Polarization charges
Now let us see what this model gives for the theory of a condenser with a dielectric. First consider a sheet of material in which there is a certain dipole moment per unit volume. Will there be on the average any charge density produced by this?  Not if $\FLPP$ is uniform. If the positive and negative charges being displaced relative to each other have the same average density, the fact that they are displaced does not produce any net charge inside the volume. On the other hand, if $\FLPP$ were larger at one place and smaller at another, that would mean that more charge would be moved into some region than away from it; we would then expect to get a volume density of charge. For the parallel-plate condenser, we suppose that $\FLPP$ is uniform, so we need to look only at what happens at the surfaces. At one surface the negative charges, the electrons, have effectively moved out a distance $\delta$; at the other surface they have moved in, leaving some positive charge effectively out a distance $\delta$. As shown in Fig. 10–5, we will have a surface density of charge, which will be called the surface polarization charge. This charge can be calculated as follows. If $A$ is the area of the plate, the number of electrons that appear at the surface is the product of $A$ and $N$, the number per unit volume, and the displacement $\delta$, which we assume here is perpendicular to the surface. The total charge is obtained by multiplying by the electronic charge $q_e$. To get the surface density of the polarization charge induced on the surface, we divide by $A$. The magnitude of the surface charge density is \begin{equation*} \sigma_{\text{pol}}=Nq_e\delta. \end{equation*}  But this is just equal to the magnitude $P$ of the polarization vector $\FLPP$, Eq. (10.4): \begin{equation} \label{Eq:II:10:5} \sigma_{\text{pol}}=P. \end{equation}  The surface density of charge is equal to the polarization inside the material. The surface charge is, of course, positive on one surface and negative on the other. Now let us assume that our slab is the dielectric of a parallel-plate capacitor. The plates of the capacitor also have a surface charge, which we will call $\sigma_{\text{free}}$, because they can move “freely” anywhere on the conductor. This is, of course, the charge that we put on when we charged the capacitor. It should be emphasized that $\sigma_{\text{pol}}$ exists only because of $\sigma_{\text{free}}$. If $\sigma_{\text{free}}$ is removed by discharging the capacitor, then $\sigma_{\text{pol}}$ will disappear, not by going out on the discharging wire, but by moving back into the material—by the relaxation of the polarization inside the material. We can now apply Gauss’ law to the Gaussian surface $S$ in Fig. 10–1. The electric field $\FLPE$ in the dielectric is equal to the total surface charge density divided by $\epsO$. It is clear that $\sigma_{\text{pol}}$ and $\sigma_{\text{free}}$ have opposite signs, so \begin{equation} \label{Eq:II:10:6} E=\frac{\sigma_{\text{free}}-\sigma_{\text{pol}}}{\epsO}. \end{equation}  Note that the field $E_0$ between the metal plate and the surface of the dielectric is higher than the field $E$; it corresponds to $\sigma_{\text{free}}$ alone. But here we are concerned with the field inside the dielectric which, if the dielectric nearly fills the gap, is the field over nearly the whole volume. Using Eq. (10.5), we can write \begin{equation} \label{Eq:II:10:7} E=\frac{\sigma_{\text{free}}-P}{\epsO}. \end{equation}  This equation doesn’t tell us what the electric field is unless we know what $P$ is. Here, however, we are assuming that $P$ depends on $E$—in fact, that it is proportional to $E$. This proportionality is usually written as \begin{equation} \label{Eq:II:10:8} \FLPP=\chi\epsO\FLPE. \end{equation}  The constant $\chi$ (Greek “khi”) is called the electric susceptibility of the dielectric. Then Eq. (10.7) becomes \begin{equation} \label{Eq:II:10:9} E=\frac{\sigma_{\text{free}}}{\epsO}\,\frac{1}{(1+\chi)}, \end{equation}  which gives us the factor $1/(1+\chi)$ by which the field is reduced. The voltage between the plates