ =
∞∑
m=1
amAmAm′
∫ β
0
f(φ) sin
(
mπφ
β
)
sin
(
m′πφ
β
)
dφ = am′A2m′
β
2
am =
2
βAm
∫ β
0
f(φ′) sin
(
mπφ′
β
)
dφ′ (15)
where, for later convenience, we have switched the primes in the second line. Inserting this expression into
the expansion of f(φ),
f(φ) =
∞∑
m=1
{
2
βAm
∫ β
0
f(φ′) sin
(
mπφ′
β
)
dφ′
}
Am sin
(
mπφ
β
)
=
∫ β
0
{ ∞∑
m=1
2
β
sin
(
mπφ′
β
)
sin
(
mπφ
β
)}
f(φ′) dφ′ ∀φ
Thus,
∞∑
m=1
2
β
sin
(
mπφ′
β
)
sin
(
mπφ
β
)
= δ(φ− φ′) , q.e.d. (16)
2.24
66
Additional information: You may read Problem 1.21, 1.24, 2.15, 2.16, and use any relevant information given
in these problems. In the comparison part of the problem, it is only required to compare the result of the
finite-element method with the exact result.
Y1 Y2
Y0 Y1
Y1
Y1
Y2 Y2
Y2
h=0.25
Figure 3: Finite element method for 2D square problem.
For symmetry, it is sufficient to write down linear equations for the three encircled grid points of the figure,
∑
k Ψk
∫
V
∇φi · ∇φk dxdy
︸ ︷︷ ︸
= ρi²0
∫
V
φidxdy
︸ ︷︷ ︸
i = 0, 1, 2
∑
k Ψk Aki =
ρi
²0
h2 i = 0, 1, 2
(37)
The first equation is Eq. 2.80 in the textbook, except that for simplicity we use only one index k instead
of k and l, and we use i instead of i and j. The Aki are 8/3, -1/3 or 0 according to the rules explained in
the textbook (see, for instance, Problem 2.29). The sum over k includes, in principle, the boundary nodes.
However, since the boundary is on zero potential, in the present problem the sums do not explicitly show
boundary points. Also, when writing down the sums we employ the fact that due to symmetry the potentials
on many nodes are equal (see figure). Further, ρi is constant and equals ρ. We find
8
3Ψ0 − 43Ψ1 − 43Ψ2 = h2 ρ²0− 13Ψ0 + 63Ψ1 − 23Ψ2 = h2 ρ²0− 13Ψ0 − 23Ψ1 + 83Ψ2 = h2 ρ²0
⇔
8Ψ0 − 4Ψ1 − 4Ψ2 = a
−Ψ0 + 6Ψ1 − 2Ψ2 = a
−Ψ0 − 2Ψ1 + 8Ψ2 = a
(38)
2.30
67
where a = 3h
2ρ
²0
. The solution is Ψ0 = 2970a, Ψ1 =
9
28a and Ψ2 =
9
35a. For a charge density of unity and
h = 0.25, we find the following numerical values listed under F.E.M.:
× F.E.M. Exact
4π²0Ψ0 0.9761 0.9258
4π²0Ψ1 0.7573 0.7205
4π²0Ψ2 0.6059 0.5691
The exact values are taken from Problem 1.24c). Considering the roughness of the grid, the accuracy achieved
by the finite element method (F.E.M.) is good. In contrast to the iteration methods, which require many
iterations to converge to a final result, the F.E.M. yields its final result after only a single calculation.
In larger problems, the complexity of the F.E.M. calculation rapidly increases with the number of grid points,
while iteration methods remain very simple.
