656 Laplace’s Equation: Cylindrical Coordinates
where A
mn
and B
mn
are constants to be determined by the remaining bound-
ary condition which states that Φ(ρ, ϕ, h)=V (ρ, ϕ)or
V (ρ, ϕ)=
∞
m=0
∞
n=1
J
m
x
mn
a
ρ
sinh
x
mn
a
h
(A
mn
cos mϕ + B
mn
sin mϕ).
(27.40)
Multiplying both sides by ρJ
m
(x
mk
a/ρ)cosjϕ and integrating from zero to
2π in ϕ, and from zero to a in ρ gives A
jk
. Changing cosine to sine and
following the same steps yields B
jk
. Switching back to m and n,thereader
may verify that
A
mn
=
2
#
2π
0
dϕ
#
a
0
dρ ρV (ρ, ϕ)J
m
x
mn
a
ρ
cos mϕ
πa
2
J
2
m+1
(x
mn
) sinh(x
mn
h/a)
,
B
mn
=
2
#
2π
0
dϕ
#
a
0
dρ ρV (ρ, ϕ)J
m
x
mn
a
ρ
sin mϕ
πa
2
J
2
m+1
(x
mn
)sinh(x
mn
h/a)
, (27.41)
where we have used Equation (27.24).
The important case of azimuthal symmetry requires special consideration.
In such a case, the potential of the top surface V (ρ, ϕ) must be independent
of ϕ. Furthermore, since S(ϕ) is constant,
5
its derivative must vanish. Hence,
the second equation in (27.1) yields μ = −m
2
= 0. This zero value for m
reduces the double summation of (27.39) to a single sum, and we get
Φ(ρ, z)=
∞
n=1
A
n
J
0
x
0n
a
ρ
sinh
x
0n
a
z
. (27.42)
The coefficients A
n
can be obtained by setting m = 0 in the first equation of
(27.41):
A
n
=
4
a
2
J
2
1
(x
0n
) sinh(x
0n
h/a)
#
a
0
ρV (ρ)J
0
x
0n
a
ρ
dρ, (27.43)
where V (ρ)istheϕ-independent potential of the top surface.
Example 27.6.1.
Suppose that the top face of a conducting cylindrical can is
held at the constant potential V
0
while the lateral surface and the bottom face are
grounded. We want to find the electrostatic potential Φ at all points inside the can.
Since the potential of the top is independent of ϕ, azimuthal symmetry prevails,
and Equation (27.43) gives
A
n
=
4V
0
a
2
J
2
1
(x
0n
) sinh(x
0n
h/a)
#
a
0
ρJ
0
x
0n
a
ρ
dρ =
4V
0
x
0n
J
1
(x
0n
) sinh(x
0n
h/a)
,
where we used Equation (27.18). The detail of calculating the integral is left as
Problem 27.15 for the reader. Therefore,
Φ(ρ, z)=4V
0
∞
n=1
J
0
(x
0n
ρ/a) sinh(x
0n
z/a)
x
0n
J
1
(x
0n
) sinh(x
0n
h/a)
.
5
S(ϕ) must be a constant. Otherwise, the potential would depend on ϕ.