Charges in dielectric medium

Now let us consider the energy associated with charges in dielectric medium. In the medium also the curl of electric field vanishes ( $\vec \nabla \times \vec E(\vec r) = 0$ ) so we can express the electric field as the negative gradiant of electric potential $\vec E(\vec r) = - \vec \nabla \phi(\vec r)$. If $\rho(\vec r)$ is the charge density in the medium, the change in the energy because of change in the charge density because of the change in the charge density $\delta \rho(\vec r)$ is

$$\delta E = \int d^3r \phi(\vec r) \delta \rho(\vec r) = \frac... ...\int d^3r \phi(\vec r) \vec \nabla \cdot \delta \vec D(\vec r).$$

Here we have used the expression of displacement vector in terms of the charge density. Now, if we do partial integration and assume that the surface at infinity does not contribute the energy we get

$$\delta E = - \frac{1}{4 \pi} \int d^3r \delta \vec D(\vec r) \cdot \vec \nabla \phi(\vec r)$$

$$= \frac{1}{4\pi} \int d^3r \delta \vec D(\vec r) \cdot \vec E(\vec r)$$ (148)

So far, we have not assumed any relation between the displacement vector and the electric field. The total energy of the system is obtained by integrating $\delta E$ starting with no charges upto the given charge density $\rho(\vec r)$. If the dielectric medium is linear, things become simpler and $\delta \vec D = \kappa \delta \vec E$ assuming that the dielectric remains same, that is the dielectric constant does not change. We can then write

$$\delta \vec D \cdot \vec E = \kappa \delta \vec E \cdot \vec... .../2 \delta \vec E^2 = \frac{1}{2} \delta (\vec D \cdot \vec E).$$

After integration, the energy of the system of charges in the dielectric medium becomes

$$E = \frac{1}{8\pi} \int d^3r \vec D(\vec r) \cdot \vec E(\vec r)$$ (149)

A natural question to ask is why does the displacement vector enter in the definition of the energy. Physically, we can understand this as follows. For changing the electric charge density we need to do work to bring the excess charge from infinity. We also need to do work to polarise the medium to produce a change in the bound charge distribution. Thus the energy of the system is due to build up of free charges as well as the reistribution of bound charges.

Let us consider the case when the distribution of ( free ) charges is fixed but the polarisability of the medium changes ( for example, a dielectric medium is introduced in the free space ). Let us consider that the dielectric constant of the medium is $\kappa_0$ initially and it changes to $\kappa$ in certain volume $V$. So $\vec D_0 = \kappa_0 \vec E_0$ and $\vec D = \kappa \vec E$. We shall assume that the media are linear. So the change in the energy of the system is

$$\Delta E = \frac{1}{8\pi}\int d^3r \Big ( \vec D(\vec r) \cdot \vec E(\vec r) - \vec D_0(\vec r) \cdot \vec E_0(\vec r) \Big )$$

$$= \frac{1}{8\pi} \int d^3r \Big ( \vec D_0(\vec r) \cdot \vec E(\vec r) - \vec D_(\vec r) \cdot \vec E_0(\vec r) \Big ) +$$

$$\frac{1}{8\pi} \int d^3r \Big ( ( \vec E(\vec r) + \vec E_0(\vec r) ) \cdot ( \vec D(\vec r) - \vec D_0(\vec r) ) \Big )$$ (150)

The second term in the expression above vanishes because $\vec \nabla \times ( \vec E + \vec E_0 ) =0$ and therefore we can write $\vec E + \vec E_0 = \vec \nabla \Phi$. Substituting this in the integral and doing a partial integration we get $\int d^3r \Phi (\vec r) \vec \nabla \cdot ( \vec D(\vec r) - \vec D_0(\vec r) ) = 0$ since we have not changed the charge distribution but only changed the dielectric constant of part of the medium. So, the change in the energy due to the change in the dielectric constant of the medium in certain region of the space is

$$\Delta E = \frac{1}{8\pi} \int d^3r \Big ( \vec D_0(\vec r) \cdot \vec E(\vec r) - \vec D_(\vec r) \cdot \vec E_0(\vec r) \Big )$$

$$= - \frac{1}{8\pi} ( \kappa - \kappa_0 ) \int d^3r \vec E(\vec r) \cdot \vec E_0(\vec r)$$ (151)

Here the integration is restricted to the region in which the dielectric constant changes. This expression tells us about the change in the energy due to an introduction of a dielectric medium. If we consider that a dielectric medium of dielectric constant $\kappa$ is introduced in vacuum ( $\kappa_0 = 0$ ), the change in the energy is

$$\Delta E = - \frac{1}{8\pi} ( \kappa - 1 ) \int d^3r \vec E(... ...) = -\frac{1}{2} \int d^3r \vec p(\vec r) \cdot \vec E(\vec r)$$

where $\vec p(\vec r)$ is the polarisation density of the medium. For $\kappa > 1$ the change is negative and increases with the increase in the magnitude of the electric field. Thus the electric field gives rise to a force on the medium.