Electronic polarization and Clausius-Mossotti equation

Electronic polarization is the displacement of the electron cloud of an atom relative to its nucleus under an applied field, producing an induced dipole $p = \alpha_eE_{loc}$, with $\alpha_e = 4\pi\varepsilon_0R^3$. It occurs in all materials and is the only mechanism in non-polar covalent solids such as Si, Ge and diamond.

Derivation

(a) Macroscopic view: from $D = \varepsilon_0E + P = \varepsilon_0\varepsilon_rE$,

$$P = \varepsilon_0(\varepsilon_r - 1)E \quad (1)$$

(b) Microscopic view: each atom feels the local field:

$$P = N\alpha_eE_{loc} \quad (2)$$

(c) Local (Lorentz) field: for cubic symmetry, the field inside a small spherical cavity around an atom is the applied field plus the field of the polarization charges on the cavity surface:

$$E_{loc} = E + \frac{P}{3\varepsilon_0} \quad (3)$$

Substituting (1) in (3):

$$E_{loc} = E + \frac{(\varepsilon_r - 1)E}{3} = \frac{\varepsilon_r + 2}{3}E$$

(d) Equate (1) and (2):

$$\begin{aligned} \varepsilon_0(\varepsilon_r - 1)E &= N\alpha_e\frac{\varepsilon_r + 2}{3}E \\ \frac{\varepsilon_r - 1}{\varepsilon_r + 2} &= \frac{N\alpha_e}{3\varepsilon_0} \end{aligned}$$

This Clausius–Mossotti equation relates the relative permittivity (measured) to the electronic polarizability (atomic). Solving for $\varepsilon_r$:

$$\varepsilon_r = \frac{1 + \dfrac{2N\alpha_e}{3\varepsilon_0}}{1 - \dfrac{N\alpha_e}{3\varepsilon_0}}$$

Example: for Si, $\varepsilon_r = 11.9$ and $N = 5\times10^{28}$ m⁻³ give $\alpha_e = \dfrac{3\varepsilon_0}{N}\cdot\dfrac{10.9}{13.9} \approx 4.2\times10^{-40}$ F m².