Clausius-Mossotti relation

The Clausius-Mossotti relation connects the relative permittivity ε_r of a dielectric to the polarizability α of the molecules constituting a slab of dielectric. The relative permittivity is a bulk (macroscopic) property and polarizability is a microsocopic property of matter.

(To be continued)

Internal electric field in dielectric

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

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

An infinitesimally small spherical cavity of radius r is made in the dielectric and inside this cavity there is vacuum with permittivity (electric constant) ε₀. Because of electric neutrality the surface of the cavity has the charge density (in absolute value),

$\sigma (\theta )=\mathbf {n} \cdot \mathbf {P} =P\cos \theta ,$

where n is a unit length vector perpendicular to the surface of the cavity at the point θ. By definition, the positive direction of the normal n is outward.

An infinitesimal element $d\Omega =r^{2}\;\sin \theta \;d\theta d\phi$ on the surface of the cavity gives a contribution to the internal electric field E_int parallel to n,

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

The z-component of this contribution to the field is

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

Integration over the whole surface gives

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

The total electric field in the cavity is in the z direction and has magnitude

$E_{\mathrm {tot} }=E+E_{\mathrm {int} }=E+{\frac {1}{3\epsilon _{0}}}P$