PHY–352 K.
Solutions for problem set #6.
Problem 3.13:
The potential in the infinite slot is explained in the textbook example 3.3 in detail; I have also
explained it in class. According to the textbook equations (3.30) and (3.34), the potential
inside the slot is
V (x, y) =
∞
X
n=1
Cn × sin
nπy a
nπx × exp −
,
a
(3.30)
where a is the slot’s width (which I have called D in class), and the coefficient Cn follow
from expanding the boundary potential Vb (y) at x = 0 into the Fourier series,
Cn
2
=
a
Za
0
Vb (y) × sin
nπy a
dy
=⇒
Vb (y) =
∞
X
n=1
Cn × sin
nπy a
.
(3.34)
For the problem at hand, Vb (y) piece-wise constant function with a jump at x = a2 ,
v0 (y) =
(
+V0
for 0 < y < a2 ,
−V0
for
a
2
> y < a.
(1)
Plugging this boundary potential into eq. (3.34), we obtain
Cn
2V0
= +
a
Za/2 Za
nπy nπy 2V0
sin
dy −
sin
dy
a
a
a
0
a/2
nπ 2V0
nπ
a a 2V0
1 − cos
−
cos
×
×
− cos(πn)
= +
a
nπ
2
a
nπ
2
2V0 =
× 1 + (−1)n − in − (−i)n
nπ
4 if n ≡ 2 (mod 4),
2V0
=
×
nπ
0 otherwise.
(2)
Thus, in the expansion (3.30) we may restrict n to n = 4m + 2 for m = 0, 1, 2, 3, . . .; for all
1
other n the coefficients Cn are zero. Consequently, the series (3.30) becomes
∞
4V0 X
(4m + 2)πy
(4m + 2)πy
2
V (x, y) =
× exp −
.
× sin
π
4m + 2
a
a
(3)
m=0
Note that this series looks just like the series in the textbook equation (3.36), except for
rescaling the coordinates, x → 2x, y → 2y. Consequently, we may sum up the series (3) in
exactly the same way as in eq. (3.37), namely
2V0
× arctan
V (x, y) =
π
sin(2πy/a)
sinh(2πx/a)
.
(4)
Graphically, it looks like the square wave (1) at small x ≪ a, but for larger X it starts looking
more and more like the sine wave sin(2πy/a) with exponentially decreasing amplitudes. Here
is the slice-plot of the potential V (y) for few fixed x ranging from 0 to 0.3 a:
V
x=0
x = 0.05 a
x = 0.1 a
x = 0.2 a
x = 0.3 a
y
2
Problem 3.14:
The electric charge density σ on a surface of a conductor is related to the electric field
immediately outside the conductor,
σ = ǫ0 × E⊥ [just outside the conductor].
(5)
For the problem at hand, the textbook eqs. (3.36–37) give us the potential inside the slot,
4V0
π
nπx nπy 1
× exp −
× sin
n
a
a
odd n≥1
2V0
sin(πy/a)
=
,
arctan
π
sinh(πx/a)
V (x, y) =
X
(3.36)
(3.37)
from which we may obtain the electric field anywhere inside the slot, and in particular just
outside the conducting strip at the slot’s left end. Consequently, eq. (5) will gives us the
charge density σ(y) on the right side of the vertical conducting strip at x = 0 — the side
facing the inside of the slot.
Presumably, these is also some other charge density on the left side of the vertical strip
facing the outside world, but to find that density we would need to know the electric field
outside the slot. That task is beyond the present homework, so I am not going to write the
solution here.
Going back to the potential (3.37) and taking the ∂/∂x derivative, we have
A
d arctan(A/B) d(A/B(x))
1
−AB ′
AB ′
d
arctan
=
×
=
×
=
−
dx
B(x)
d(A/B)
dx
1 + (A/B)2
B2
A2 + B 2
(6)
hence for A = sin(πy/a) and B(x) = sinh(πx/a),
∂
sin(πy/a) × (π/a) cosh(πx/a)
sin(πy/a)
= −
arctan
∂x
sinh(πx/a)
sin2 (πy/a) + sinh2 (πx/a)
(7)
and therefore
Ex (x, y) = −
∂V
2V0
sin(πy/a) × cosh(πx/a)
.
