Required math: calculus
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problem 4.24.
As an exercise in applying Laplace’s equation to a problem with dielectrics, suppose we have a conducting sphere of radius surrounded by a spherical shell of dielectric of outer radius and susceptibility , with the whole system in a uniform electric field .
The general solution to Laplace’s equation for the potential in spherical coordinates is
where is the degree- Legendre polynomial.
Inside the sphere, the potential is constant (since the field is zero inside a conductor), so we might as well take it to be zero.
In the region , the solution is the general one given above. For , in order to avoid an infinite field for large , we have to drop the terms except for since we need a potential that gives a constant field. If we take the field to lie in the direction, then since we have for this region
Note that we’ve used a different set of coefficients here, since the solution in this region is distinct from that for the region . That is, the coefficients .
We can obtain conditions on the coefficients by equating terms of the various Legendre polynomials in the boundary conditions. Continuity of the potential at gives the condition
Continuity at gives
Finally, we can use the condition at the boundary of two dielectrics to get, since there is no free charge at the boundary:
Consider first the terms for . We get
Since both and are proportional to , the only way the third boundary condition (at the dielectric/vacuum boundary) above can be satisfied is if for . We are therefore left with the terms. For these we get
Plugging these into the third boundary condition gives
In terms of the dielectric constant we get
The potential inside the dielectric shell is therefore
The field can be found from the gradient
This reduces to the situation of a conducting sphere in a uniform field if we set (effectively replacing the dielectric by a vacuum). In that case, the potential reduces to
The first term is just the applied field, and the second term arises from the induced charge on the conductor.