21.3 SEPARATION OF VARIABLES IN POLAR COORDINATES
+ + + + + + + +
+
+
+
+
−
−
−
−
−
−
−
−
−
−
−
−
θ
a
Figure 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)=
∑∞
=0
AaP(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
− 1
v(a, θ, φ)P(μ)dμ
=
2 +1
2
v 0
∫ 1
0
P(μ)dμ,