# 5.14: Mixed Dielectrics

- Page ID
- 6022

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

- Whether by “electric field” you mean \(E\) or \(D\);
- 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:

- The component of \(\textbf{D}\) perpendicular to a boundary is continuous;
- 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:

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

\(\text{FIGURE V.18}\)