PDES: SEPARATION OF VARIABLES AND OTHER METHODS
x
z
y
a
−a
φ
θ
v=0
v=v 0
r
Figure 21.7 A hollow split conducting sphere with its top half charged to a
potentialv 0 and its bottom half at zero potential.
where in the last line we have used (21.50). The integrals of the Legendre polynomials are
easily evaluated (see exercise 17.3) and we find
A 0 =
v 0
2
,A 1 =
3 v 0
4 a
,A 2 =0,A 3 =−
7 v 0
16 a^3
, ···,
so that the required solution inside the sphere is
v(r, θ, φ)=
v 0
2
[
1+
3 r
2 a
P 1 (cosθ)−
7 r^3
8 a^3
P 3 (cosθ)+···
]
.
Outside the sphere (forr>a) we require the solution to be bounded asrtends to
infinity and so in (21.51) we must haveA=0forall. In this case, by imposing the
boundary condition atr=awe require
v(a, θ, φ)=
∑∞
=0
Ba−(+1)P(cosθ),
wherev(a, θ, φ) is given by (21.50). Following the above argument the coefficients in the
expansion are given by
Ba−(+1)=
2 +1
2
v 0
∫ 1
0
P(μ)dμ,
so that the required solution outside the sphere is
v(r, θ, φ)=
v 0 a
2 r
[
1+
3 a
2 r
P 1 (cosθ)−
7 a^3
8 r^3
P 3 (cosθ)+···