The energy density within a dielectric medium

Consider a system of free charges embedded in a dielectric medium. The increase in the total energy when a small amount of free charge $\delta\rho_f$ is added to the system is given by

$$\delta U = \int \phi \,\delta\rho_f \,d^3{\bfm r},$$ (481)

where the integral is taken over all space and $\phi({\bfm r})$ is the electrostatic potential. Here, it is assumed that the original charges and the dielectric are held fixed, so that no mechanical work is performed. It follows from Eq. (3.5) that

$$\delta U = \int \phi\,\nabla\!\cdot\!\delta {\bfm D}\,d^3{\bfm r},$$ (482)

where $\delta{\bfm D}$ is the change in the electric displacement associated with the charge increment. Now the above equation can also be written

$$\delta U = \int \nabla\!\cdot\!(\phi\,\delta{\bfm D})\,d^3{\bfm r} - \int \nabla\phi\!\cdot\!\delta{\bfm D}\,d^3{\bfm r},$$ (483)

giving

$$\delta U = \int \phi \,\delta{\bfm D}\!\cdot\!d{\bfm S} - \int \nabla\phi\!\cdot\!\delta{\bfm D}\,d^3{\bfm r},$$ (484)

where use has been made of Gauss's theorem. If the dielectric medium is of finite spatial extent then we can neglect the surface term to give

$$\delta U = -\int\nabla\phi \!\cdot\!\delta{\bfm D}\,d^3{\bfm r} = \int {\bfm E}\!\cdot\! \delta {\bfm D}\,d^3{\bfm r}.$$ (485)

This energy increment cannot be integrated unless ${\bfm E}$ is a known function of ${\bfm D}$. Let us adopt the conventional approach and assume that ${\bfm D} =\epsilon_0 \epsilon {\bfm E}$, where the dielectric constant $\epsilon$ is independent of the electric field. The change in energy associated with taking the displacement field from zero to ${\bfm D}({\bfm r})$ at all points in space is given by

$$U = \int_{\bf0}^{\bfm D} \delta U = \int_{\bf0}^{\bfm D }\int {\bfm E}\!\cdot\! \delta{\bfm D}\,d^3{\bfm r},$$ (486)

or

$$U = \int \int_0^E \frac{\epsilon_0\epsilon \,\delta (E^2)}{2... ... r} = \frac{1}{2} \int \epsilon_0\epsilon E^2\,d^{3} {\bfm r},$$ (487)

which reduces to

$$U = \frac{1}{2} \int {\bfm E}\!\cdot\!{\bfm D}\,d^3{\bfm r}.$$ (488)

Thus, the electrostatic energy density inside a dielectric is given by

$$W = \frac{{\bfm E}\!\cdot\!{\bfm D}}{2}.$$ (489)

This is a standard result which is often quoted in textbooks. Nevertheless, it is important to realize that the above formula is only valid in dielectric media in which the electric displacement ${\bfm D}$ varies linearly with the electric field ${\bfm E}$.