68
The system is picturedin the following figure:
∫
0
1
Plxdx
The problem is symmetric around the z axis so
φr,θ = ∑
l
A lr l + B lr−l−1 Plcosθ
The A l and B l are determined by the conditions
1)
∫
−1
1
φa,xPlxdx = 22l + 1 A la
l + B la−l−1 
2)
∫
−1
1
φb,xPlxdx = 22l + 1 A lb
l + B lb−l−1 
Solving these two equations gives
A l = 2l + 1
2a2l+1 − b2l+1 
a l+1 ∫
−1
1
φa,xPlxdx − b l+1 ∫
−1
1
φb,xPlxdx
B l = a l+1 2l + 12 ∫−1
1
φa,xPlxdx − A la2l+1
Using
∫
−1
1
φa,xPlxdx = V ∫
0
1
Plxdx
∫
−1
1
φb,xPlxdx = V ∫
−1
0
Plxdx = V−1 l ∫
0
1
Plxdx
So
A l = 2l + 1
2a2l+1 − b2l+1 
Va l+1 − b l+1−1 l  ∫
0
1
Plxdx
B l = a l+1 2l + 12
V ∫
0
1
Plxdx − A la2l+1
3.1
69
Note that
∫
0
1
Plxdx = 12 ∫−1
1
Plxdx
for l even. For even l ∑ 0, ∫
0
1
Plxdx = 0. Thus we have
∫
0
1
P0xdx = 1; ∫
0
1
P1xdx = 12
, ∫
0
1
P3xdx = − 18
and
A0 = V2
, A1 = 3
4a3 − b3 
Va2 + b2 , A3 = − 7
16a7 − b7 
Va4 + b4 
B0 = 12
Va − 1
2
Va = 0
B1 = 34
a2V − 3
4a3 − b3 
Va2 + b2 a3 = 3
4
Va2b2 b + a
−a3 + b3
B3 = − 716
a4V − a7 − 7
16a7 − b7 
Va4 + b4  = − 7
16
Va4b4 b
3 + a3
−a7 + b7
As b → ∞, only the B l terms (and A0 survive. Thus using the general expression for ∫
0
1
Plxdx
given by (3.26)
φr,θ = V
2
P0x + 32
a3
r2
P1x − 78
a4
r4
P3x +. . . . .
Let’s now solve the problem neglecting the outer sphere (since b → ∞ using the Green’s function
result
(2.19) this integral give, for cosθ = 1
φr,θ = V
2
1 − ρ2  1
1 − ρ
− 1
1 + ρ2 
with ρ = a/r. Expanding the above,
φr,θ = V
2
a
r +
3
2
a2
r2
− 7
8
a4
r4
+. . . . . .
Comparing with our previous solution with x = 1, we see the Green’s function solution differs by
having a B0 term and by not having an A0 term. All the other higher power terms agree in the series.
This difference is due to having a potential at ∞ in the original problem.
70
3.2
71
72
73
3.3
74
75
Slice the sphere equally by n planes slicing through the z axis, subtending angle Δφ about
this axis with the surface of each slice of the pie alternating as ±V.
φr,θ,φ = >
l,m
A lmr lYl
mθ,φ
so
A lm = 1
a l
∫ dΩYlmθ,φ∗φa,θ,φ
Symmetries:
A l−m = −1mA lm ∗
φr,θ,φ + 2Δφ = φr,θ,φ
where
Δφ = 2π
2n
Thus
m = ±n, and integral multiples thereof
φ−r⃗ = −φr⃗, n = 1
φ−r⃗ = φr⃗, n > 1
Since
PYl
mθ,φ = −1 lYlmθ,φ
Then
l is odd for n = 1; l is even for n > 1
Thus we only have contributions of l ≥ n. Using
A lm = 1
a l
∫ dΩYlmθ,φ∗φa,θ,φ
The integral over φ can be done trivially, since the integrand is just e−imφ leaving the desired answer
in terms of an integral over cosθ.
n = 1 case: I am going to keep only the lowest novanishing terms, involving A11 and A1−1.
φ = rA11Y11 + A1−1Y1−1  = rA11Y11 + A11Y11 
∗  = 2r ReA11Y11 
Y1
1 = − 3
8π
1 − x21/2e iφ
A11 = − 1a
3
8π
V ∫
−1
1
1 − x21/2dx ∫
0
π
e−iφdφ − ∫
π
2π
e−iφdφ
3.4
76
A11 = 2iπa
3
8π
V
φ = 2r Re 2iπa
3
8π
V − 3
8π
sinθe iφ = 3r
2a
Vsinθ sinφ
From the figure
we see
sinθ sinφ = cosθ ′
So
φ = 3r
2a
Vcosθ ′ = V 3
2
r
a P1cosθ
′ +. . . . . .
