Energy Density Within 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

$$\delta U = \int_V \phi \,\delta\rho_f \,dV,$$ (559)

where the integral is taken over all space, and $\phi({\bf r})$ 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 Equation (501) that

$$\delta U = \int_V \phi\,\nabla \cdot\delta {\bf D}\,dV,$$ (560)

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

$$\delta U = \int_V \nabla \cdot (\phi\,\delta{\bf D})\,dV - \int_V \nabla\phi \cdot\delta{\bf D}\,dV,$$ (561)

giving

$$\delta U = \int_S \phi\, \delta{\bf D}\cdot d{\bf S} - \int_V \nabla\phi\cdot\delta{\bf D}\,dV,$$ (562)

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

$$\delta U = -\int_V\nabla\phi \cdot \delta{\bf D}\,dV = \int_V {\bf E}\cdot \delta {\bf D}\,dV.$$ (563)

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

$$U = \int_{\bf0}^{\bf D} \delta U = \int_{\bf0}^{\bf D }\int_V {\bf E} \cdot \delta{\bf D}\,dV,$$ (564)

or

$$U = \int_V \int_0^E \frac{\epsilon_0\,\epsilon \,\delta (E^{\,2})}{2}\,dV = \frac{1}{2} \int_V \epsilon_0\,\epsilon\, E^{\,2}\,dV,$$ (565)

which reduces to

$$U = \frac{1}{2} \int_V {\bf E}\cdot{\bf D}\,dV.$$ (566)

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

$$W = \frac{1}{2}\,{\bf E}\cdot{\bf D}.$$ (567)

This is a standard result that 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, ${\bf D}$ , varies linearly with the electric field, ${\bf E}$ .