Boundary value problems with dielectrics - II

Consider a plane slab of dielectric $\epsilon$ lying between $z=0$ and $z=b$. Suppose that this slab is placed in a uniform $z$-directed electric field of strength $E_0$. What is the field strength $E_1$ inside the slab?

Since there are no free charges and this is a one-dimensional problem, it is clear from Eq. (3.5) that the electric displacement $D$ is the same in both the dielectric slab and the vacuum region which surrounds it. In the vacuum region $D= \epsilon_0 E_0$, whereas $D= \epsilon_0\epsilon\, E_1$ in the dielectric. It follows that

$$E_1=\frac{E_0}{\epsilon}.$$ (451)

In other words, the electric field inside the slab is reduced by polarization charges. As before, there is zero polarization charge density inside the dielectric. However, there is a uniform polarization charge sheet on both surfaces of the slab. It is easily demonstrated that

$$\sigma_{\rm pol}(z=b)=-\sigma_{\rm pol} (z=0) = \epsilon_0 \,E_0\, \frac{\epsilon-1}{\epsilon}.$$ (452)

In the limit $\epsilon\gg 1$, the slab acts like a conductor and $E_1\rightarrow 0$.

Let us now generalize this result. Consider a dielectric medium whose dielectric constant $\epsilon$ varies with $z$. The medium is assumed to be of finite extent and is surrounded by a vacuum, so that $\epsilon(z)\rightarrow 1$ as $\vert z\vert\rightarrow \infty$. Suppose that this dielectric is placed in a uniform $z$-directed electric field $E_0$. What is the field $E(z)$ inside the dielectric?

We know that the electric displacement inside the dielectric is given by $D(z) = \epsilon_0\, \epsilon(z)\,E(z)$. We also know from Eq. (3.5) that, since there are no free charges and this is a one-dimensional problem,

$$\frac{d D(z)}{dz} = \epsilon_0\, \frac{d [\epsilon(z) E(z)]}{d z} = 0.$$ (453)

Furthermore, $E(z)\rightarrow E_0$ as $\vert z\vert\rightarrow \infty$. It follows that

$$E(z) = \frac{E_0}{\epsilon(z)}.$$ (454)

Thus, the electric field is inversely proportional to the dielectric constant inside the dielectric medium. The polarization charge density inside the dielectric is given by

$$\rho_b = \epsilon_0 \,\frac{d E(z)}{d z} = \epsilon_0\, E_0\frac{d}{dz} \!\left[\frac{1}{\epsilon(z)}\right].$$ (455)