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

# Polarization and Clausius-Mossotti equation

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2075 Chaitra, 8 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/polarization-clausius-mossotti
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

Define polarization. Derive the Clausius-Mossotti equation showing the relation between relative permittivity and electronic polarizability.

## Short answer

Polarization P is the net dipole moment per unit volume, P = epsilon0 (epsilonr - 1) E. Using the Lorentz local field E_loc = E + P/(3 epsilon0) and P = N alpha_e E_loc, equating gives the Clausius-Mossotti equation (epsilonr - 1)/(epsilonr + 2) = N alpha_e/(3 epsilon0), relating macroscopic relative permittivity epsilonr to electronic polarizability alpha_e. Valid for non-polar materials with cubic symmetry.

## Answer

**Polarization** $\vec P$ is the net induced dipole moment per unit volume of a dielectric: $P = Np$, where $N$ is the number of molecules per m³ and $p$ the average dipole moment of each. It is related to the field by $P = \varepsilon_0(\varepsilon_r - 1)E$.

#### Step 1: Polarization in terms of $\varepsilon_r$
From $D = \varepsilon_0 E + P = \varepsilon_0\varepsilon_r E$:
$$
P = \varepsilon_0(\varepsilon_r - 1)E \quad (1)
$$

#### Step 2: Polarization in terms of polarizability
A molecule feels the **local field** $E_{loc}$, not the applied field, so
$$
P = N\alpha_e E_{loc} \quad (2)
$$

#### Step 3: Local (Lorentz) field
Imagine a small spherical cavity around a molecule. The field at its centre is
$$
E_{loc} = E_1 + E_2 + E_3 + E_4
$$
- $E_1$: field of the free charges on the plates
- $E_2$: field of the polarization charges on the dielectric surfaces facing the plates; together $E_1 + E_2 = E$, the applied macroscopic field
- $E_3$: field of the polarization charges on the cavity surface $= \dfrac{P}{3\varepsilon_0}$ (Lorentz field)
- $E_4$: field of the dipoles inside the cavity $= 0$ for cubic symmetry

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

```text
   +  +  +  +  +  +   positive plate      |
       + + + +                            |  E
     (   mol   )      spherical cavity    v
       - - - -        (+ on upper wall, - on lower wall;
   -  -  -  -  -  -   negative plate      field P/(3 e0) adds to E)
```

#### Step 4: Combine
Put (1) in (3):
$$
\begin{aligned}
E_{loc} &= E + \frac{\varepsilon_0(\varepsilon_r - 1)E}{3\varepsilon_0} = E\,\frac{\varepsilon_r + 2}{3}
\end{aligned}
$$
Equate (1) and (2):
$$
\begin{aligned}
\varepsilon_0(\varepsilon_r - 1)E &= N\alpha_e\, E\,\frac{\varepsilon_r + 2}{3} \\
\frac{\varepsilon_r - 1}{\varepsilon_r + 2} &= \frac{N\alpha_e}{3\varepsilon_0}
\end{aligned}
$$

This is the **Clausius–Mossotti equation**. It links the macroscopic quantity $\varepsilon_r$ with the microscopic quantity $\alpha_e$.

#### Notes
- Valid for non-polar materials with cubic symmetry (gases, elemental solids like Si, Ge, diamond).
- If $N\alpha_e$ is known, $\varepsilon_r = \dfrac{1 + 2N\alpha_e/3\varepsilon_0}{1 - N\alpha_e/3\varepsilon_0}$.
- At optical frequency $\varepsilon_r = n^2$, giving the Lorentz–Lorenz equation.

## Check yourself

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

1. Polarization P is:
   A. The field inside a dielectric
   B. Net dipole moment per unit volume, P = Np
   C. The charge on the plates
   D. The dipole moment of one atom
   Answer: B. N is the number of molecules per m^3 and p their average dipole moment. (Key point: Definition of polarization)

2. From D = eps0 E + P = eps0 eps_r E, P equals:
   A. eps0 eps_r E
   B. eps0 E/(eps_r - 1)
   C. (eps_r - 1)E/eps0
   D. eps0(eps_r - 1)E
   Answer: D. Subtracting eps0 E from D gives P = eps0(eps_r - 1)E. (Key point: Macroscopic relation)

3. Substituting P into E_loc = E + P/(3 eps0) gives E_loc =
   A. E(eps_r - 1)/3
   B. E(eps_r + 2)/3
   C. E(eps_r + 2)
   D. 3E/(eps_r + 2)
   Answer: B. E + (eps_r - 1)E/3 = E(eps_r + 2)/3. (Key point: Local field in terms of eps_r)

4. The Clausius-Mossotti equation is valid for:
   A. Polar liquids such as water
   B. Non-polar materials with cubic symmetry such as Si or gases
   C. Ferroelectrics
   D. Any material
   Answer: B. It fails where permanent dipoles interact strongly. (Key point: Validity)

## Close variants set in other exams

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

## Related in Electrical Engineering Material

- [Clausius-Mossotti equation for ionic polarization](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aclausius-mossotti-ionic) — Electrical Engineering Material, chapter 3, asked 5×
- [Dipole moment and Clausius-Mossotti equation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Adipole-moment-clausius-mossotti) — Electrical Engineering Material, chapter 3, asked 2×
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