---
title: "Electronic polarization and Clausius-Mossotti equation"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:electronic-polarization-clausius-mossotti"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/electronic-polarization-clausius-mossotti
site: BCT Engineering
---

# Electronic polarization and Clausius-Mossotti equation

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2069 Asar, 6 marks
- Video: [Derive Clausius Mossotti relation](https://www.youtube.com/watch?v=OZchTd4fxo0) — Physics mee
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/electronic-polarization-clausius-mossotti
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

What is electronic polarization? Derive the mathematical relation showing the relation between electronic polarization and relative permittivity, using Clausius-Mossotti equation.

## Short answer

Electronic polarization is the displacement of electron cloud relative to nucleus, with polarizability α_e = 4πε₀R³. The Clausius–Mossotti relation linking it to relative permittivity is (ε_r−1)/(ε_r+2) = Nα_e/(3ε₀).

## Answer

**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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is electronic polarization?
   A. Displacement of electron cloud relative to the nucleus under a field
   B. Alignment of permanent dipoles
   C. Shift of positive and negative ions
   D. Charge build-up at grain boundaries
   Answer: A. The cloud shifts one way and the nucleus the other, forming an induced dipole. (Key point: Electronic polarization)

2. For cubic symmetry the local field in terms of applied E and P is:
   A. E_loc = E - P/(3ε0)
   B. E_loc = E + P/(3ε0)
   C. E_loc = E + P/ε0
   D. E_loc = E + 3P/ε0
   Answer: B. The polarization charges on the cavity surface add P/3ε0 to the applied field. (Key point: Lorentz local field)

3. Substituting P = ε0(εr-1)E into E_loc gives:
   A. E_loc = (εr + 1)E/2
   B. E_loc = (εr + 2)E/3
   C. E_loc = (εr - 1)E/3
   D. E_loc = 3E/(εr + 2)
   Answer: B. E + (εr-1)E/3 = (εr+2)E/3. (Key point: Local field in εr)

4. The Clausius-Mossotti equation is:
   A. (εr+2)/(εr-1) = Nα_e/(3ε0)
   B. (εr-1)/(εr+2) = 3ε0/(Nα_e)
   C. (εr-1)/(εr+1) = Nα_e/ε0
   D. (εr-1)/(εr+2) = Nα_e/(3ε0)
   Answer: D. Equating ε0(εr-1)E with Nα_e(εr+2)E/3 gives the equation. (Key point: Clausius-Mossotti)

## Close variants set in other exams

- [Polarization and Clausius-Mossotti equation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Apolarization-clausius-mossotti) — Electrical Engineering Material, chapter 3, asked 2×

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