The other terms, for l = 2,3, can be obtained in the same way in agreement with the result of
(3.36)
77
3.6
78
79
3.7
80
81
3.9
82
83
This problem is described by
a) From the class notes
Φρ, z,φ = ∑
nν
Anν sinνφ + Bnν cosνφIν
nπρ
L
sin nπz
L
where
Anν = 2πL
1
Iν nπbL
∫
0
L ∫
0
2π
Vφ, z sin nπa
L
 sinνφdφdz, ν ∑ 0
Bnν = 2πL
1
Iν nπbL
∫
0
L ∫
0
2π
Vφ, z sin nπa
L
cosνφdφdz, ν Φ 0
Bnν = 1πL
1
Iν nπbL
∫
0
L ∫
0
2π
Vφ, z sin nπa
L
dφdz, ν = 0
Noting
∫
− π2
π
2 sinνφdφ − ∫
π
2
3π
2 sinνφdφ = 0
we conclude Anν = 0. Similarly, noting
∫
− π2
π
2
cosνφdφ − ∫
π
2
3π
2
cosνφdφ = 4−1
m
2m + 1
, m = 0,1,2, . . . .
where I’ve recognized that ν must be odd, ie, ν = 2m + 1. Also
∫
0
L
sin nπz
L
dz = 2
2l + 1π
, l = 0,1,2, . . . . .
where again I’ve recognized that n must be odd, ie, n = 2l + 1. Thus
Bnν =
16−1mV
π2I2m+1 nπbL 2l + 12m + 1
b) Now z = L/2, L ∑∑ b, L ∑∑ ρ. Then from the class notes
3.10
84
I2m+1
2l + 1πρ
L
  1
Γ2m + 2
2l + 1πρ
2L
m+1
Also
sin 2l + 1π
L
= −1 l
so
Φρ, z,φ = ∑
l,m
16−1 l+mV
π22l + 12m + 1
ρ
b
2m+1
cos2m + 1φ
Using
tan−1x = ∑
l=0
∞
x2l+1
1l + 1
−1 l
π
4
= tan−11 = ∑
l=0
∞
−1 l
2l + 1
so
Φρ, z,φ = 4Vπ ∑
m
−1m
2m + 1
ρ
b
2m+1
cos2m + 1φ
Remembering from problem 2.13 that
∑
m
−1m
2m + 1
ρ
b
2m+1
cos2m + 1φ = 1
2
tan−1
2 ρ
b
cosφ
1 − ρ
2
b2
we find
Φρ, z,φ = 2Vπ tan
−1 2
ρ
b
cosφ
1 − ρ
2
b2
which is the answer for problem 2.13.
85
3.12
86
87
3.13 Solve for the potential in Problem 3.1, using the appropriate Green function obtained
in the text, and verify that the answer obtained in this way agrees with the direct
solution from the differential equation.
Recall that Problem 3.1 asks for the potential between two concentric spheres
of radii a and b (with b > a), where the upper hemisphere of the inner sphere
and the lower hemisphere of the outer sphere are maintained at potential V , and
where the other hemispheres are at zero potential. Since this problem involves
the potential between two spheres, we use the Dirichlet Green’s function
G(~x, ~x ′)
=
∑
l,m
4π
2l + 1
1
1−
(
a
b
)2l+1 (rl< − a2l+1rl+1<
)(
1
rl+1>
−
rl>
b2l+1
)
Y ml (Ω)Y
m ∗
l (Ω
′)
(1)
Because there are no charges between the spheres, the Green’s function solution
for the potential only involves the surface integral
Φ(~x ) = − 1
4π
∫
S
Φ(~x ′)
∂G
∂n′
da′
Here, we note a subtlety in that the boundary surface is actually disconnected,
and includes both the inner sphere of radius a and the outer sphere of radius b.
