Capacitance of Spherical Capacitors: Solid and Hollow

Beyond the flat-plate geometry, capacitors can equally well be built from spherical conductors, and the WBCHSE syllabus asks specifically for two related spherical cases: a single isolated conducting sphere (treated as a "solid" sphere capacitor) and a "hollow" spherical capacitor made from two concentric spherical shells.

Capacitance of an isolated (solid) conducting sphere. Consider a single conducting sphere of radius $R$, carrying charge $Q$, isolated in space with no other conductor anywhere nearby. From Section 2.4 (a uniformly charged sphere behaves, from outside itself, exactly like a point charge of the same total charge located at its centre), the potential at the sphere's own surface is

Treating the isolated sphere as a capacitor whose "other plate" is an imaginary conducting shell at infinity (where $V=0$ by the usual convention), its capacitance is

using $k = 1/(4\pi\epsilon_0)$. This is a genuinely useful, self-contained result: capacitance depends on the sphere's radius ALONE, growing larger for a physically larger sphere -- a bigger sphere can hold more charge for the very same surface potential, exactly the geometric intuition capacitance is meant to capture.

Capacitance of a hollow spherical capacitor (two concentric shells). Now consider two concentric conducting spherical shells, an inner shell of radius $a$ carrying charge $+Q$ and an outer, concentric shell of radius $b > a$ carrying charge $-Q$. Between the two shells (in the region $a < r < b$), the field is exactly that of a point charge $Q$ at the centre (by the same reasoning as above, applied now only to the enclosed inner shell); outside the outer shell ($r > b$), the field is exactly zero, since the net enclosed charge is $Q + (-Q) = 0$. The potential difference between the two shells is therefore

giving a capacitance

$C = \frac{Q}{V} = \frac{ab}{k(b-a)} = \frac{4\pi\epsilon_0\, ab}{b - a}$ …