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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2021/paper-1/7/iii/b/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2021 / Paper 1 / 7 / iii / b / Solution by

Codex  0 2026-10-20

Treat the film as a parallel-plate capacitor, initially compensated and then isolated electrically. Its capacitance and temperature-induced electric polarization charge are

C = \frac{ϵ _{0} ϵ A}{d}, Δ Q = A p Δ T .

(1)

The corresponding ideal open-circuit signal, with the electrode polarity defined consistently with this charge convention, is

V_{oc} = \frac{Δ Q}{C} = \frac{p d Δ T}{ϵ _{0} ϵ} .

(2)

Area cancels because both charge and capacitance are proportional to it. Substitution gives

V_{oc} = \frac{( - 35 \times 1 0 ^{- 6} ) ( 2 \times 1 0 ^{- 6} ) ( 1 )}{( 8.854 \times 1 0 ^{- 12} ) ( 8 )} = - 0.988 V .

(3)

The voltage magnitude is approximately $0.99 V$ ; interchanging the voltmeter leads reverses its sign. This is an ideal capacitive result, before leakage, mobile-charge screening or loading by the measurement circuit reduces the signal.

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