---
title: "Si electronic polarizability and local field"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:si-polarizability-local-field"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-polarizability-local-field
site: BCT Engineering
---

# Si electronic polarizability and local field

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2081 Baisakh, 6 marks; 2073 Chaitra, 4 marks (asked 2 times)
- Video: [Electronic Polarization- Derivation in Dielectrics](https://www.youtube.com/watch?v=xLzioXSW1Gw) — Physics World
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-polarizability-local-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

A pure Si crystal that has εr = 11.9. (i) What is the electronic polarizability due to valence electrons per Si atom? (ii) Suppose a voltage is applied across Si crystal sample, by how much is the local field greater than the applied field. Given that the density of Si atoms; N = 5×10²⁸ atoms per m³, ε₀ = 8.85×10⁻¹² Fm⁻¹.

## Short answer

α_e = (3ε0/N)(εr-1)/(εr+2) = 4.16×10⁻⁴⁰ F m². Local field E_loc = (εr+2)/3 E = 4.63E, i.e., 4.63 times the applied field.

## Answer

Pure Si has only electronic polarization (covalent, no permanent dipoles, cubic structure), so the Clausius-Mossotti equation with the Lorentz local field applies.

**Given:** $\varepsilon_r = 11.9$, $N = 5 \times 10^{28}\ \text{m}^{-3}$, $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$.

#### (i) Electronic polarizability per Si atom

Clausius-Mossotti equation:

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

$$
\begin{aligned}
\alpha_e &= \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2} \\
&= \frac{3 \times 8.85 \times 10^{-12}}{5 \times 10^{28}} \times \frac{11.9 - 1}{11.9 + 2} \\
&= 5.31 \times 10^{-40} \times 0.7842 = 4.16 \times 10^{-40}\ \text{F m}^2
\end{aligned}
$$

(If the local field were ignored, $\alpha_e = \varepsilon_0(\varepsilon_r - 1)/N = 1.93 \times 10^{-39}$ F m², about 4.6 times too large; this shows why the local field matters.)

#### (ii) Local field compared with applied field

The Lorentz local field is

$$
E_{loc} = E + \frac{P}{3\varepsilon_0}, \qquad P = \varepsilon_0(\varepsilon_r - 1)E
$$

$$
E_{loc} = E\left(1 + \frac{\varepsilon_r - 1}{3}\right) = \frac{\varepsilon_r + 2}{3}E = \frac{13.9}{3}E = 4.63E
$$

So the field acting on each Si atom is about 4.63 times the applied field, i.e. greater than it by $3.63E$ (363%).

**Answer:** (i) $\alpha_e \approx 4.16 \times 10^{-40}\ \text{F m}^2$; (ii) $E_{loc} = 4.63E$.

## Check yourself

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

1. Why can the Clausius-Mossotti equation with only alpha_e be applied to pure Si?
   A. Si is covalent, non-polar and cubic, so only electronic polarization exists
   B. Si is an ionic crystal
   C. Si has permanent dipoles that cancel
   D. Si is a gas
   Answer: A. Covalent Si has no ions or permanent dipoles, and its cubic structure makes the Lorentz field valid. (Key point: Why only electronic polarization)

2. Using (eps_r - 1)/(eps_r + 2) = 0.784 and 3 eps0/N = 5.31 x 10^-40, alpha_e is about:
   A. 4.16 x 10^-40 F m^2
   B. 5.31 x 10^-40 F m^2
   C. 1.93 x 10^-39 F m^2
   D. 6.8 x 10^-40 F m^2
   Answer: A. 5.31e-40 x 0.784 = 4.16e-40 F m^2. (Key point: Value of alpha_e)

3. If the local field were ignored (P = N alpha E), the computed alpha_e would be:
   A. About 4.6 times too small
   B. The same
   C. About 4.6 times too large
   D. Zero
   Answer: C. Ignoring E_loc > E gives eps0(eps_r - 1)/N = 1.93e-39, about 4.6 times the correct value. (Key point: Ignoring local field)

4. The local field in Si compared with the applied field is:
   A. E_loc = 11.9E
   B. E_loc = 3.63E
   C. E_loc = 13.9E
   D. E_loc = 4.63E
   Answer: D. E_loc = (eps_r + 2)E/3 = 13.9E/3 = 4.63E. (Key point: Local field ratio)

## Related in Electrical Engineering Material

- [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 polarizability with 6×10²² electrons/cm³](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-polarizability-6e22) — Electrical Engineering Material, chapter 3
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