= +
×
∂x
a
sin2 (πy/a) + sinh2 (πx/a)
3
(8)
For x → 0 this formula simplifies to
2V0
1
×
,
a
sin(πy/a)
Ex (0, y) =
(9)
so according to eq. (5), the charge density on the right side of the vertical strip is
σ(y) =
1
2ǫ0 V0
×
.
a
sin(πy/a)
(10)
Alternative derivation of eq. (9):
Suppose we did not have the analytical formula (3.37) for the potential inside the slot but
only the un-summed series (3.36). In this case, we may take the x–derivative term-by-term
to get a series for the electric field, then simplify that series for x = 0, and finally try to sum
it up. Thus, starting with the potential (3,36), we have
Ex (x, y) = −

∂  4V0
∂x π
X
odd n≥1
1
× sin
n
nπy a

nπx 
× exp −
a
nπy d
nπx nπx 4V0 X 1
nπ
= −
×
= −
× sin
exp exp −
× exp −
π
n
a
dx
a
a
a
odd n≥1
nπy nπx 4v0 X
sin
× exp −
,
= +
a
a
a
odd n≥1
(11)
which for x = 0 simplifies to
4V0
Ex (0, y) =
a
X
sin
odd n≥1
nπy a
.
(12)
To sum up this Fourier series we use
inξ
sin(nξ) = Im e
= Im
4
iξ n
e
for ξ = π(y/a).
(13)
Consequently,
X
odd n≥1

sin(nξ) = Im 
X
odd n≥1
= Im
∞
X
m=0
=

n
eiξ 
2m+1
eiξ
=
eiξ
1 − eiξ
+1
,
2 sin(ξ)
2 =
e−iξ
1
1
=
+iξ
−2i sin(ξ)
−e
!
(14)
and therefore
Ex (0, y) =
2V0
1
×
,
a
sin(πy/a)
(15)
in perfect agreement with eq. (9).
Problem 3.16:
First, read through the textbook example 3.5, the rectangular pipe where Vb = 0 on the 4
walls of the pipe.
For the problem at hand, we follows the same strategy: Start by looking for the solution
of the 3D Laplace equation in the form
V (x, y, z) = f (x) × g(y) × h(z)
(16)
and demand that it vanishes on five sides of the cube without specifying V (x, y) on the top
side at z = a. Consequently, we arrive at the equations
f ′′ (x) + C1 × F (x) = 0,
g ′′ (y) + C2 × g(y) = 0,
h′′ (z) + C3 × h(z) = 0,
for some constants C1 , C2 , and C3 ,
(17)
C1 + C2 + C3 = 0,
(18)
f (0) = f (a) = g(0) = g(a) = h(0) = 0.
(19)
5
Solving these equations for f (x) and g(y), we find
f (x) = sin(nπx/a),
g(y) = sin(mπy/a)
(20)
for some positive integers n and m. As to the g(z), the general solution has form
+kz
g(z) = E × e
−kx
+ F ×e
√
π n2 + m2
for k =
a
(21)
for some constants E and F . In the textbook example of the pipe, the boundary condition
was g → 0 for z → +∞, which required E = 0. But for the cube at hand, the boundary
condition is g(0) = 0, which requires E = −F , hence
g(z) = E × e+kz − E × e−kz = 2E × sinh(kz).
(22)
Altogether, a separable solution for 5 sides of the cube looks like
p
V (x, y, z) = const × sin(nπx/a) × sin(mπy/a) × sinh (π/a) n2 + m2 × z
(23)
for some positive integers n, m, so the general solution is the double series
V (x, y, z) =
∞ X
∞
X
n=1 m=1
Cn,m × sin(nπx/a) × sin(mπy/a) × sinh (π/a)
p
n2 + m2 × z . (24)
The coefficients Cn,m on this series follow from the expansion of the boundary potential
at the top side of the cube into a (double) Fourier series:
Vb (x, y) = V (x, y, z = a) =
∞ X
∞
X
n=1 m=1
Cn,m × sin(nπx/a) × sin(mπy/a) × sinh π
p
n2 + m2 .
