Clausius-Mossotti relation

The Clausius-Mossotti relation connects the relative permittivity &epsilon;r to the polarizability &alpha; of the molecules constituting a chunk of dielectric.

Internal electric field in dielectric


A macroscopic piece of dielectric is placed in an outer electric field E (the z-direction), which polarizes the dielectric, so that a surface charge density is created. Since E "pushes" positive charge and "pulls" negative charge, the sign of the charge density is as is indicated in the figure on the outer surface. The polarization vector P points by definition from negative to positive charge. The surface charge density is in absolute value equal to the magnitude P of the polarization vector P.

A microscopic spherical cavity of radius r is made inside the dielectric; inside the cavity there is vacuum with permittivity (electric constant) &epsilon;0. Because of electric neutrality the cavity has the same surface charge density as the outer surface:

\sigma(\theta) = P\;\cos\theta. $$ A surface element $$d\Omega = r^2\;\sin\theta\; d\theta d\phi$$ inside the cavity gives a contribution to the internal electric field Ei

$$ The z-component of this field is
 * d\mathbf{E}_i| = \frac{\sigma(\theta) d\Omega}{4\pi \epsilon_0 r^2} = \frac{P\cos\theta d\Omega}{4\pi \epsilon_0 r^2} = \frac{P \cos\theta\sin\theta d\theta d\phi}{4\pi \epsilon_0 }

dE_{i,z} = \frac{P}{4\pi \epsilon_0} (\cos\theta)^2\sin\theta d\theta d\phi $$ Integration over the whole surface gives

E_{i,z} = \iint dE_{i,z} = - \frac{P}{4\pi \epsilon_0} \int_{0}^{\pi} (\cos\theta)^2 d\cos\theta \int_{0}^{2\pi} d\phi = \frac{1}{3\epsilon_0} P. $$