PDES: SEPARATION OF VARIABLES AND OTHER METHODS
xzya−aφθv=0v=v 0rFigure 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 aP 1 (cosθ)−7 r^3
8 a^3P 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)=∑∞
=0Ba−(+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
0P(μ)dμ,so that the required solution outside the sphere is
v(r, θ, φ)=v 0 a
2 r[
1+
3 a
2 rP 1 (cosθ)−7 a^3
8 r^3P 3 (cosθ)+···