(25)
Reversing this Fourier transforms gives us
Cn,m
4
1
√
=
× 2
a
sinh π n2 + m2
Za
0
dx
Za
0
dy Vb (x, y) × sin(nπx/a) × sin(mπy/a).
(26)
For the Vb being a constant but non-zero V0 over the whole top surface, the double integral
6
here evaluates to
Za
dx
0
Za
0
dy V0 × sin(nπx/a) × sin(mπy/a) = V0 ×
Za
= V0 ×
16V0
1
1
√
=
×
×
×
2
π
nm sinh π n2 + m2
sin(mπy/a)
0
1 − (−1)n 1 − (−1)m
×
nπ/a
mπ/a

2

 4V0 a
nmπ 2
=


0
hence
Cn,m
0
sin(nπx/a) dx ×
Za
when both n and m are odd,
otherwise,
(27)
1 when both n and m are odd,
0 otherwise.
(28)
Finally, plugging these coefficients into the series (24), we arrive at the potential inside the
cube,
16V0
×
V (x, y, z) =
π2
X
odd n,m≥1
√
sinh π n2 + m2 × (z/a)
1
√
.
× sin(nπx/a) × sin(nπy/a) ×
nm
sinh π n2 + m2
(29)
As a second part of this problem, consider the potential at the cube’s center x = y =
z =
a
2.
Using the superposition principle and the symmetry between the six sides of the
cube, we expect
V (center) =
1X
V0 + 5 × 0
V0
V (side) =
=
.
6
6
6
(30)
sides
At the same time, evaluating the potential (29) for the center, we have
sin(nπx/a) → sin(nπ/2) = (−1)(n−1)/2
sin(mπy/a) → sin(mπ/2) = (−1)(m−1)/2
sinh(A/2)
1
sinh(Az/a)
→
=
sinh(A)
sinh(A)
2 cosh(A/2)
7
hh for an odd n ii,
hh for an odd m ii,
for A = π
p
n2 + m2 ,
(31)
hence
V (center) =
8V0
×
π2
1
(−1)(n+m−2)/2
√
×
.
nm
cosh π n2 + m2 /2
odd n,m≥1
X
(32)
This series is too painful to evaluate analytically, but thanks to the cosh in the denominator
it converges very rapidly. Indeed, the terms with n or m greater than about 21 are smaller
than
1
∼ 4 · 10−16
21 cosh(21π/2)
so one may sum the series numerically with 15 digits accuracy by summing over odd n, m ≤
21 only. I did this using Mathematica, and got
1
V (center
=
± 2 · 10−17 .
V0
6
(33)
This agrees with eq. (30) so well that there is no doubt that exact answer is indeed 61 .
Problem 3.17:
The Rodrigues for the Legendre polynomials is
Pℓ (x) =
1 dℓ
(x2 − 1)ℓ .
ℓ
ℓ
2 ℓ! dx
(34)
For ℓ = 3 it evaluates to
1 d3 2
1
3
6
4
2
3
P3 (x) =
(x
−1)
=
x
−3x
+3x
−1
=
×
120x
−
3×24x
=
8 × 6 dx3
48
5 3
2x
− 32 x.
(35)
Let’s check that this polynomial obeys the Legendre equation for ℓ = 3:
(1 − x2 ) × P3′′ (x) − 2x × P3′ (x) + 3(3 + 1) × P3 (x) =
15x2 − 3
5x3 − 3x
+ 12 ×
2
2
3
3
3
= 15x − 15x − 15x + 3x + 30x − 18x
= (1 − x2 ) × 15x − 2x ×
= 0.
