Page 513 - Electromagnetics

Next we solve the boundary value problem
2
∇ V (r,θ,φ) = 0,
V (a,θ,φ) =−V 0 , π/2 ≤ θ< π, −π ≤ φ ≤ π,
V (a,θ,φ) =+V 0 , 0 <θ ≤ π/2, −π ≤ φ ≤ π,

for both r > a and r < a. This yields the potential field of a conducting sphere split
into top and bottom hemispheres and held at a potential difference of 2V 0 . Azimuthal
symmetry gives µ = 0. The two possible solutions for %(θ) are

A θ + B θ ln cot(θ/2), n = 0,
%(θ) =
A θ P n (cos θ), n = 0,
0
where we have discarded Q (cos θ) because the region of interest contains the z-axis. The
0
n = 0 solution cannot match the boundary conditions; neither can a single term of the
type A θ P n (cos θ), but a series of these latter terms can. We use
∞ ∞
n −(n+1)
V (r,θ) = V n (r,θ) = [A r r + B r r ]P n (cos θ). (A.147)
n=0 n=0
n
The terms r −(n+1) and r are not allowed, respectively, for r < a and r > a.For r < a
then,
∞
n
V (r,θ) = A n r P n (cos θ).
n=0
Letting V 0 (θ) be the potential on the surface of the split sphere, we impose the boundary
condition:
∞
n
V (a,θ) = V 0 (θ) = A n a P n (cos θ), 0 ≤ θ ≤ π.
n=0
This is a Fourier–Legendre expansion of V 0 (θ). The A n are evaluated by orthogonality.
Multiplying by P m (cos θ) sin θ and integrating from θ = 0 to π, we obtain

∞ π π
n
A n a P n (cos θ)P m (cos θ) sin θ dθ = V 0 (θ)P m (cos θ) sin θ dθ.
n=0 0 0
Using orthogonality relationship (A.93) and the given V 0 (θ) we have
2 π/2 π
m
A m a = V 0 P m (cos θ) sin θ dθ − V 0 P m (cos θ) sin θ dθ.
2m + 1 0 π/2
The substitution η = cos θ gives

2 1 0
m
A m a = V 0 P m (η) dη − V 0 P m (η) dη
2m + 1 0 −1
1 1
= V 0 P m (η) dη − V 0 P m (−η) dη;
0 0
m
then P m (−η) = (−1) P m (η) gives
2m + 1 m 1
−m
A m = a V 0 [1 − (−1) ] P m (η) dη.
2 0

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