Required math: calculus
Required physics: electrostatics
Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 4.2.1; Problem 4.10.
The potential of an ideal dipole is
where is the vector from the dipole to the observation point.
If we now consider an object consisting of polarized dielectric, we can define the polarization density, or polarization per unit volume as . The potential due to such an object is then
Note that this is a volume integral over the primed coordinates , that is, over the location of the volume element containing the polarized material.
We can transform this integral by doing a calculation of a gradient:
where we’ve omitted the terms in and in the first two lines since they have the same form. We can therefore write the original integral as
A standard theorem from vector calculus says, for a function and vector field :
so we can transform the integral to get
The first integral is a volume integral of a divergence so we can apply the divergence theorem to transform this to a surface integral, so we finally get
where the first integral is over the surface of the polarized object, and the second integral is over its volume. That is, the potential of a polarized object can be expressed as the sum of the potential of a surface charge density and a volume charge density, where we have
These charge distributions are known as bound charges, which is why we’ve used a subscript . If we’re interested in the electric field of a dipole distribution, we can work out these integrals and then take the negative gradient to get , or, if the problem has the right symmetry, we can use Gauss’s law to work out the fields of the two charge distributions and then add them together.
As an example, suppose we have a sphere of radius with a polarization density given by
where is a constant. Then
where the last line can be found by expressing in rectangular coordinates. Note that the total charge in the shell is equal and opposite to the total volume charge .
Rather than working out the integrals above, we can use Gauss’s law to find the electric field from a spherical shell and a uniformly charged sphere. We’ve already solved this problem in examples 1 and 2 here, so we can just quote the results. Inside the sphere, the spherical shell contributes nothing, and the volume charge gives a radial field of
Outside the sphere, the shell contributes
and the volume charge contributes
so the total field is .