Electric fields in cavities of a dielectric
We now turn to an interesting but complicated question—the problem of the dielectric constant in dense materials. Suppose that we take liquid helium or liquid argon or some other nonpolar material. We still expect electronic polarization. But in a dense material, $\FLPP$ can be large, so the field on an individual atom will be influenced by the polarization of the atoms in its close neighborhood. The question is, what electric field acts on the individual atom? Imagine that the liquid is put between the plates of a condenser. If the plates are charged they will produce an electric field in the liquid. But there are also charges in the individual atoms, and the total field $\FLPE$ is the sum of both of these effects. This true electric field varies very, very rapidly from point to point in the liquid. It is very high inside the atoms—particularly right next to the nucleus—and relatively small between the atoms. The potential difference between the plates is the line integral of this total field. If we ignore all the fine-grained variations, we can think of an average electric field $E$, which is just $V/d$. (This is the field we were using in the last chapter.) We should think of this field as the average over a space containing many atoms. Now you might think that an “average” atom in an “average” location would feel this average field. But it is not that simple, as we can show by considering what happens if we imagine different-shaped holes in a dielectric. For instance, suppose that we cut a slot in a polarized dielectric, with the slot oriented parallel to the field, as shown in part (a) of Fig. 11–5. Since we know that $\FLPcurl{\FLPE}=\FLPzero$, the line integral of $\FLPE$ around the curve, $\Gamma$, which goes as shown in (b) of the figure, should be zero. The field inside the slot must give a contribution which just cancels the part from the field outside. Therefore the field $E_0$ actually found in the center of a long thin slot is equal to $E$, the average electric field found in the dielectric. Now consider another slot whose large sides are perpendicular to $E$, as shown in part (c) of Fig. 11–5. In this case, the field $E_0$ in the slot is not the same as $E$ because polarization charges appear on the surfaces. If we apply Gauss’ law to a surface $S$ drawn as in (d) of the figure, we find that the field $E_0$ in the slot is given by \begin{equation} \label{Eq:II:11:22} E_0=E+\frac{P}{\epsO}, \end{equation}  where $E$ is again the electric field in the dielectric. (The Gaussian surface contains the surface polarization charge $\sigma_{\text{pol}}=P$.) We mentioned in Chapter 10 that $\epsO E+P$ is often called $D$, so $\epsO E_0=D_0$ is equal to $D$ in the dielectric. Earlier in the history of physics, when it was supposed to be very important to define every quantity by direct experiment, people were delighted to discover that they could define what they meant by $E$ and $D$ in a dielectric without having to crawl around between the atoms. The average field $\FLPE$ is numerically equal to the field $\FLPE_0$ that would be measured in a slot cut parallel to the field. And the field $\FLPD$ could be measured by finding $E_0$ in a slot cut normal to the field. But nobody ever measures them that way anyway, so it was just one of those philosophical things. For most liquids which are not too complicated in structure, we could expect that an atom finds itself, on the average, surrounded by the other atoms in what would be a good approximation to a spherical hole. And so we should ask: “What would be the field in a spherical hole?” We can find out by noticing that if we imagine carving out a spherical hole in a uniformly polarized material, we are just removing a sphere of polarized material. (We must imagine that the polarization is “frozen in” before we cut out the hole.) By superposition, however, the fields inside the dielectric, before the sphere was removed, is the sum of the fields from all charges outside the spherical volume plus the fields from the charges within the polarized sphere. That is, if we call $E$ the field in the uniform dielectric, we can write \begin{equation} \label{Eq:II:11:23} E=E_{\text{hole}}+E_{\text{plug}}, \end{equation}  where $E_{\text{hole}}$ is the field in the hole and $E_{\text{plug}}$ is the field inside a sphere which is uniformly polarized (see Fig. 11–6). The fields due to a uniformly polarized sphere are shown in Fig. 11–7. The electric field inside the sphere is uniform, and its value is \begin{equation} \label{Eq:II:11:24} E_{\text{plug}}=-\frac{P}{3\epsO}. \end{equation}  Using (11.23), we get \begin{equation} \label{Eq:II:11:25} E_{\text{hole}}=E+\frac{P}{3\epsO}. \end{equation}  The field in a spherical cavity is greater than the average field by the amount $P/3\epsO$. (The spherical hole gives a field $1/3$ of the way between a slot parallel to the field and a slot perpendicular to the field.)