This means that the potential may be written as a sum of two contributions
Φ(~x ) = − 1
4π
∫
r′=a
Φ(~x ′)
∂G
∂n′
a2dΩ′ − 1
4π
∫
r′=b
Φ(~x ′)
∂G
∂n′
b2dΩ′ (2)
We now compute the normal derivatives of the Green’s function (1)
∂G
∂n′
∣∣∣
r′=a
= −∂G
∂r′
∣∣∣
r<=r′=a
=
= −4π
a2
∑
l,m
1
1−
(
a
b
)2l+1 [(ar)l+1 − (ab)l+1 (rb)l
]
Y ml (Ω)Y
m ∗
l (Ω
′)
and
∂G
∂n′
∣∣∣
r′=b
=
∂G
∂r′
∣∣∣
r>=r′=b
=
= −4π
b2
∑
l,m
1
1−
(
a
b
)2l+1 [(rb)l − (ab)l (ar)l+1
]
Y ml (Ω)Y
m ∗
l (Ω
′)
88
Inserting these expressions into (2) yields
Φ(~x )
=
∑
l,m
[∫
Va(Ω′)Y m ∗l (Ω
′)dΩ′
]
1
1−
(
a
b
)2l+1 [(ar)l+1 − (ab)l+1 (rb)l
]
Y ml (Ω)
+
∑
l,m
[∫
Vb(Ω′)Y m ∗l (Ω
′)dΩ′
]
1
1−
(
a
b
)2l+1 [(rb)l − (ab)l (ar)l+1
]
Y ml (Ω)
This is the general expression for the solution to the boundary value problem
where Va(Ω) is the potential on the inner sphere and Vb(Ω) is the potential on
the outer sphere.
For the upper/lower hemispheres problem, we note that azimuthal symmetry
allows us to restrict the m values to m = 0 only. In this case, the spherical
harmonic expansion reduces to a Legendre polynomial expansion
Φ(~x )
=
∑
l
[∫ 1
−1
Va(ζ)Pl(ζ)dζ
]
2l + 1
2
(
1−
(
a
b
)2l+1) [(ar)l+1 − (ab)l+1 (rb)l
]
Pl(cos θ)
+
∑
l
[∫
Vb(ζ)Pl(ζ)dζ
]
2l + 1
2
(
1−
(
a
b
)2l+1) [(rb)l − (ab)l (ar)l+1
]
Pl(cos θ)
where ζ = cos θ′. Since Va = V for ζ > 0 and Vb = V for ζ < 0, this above
expression reduces to
Φ(~x ) =
∑
l
(2l + 1)V Nl
2
(
1−
(
a
b
)2l+1) [(ar)l+1 − (ab)l+1 (rb)l
]
Pl(cos θ)
+
∑
l
(2l + 1)(−1)lV Nl
2
(
1−
(
a
b
)2l+1) [(rb)l − (ab)l (ar)l+1
]
Pl(cos θ)
=
∑
l
(2l + 1)V Nl
2
(
1−
(
a
b
)2l+1)[((ar)l+1 − (ab)l+1 (rb)l
)
+ (−1)l
((r
b
)l
−
(a
b
)l (a
r
)l+1)]
Pl(cos θ)
where
Nl =
∫ 1
0
Pl(ζ)dζ =
{ 1 l = 0
(−1)j+1
Γ(j − 12 )
2
√
πj!
l = 2j − 1 odd
89
If desired, the potential may be rearranged to read
Φ(~x ) =
∑
l
(2l + 1)V Nl
2
(
1−
(
a
b
)2l+1)[(1 + (−1)l+1 (ab)l
)(a
r
)l+1
+ (−1)l
(
1 + (−1)l+1
(a
b
)l+1)(r
b
)l]
Pl(cos θ)
=
V
2
+ V
∞∑
j=1
(−1)j+1(4j − 1)Γ(j − 12 )
4
√
πj!
(
1−
(
a
b
)4j−1)
[(
1 +
(a
b
)2j−1)(a
r
)2j
−
(
1 +
(a
b
)2j)(r
b
)2j−1]
P2j−1(cos θ)
which agrees with the solution to Problem 3.1 that wehave found earlier.
90
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92
3.17
To demonstrate the general method, this problem is worked out in a verbose manner.