8
(36)
Or in terms of g(θ) = P3 (cos θ),
g(θ) =
5
2
cos3 θ −
3
2
cos θ,
d
2
3
g(θ) = − 15
2 cos θ sin θ + 2 sin θ,
dθ
d2
3
g(θ) = = +15 cos θ sin2 θ − 15
2 cos θ +
dθ2
(37)
3
2
cos θ,
hence
d2 g
cos θ dg
+
×
+ 3(3 + 1) × g =
dθ2
sin θ
dθ
= 15 cos θ sin2 θ −
−
15
2
cos3 θ +
15
2
3
2
cos3 θ +
3
2
cos θ
cos θ + 12 × 52 cos3 θ − 12 × 23 cos θ
(38)
= 15 cos θ sin2 θ + 15 cos3 θ − 15 cos θ
= 15 cos θ × sin2 θ + cos2 θ − 1
= 0.
Finally, let’s check the orthogonality of the Legendre polynomials P3 (x) and P1 (x) = x:
Z+1 4
Z+1
Z+1 3
5x − 3x2
5x − 3x
× x dx =
dx
P3 (x) × P1 (x) dx =
2
2
−1
=
−1
Z+1
−1
−1
5
x5 − x3
x − x3
d
=
2
2
x=+1
(39)
x=−1
= 0 − 0 = 0.
Problem 3.19:
The problem gives us the potential on the sphere as
Vb (θ) = k × cos(3θ)
(40)
for some constant k. Let’s re-express this potential in terms of the Legendre polynomials of
9
cos θ. First, a bit of trigonometry:
cos(3θ) = cos3 θ − 3 cos θ sin2 θ = 4 cos3 θ − 3 cos θ.
(41)
This is a cubic polynomial in x = cos θ, so it’s linear combination of Pℓ (x) for ℓ ≤ 3 only.
Moreover, since the polynomial (41) is odd in x = cos θ, only the odd ℓ’s are involved in its
expansion. This lives us with just the P3 (x) = 25 x3 − 32 x and the P1 (x) = x, and the right
combination is
8
5 P3 (x)
−
3
5 P1 (x)
= 4x3 −
12
5x
+
3
5x
= 4x3 − 3x.
(42)
Thus, the boundary potential on the sphere amounts to
Vb (θ) =
8k
3k
× P3 (cos θ) −
× P1 (cos θ).
5
5
(43)
For each Legendre polynomial Pℓ (cos θ), the potential behaves like r ℓ inside the sphere
and r −ℓ−1 outside the sphere, cf. equations (55–57) of my notes on separation of variables
(pages 9–10). Thus,
inside the sphere V (r, θ) =
outside the sphere V (r, θ) =
r3
3k
r
8k
× P3 (cos θ) × 3 −
× P1 (cos θ) × ,
5
R
5
R
(44)
R4
3k
R2
8k
× P3 (cos θ) × 4 =
× P1 (cos θ) × 2 . (45)
5
r
5
r
Problem 3.26:
This problem calls for separation of variables in cylindrical coordinates. It is not one of the
textbook examples, but it is discussed in detail in my notes on separation of variables (pages
1–6). According to eqs. (17) and (29–31) of my notes,
inside the cylinder V (s, φ) =
+∞
X
m=−∞
imφ
Cm × e
×
r |m|
R
,
(46)
+∞
X
|m|
R
imφ
outside the cylinder V (s, φ) =
Cm × e
×
,
r
m=−∞
where the coefficients Cm follow from expanding the boundary potential Vb (φ) on the cylin10
der’s surface into a Fourier series:
Vb (φ) =
+∞
X
m=−∞
Cm × eimφ ,
Cm
1
=
2π
Z2π
Vb (φ) × e−imφ dφ.
(47)
0
The potential (46) is continuous across the cylinder’s surface, but the radial electric field has
a discontinuity:
Inside the cylinder
Es (s, φ) =
+∞
X
m=−∞
−−→
s→R
imφ
Cm × e
|m| r |m|−1
×
R|m|
+∞
+1 X
|m| × Cm × eimφ ,
R m=−∞
(48)
Outside the cylinder
Es (s, φ) =
+∞
X
m=−∞
Cm × eimφ ×
−|m| R|m|
r |m|+1
+∞
−1 X
−−→
|m| × Cm × eimφ ,
s→R R
m=−∞
(49)
Hence the discontinuity at s = R is
disc(Es ) = −
+∞
2 X
|m| × Cm × eimφ .
