Capacitance of coaxial capacitor

Aim: Solve Laplace's equation for the region between coaxial cylinders (inner radius a at potential V₀, outer radius b at potential 0, length L, dielectric ε) and find the capacitance.

        ╭──────────────╮
       ╱  ╭──────────╮  ╲    inner: ρ = a, V = V₀
      │  │   ●──a    │  │    outer: ρ = b, V = 0
       ╲  ╰──────────╯  ╱    length L (edge effects ignored)
        ╰──────b───────╯

Step 1: Laplace's equation

There is no charge between the cylinders, so ∇²V = 0. V depends only on ρ, so:

(1/ρ) d/dρ ( ρ dV/dρ ) = 0

Step 2: Integrate twice

ρ dV/dρ = A          →   dV/dρ = A/ρ
V = A ln ρ + B

Step 3: Boundary conditions

V(b) = 0:    A ln b + B = 0    →  B = −A ln b
V(a) = V₀:   A ln a − A ln b = V₀   →  A = −V₀ / ln(b/a)

V(ρ) = V₀ ln(b/ρ) / ln(b/a)

Step 4: Field and charge

E = −dV/dρ a_ρ = ( V₀ / (ρ ln(b/a)) ) a_ρ

On the inner conductor (ρ = a): ρ_s = εE_ρ(a) = ε V₀ / (a ln(b/a)).

Q = ρ_s × (2πaL) = 2π ε L V₀ / ln(b/a)

Step 5: Capacitance

C = Q / V₀ = 2π ε L / ln(b/a)      farads

Result: C = 2πεL / ln(b/a), or per unit length C/L = 2πε/ln(b/a) F/m.