Energy density in electrostatic field

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:

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

W_E = (1/2) ∫_v (∇ · D) V dv

Step 2: Use a vector identity

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

Therefore

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

Step 3: Apply the divergence theorem to the first term

(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

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

Since D = εE:

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:

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.