Clausius-Mossotti relation

The Clausius-Mossotti relation connects the relative permittivity ε_r to the polarizability α of the molecules constituting a chunk of dielectric.

Internal electric field in dielectric

Chunk of dielectric (green) in external electric field E. Microscopic spherical cavity (size greatly exaggerated in drawing) of radius r 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) ε₀. 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 E_i

$|d\mathbf {E} _{i}|={\frac {\sigma (\theta )d\Omega }{4\pi \epsilon _{0}r^{2}}}={\frac {P\cos \theta d\Omega }{4\pi \epsilon _{0}r^{2}}}={\frac {P\cos \theta \sin \theta d\theta d\phi }{4\pi \epsilon _{0}}}$

The z-component of this field is

$dE_{i,z}={\frac {P}{4\pi \epsilon _{0}}}(\cos \theta )^{2}\sin \theta d\theta d\phi$

Integration over the whole surface gives

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