Potential Between Concentric Spheres
We want to find the potential V(r, θ) in the region between a grounded sphere of radius aand a concentric
spherical shell of radius bheld at potential V(b, θ) = 3
2sin2θ. The region of interest is a < r < b.
1 Governing Equation and Symmetry
In the region between the conductors (a < r < b), there is no charge density (ρ= 0). Therefore, the electric
potential Vmust satisfy Laplace’s equation:
∇2V= 0
Due to the spherical geometry and the θ-dependent boundary condition on the outer shell, we use spherical
coordinates (r, θ,φ). The potential Vwill be independent of the azimuthal angle φdue to the symmetry of
the problem (azimuthal symmetry), so V=V(r, θ). Laplace’s equation in spherical coordinates with ∂V
∂φ = 0
simplifies to:
1
r2
∂
∂rr2∂V
∂r+1
r2sin θ
∂
∂θ sin θ∂V
∂θ = 0
Multiplying by r2gives:
∂
∂rr2∂V
∂r+1
sin θ
∂
∂θ sin θ∂V
∂θ = 0
2 General Solution using Separation of Variables
The general solution to Laplace’s equation in spherical coordinates with azimuthal symmetry is found using
the method of separation of variables. It is given by an infinite series involving Legendre polynomials,
Pl(cos θ):
V(r, θ) =
∞
l=0 Alrl+Blr−(l+1)Pl(cos θ) (1)
where Aland Blare constants determined by the boundary conditions.
3 Applying Boundary Conditions
3.1 Inner Boundary (r = a)
The inner sphere is grounded, meaning its potential is zero:
V(a, θ) = 0
Substituting r=ainto the general solution (1):
∞
l=0 Alal+Bla−(l+1)Pl(cos θ) = 0
Since this must hold for all values of θ, and the Legendre polynomials are linearly independent (orthogonal),
the coefficient of each Pl(cos θ) must be zero:
Alal+Bla−(l+1) = 0
1
