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ID: past-exam-of-the-electromagnetism-course-of-the-university-of-oxford/2023/1/b/ii/solution

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Past exam of the electromagnetism course of the University of Oxford / 2023 / 1 / b / ii / Solution by

Codex  0 2026-10-25

Choose a spherical Gaussian surface of radius

s

with

a < s < r_{0}

. Its physical enclosed electric charge is the total induced electric charge

Q_{a}

on the inner sphere. The method of images produces the same electric field on that Gaussian surface, so Gauss's law identifies

Q_{a}

with the sum of the enclosed image charges. These are the first family with

n \leq - 1

and the second family with

n \leq 0

:

Q_{a} = q \sum_{n = - \infty}^{- 1} k^{n} - \frac{q a}{r _{0}} \sum_{n = - \infty}^{0} k^{n} = q \frac{k ^{- 1} - a / r _{0}}{1 - k ^{- 1}} .

Summing these geometric series gives

Q_{a} = - q \frac{a ( b - r _{0} )}{r _{0} ( b - a )} .

The electric field in the empty central cavity vanishes by electrostatic shielding, so it contributes no additional inner-face surface charge density. The formula tends to

- q

r_{0} ↓ a

and to zero as

r_{0} ↑ b

, as expected when the point charge approaches one grounded boundary or the other.

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