Preparation: We use the completeness (=closure) relations
δ(φ− φ′) = 1
2π
∞∑
m=−∞
exp(−imφ′) exp(imφ)
δ(z − z′) = 2
L
∞∑
n=1
sin
(nπz
L
)
sin
(
nπz′
L
)
δ(ρ− ρ′)
ρ′
=
∫ ∞
k=0
kJm(kρ′)Jm(kρ)dk ∀m , (1)
where the last one is obtained by treating Eq. 3.108 in the textbook with the replacements explained after
Eq. 3.112. In doubt, for any complete set of functions with given orthogonality relation the appropriate
closure relation can be obtained using the procedure outlined in class. The procedure is demonstrated again
in the following:
Completeness of the set {Jm(kρ), k ∈ [0,∞[ ,m fixed } ⇒ any f(ρ) can be expanded as
f(ρ) =
∫ ∞
k=0
A(k)Jm(kρ)dk (2)
Obtain expansion coefficient function A(k) using the orthogonality relation Eq. 3.108 of the textbook:
∫ ∞
ρ=0
f(ρ)Jm(k′ρ)ρdρ =
∫ ∞
k=0
A(k)
∫ ∞
ρ=0
Jm(kρ)Jm(k′ρ)ρdρdk
∫ ∞
ρ=0
f(ρ)Jm(k′ρ)ρdρ =
∫ ∞
k=0
A(k)
1
k
δ(k − k′)dk
A(k′) = k′
∫ ∞
ρ=0
f(ρ)Jm(kρ)ρdρ
To simplify writing in what follows, swap primes from k to ρ:
A(k) = k
∫ ∞
ρ=0
f(ρ′)Jm(kρ′)ρ′dρ′
Insert into Eq. 2:
f(ρ) =
∫ ∞
k=0
k
∫ ∞
ρ=0
f(ρ′)Jm(kρ′)ρ′dρ′Jm(kρ)dk
93
f(ρ) =
∫ ∞
ρ=0
{∫ ∞
k=0
kJm(kρ′)Jm(kρ)dk
}
ρ′f(ρ′)dρ′
From the last line, it is obvious that
δ(ρ− ρ′)
ρ′
=
∫ ∞
k=0
kJm(kρ′)Jm(kρ)dk ∀m
a): Obtain given expansion of Green’s function.
Step 1: Write down Equation for Green’s function with δ-function in cylindrical coordinates,
∆G(x,x′) = −4πδ(x− x′) = −4π δ(ρ− ρ
′)
ρ′
δ(φ− φ′)δ(z − z′)
Step 2: On right side, use completeness relations for two out of the three δ-functions. From the given result,
we suspect that we need to use the completeness relations for δ(z − z′) and δ(φ− φ′):
∆G(x,x′) = − 4
L
∞∑
m=−∞
∞∑
n=1
exp(−imφ′) exp(imφ) sin
(
nπz′
L
)
sin
(nπz
L
) δ(ρ− ρ′)
ρ′
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
{
− 4
L
exp(−imφ′) sin
(
nπz′
L
)
δ(ρ− ρ′)
ρ′
}
exp(imφ) sin
(nπz
L
)
(3)
Step 3: On left side, expand the Green’s function using the orthogonal function sets that have also been used
in Step 2. Note that x′ only enters as a parameter of the calculation; ∆ acts on x.
∆G(x,x′) = ∆
∞∑
m=−∞
∞∑
n=1
exp(imφ) sin
(nπz
L
)
Amn(ρ|ρ′, z′, φ′)
Step 4: Write Laplacian in the proper coordinates and take derivatives of the orthogonal functions:
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
{[
1
ρ
d
dρ
ρ
d
dρ
− m
2
ρ2
− n
2π2
L2
]
Amn(ρ|ρ′, z′, φ′)
}
exp(imφ) sin
(nπz
L
)
(4)
Step 5: Expansions in orthogonal sets of functions are unique. Thus, we can separately equate the co-
efficients of the exp(imφ) sin
(
nπz
L
)
in Eq. 3 and Eq. 4. Dividing by exp(−imφ′) sin
(
nπz′
L
)
and using
Dρ =
[
1
ρ
d
dρ ρ
d
dρ − m
2
ρ2 − n
2π2
L2
]
as abbreviation for the involved linear differential operator in ρ, we find
Dρ
{
Amn(ρ|ρ′, z′, φ′)
exp(−imφ′) sin (nπz′L
)
}
= − 4
L
δ(ρ− ρ′)
ρ′
94
Step 6: Noticing that the expression in the curly brackets of the last equation can only depend on the
parameters of Dρ (which are m and n), on ρ and on ρ′, we define the reduced Green’s function
gmn(ρ, ρ′) =
Amn(ρ|ρ′, z′, φ′)
exp(−imφ′) sin (nπz′L
)
and proceed to solve
[
1
ρ
∂
∂ρ
ρ
∂
∂ρ
− m
2
ρ2
− n
2π2
L2
]
gmn(ρ, ρ′) = − 4
L
δ(ρ− ρ′)
ρ′
(5)
Inspection of Eq. 3.98ff of the textbook shows that the differential equation is the modified Bessel differ-
ential equation (except the δ-function inhomogeneity), and that the linearly independent solutions of the
homogeneous equation are Im(kρ) and Km(kρ), with k = nπL . In the domain ρ < ρ
′ only Im(kρ) is regular,
while in the domain ρ > ρ′ only Km(kρ) is regular. Thus, for the reduced Green’s function to be regular,
symmetric in ρ and ρ′, and continuous at ρ = ρ′, it must be of the form
gmn(ρ, ρ′) = C Im(kρ<) Km(kρ>) ,
with a constant C to be determined to match the inhomogeneity. Also, ρ< = min(ρ, ρ′) and ρ> = max(ρ, ρ′).
