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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2022/9/c/solution

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Past exam of the physics 2 paper of the University of Oxford / 2022 / 9 / c / Solution by

Codex  0 2026-10-25

The electric field within a conductor in electrostatic equilibrium vanishes. The normal electric field boundary condition, with outward normal vector

+ \hat{z}

, therefore gives

σ = ϵ_{0} E_{z} (0^{+})

. Directly below the monopole,

a = 0

:

σ_{mono} (0) = - \frac{q}{2 π h ^{2}} .

For the electric dipole,

σ_{dip} (0) = - \frac{q d}{π h ^{3}} + O (q d^{2} / h^{4}) .

The exact finite-separation answer is

σ (0) = \frac{q}{2 π} [\frac{1}{( h + d ) ^{2}} - \frac{1}{h ^{2}}],

whose Taylor series agrees with the boxed leading term. Although the real point charges have zero total electric charge, their unequal distances leave negative induced surface charge density immediately beneath them.

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