**Required math: calculus**

**Required physics: electrostatics**

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Section 4.4.3 & Problem 4.26.

We’ve seen that the energy in a system containing dielectrics can be written as

This is the energy required to place the free charges, and includes the energy needed to polarize the dielectric.

As a simple example of this formula, suppose we have a spherical conductor of radius that has a free charge on it, and we surround this conductor with a spherical shell of linear dielectric that extends from to . The free charge on the conductor will polarize the dielectric, resulting in surface charges on the inner and outer surfaces of the dielectric.

We can find the displacement from its relation to the free charge. That is,

where is the free charge (excluding the bound charge) that is contained within the surface of integration.

In this case, and because the system has spherical symmetry, we can take the surface of integration to be a sphere. Since the enclosed free charge is for any surface with , we get in this region:

For a linear dielectric . We therefore have for the electric field:

The energy is then

If we remove the dielectric, this is the same as setting , so the energy stored in the field of a charged conducting sphere is

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georgianajackalgot this easily 🙂