Step 7: Find C. Integrating Eq. 5 from ρ′ − ² to ρ′ + ² with ² → 0 and dropping vanishing terms we find
d
dρ
gmn(ρ, ρ′)|ρ=ρ′+² − d
dρ
gmn(ρ, ρ′)|ρ=ρ′−² = − 4
ρ′L
The left side equals C k W [Im(x), Km(x)] with x = kρ′ and the Wronski determinant W =
Im(x)
(
d
dxKm(x)
) − ( ddxIm(x)
)
Km(x). The Wronsky determinant can be evaluated in the asymptotic re-
gion, where both Im(x) and Km(x) have simple forms given in Eqs. 3.102ff of the textbook, leading to
W [Im(x),Km(x)] = − 1x (see Eq. 3.147 in textbook). We thus find
−Ck 1
x
= −Ck 1
kρ′
= − 4
ρ′L
and therfore C = 4L .
Step 8:
Going backward through all steps, the Green’s function expansion is assembled into the final result:
gmn(ρ, ρ′) =
4
L
Im(kρ<) Km(kρ>) ,
Amn(ρ|ρ′, z′, φ′) = 4
L
Im(kρ<)Km(kρ>) exp(−imφ′) sin
(
nπz′
L
)
95
and, using k = nπL ,
G(x,x′) =
4
L
∞∑
m=−∞
∞∑
n=1
exp(−imφ′) exp(imφ) sin
(
nπz′
L
)
sin
(nπz
L
)
Im(
nπ
L
ρ<) Km(
nπ
L
ρ>) , q.e.d.
b:
Step 1: Write down Equation for Green’s function with δ-function in cylindrical coordinates,
∆G(x,x′) = −4πδ(x− x′) = −4π δ(ρ− ρ
′)
ρ′
δ(φ− φ′)δ(z − z′)
Step 2: Use completeness relations for δ(ρ−ρ
′)
ρ′ and δ(φ− φ′):
∆G(x,x′) = −2
∞∑
m=−∞
∫ ∞
k=0
exp(−imφ′) exp(imφ) Jm(kρ) Jm(kρ′)δ(z − z′) k dk
∆G(x,x′) =
∞∑
m=−∞
∫ ∞
k=0
{−2k exp(−imφ′)Jm(kρ′)δ(z − z′)} exp(imφ) Jm(kρ) dk (6)
Step 3: Expand the Green’s function using the orthogonal functionsets that have also been used in Step 2.
∆G(x,x′) = ∆
∞∑
m=−∞
∫ ∞
k=0
exp(imφ) Jm(kρ) Amk(z|ρ′, z′, φ′)dk
Step 4: Write Laplacian in the proper coordinates and take the obvious derivatives of the orthogonal func-
tions:
∆G(x,x′) =
∞∑
m=−∞
∫ ∞
k=0
[
∂2
∂z2
− m
2
ρ2
+
1
ρ
∂
∂ρ
ρ
∂
∂ρ
]
exp(imφ) Jm(kρ) Amk(z|ρ′, z′, φ′)dk
The ordinary Bessel diff. equation reads
[
−m2ρ2 + 1ρ ∂∂ρρ ∂∂ρ
]
Jm(kρ) = −k2Jm(kρ), and thus:
∆G(x,x′) =
∞∑
m=−∞
∫ ∞
k=0
{[
d2
dz2
− k2
]
Amk(z|ρ′, z′, φ′)
}
exp(imφ) Jm(kρ)dk (7)
Step 5: We equate coefficients in Eqs. 6 and Eq. 7. Dividing by exp(−imφ′) Jm(kρ′) we find
[
d2
dz2
− k2
]{
Amk(z|ρ′, z′, φ′)
exp(−imφ′)Jm(kρ′)
}
= −2k δ(z − z′)
Step 6: Noting that the expression in the curly brackets of the last equation can only depend on k, z and z′,
we define the reduced Green’s function
96
gk(z, z′) =
Amk(z|ρ′, z′, φ′)
exp(−imφ′)Jm(kρ′)
and proceed to solve
[
d2
dz2
− k2
]
gk(z, z′) = −2k δ(z − z′)
A reduced Green’s function that matches the boundary conditions at z = 0 and z = L, that is symmetric in
z and z′ and that is continuous at z = z′ must be of the form
gk(z, z′) = C sinh(k z<) sinh(k (L− z>)) ,
where the constant C needs to be determined to match the inhomogeneity at z = z′. Also, z< = min(z, z′)
and z> = max(z, z′).
