5.14: Mixed Dielectrics

This section addresses the question: If there are two or more dielectric media between the plates of a capacitor, with different permittivities, are the electric fields in the two media different, or are they the same? The answer depends on

1. Whether by “electric field” you mean $E$ or $D$;
2. The disposition of the media between the plates – i.e. whether the two dielectrics are in series or in parallel.

Let us first suppose that two media are in series (Figure $V.$16).

$\text{FIGURE V.16}$

Our capacitor has two dielectrics in series, the first one of thickness $d_1$ and permittivity $\epsilon_1$ and the second one of thickness $d_2$ and permittivity $\epsilon_2$. As always, the thicknesses of the dielectrics are supposed to be small so that the fields within them are uniform. This is effectively two capacitors in series, of capacitances $\epsilon_1A/d_1 \text{ and }\epsilon_2A/d_2$. The total capacitance is therefore

$$C=\frac{\epsilon_1\epsilon_2A}{\epsilon_2d_1+\epsilon_1d_2}.\label{5.14.1}$$

Let us imagine that the potential difference across the plates is $V_0$. Specifically, we’ll suppose the potential of the lower plate is zero and the potential of the upper plate is $V_0$. The charge $Q$ held by the capacitor (positive on one plate, negative on the other) is just given by $Q = CV_0$, and hence the surface charge density $\sigma$ is $CV_0/A$. Gauss’s law is that the total $D$-flux arising from a charge is equal to the charge, so that in this geometry $D = \sigma$, and this is not altered by the nature of the dielectric materials between the plates. Thus, in this capacitor, $D = CV_0/A = Q/A$ in both media. Thus $D$ is continuous across the boundary.

Then by application of $D = \epsilon E$ to each of the media, we find that the $E$-fields in the two media are $E_1$=$Q$/$(\epsilon_1A$) and $E_2$=$Q$/$(\epsilon_2A$), the $E$-field (and hence the potential gradient) being larger in the medium with the smaller permittivity.

The potential V at the media boundary is given by $V/d_2=E_2$. Combining this with our expression for $E_2$, and $Q = CV$and Equation $\ref{5.14.1}$, we find for the boundary potential:

$$V=\frac{\epsilon_1d_2}{\epsilon_2d_1+\epsilon_1d_2}V_0.\label{5.14.2}$$

Let us now suppose that two media are in parallel (Figure $V.$17).

$\text{FIGURE V.17}$

This time, we have two dielectrics, each of thickness $d$, but one has area $A_1$ and permittivity $\epsilon_1$ while the other has area $A_2$ and permittivity $\epsilon_2$. This is just two capacitors in parallel, and the total capacitance is

$$C=\frac{\epsilon_1A_1}{d}+\frac{\epsilon_2A_2}{d}\label{5.14.3}$$

The $E$-field is just the potential gradient, and this is independent of any medium between the plates, so that $E = V/d$. in each of the two dielectrics. After that, we have simply that $D_1=\epsilon_1E \text{ and }D_2=\epsilon_2E$. The charge density on the plates is given by Gauss’s law as $\sigma = D$, so that, if $\epsilon_1 < \epsilon_2$, the charge density on the left hand portion of each plate is less than on the right hand portion – although the potential is the same throughout each plate. (The surface of a metal is always an equipotential surface.) The two different charge densities on each plate is a result of the different polarizations of the two dielectrics – something that will be more readily understood a little later in this chapter when we deal with media polarization.

We have established that:

1. The component of $\textbf{D}$ perpendicular to a boundary is continuous;
2. The component of $\textbf{E}$ parallel to a boundary is continuous.

In Figure $V.$18 we are looking at the $D$-field and at the $E$-field as it crosses a boundary in which $\epsilon_1 < \epsilon_2$. Note that $D_y$ and $E_x$are the same on either side of the boundary. This results in:

$$\frac{\tan \theta_1}{\tan \theta_2}=\frac{\epsilon_1}{\epsilon_2}.\label{5.14.4}$$

$\text{FIGURE V.18}$