---
title: "Energy density in electrostatic field"
subject: "Electromagnetism"
chapter: 2
key: "semester_3/electromagnetism:energy-density-electrostatic-field"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/energy-density-electrostatic-field
site: BCT Engineering
---

# Energy density in electrostatic field

Electromagnetism · Chapter 2: Electric field

- Asked in: 2076 Asoj, 7 marks; 2068 Baisakh, 8 marks (asked 2 times)
- Video: [Electric field energy density](https://www.youtube.com/watch?v=ZCCc-qSiRDM) — Physics with Professor Matt Anderson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/energy-density-electrostatic-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-2

## Question

Derive the expression for energy density in an electrostatic field.

## Short answer

The energy stored in an electrostatic field per unit volume is w_E = (1/2) D·E = (1/2) ε E² = D²/(2ε) J/m³. It is derived by expressing work W_E = (1/2) ∫ ρ_v V dv, replacing ρ_v with ∇·D, using vector identity and divergence theorem, and recognizing E = −∇V.

## Answer

**Aim:** Find the energy stored in an electrostatic field per unit volume.

#### Energy of a charge distribution

The work done in assembling a continuous volume charge distribution ρ_v in space is stored as potential energy:

```text
W_E = (1/2) ∫_v ρ_v V dv
```

(The factor 1/2 avoids counting each pair of charges twice.)

#### Step 1: Replace ρ_v by the divergence of D

Gauss's law in point form gives ρ_v = ∇ · D, so

```text
W_E = (1/2) ∫_v (∇ · D) V dv
```

#### Step 2: Use a vector identity

```text
∇ · (V D) = V (∇ · D) + D · ∇V
 →  V (∇ · D) = ∇ · (V D) − D · ∇V
```

Therefore

```text
W_E = (1/2) ∫_v ∇ · (V D) dv − (1/2) ∫_v D · ∇V dv
```

#### Step 3: Apply the divergence theorem to the first term

```text
(1/2) ∫_v ∇ · (V D) dv = (1/2) ∮_S V D · dS
```

Take the volume v large enough to include all the charges (take S as a very large sphere of radius R). For a bounded charge distribution at large distance, V varies as 1/R, D as 1/R², and the area S as R². Hence the surface integral varies as 1/R and tends to zero as R → ∞.

#### Step 4: Use E = −∇V

```text
W_E = −(1/2) ∫_v D · ∇V dv = (1/2) ∫_v D · E dv
```

Since D = εE:

```text
W_E = (1/2) ∫_v D · E dv = (1/2) ∫_v ε E² dv = (1/2) ∫_v (D²/ε) dv
```

#### Energy density

The energy stored per unit volume is the integrand:

```text
w_E = dW_E/dv = (1/2) D · E = (1/2) ε E² = D² / (2ε)     J/m³
```

**Check with a parallel-plate capacitor:** E = V/d, volume = Ad: W = (1/2) ε (V/d)² A d = (1/2)(εA/d) V² = (1/2) C V². ✓

**Notes:** The energy is considered to be stored in the field itself, spread through space with density (1/2) ε E². Where E is zero, no energy is stored.

## Check yourself

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

1. The derivation starts from W_E = (1/2) ∫ ρ_v V dv. Why is the factor 1/2 present?
   A. It avoids counting the interaction of each pair of charges twice
   B. Because only half of the charges contribute to the potential
   C. Because V is the average of the potential and zero
   D. Because energy is shared equally between E and D
   Answer: A. Each pair of charges is counted twice when summing over all charges, so the factor 1/2 corrects this. (Key point: Starting energy expression)

2. Which relation is used in Step 1 to replace ρ_v in the energy integral?
   A. E = -∇V
   B. Continuity equation, ∇ · J = -∂ρ_v/∂t
   C. Ohm's law, J = σE
   D. Gauss's law in point form, ρ_v = ∇ · D
   Answer: D. Gauss's law in point form gives ρ_v = ∇·D, so the integrand can be written in terms of D and V. (Key point: Replacing charge density)

3. Why does the surface integral (1/2)∮ V D · dS vanish when the surface is taken as a very large sphere of radius R?
   A. Because D is zero everywhere in free space
   B. Because dS is perpendicular to V
   C. Because V varies as 1/R, D as 1/R² and S as R², so the integral varies as 1/R and tends to zero
   D. Because the divergence theorem makes every surface integral zero
   Answer: C. V ~ 1/R, D ~ 1/R² and area ~ R² give an integral ~ 1/R, which goes to zero as R tends to infinity. (Key point: Surface integral vanishes)

4. What is the final expression for energy density in the electrostatic field?
   A. w_E = ε E
   B. w_E = (1/2) ε E² = (1/2) D·E = D²/(2ε) in J/m³
   C. w_E = ε E² in J/m
   D. w_E = (1/2) E²/ε in J/m³
   Answer: B. The integrand of W_E = (1/2)∫D·E dv is the energy per unit volume, (1/2) D·E = (1/2)εE² = D²/(2ε). (Key point: Final energy density)

## Related in Electromagnetism

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