Step 7: Find C. Integrating the differential equation for gk(z, z′) from z′−² to z′+² with ² → 0 and dropping
vanishing terms we find
d
dz
gk(z, z′)|z=z′+² − d
dz
gk(z, z′)|z=z′−² = −2k
Direct calculation yields
d
dz
gk(z, z′)|z=z′+² − d
dz
gk(z, z′)|z=z′−² = −kC [sinh(kz′) cosh(k (L− z>)) + cosh(kz′) sinh(k (L− z>))]
= −kC sinh(kL)
and thus, with the equation before, C = 2sinh(kL) .
Step 8:
Going backward through all steps, the Greens function expansion is assembled into the final result
G(x,x′) = 2
∞∑
m=−∞
∫ ∞
k=0
exp(−imφ′) exp(imφ) Jm(kρ) Jm(kρ′) sinh(k z<) sinh(k (L− z>))sinh(kL) dk , q.e.d.
97
3.22
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99
3.23
Hint: Use Green’s function expansion techniques. For the second line, note Eq. 3.147.
We use the completeness relations
δ(φ− φ′) = 1
2π
∞∑
m=−∞
exp(−imφ′) exp(imφ)
δ(z − z′) = 2
L
∞∑
n=1
sin
(nπz
L
)
sin
(
nπz′
L
)
δ(ρ− ρ′)
ρ′
=
∞∑
n=1
2
a2J2m+1(xmn)
Jm(xmn
ρ
a
)Jm(xmn
ρ′
a
) ∀m . (8)
a):
Step 1: Write down Equation for Green’s function with δ-function in cylindrical coordinates,
∆G(x,x′) = −4πδ(x− x′) = −4π δ(ρ− ρ
′)
ρ′
δ(φ− φ′)δ(z − z′)
Step 2: We use the completeness relations for δ(φ− φ′) and δ(ρ−ρ′)ρ′ to write:
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
{
− 4
a2J2m+1(xmn)
exp(−imφ′) Jm(xmn ρ
′
a
) δ(z − z′)
}
exp(imφ) Jm(xmn
ρ
a
) (9)
Step 3: Expansion of the Green’s function in exp(imφ) Jm(xmn ρa ):
∆G(x,x′) = ∆
∞∑
m=−∞
∞∑
n=1
exp(imφ) Jm(
xmn
a
ρ)Amn(z|ρ′, z′, φ′)
Step 4: Apply Laplacian:
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
[
∂2
∂z2
− m
2
ρ2
+
1
ρ
∂
∂ρ
ρ
∂
∂ρ
]
exp(imφ) Jm(
xmn
a
ρ)Amn(z|ρ′, z′, φ′)
The ordinary Bessel differential equation reads, in the present case,
[
1
ρ
∂
∂ρρ
∂
∂ρ − m
2
ρ2
]
Jm(xmna ρ) =
− (xmna
)2
Jm(xmna ρ), and thus:
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
{[
d2
dz2
−
(xmn
a
)2]
Amn(z|ρ′, z′, φ′)
}
exp(imφ) Jm(
xmn
a
ρ) (10)
Step 5: We equate coefficients in Eqs. 9 and Eq. 10. Dividing by − 4
a2J2
m+1(xmn)
exp(−imφ′)Jm(kρ′) yields
100
[
d2
dz2
−
(xmn
a
)2]
gmn(z, z′) = δ(z − z′)
with the reduced Green’s function
gmn(z, z′) = −
a2 J2m+1(xmn)Amn(z|ρ′, z′, φ′)
4 exp(−imφ′)Jm(xmna ρ′)
Step 6: A reduced Green’s function that matches the boundary conditions at z = 0 and z = L, that is
symmetric in z and z′ and that is continuous at z = z′ must be of the form
gmn(z, z′) = C sinh(
xmn
a
z<) sinh(
xmn
a
(L− z>)) ,
where the constant C needs to be determined to match the inhomogeneity at z = z′. Also, z< = min(z, z′)
and z> = max(z, z′).
