Capacitance of spherical capacitor

Aim: Capacitance C = Q/V₀ of two concentric spherical conductors: inner radius a at potential V₀, outer radius b at potential 0, with a dielectric of permittivity ε between them.

     ╭─────────╮
    ╱  ╭─────╮  ╲      inner sphere: radius a, V = V₀
   │  │  ●─a │   │     outer sphere: radius b, V = 0
    ╲  ╰─────╯  ╱      region a < r < b: ρ_v = 0
     ╰─────b───╯

Step 1: Laplace's equation

The region between the spheres has no free charge, so ∇²V = 0. By spherical symmetry V depends only on r:

(1/r²) d/dr ( r² dV/dr ) = 0

Step 2: Integrate twice

r² dV/dr = A                    (first integration; A is a constant)
dV/dr = A / r²
V = −A / r + B                  (second integration)

Write the constants as C₁ = −A and C₂ = B: V = C₁/r + C₂.

Step 3: Boundary conditions

V(r = b) = 0:      C₁/b + C₂ = 0     →  C₂ = −C₁/b
V(r = a) = V₀:     C₁/a + C₂ = V₀    →  C₁ (1/a − 1/b) = V₀

C₁ = V₀ a b / (b − a)

V(r) = V₀ ( ab / (b − a) ) ( 1/r − 1/b )

Step 4: Electric field and charge

E = −∇V = −(dV/dr) a_r = ( V₀ ab / (b − a) ) (1/r²) a_r

Surface charge density on the inner sphere (r = a): ρ_s = D_n = ε E_r(a) = ε V₀ b / ( a (b − a) ).

Q = ρ_s (4πa²) = 4π ε a b V₀ / (b − a)

Step 5: Capacitance

C = Q / V₀ = 4π ε a b / (b − a)      farads

Result: C = 4πε ab / (b − a), with ε = ε₀ε_r.

Special case: For b → ∞ (an isolated sphere), C = 4πεa.