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

# Clausius-Mossotti equation for ionic polarization

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2082 Kartik (new course), 4 marks; 2076 Asoj, 8 marks; 2073 Shrawan, 6 marks; 2072 Kartik, 6 marks; 2068 Baisakh, 10 marks (asked 5 times)
- 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/clausius-mossotti-ionic
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

Define local field in relation to polarization. Derive the Clausius-Mossotti Equation for ionic polarization, relating polarizability with the permittivity.

## Short answer

The local field is E_loc = E + P/(3 epsilon_0). For ionic crystals, the Clausius-Mossotti equation is (epsilon_r - 1)/(epsilon_r + 2) = N_i (alpha_i + alpha_e+ + alpha_e-) / (3 epsilon_0), relating total polarizability per ion pair to static permittivity.

## Answer

#### Local field and polarization

**Polarization** $P$ is the net induced dipole moment per unit volume of a dielectric: $P = N p$, where $N$ is the number of dipoles per unit volume and $p$ the average dipole moment.

The **local field** $E_{loc}$ is the field that actually acts on an ion or atom. It equals the applied field $E$ plus the field produced by all the surrounding polarized ions. Since the dipoles of the neighbours point the same way, $E_{loc}$ is greater than $E$, and the dipole moment of each unit is $p = \alpha E_{loc}$, not $\alpha E$.

#### Ionic polarization and ionic polarizability

In an ionic crystal such as NaCl, an applied field pushes the cations along the field and the anions against it. Each ion pair is displaced from its equilibrium spacing by a small amount $x$, which induces a dipole.

```text
  No field:   Na+  ---a---  Cl-
  With field E  -->
              Na+ -->   <-- Cl-
              (spacing changes by x)
```

The restoring force behaves like a spring of constant $\beta$: $F = \beta x = eE_{loc}$. The induced dipole is

$$
p_i = e x = \frac{e^2}{\beta}E_{loc} = \alpha_i E_{loc}, \qquad \alpha_i = \frac{e^2}{\beta}
$$

Each ion is also electronically polarized, so the total polarizability per ion pair is

$$
\alpha = \alpha_i + \alpha_{e+} + \alpha_{e-}
$$

#### Lorentz local field

Take a spherical cavity around the reference ion. The field at its centre is

$$
E_{loc} = E + E_{sphere} + E_{inside}
$$

The polarization charge density on the cavity wall at angle $\theta$ is $P\cos\theta$. Summing the axial field of all rings:

$$
E_{sphere} = \int_0^{\pi}\frac{P\cos\theta \cdot 2\pi r^2\sin\theta\, d\theta}{4\pi\varepsilon_0 r^2}\cos\theta = \frac{P}{3\varepsilon_0}
$$

For a cubic crystal the dipoles inside the sphere give $E_{inside} = 0$. Hence

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

#### Clausius-Mossotti equation

With $N_i$ ion pairs per unit volume:

$$
P = N_i\alpha E_{loc} = N_i\alpha\left(E + \frac{P}{3\varepsilon_0}\right)
\;\Rightarrow\; P = \frac{N_i\alpha E}{1 - \dfrac{N_i\alpha}{3\varepsilon_0}}
$$

Macroscopically $P = \varepsilon_0(\varepsilon_r - 1)E$. Equating:

$$
\begin{aligned}
\varepsilon_0(\varepsilon_r - 1) &= \frac{N_i\alpha}{1 - N_i\alpha/3\varepsilon_0} \\
(\varepsilon_r - 1)\left(1 - \frac{N_i\alpha}{3\varepsilon_0}\right) &= \frac{N_i\alpha}{\varepsilon_0} \\
\varepsilon_r - 1 &= \frac{N_i\alpha}{3\varepsilon_0}\left(3 + \varepsilon_r - 1\right) = \frac{N_i\alpha}{3\varepsilon_0}(\varepsilon_r + 2)
\end{aligned}
$$

$$
\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N_i(\alpha_i + \alpha_{e+} + \alpha_{e-})}{3\varepsilon_0}}
$$

#### Low and high frequency forms

- At low frequency (DC up to infrared), both ionic and electronic polarizations follow the field: $\varepsilon_r = \varepsilon_r(0)$ and the full $\alpha$ is used.
- At optical frequencies the heavy ions cannot follow, so only the electronic terms remain, and $\varepsilon_r = \varepsilon_{op} = n^2$ (refractive index squared):

$$
\frac{\varepsilon_{op} - 1}{\varepsilon_{op} + 2} = \frac{N_i(\alpha_{e+} + \alpha_{e-})}{3\varepsilon_0}
$$

Subtracting the two equations gives $\alpha_i$ alone. Example: NaCl has $\varepsilon_r(0) \approx 5.9$ and $\varepsilon_{op} \approx 2.3$ ($n \approx 1.54$), showing that ionic polarization contributes more than half of its static permittivity.

#### Significance

The equation relates the microscopic polarizability of each ion pair to the measurable permittivity, explains why $E_{loc} = \frac{\varepsilon_r + 2}{3}E$, and shows why ionic crystals have a higher static permittivity than optical permittivity. It holds for cubic ionic crystals; it fails for polar liquids with permanent dipoles, where dipole interactions are strong.

## Check yourself

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

1. Why is the dipole moment of each ion pair written as p = alpha*E_loc rather than alpha*E?
   A. Because the applied field is zero inside a crystal
   B. Because the ion feels the applied field plus the field of surrounding polarized ions
   C. Because alpha already contains the applied field
   D. Because E_loc is smaller than E in a dielectric
   Answer: B. The field acting on an ion is the applied field plus the field from the neighbouring dipoles, which point the same way, so E_loc > E. (Key point: Local field)

2. In the spring model of an ionic crystal, F = beta*x = e*E_loc. What is the ionic polarizability?
   A. alpha_i = beta/e^2
   B. alpha_i = e*beta
   C. alpha_i = e^2/beta
   D. alpha_i = e/beta
   Answer: C. p_i = e*x = e*(eE_loc/beta) = (e^2/beta)E_loc, so alpha_i = e^2/beta. (Key point: Ionic polarizability)

3. For a cubic crystal, the Lorentz local field is:
   A. E_loc = E + P/(3*eps0)
   B. E_loc = E - P/(3*eps0)
   C. E_loc = E + P/eps0
   D. E_loc = E + 3P/eps0
   Answer: A. The cavity surface charge gives P/(3 eps0) and the dipoles inside a cubic cavity give zero. (Key point: Lorentz field)

4. Which is the Clausius-Mossotti equation for an ionic crystal, with total alpha = alpha_i + alpha_e+ + alpha_e-?
   A. (eps_r + 2)/(eps_r - 1) = N alpha/(3 eps0)
   B. (eps_r - 1)/(eps_r + 2) = N alpha/(3 eps0)
   C. (eps_r - 1)/(eps_r + 2) = N alpha/eps0
   D. (eps_r - 1)/(eps_r + 2) = 3 N alpha/eps0
   Answer: B. Equating P = N alpha E_loc with P = eps0(eps_r - 1)E gives (eps_r - 1)/(eps_r + 2) = N alpha/(3 eps0). (Key point: Result of the derivation)

## Related in Electrical Engineering Material

- [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×
- [Local field and Clausius-Mossotti equation](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Aclausius-mossotti-electronic) — Electrical Engineering Material, chapter 3, asked 8×
- [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×
- [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×
- [Si polarizability, local field and resonant frequency](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-polarizability-resonant-frequency) — Electrical Engineering Material, chapter 3, asked 2×
- [Si electronic polarizability and local field](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-polarizability-local-field) — Electrical Engineering Material, chapter 3, asked 2×