---
title: "Average dipole energy of HCl molecule"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:hcl-average-dipole-energy"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/hcl-average-dipole-energy
site: BCT Engineering
---

# Average dipole energy of HCl molecule

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2071 Shrawan, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/hcl-average-dipole-energy
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

Derive the relation for average dipole energy of HCl molecule when it is applied with electric field of magnitude E.

## Short answer

For a permanent dipole p₀ in field E, average of cosθ is ⟨cosθ⟩ = p₀E/(3kT) when p₀E≪kT. The average dipole energy is then ⟨U⟩ = −p₀E⟨cosθ⟩ = −p₀²E²/(3kT).

## Answer

HCl is a polar molecule with a permanent dipole moment $p_0$. In a field $E$, a dipole at angle $\theta$ to the field has potential energy
$$
U = -p_0E\cos\theta
$$
Thermal motion randomizes the dipoles, so the number of dipoles in an energy state follows the **Boltzmann distribution**, $\propto e^{-U/kT} = e^{p_0E\cos\theta/kT}$.

#### Average of $\cos\theta$
The number of dipoles in the solid angle between $\theta$ and $\theta + d\theta$ is proportional to $2\pi\sin\theta\,d\theta\, e^{x\cos\theta}$, where $x = p_0E/kT$.
$$
\langle\cos\theta\rangle = \frac{\int_0^{\pi}\cos\theta\, e^{x\cos\theta}\sin\theta\,d\theta}{\int_0^{\pi} e^{x\cos\theta}\sin\theta\,d\theta}
$$
Put $u = \cos\theta$:
$$
\langle\cos\theta\rangle = \frac{\int_{-1}^{1}u\,e^{xu}du}{\int_{-1}^{1}e^{xu}du} = \coth x - \frac{1}{x} = L(x)
$$
$L(x)$ is the **Langevin function**.

#### Small-field approximation
At normal fields $p_0E \ll kT$ ($x \ll 1$), so $L(x) \approx x/3$:
$$
\langle\cos\theta\rangle = \frac{p_0E}{3kT}
$$

#### Average dipole energy
$$
\begin{aligned}
\langle U\rangle &= -p_0E\langle\cos\theta\rangle \\
&= -p_0E\cdot\frac{p_0E}{3kT} \\
&= -\frac{p_0^2E^2}{3kT}
\end{aligned}
$$

#### Related results
- Average dipole moment along the field: $p_{av} = p_0\langle\cos\theta\rangle = \dfrac{p_0^2E}{3kT}$
- Orientational polarizability: $\alpha_d = \dfrac{p_0^2}{3kT}$

The average energy is negative (alignment lowers energy), grows as $E^2$, and falls as $1/T$ because heat opposes alignment.

## Check yourself

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

1. What is the potential energy of a dipole p0 at angle θ to field E?
   A. U = -p0 E cosθ
   B. U = +p0 E cosθ
   C. U = -p0 E sinθ
   D. U = p0 E^2 cosθ
   Answer: A. A dipole aligned with the field has lowest energy, so U = -p0E cosθ. (Key point: Dipole energy)

2. How are the dipoles distributed over orientations in the derivation?
   A. All dipoles align exactly with E
   B. Boltzmann distribution, proportional to e^(-U/kT)
   C. Uniformly at all angles regardless of field
   D. Only at θ = 90°
   Answer: B. Thermal motion randomizes the dipoles, giving a Boltzmann weighting e^(p0E cosθ/kT). (Key point: Boltzmann weighting)

3. What does the average of cosθ evaluate to before approximation?
   A. tanh x
   B. x/3 exactly for all x
   C. coth x - 1/x (Langevin function)
   D. 1/x
   Answer: C. With u = cosθ the integrals give L(x) = coth x - 1/x, with x = p0E/kT. (Key point: Langevin result)

4. What is the final average dipole energy for p0E << kT?
   A. -p0 E / (3kT)
   B. +p0^2 E^2 / (3kT)
   C. -p0^2 E / (3kT)
   D. -p0^2 E^2 / (3kT)
   Answer: D. <U> = -p0E<cosθ> = -p0E(p0E/3kT) = -p0^2E^2/3kT, negative and proportional to E^2 and 1/T. (Key point: Average energy)

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