The Clausius-Mossotti relation

Let us now investigate what a dielectric equation of state actually looks like. Suppose that a dielectric medium is made up of identical molecules which develop a dipole moment

$${\bfm p} = \alpha \epsilon_0 {\bfm E}$$ (507)

when placed in an electric field ${\bfm E}$. The constant $\alpha$ is called the molecular polarizability. If $N$ is the number density of such molecules then the polarization of the medium is

$${\bfm P} = N {\bfm p} = N\alpha \epsilon_0 {\bfm E},$$ (508)

or

$${\bfm P} = \frac{N_A\, \rho_m \alpha}{M} \,\epsilon_0 {\bfm E},$$ (509)

where $\rho_m$ is the mass density, $N_A$ is Avogadro's number, and $M$ is the molecular weight. But, how does the electric field experienced by an individual molecule relate to the average electric field in the medium? This is not a trivial question since we expect the electric field to vary strongly (on atomic length-scales) inside the dielectric.

Suppose that the dielectric is polarized with a mean electric field ${\bfm E}_0$ which is uniform (on macroscopic length-scales) and directed along the $z$-axis. Consider one of the molecules which constitute the dielectric. Let us draw a sphere of radius $a$ about this particular molecule. This is intended to represent the boundary between the microscopic and the macroscopic range of phenomena affecting the molecule. We shall treat the dielectric outside the sphere as a continuous medium and the dielectric inside the sphere as a collection of polarized molecules. According to Eq. (3.29) there is a polarization surface charge of magnitude

$$\sigma_{\rm pol} = -P \cos\theta$$ (510)

on the inside of the sphere, where $(r,\theta, \varphi)$ are spherical polar coordinates, and ${\bfm P} = P\,\hat{\bfm z} = \epsilon_0(\epsilon-1) E_0\,\hat{\bfm z}$ is the uniform polarization of the dielectric. The magnitude of $E_z$ at the molecule due to the surface charge is

$$E_z = -\frac{1}{4\pi\epsilon_0} \int \frac{\sigma_{\rm pol} \cos\theta}{a^2}\,dS,$$ (511)

where $dS = 2\pi a^2\,\sin\theta\,d\theta$ is a surface element of the sphere. It follows that

$$E_z = \frac{P }{2\epsilon_0} \int_0^\pi \cos^2\theta\,\sin\theta\,d\theta = \frac{P}{3\epsilon_0}.$$ (512)

It is easily demonstrated that $E_\theta =E_\varphi=0$ at the molecule. Thus, the field at the molecule due to the surface charges on the sphere is

$${\bfm E} = \frac{\bfm P}{3\epsilon_0}.$$ (513)

The field due to the individual molecules within the sphere is obtained by summing over the dipole fields of these molecules. The electric field at a distance ${\bfm r}$ from a dipole ${\bfm p}$ is

$${\bfm E} = -\frac{1}{4\pi\epsilon_0}\left[\frac{\bfm p}{r^3} - \frac{3({\bfm p}\!\cdot\!{\bfm r}){\bfm r}}{r^5}\right].$$ (514)

It is assumed that the dipole moment of each molecule within the sphere is the same, and also that the molecules are evenly distributed throughout the sphere. This being the case, the value of $E_z$ at the molecule due to all of the other molecules within in the sphere,

$$E_z = -\frac{1}{4\pi\epsilon_0}\sum_{\rm mols}\left[\frac{ p_z}{r^3} - \frac{3(p_x\, xz + p_y\, yz + p_z\, z^2)}{r^5}\right],$$ (515)

is zero, since

$$\sum_{\rm mols} x^2= \sum_{\rm mols} y^2 = \sum_{\rm mols} z^2 = \frac{1}{3} \sum_{\rm mols} r^2$$ (516)

and

$$\sum_{\rm mols} xy = \sum_{\rm mols} yz = \sum_{\rm mols} zx = 0.$$ (517)

It is easily seen that $E_\theta =E_\varphi=0$. Hence, the electric field at the molecule due to the other molecules within the sphere vanishes.

It is clear that the net electric field seen by an individual molecule is

$${\bfm E} = {\bfm E}_0 + \frac{\bfm P}{3\epsilon_0}.$$ (518)

This is larger than the average electric field ${\bfm E}_0$ in the dielectric. The above analysis indicates that this effect is ascribable to the long range (rather than the short range) interactions of the molecule with the other molecules in the medium. Making use of Eq. (3.88) and the definition ${\bfm P} =\epsilon_0 (\epsilon-1) {\bfm E}_0$, we obtain

$$\frac{\epsilon -1}{\epsilon+2} = \frac{N_A\,\rho_m \alpha}{3 M}.$$ (519)

This is called the Clausius-Mossotti relation. This formula is found to work pretty well for a wide class of dielectric liquids and gases. The Clausius-Mossotti relation yields

$$\frac{d\epsilon}{d\rho_m} = \frac{(\epsilon-1)(\epsilon+2)}{3\rho_m}.$$ (520)