21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES
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aFigure 21.6 Induced charge and field lines associated with a conducting
sphere placed in an initially uniform electrostatic field.A hollow split conducting sphere of radiusais placed at the origin. If one half of its
surface is charged to a potentialv 0 and the other half is kept at zero potential, find the
potentialvinside and outside the sphere.Let us choose the top hemisphere to be charged tov 0 and the bottom hemisphere to be
at zero potential, with the plane in which the two hemispheres meet perpendicular to the
polar axis; this is shown in figure 21.7. The boundary condition then becomes
v(a, θ, φ)={
v 0 for 0<θ<π/2(0<cosθ<1),
0forπ/ 2 <θ<π (− 1 <cosθ<0). (21.50)The problem is clearly axially symmetric and so we may setm= 0. Also, we require the
solution to be finite on the polar axis and so it cannot containQ(cosθ). Therefore the
general form of the solution to (21.38) is
v(r, θ, φ)=∑∞
=0(Ar+Br−(+1))P(cosθ). (21.51)Inside the sphere (forr<a) we require the solution to be finite at the origin and so
B=0forallin (21.51). Imposing the boundary condition atr=awe must then have
v(a, θ, φ)=∑∞
=0AaP(cosθ),wherev(a, θ, φ) is also given by (21.50). Exploiting the mutual orthogonality of the Legendre
polynomials, the coefficients in the Legendre polynomial expansion are given by (18.14)
as (writingμ=cosθ)
Aa=2 +1
2
∫ 1
− 1v(a, θ, φ)P(μ)dμ=
2 +1
2
v 0∫ 1
0P(μ)dμ,