is the integral of the electric field. Since the field is uniform, the integral is just the product of $E$ and the plate separation $d$. We have that \begin{equation*} V=Ed=\frac{\sigma_{\text{free}}d}{\epsO(1+\chi)}. \end{equation*}  The total charge on the capacitor is $\sigma_{\text{free}}A$, so that the capacitance defined by (10.2) becomes \begin{equation} \label{Eq:II:10:10} C=\frac{\epsO A(1+\chi)}{d}=\frac{\kappa\epsO A}{d}. \end{equation}  We have explained the observed facts. When a parallel-plate capacitor is filled with a dielectric, the capacitance is increased by the factor \begin{equation} \label{Eq:II:10:11} \kappa=1+\chi, \end{equation}  which is a property of the material. Our explanation, of course, is not complete until we have explained—as we will do later—how the atomic polarization comes about. Let’s now consider something a little bit more complicated—the situation in which the polarization $\FLPP$ is not everywhere the same. As mentioned earlier, if the polarization is not constant, we would expect in general to find a charge density in the volume, because more charge might come into one side of a small volume element than leaves it on the other. How can we find out how much charge is gained or lost from a small volume? First let’s compute how much charge moves across any imaginary surface when the material is polarized. The amount of charge that goes across a surface is just $P$ times the surface area if the polarization is normal to the surface. Of course, if the polarization is tangential to the surface, no charge moves across it. Following the same arguments we have already used, it is easy to see that the charge moved across any surface element is proportional to the component of $\FLPP$ perpendicular to the surface. Compare Fig. 10–6 with Fig. 10–5. We see that Eq. (10.5) should, in the general case, be written \begin{equation} \label{Eq:II:10:12} \sigma_{\text{pol}}=\FLPP\cdot\FLPn. \end{equation}  If we are thinking of an imagined surface element inside the dielectric, Eq. (10.12) gives the charge moved across the surface but doesn’t result in a net surface charge, because there are equal and opposite contributions from the dielectric on the two sides of the surface. The displacements of the charges can, however, result in a volume charge density. The total charge displaced out of any volume $V$ by the polarization is the integral of the outward normal component of $\FLPP$ over the surface $S$ that bounds the volume (see Fig. 10–7). An equal excess charge of the opposite sign is left behind.  Denoting the net charge inside $V$ by $\Delta Q_{\text{pol}}$ we write \begin{equation} \label{Eq:II:10:13} \Delta Q_{\text{pol}}=-\int_S\FLPP\cdot\FLPn\,da. \end{equation}  We can attribute $\Delta Q_{\text{pol}}$ to a volume distribution of charge with the density $\rho_{\text{pol}}$, and so \begin{equation} \label{Eq:II:10:14} \Delta Q_{\text{pol}}=\int_V\rho_{\text{pol}}\,dV. \end{equation}  Combining the two equations yields \begin{equation} \label{Eq:II:10:15} \int_V\rho_{\text{pol}}\,dV=-\int_S\FLPP\cdot\FLPn\,da. \end{equation}  We have a kind of Gauss’ theorem that relates the charge density from polarized materials to the polarization vector $\FLPP$. We can see that it agrees with the result we got for the surface polarization charge or the dielectric in a parallel-plate capacitor. Using Eq. (10.15) with the Gaussian surface of Fig. 10–1, the surface integral gives $P\,\Delta A$, and the charge inside is $\sigma_{\text{pol}}\,\Delta A$, so we get again that $\sigma_{\text{pol}}=P$. Just as we did for Gauss’ law of electrostatics, we can convert Eq. (10.15) to a differential form—using Gauss’ mathematical theorem: \begin{equation} \int_S\FLPP\cdot\FLPn\,da=\int_V\FLPdiv{\FLPP}\,dV.\notag \end{equation}  We get \begin{equation} \label{Eq:II:10:16} \rho_{\text{pol}}=-\FLPdiv{\FLPP}. \end{equation}  If there is a nonuniform polarization, its divergence gives the net density of charge appearing in the material. We emphasize that this is a perfectly real charge density; we call it “polarization charge” only to remind ourselves how it got there.