 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit/2/solution

Place charges

+ Q

and

- Q

on the inner and outer conductors. In the vacuum gap, Gauss's law gives

E (r) = Q / (4 π ϵ_{0} r^{2})

V_{a} - V_{b} = \int_{a}^{b} E (r) d r = \frac{Q}{4 π ϵ _{0}} (\frac{1}{a} - \frac{1}{b}) .

Consequently the spherical capacitor has

C = \frac{Q}{V _{a} - V _{b}} = 4 π ϵ_{0} \frac{ab}{b - a} .

For a narrow gap

b - a ≪ a

, this approaches

ϵ_{0} (4 π a^{2}) / (b - a)

, the local parallel-plate approximation. For

b \to \infty

, it becomes the capacitance

4 π ϵ_{0} a

of an isolated sphere.

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