= Solution
Place charges $+Q$ and $-Q$ on the inner and outer conductors. In the vacuum gap, <Gauss's law> gives $E(r)=Q/(4\pi\epsilon_0r^2)$. Their <electric potential difference> is
$$
V_a-V_b=\int_a^bE(r)\,dr=\frac{Q}{4\pi\epsilon_0}\left(\frac1a-\frac1b\right).
$$
Consequently the <spherical capacitor> has
$$
\boxed{C=\frac{Q}{V_a-V_b}=4\pi\epsilon_0\frac{ab}{b-a}.}
$$
For a narrow gap $b-a\ll a$, this approaches $\epsilon_0(4\pi a^2)/(b-a)$, the local parallel-plate approximation. For $b\to\infty$, it becomes the capacitance $4\pi\epsilon_0a$ of an isolated sphere.
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