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Capacitance of parallel plate capacitor

Short answer

Solving Laplace's equation d²V/dy²=0 with V(0)=0, V(d)=V₀ yields V = V₀ y/d. Then E = −V₀/d a_y, D = εE, surface charge ρ_s = ε V₀/d, so Q = ε V₀ S/d. Capacitance C = ε S/d.

By solving Laplace's equation, find the capacitance of a parallel plate capacitor with potential difference between the plates equal to V0.

Setup. Two large parallel plates of area S are separated by distance d, with the dielectric of permittivity ε between them. Fringing is neglected, so the potential V depends only on y. Let the lower plate (y = 0) be at V = 0 and the upper plate (y = d) at V = V₀.

 y = d  ======================  V = V0
          ↓  E = −(V0/d) a_y
 y = 0  ======================  V = 0

Solving Laplace's equation

Between the plates there is no free charge, so ∇²V = 0, which reduces to

d²V/dy² = 0
dV/dy   = A
V       = A y + B

Boundary conditions:

  • V = 0 at y = 0 gives B = 0.
  • V = V₀ at y = d gives A = V₀/d.
V = V₀ y / d

Field, charge and capacitance

E = −∇V = −(V₀/d) a_y
D = εE  = −(εV₀/d) a_y

The field leaves the plate at the higher potential, so the upper plate (V₀) carries +Q and the lower plate carries −Q. The magnitude of the surface charge density is

|ρ_s| = εV₀/d
Q = ρ_s S = εV₀S/d

Capacitance C = Q/V₀:

C = εS/d = ε₀ε_r S/d   farads

Answer: C = εS/d, independent of V₀. Energy stored is W = ½CV₀² = εSV₀²/(2d).

✓ Answer checked by Claude Opus 5.5 (high effort)

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