Electronic polarization
We will first discuss the polarization of nonpolar molecules. We can start with the simplest case of a monatomic gas (for instance, helium). When an atom of such a gas is in an electric field, the electrons are pulled one way by the field while the nucleus is pulled the other way, as shown in Fig. 10–4. Although the atoms are very stiff with respect to the electrical forces we can apply experimentally, there is a slight net displacement of the centers of charge, and a dipole moment is induced.  For small fields, the amount of displacement, and so also the dipole moment, is proportional to the electric field. The displacement of the electron distribution which produces this kind of induced dipole moment is called electronic polarization. We have already discussed the influence of an electric field on an atom in Chapter 31 of Vol. I, when we were dealing with the theory of the index of refraction. If you think about it for a moment, you will see that what we must do now is exactly the same as we did then. But now we need worry only about fields that do not vary with time, while the index of refraction depended on time-varying fields. In Chapter 31 of Vol. I we supposed that when an atom is placed in an oscillating electric field the center of charge of the electrons obeys the equation \begin{equation} \label{Eq:II:11:2} m\,\frac{d^2x}{dt^2}+m\omega_0^2x=q_eE. \end{equation}  The first term is the electron mass times its acceleration and the second is a restoring force, while the right-hand side is the force from the outside electric field. If the electric field varies with the frequency $\omega$, Eq. (11.2) has the solution \begin{equation} \label{Eq:II:11:3} x=\frac{q_eE}{m(\omega_0^2-\omega^2)}, \end{equation}  which has a resonance at $\omega=\omega_0$. When we previously found this solution, we interpreted it as saying that $\omega_0$ was the frequency at which light (in the optical region or in the ultraviolet, depending on the atom) was absorbed. For our purposes, however, we are interested only in the case of constant fields, i.e., for $\omega=0$, so we can disregard the acceleration term in (11.2), and we find that the displacement is \begin{equation} \label{Eq:II:11:4} x=\frac{q_eE}{m\omega_0^2}. \end{equation}   From this we see that the dipole moment $p$ of a single atom is \begin{equation} \label{Eq:II:11:5} p=q_ex=\frac{q_e^2E}{m\omega_0^2}. \end{equation}  In this theory the dipole moment $p$ is indeed proportional to the electric field. People usually write \begin{equation} \label{Eq:II:11:6} \FLPp=\alpha\epsO\FLPE. \end{equation}  (Again the $\epsO$ is put in for historical reasons.) The constant $\alpha$ is called the polarizability of the atom, and has the dimensions $L^3$. It is a measure of how easy it is to induce a moment in an atom with an electric field. Comparing (11.5) and (11.6), our simple theory says that \begin{equation} \label{Eq:II:11:7} \alpha=\frac{q_e^2}{\epsO m\omega_0^2}=\frac{4\pi e^2}{m\omega_0^2}. \end{equation}  If there are $N$ atoms in a unit volume, the polarization $P$—the dipole moment per unit volume—is given by \begin{equation} \label{Eq:II:11:8} \FLPP=N\FLPp=N\alpha\epsO\FLPE. \end{equation}  Putting (11.1) and (11.8) together, we get \begin{equation} \label{Eq:II:11:9} \kappa-1=\frac{P}{\epsO E}=N\alpha \end{equation}  or, using (11.7), \begin{equation} \label{Eq:II:11:10} \kappa-1=\frac{4\pi Ne^2}{m\omega_0^2}. \end{equation}  From Eq. (11.10) we would predict that the dielectric constant $\kappa$ of different gases should depend on the density of the gas and on the frequency $\omega_0$ of its optical absorption. Our formula is, of course, only a very rough approximation, because in Eq. (11.2) we have taken a model which ignores the complications of quantum mechanics. For example, we have assumed that an atom has only one resonant frequency, when it really has many. To calculate properly the polarizability $\alpha$ of atoms we must use the complete quantum-mechanical theory, but the classical ideas above give us a reasonable estimate. Let’s see if we can get the right order of magnitude for the dielectric constant of some substance. Suppose we try hydrogen. We have once estimated (Chapter 38, Vol. I) that the energy needed to ionize the hydrogen atom should be approximately \begin{equation} \label{Eq:II:11:11} E\approx\frac{1}{2}\,\frac{me^4}{\hbar^2}. \end{equation}  For an estimate of the natural frequency $\omega_0$, we can set this energy equal to $\hbar\omega_0$—the energy of an atomic oscillator whose natural frequency is $\omega_0$. We get \begin{equation*} \omega_0\approx\frac{1}{2}\,\frac{me^4}{\hbar^3}. \end{equation*}  If we now use this value of $\omega_0$ in Eq. (11.7), we find for the electronic polarizability \begin{equation} \label{Eq:II:11:12} \alpha\approx16\pi\biggl[\frac{\hbar^2}{me^2}\biggr]^3. \end{equation}  The quantity $(\hbar^2/me^2)$ is the radius of the ground-state orbit of a Bohr atom (see Chapter 38, Vol. I) and equals $0.528$ angstroms. In a gas at standard pressure and temperature ($1$ atmosphere, $0^\circ$C) there are $2.69\times10^{19}$ atoms/cm$^3$, so Eq. (11.9) gives us  \begin{equation} \label{Eq:II:11:13} \kappa=1+(2.69\times10^{19})16\pi(0.528\times10^{-8})^3=1.00020. \end{equation} 
 \begin{align} \kappa&=1+(2.69\times10^{19})16\pi(0.528\times10^{-8})^3\notag\\[1ex] \label{Eq:II:11:13} &=1.00020. \end{align} 
 The dielectric constant for hydrogen gas is measured to be \begin{equation*} \kappa_{\text{exp}}=1.00026. \end{equation*}  We see that our theory is about right. We should not expect any better, because the measurements were, of course, made with normal hydrogen gas, which has diatomic molecules, not single atoms. We should not be surprised if the polarization of the atoms in a molecule is not quite the same as that of the separate atoms.  The molecular effect, however, is not really that large. An exact quantum-mechanical calculation of $\alpha$ for hydrogen atoms gives a result about $12\%$ higher than (11.12) (the $16\pi$ is changed to $18\pi$), and therefore predicts a dielectric constant somewhat closer to the observed one. In any case, it is clear that our model of a dielectric is fairly good. Another check on our theory is to try Eq. (11.7) on atoms which have a higher frequency of excitation. For instance, it takes about $24.6$ electron volts to pull the electron off helium, compared with the $13.6$ electron volts required to ionize hydrogen.  We would, therefore, expect that the absorption frequency $\omega_0$ for helium would be about twice as big as for hydrogen and that $\alpha$ would be one-quarter as large. So, from (11.13) we expect that \begin{equation*} \kappa_{\text{helium}}\approx1.000050. \end{equation*}  Experimentally, \begin{equation*} \kappa_{\text{helium}}=1.000068, \end{equation*}  so you see that our rough estimates are coming out on the right track. So we have understood the dielectric constant of nonpolar gas, but only qualitatively, because we have not yet used a correct atomic theory of the motions of the atomic electrons.