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

# Local field and Clausius-Mossotti equation

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2081 Chaitra (new course), 4 marks; 2081 Bhadra, 2+6 marks; 2079 Bhadra, 6 marks; 2075 Asoj, 8 marks; 2074 Asoj, 6 marks; 2071 Shrawan, 6 marks; 2070 Chaitra, 6 marks; 2069 Chaitra, 6 marks (asked 8 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-electronic
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

What is Local Field? Derive Clausius-Mossotti equation.

## Short answer

The local field is E_loc = E + P/(3 epsilon_0). Combining P = N alpha_e E_loc and P = epsilon_0 (epsilon_r - 1) E yields the Clausius-Mossotti equation: (epsilon_r - 1)/(epsilon_r + 2) = N alpha_e/(3 epsilon_0), linking electronic polarizability to permittivity.

## Answer

#### Local field

The **local field** $E_{loc}$ is the actual electric field acting on an individual atom or molecule inside a dielectric. It is not equal to the applied (macroscopic) field $E$, because the induced dipoles of all the surrounding atoms also produce a field at that point. In a polarised solid, $E_{loc} > E$.

#### Lorentz method for the local field

Imagine a small spherical cavity around the chosen atom, inside a dielectric between capacitor plates. The field at the centre is

$$
E_{loc} = E_1 + E_2 + E_3
$$

- $E_1$: field due to charges on the plates and the polarization charges on the outer dielectric surfaces; together this is the applied macroscopic field $E$.
- $E_2$: field due to polarization charges on the surface of the spherical cavity.
- $E_3$: field of the dipoles inside the sphere; it is zero for a cubic (symmetric) arrangement.

```text
  +  +  +  +  +  +  +   plate
  ------------------
      .--------.
     /  - - -  \     surface charge
    |  (atom)   |    on cavity: E2
     \  + + +  /
      '--------'
  ------------------
  -  -  -  -  -  -  -   plate
```

**Field of cavity surface charge ($E_2$):** on a ring of the sphere at angle $\theta$ to the field, the surface charge density is $P\cos\theta$. The ring of radius $r\sin\theta$, width $r\,d\theta$ has area $2\pi r^2\sin\theta\, d\theta$, and its field component along $E$ at the centre is

$$
dE_2 = \frac{(P\cos\theta)(2\pi r^2 \sin\theta\, d\theta)}{4\pi\varepsilon_0 r^2}\cos\theta
$$

$$
E_2 = \frac{P}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\, d\theta = \frac{P}{2\varepsilon_0}\cdot\frac{2}{3} = \frac{P}{3\varepsilon_0}
$$

So the **Lorentz local field** is

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

#### Derivation of Clausius-Mossotti equation

For electronic polarization, each atom acquires an induced dipole $p = \alpha_e E_{loc}$. With $N$ atoms per unit volume:

$$
P = N\alpha_e E_{loc} = N\alpha_e\left(E + \frac{P}{3\varepsilon_0}\right)
$$

Also, from the macroscopic definition, $P = \varepsilon_0(\varepsilon_r - 1)E$, so $E = \dfrac{P}{\varepsilon_0(\varepsilon_r - 1)}$. Then

$$
\begin{aligned}
P &= N\alpha_e\left[\frac{P}{\varepsilon_0(\varepsilon_r - 1)} + \frac{P}{3\varepsilon_0}\right] \\
1 &= \frac{N\alpha_e}{\varepsilon_0}\left[\frac{3 + \varepsilon_r - 1}{3(\varepsilon_r - 1)}\right] = \frac{N\alpha_e}{3\varepsilon_0}\cdot\frac{\varepsilon_r + 2}{\varepsilon_r - 1}
\end{aligned}
$$

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

This is the **Clausius-Mossotti equation**. It links a microscopic property (polarizability $\alpha_e$ of one atom) to a macroscopic property (relative permittivity $\varepsilon_r$).

**Notes:**
- It is valid for non-polar solids with cubic symmetry and for gases and liquids where $E_3 = 0$.
- From it, $E_{loc} = \dfrac{\varepsilon_r + 2}{3}E$. Example: Si ($\varepsilon_r = 11.9$) has $E_{loc} = 4.63E$.
- For a dilute gas, $\varepsilon_r \approx 1$, and it reduces to $\varepsilon_r = 1 + N\alpha_e/\varepsilon_0$.

## Check yourself

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

1. The local field is
   A. The applied field E
   B. The actual field at an atom, including fields of other induced dipoles; E_loc > E
   C. The field inside the atom only
   D. The depolarising field
   Answer: B. Neighbouring dipoles add to the applied field. (Key point: Local field)

2. Lorentz local field 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: A. E2 from cavity surface charge integrates to P/(3ε0), E3 = 0 for cubic. (Key point: Lorentz field)

3. The field E3 of dipoles in the sphere is zero for
   A. Any arrangement
   B. Polar liquids only
   C. Metals only
   D. Cubic (symmetric) arrangement
   Answer: D. Symmetry cancels it. (Key point: E3)

4. The Clausius-Mossotti equation is
   A. (ε_r + 2)/(ε_r - 1) = N α_e/(3ε0)
   B. (ε_r - 1)/(ε_r + 2) = N α_e/ε0
   C. (ε_r - 1)/(ε_r + 2) = N α_e/(3ε0)
   D. (ε_r - 1)/ε_r = N α_e/(3ε0)
   Answer: C. Combining P = Nα E_loc and P = ε0(ε_r -1)E. (Key point: Clausius-Mossotti)

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