R m=−∞
(50)
Physically, this discontinuity stems from the electric charge density on the surface of the
cylinder,
σ
,
ǫ0
disc(Es ) =
(51)
hence the φ-dependence of this charge density is related to the coefficients Cm according to
+∞
X
2ǫ0
|m| × Cm × eimφ ,
×
σ(φ) = −
R
m=−∞
Cm
1
R
= −
×
×
|m| 4πǫ0
11
Z2π
σ(φ) × e−imφ dφ.
0
(52)
(53)
For the problem at hand, the surface charge density has angular dependence
a
a
× e+5iφ + × e−5iφ
2
2
σ(φ) = a × cos(5φ) =
(54)
for some constant a. Comparing this charge density to eq. (52), we immediately identify
C+5 = C−5 = −
aR
,
20ǫ0
all other Cm = 0.
(55)
Consequently, plugging these coefficients into the potential (46) gives us
Inside the cylinder
r5
aR
+imφ
−5imφ
× e
+e
× 5
V (s, φ) = −
20ǫ0
R
5
aR
r
= −
× cos(5φ) × 5 ,
10ǫ0
R
(56)
Outside the cylinder
R5
aR
× e+imφ + e−5imφ × 5
20ǫ0
r
5
R
aR
× cos(5φ) × 5 .
= −
10ǫ0
r
V (s, φ) = −
(57)
Problem 3.43:
(a) Separating variables in spherical coordinates (r, θ, φ) and assuming φ–independent potentials and charges, we get a solution of the form
V (r, θ) = Pℓ (cos θ) × A × r ℓ +
B
r ℓ+1
(58)
for some constant coefficients A and B which depend on the boundary conditions. Thus in
general,
Outside the outer shell (r > b)
V (r, θ) =
∞
X
ℓ=0
12
Bℓ
Pℓ (cos θ) × Aℓ × r + ℓ+1
r
ℓ
,
(59)
Between the inner ball and the outer shell (a < r < b)
∞
X
Dℓ
ℓ
V (r, θ) =
Pℓ (cos θ) × Cℓ × r + ℓ+1 ,
r
(60)
ℓ=0
and all we need to do is to determine the coefficients Aℓ , Bℓ , Cℓ , and Dℓ from the boundary
conditions:
• Far away from the outer shell, the potential must vanish as we go to infinity. To satisfy
this boundary condition, we must have all Aℓ = 0.
• At the surface of the inner ball, the potential is a given constant V0 . In terms of the
expansion (60), this means
@r = a,
V (θ) = V0 × P0 (cos θ) + 0
(61)
and therefore
D
Cℓ + ℓ =
a
V0
for ℓ = 0,
0
for all other ℓ.
(62)
• The potential must be continuous across the outer shell at r = b, hence
Aℓ × bℓ +
Bℓ
Dℓ
ℓ
=
C
×
b
+
.
ℓ
bℓ+1
bℓ+1
(63)
• Finally, the discontinuity of the radial electric field at the outer shell is related to the
charge density σ at that shell, disc(Er ) = σ/ǫ0 . For the problem at hand σ(θ) = k cos θ
for some constant k, hence
disc(Er ) =
k
k
× cos θ =
× P1 (cos θ).
ǫ0
ǫ0
(64)
At the same time, for the potentials (59) and (60),
for r > b,
Er (r, θ) =
∞
X
ℓ=0
(ℓ + 1)Bℓ
ℓ−1
, (65)
Pℓ (cos θ) × −ℓAℓ × r
+
r ℓ+2
13
for a < r < b,
Er (r, θ) =
∞
X
ℓ=0
(ℓ + 1)Dℓ
ℓ−1
, (66)
Pℓ (cos θ) × −ℓCℓ × r
+
r ℓ+2
hence the discontinuity at r = b is
disc(Er ) =
∞
X
ℓ=0
Pℓ (cos θ) × ℓ × (Cℓ − Aℓ ) × bℓ−1 + (ℓ + 1) × (Bℓ − Dℓ ) × b−ℓ−2 . (67)
Comparing this discontinuity to eq. (64), we obtain
ℓ−1
ℓ × (Cℓ − Aℓ ) × b
−ℓ−2
+ (ℓ + 1) × (Bℓ − Dℓ ) × b
=
(k/ǫ0 ) for ℓ = 1,
0
for all other ℓ.
