---
title: "Si polarizability with 6×10²² electrons/cm³"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:si-polarizability-6e22"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-polarizability-6e22
site: BCT Engineering
---

# Si polarizability with 6×10²² electrons/cm³

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2074 Asoj, 4 marks
- Video: [AP Unit 5 Lecture 03 Derivation  Electronic and Ionic Polarization](https://www.youtube.com/watch?v=6gtZOozFF0o) — Dr. Gratian Patrick
- 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-6e22
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

The number of electrons per unit volume of Silicon is 6×10²² cm⁻³. Calculate: i) Electronic polarizability due to valence electrons per Silicon atom. ii) If the Silicon crystal sample is electrode on opposite faces, by how many times the local field is greater than the applied field?

## Short answer

Using Clausius–Mossotti with ε_r=11.9, α_e = 3ε₀(ε_r−1)/[N(ε_r+2)] ≈ 3.47×10⁻⁴⁰ F m² per Si atom (or 1.39×10⁻³⁹ F m² per valence electron). Local field ratio E_loc/E = (ε_r+2)/3 = 4.63.

## Answer

Use the Clausius–Mossotti relation for a covalent (non-polar) solid, where only electronic polarization exists.

**Data and assumptions**
- $N = 6\times10^{22}\ \text{cm}^{-3} = 6\times10^{28}\ \text{m}^{-3}$, taken as the number of polarizable Si units per m³ (as given).
- Relative permittivity of Si, $\varepsilon_r = 11.9$ (standard value, not given in the question).
- $\varepsilon_0 = 8.854\times10^{-12}$ F/m.

#### i) Electronic polarizability
$$
\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}
\;\Rightarrow\;
\alpha_e = \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2}
$$
$$
\begin{aligned}
\frac{3\varepsilon_0}{N} &= \frac{3\times8.854\times10^{-12}}{6\times10^{28}} = 4.427\times10^{-40} \\
\frac{\varepsilon_r - 1}{\varepsilon_r + 2} &= \frac{10.9}{13.9} = 0.7842 \\
\alpha_e &= 4.427\times10^{-40}\times0.7842 = 3.47\times10^{-40}\ \text{F m}^2
\end{aligned}
$$

**Answer (i): $\alpha_e \approx 3.47\times10^{-40}\ \text{F m}^2$ per Si atom.**

(If the given figure is read strictly as valence electrons, the atom density is $6\times10^{28}/4 = 1.5\times10^{28}$ m⁻³ and the polarizability per atom becomes $1.39\times10^{-39}$ F m².)

#### ii) Local field compared with applied field
For a sample with electrodes on opposite faces, the Lorentz local field is
$$
E_{loc} = E + \frac{P}{3\varepsilon_0} = E + \frac{(\varepsilon_r - 1)E}{3}
\;\Rightarrow\;
\frac{E_{loc}}{E} = \frac{\varepsilon_r + 2}{3}
$$
$$
\frac{E_{loc}}{E} = \frac{11.9 + 2}{3} = \frac{13.9}{3} = 4.63
$$

**Answer (ii): the local field is about 4.63 times the applied field.**

## Check yourself

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

1. N = 6 x 10^22 cm^-3 equals:
   A. 6 x 10^25 m^-3
   B. 6 x 10^19 m^-3
   C. 6 x 10^16 m^-3
   D. 6 x 10^28 m^-3
   Answer: D. 1 cm^-3 = 10^6 m^-3, so 6e22 cm^-3 = 6e28 m^-3. (Key point: Unit conversion)

2. Which equation is used to find alpha_e for Si?
   A. alpha_e = Ne/E
   B. alpha_e = (3 eps0/N)(eps_r - 1)/(eps_r + 2)
   C. alpha_e = eps_r/N
   D. alpha_e = 4 pi eps0 N
   Answer: B. The Clausius-Mossotti relation applies to non-polar covalent Si. (Key point: Method for alpha_e)

3. With 3 eps0/N = 4.427 x 10^-40 and the factor 0.7842, alpha_e is about:
   A. 3.47 x 10^-40 F m^2
   B. 4.43 x 10^-40 F m^2
   C. 5.6 x 10^-40 F m^2
   D. 1.39 x 10^-39 F m^2
   Answer: A. 4.427e-40 x 0.7842 = 3.47e-40 F m^2. (Key point: Result)

4. E_loc/E for Si (eps_r = 11.9) is:
   A. 3.63
   B. 4.63
   C. 11.9
   D. 13.9
   Answer: B. (eps_r + 2)/3 = 13.9/3 = 4.63. (Key point: Local field ratio)

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