# Energy of conducting sphere in a dielectric shell

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

$\displaystyle W=\frac{1}{2}\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r} \ \ \ \ \ (1)$

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 ${a}$ that has a free charge ${Q}$ on it, and we surround this conductor with a spherical shell of linear dielectric that extends from ${r=a}$ to ${r=b}$. 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 ${\mathbf{D}}$ from its relation to the free charge. That is,

$\displaystyle \int_{A}\mathbf{D}\cdot d\mathbf{a}=Q_{f} \ \ \ \ \ (2)$

where ${Q_{f}}$ is the free charge (excluding the bound charge) that is contained within the surface of integration.

In this case, ${Q_{f}=Q}$ and because the system has spherical symmetry, we can take the surface of integration to be a sphere. Since the enclosed free charge is ${Q}$ for any surface with ${r>a}$, we get in this region:

 $\displaystyle 4\pi r^{2}D$ $\displaystyle =$ $\displaystyle Q\ \ \ \ \ (3)$ $\displaystyle \mathbf{D}$ $\displaystyle =$ $\displaystyle \frac{Q}{4\pi r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (4)$

For a linear dielectric ${\mathbf{D}=\epsilon\mathbf{E}=\epsilon_{0}\left(1+\chi_{e}\right)\mathbf{E}}$. We therefore have for the electric field:

$\displaystyle \mathbf{E}=\begin{cases} \frac{Q}{4\pi\epsilon_{0}\left(1+\chi_{e}\right)r^{2}}\hat{\mathbf{r}} & ab \end{cases} \ \ \ \ \ (5)$

The energy is then

 $\displaystyle W$ $\displaystyle =$ $\displaystyle \frac{1}{2}\int\left(\mathbf{D}\cdot\mathbf{E}\right)d^{3}\mathbf{r}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{4\pi Q^{2}}{2\left(4\pi\right)^{2}\epsilon_{0}}\left[\int_{a}^{b}\frac{r^{2}dr}{\left(1+\chi_{e}\right)r^{4}}+\int_{b}^{\infty}\frac{r^{2}dr}{r^{4}}\right]\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Q^{2}}{8\pi\epsilon_{0}}\left[\frac{1}{1+\chi_{e}}\left(\frac{1}{a}-\frac{1}{b}\right)+\frac{1}{b}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{Q^{2}}{8\pi\epsilon_{0}\left(1+\chi_{e}\right)}\left(\frac{1}{a}+\frac{\chi_{e}}{b}\right) \ \ \ \ \ (9)$

If we remove the dielectric, this is the same as setting ${\chi_{e}=0}$, so the energy stored in the field of a charged conducting sphere is

$\displaystyle W=\frac{Q^{2}}{8\pi\epsilon_{0}a} \ \ \ \ \ (10)$

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