Step 7: Find C. Integrating the differential equation for gmn(z, z′) from z′ − ² to z′ + ² with ² → 0 and
dropping vanishing terms we find
d
dz
gmn(z, z′)|z=z′+² − d
dz
gmn(z, z′)|z=z′−² = 1
Direct calculation similar to Problem 3.17 b) yields
C = − a
xmn sinh(xmnLa )
.
Step 8:
Going backward through all steps, the Greens function expansion is assembled into
G(x,x′) =
4
a
∞∑
m=−∞
∞∑
n=1
exp(−imφ′) exp(imφ) Jm(xmn
a
ρ) Jm(
xmn
a
ρ′)
sinh(xmna z<) sinh(
xmn
a (L− z>))
xmn J2m+1(xmn) sinh(
xmnL
a )
.
The potential of a point charge q at x′ follows from its charge density, ρ(x′′) = qδ(x′′ − x′), and
Φ(x,x′) =
1
4π²0
∫
V
ρ(x′′)G(x,x′′)d3x′′ =
q
4π²0
∫
V
δ(x′′ − x′)G(x,x′′)d3x′′ = q
4π²0
G(x,x′)
Thus, with the above obtained G(x,x′) it is
Φ(x,x′) =
q
aπ²0
∞∑
m=−∞
∞∑
n=1
exp(−imφ′) exp(imφ) Jm(xmn
a
ρ) Jm(
xmn
a
ρ′)
sinh(xmna z<) sinh(
xmn
a (L− z>))
xmn J2m+1(xmn) sinh(
xmnL
a )
, q.e.d.
101
b):
Step 1: Write down Equation for Green’s function with δ-function in cylindrical coordinates,
∆G(x,x′) = −4πδ(x− x′) = −4π δ(ρ− ρ
′)
ρ′
δ(φ− φ′)δ(z − z′)
Step 2: We use the completeness relations for δ(φ− φ′) and δ(z − z′) to write:
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
{
− 4
L
exp(−imφ′) sin
(
nπz′
L
)
δ(ρ− ρ′)
ρ′
}
exp(imφ) sin
(nπz
L
)
(11)
Step 3: Expansion of the Green’s function in exp(imφ) sin
(
nπz
L
)
:
∆G(x,x′) = ∆
∞∑
m=−∞
∞∑
n=1
exp(imφ) sin
(nπz
L
)
Amn(ρ|ρ′, z′, φ′)
Step 4: Apply Laplacian:
∆G(x,x′) =
∞∑
m=−∞
∞∑
n=1
{[
1
ρ
d
dρ
ρ
d
dρ
−
(nπ
L
)2
− m
2
ρ2
]
Amn(ρ|ρ′, z′, φ′)
}
exp(imφ) sin
(nπz
L
)
(12)
Step 5: We equate coefficients in Eqs. 11 and Eq. 12, and divide by − 4L exp(−imφ′) sin
(
nπz′
L
)
. We find
[
1
ρ
d
dρ
ρ
d
dρ
−
(nπ
L
)2
− m
2
ρ2
]
gmn(ρ, ρ′) =
δ(ρ− ρ′)
ρ′
for the reduced Green’s function
gmn(ρ, ρ′) = − L Amn(ρ|ρ
′, z′, φ′)
4 exp(−imφ′) sin (nπz′L
)
Step 6: The differential equation for gmn is the modified Bessel equation (except the inhomogeneity). Thus,
a reduced Green’s function that solves the homogeneous equation, is symmetric in ρ and ρ′, is continuous at
ρ = ρ′, is regular at ρ = 0 and is vanishing at ρ = a must be of the form
gmn(ρ, ρ′) = C Im(
nπ
L
ρ<)
(
Im(
nπ
L
ρ>)−
Im(nπaL )
Km(nπaL )
Km(
nπ
L
ρ>)
)
where