Academy

Spherical Capacitor

Capacitance of two concentric spherical shells of radii a (inner) and b (outer). For isolated sphere (b → ∞): C = 4πε₀a.
Class 12
Derivation

Setup

Inner shell radius aa, outer shell radius bb, charge +Q+Q on inner and Q-Q on outer.

Field in the gap (a<r<ba < r < b)

By Gauss's law (enclosed charge = +Q+Q):

E=Q4πε0r2E = \frac{Q}{4\pi\varepsilon_0 r^2}

Field is zero for r<ar < a (inner shell) and r>br > b (outer shell shields it).

Potential difference

V=VaVb=abEdr=Q4πε0abdrr2=Q4πε0(1a1b)=Q(ba)4πε0abV = V_a - V_b = \int_a^b E\, dr = \frac{Q}{4\pi\varepsilon_0}\int_a^b \frac{dr}{r^2} = \frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{a} - \frac{1}{b}\right) = \frac{Q(b-a)}{4\pi\varepsilon_0 ab}

Capacitance

C=QV=4πε0abbaC = \frac{Q}{V} = \frac{4\pi\varepsilon_0 ab}{b - a} C=4πε0abba\boxed{C = 4\pi\varepsilon_0\frac{ab}{b-a}}

Limiting case: isolated sphere (bb \to \infty)

C=4πε0abbab4πε0aC = 4\pi\varepsilon_0 \frac{a \cdot b}{b - a} \xrightarrow{b\to\infty} 4\pi\varepsilon_0 a

An isolated sphere of radius aa has capacitance 4πε0a4\pi\varepsilon_0 a.