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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013/11/ii/solution

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Past exam of the physics 2 paper of the University of Oxford / 2013 / 11 / ii / Solution by

Codex  0 2026-10-25

Take

x \in [- s /2, s /2]

along the tilt, with local separation

h (x) = a + 2 δ x / s

. Since

s ≫ a

and the tilt is small, each narrow strip is a local parallel-plate capacitor at the same voltage

V

. Its surface charge densities are approximately

σ_{\pm} (x) = \pm \frac{ϵ _{0} V}{a + 2 δ x / s} .

Parallel strip capacitances add, so for

∣ δ ∣ < a

,

C (δ) = ϵ_{0} s \int_{- s /2}^{s /2} \frac{d x}{a + 2 δ x / s} = \frac{ϵ _{0} s ^{2}}{2 δ} lo g \frac{a + δ}{a - δ} .

Using

lo g [(1 + z) / (1 - z)] = 2 (z + z^{3} /3 + z^{5} /5 + \dots)

gives

C (δ) = \frac{ϵ _{0} s ^{2}}{a} [1 + \frac{δ ^{2}}{3 a ^{2}} + O (\frac{δ ^{4}}{a ^{4}})] .

The denser charge at the smaller gap explains why tilting increases the capacitance. The local-strip approximation neglects fringing and slope corrections of order

(δ / s)^{2}

, smaller than the retained

(δ / a)^{2}

term.

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