= Solution
Put <electric charge> $+Q$ on the inner <electrical conductor> and $-Q$ on the outer one. In the gap, <spherical symmetry> makes the <electric field> radial and constant on a spherical <Gaussian surface>. <Gauss's law> gives
$$
4\pi r^2E(r)=\frac Q{\epsilon_0},\qquad a<r<b.
$$
The inner-to-outer <electric potential difference> is therefore
$$
V=\int_a^b\frac Q{4\pi\epsilon_0r^2}\,dr
=\frac Q{4\pi\epsilon_0}\left(\frac1a-\frac1b\right).
$$
Thus the <capacitance> of the <spherical capacitor> is
$$
\boxed{C=4\pi\epsilon_0\frac{ab}{b-a}.}
$$
As $b\to\infty$ this gives the <capacitance of an isolated conducting sphere>, $4\pi\epsilon_0a$; for $b-a\ll a$ it approaches the <parallel-plate capacitor> result with <area> $4\pi a^2$.
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