(68)
Now, let’s solve all these boundary conditions for the coefficients Aℓ , Bℓ , Cℓ and Dℓ . For
ℓ > 1, all the coefficients turn out to be zero. Indeed, combining eqs. (68) and (63) for ℓ 6= 1
gives us
ℓ
Bℓ − Dℓ
= −(Cℓ − Aℓ )
× (Cℓ − Aℓ ) =
ℓ+1
b2ℓ+1
=⇒
Bℓ = Dℓ
and Cℓ = Aℓ . (69)
But Aℓ = 0 by the asymptotic condition at r → ∞, so we must also have Cℓ = 0. Finally,
at the inner ball, eq. (62) gives Dℓ = −a2ℓ+1 Cℓ = 0 and hence Bℓ = 0 as well.
For ℓ = 0 we also have D0 = B0 and C0 = A0 = 0, but then eq. (62) becomes
D0
= V0 ,
a
(70)
hence B0 = D0 = a × V0 .
Finally, for ℓ = 1 the relevant equations are
A1 = 0,
a3 C1 + D1 = 0,
b3 (C1 − A1 ) + (D1 − B1 ) = 0,
−b3 (C1 − A1 ) + 2(D1 − B1 ) = b3 × (k/ǫ0 ),
14
(71)
and solving these linear equations yields
A1 = 0,
B1 = +
k
× (b3 − a3 ),
3ǫ0
C1 = +
k
k
, D1 = −
× a3 .
3ǫ0
3ǫ0
(72)
It remains to plug all these coefficients into eqs. (59) and (60):
Outside the outer shell (r > b)
V (r, θ) =
aV0
k
b3 − a3
+
×
× cos θ,
r
3ǫ0
r2
Between the inner ball and the outer shell (a < r < b)
aV0
k
a3
V (r, θ) =
+
× r − 2 × cos θ.
r
3ǫ0
r
(73)
(74)
This completes part (a) of the problem.
(b) The density of the induced charge on the surface of the conducting inner ball obtains
from the radial electric field immediately outside this surface,
σ(θ) = ǫ0 × Er (a, θ) = −ǫ0 ×
∂V
∂r
.
(75)
r=a
Taking the derivative of the potential (74) (for a < r < b) yields
k
2a3
aV0
× −1 − 3 × cos θ.
Er (r, θ) = 2 +
r
3ǫ0
r
(76)
Hence, taking the limit r → a and multiplying by the ǫ0 yields the charge density on the
conducting ball,
σ(θ) =
ǫ0 V0
k
−
× 3 × cos θ .
a
3
(77)
Physically, the first term here stems from the net electric charge on the ball while the second
terms is induced by the charges on the outer shell.
15
(c) There are two ways two find the net electric charge of the system. One way is to look at
the leading 1/r term in the potential at large distances from the system,
Qnet
= lim r × v(r) .
r→∞
4πǫ0
(78)
A quick glance on the solution (73) for the potential outside the outer shell shows that
lim r × v(r)
= aV0
(79)
Qnet = 4πǫ0 × a × V0 .
(80)
r→∞
and therefore
The other way to get the net charge is to simply integrate the given σ = k cos θ over the
outer shell and the solution to part (b) over the surface of the inner ball. Thus
Qouter shell =
Zπ
0
k cos θ × 2πb2 sin θ dθ = 0,
(81)
while
inner ball
Q
=
Zπ 0
ǫ0 V0
ǫ0 V0
− k cos θ × 2πa2 sin θ dθ =
× 4πa2 + 0 = 4πǫ0 aV0 , (82)
a
a
hence
Qnet = Qouter shell + Qinner ball = 4πǫ0 aV0 .
(83)
We see that both methods yield the same net charge, and this gives us a useful consistency
check for our calculations.
16