 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1/section-a/a2/solution

+ λ

and

- λ

on the two axes. For

d ≫ a_{1}, a_{2}

, the surface potentials differ by

V_{1} - V_{2} ≃ \frac{λ}{2 π ϵ _{0}} (ln \frac{d}{a _{1}} + ln \frac{d}{a _{2}}) .

C^{'} = \frac{2 π ϵ _{0}}{ln [ d ^{2} / ( a _{1} a _{2} )]} = \frac{π ϵ _{0}}{ln [ d / a _{1} a _{2} ]} .

The two logarithms belong to one potential difference, not two independent capacitances in parallel. A homogeneous dielectric replaces

ϵ_{0}

by its absolute permittivity; finite radii and proximity corrections are neglected.

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