The dielectric constant of liquids; the Clausius-Mossotti equation
In a liquid we expect that the field which will polarize an individual atom is more like $E_{\text{hole}}$ than just $E$. If we use the $E_{\text{hole}}$ of (11.25) for the polarizing field in Eq. (11.6), then Eq. (11.8) becomes \begin{equation} \label{Eq:II:11:26} P=N\alpha\epsO\biggl(E+\frac{P}{3\epsO}\biggr), \end{equation}  or \begin{equation} \label{Eq:II:11:27} P=\frac{N\alpha}{1-(N\alpha/3)}\,\epsO E. \end{equation}  Remembering that $\kappa-1$ is just $P/\epsO E$, we have \begin{equation} \label{Eq:II:11:28} \kappa-1=\frac{N\alpha}{1-(N\alpha/3)}, \end{equation}  which gives us the dielectric constant of a liquid in terms of $\alpha$, the atomic polarizability. This is called the Clausius-Mossotti equation. Whenever $N\alpha$ is very small, as it is for a gas (because the density $N$ is small), then the term $N\alpha/3$ can be neglected compared with $1$, and we get our old result, Eq. (11.9), that \begin{equation} \label{Eq:II:11:29} \kappa-1=N\alpha. \end{equation}  Let’s compare Eq. (11.28) with some experimental results. It is first necessary to look at gases for which, using the measurement of $\kappa$, we can find $\alpha$ from Eq. (11.29). For instance, for carbon disulfide at zero degrees centigrade the dielectric constant is $1.0029$, so $N\alpha$ is $0.0029$. Now the density of the gas is easily worked out and the density of the liquid can be found in handbooks. At $20^\circ$C, the density of liquid CS$_2$ is $381$ times higher than the density of the gas at $0^\circ$C. This means that $N$ is $381$ times higher in the liquid than it is in the gas, so that—if we make the approximation that the basic atomic polarizability of the carbon disulfide doesn’t change when it is condensed into a liquid—$N\alpha$ in the liquid is equal to $381$ times $0.0029$, or $1.11$. Notice that the $N\alpha/3$ term amounts to almost $0.4$, so it is quite significant. With these numbers we predict a dielectric constant of $2.76$, which agrees reasonably well with the observed value of $2.64$. In Table 11–1 we give some experimental data on various materials (taken from the Handbook of Chemistry and Physics), together with the dielectric constants calculated from Eq. (11.28) in the way just described. The agreement between observation and theory is even better for argon and oxygen than for CS$_2$—and not so good for carbon tetrachloride. On the whole, the results show that Eq. (11.28) works very well. Our derivation of Eq. (11.28) is valid only for electronic polarization in liquids.  It is not right for a polar molecule like H$_2$O. If we go through the same calculations for water, we get $13.2$ for $N\alpha$, which means that the dielectric constant for the liquid is negative, while the observed value of $\kappa$ is $80$. The problem has to do with the correct treatment of the permanent dipoles, and Onsager has pointed out the right way to go. We do not have the time to treat the case now, but if you are interested it is discussed in Kittel’s book, Introduction to Solid State Physics.