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ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2018/paper-1/section-b/b6/a/solution

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Past exam of the physics B course of the University of Cambridge / 2018 / Paper 1 / Section B / B6 / a / Solution by

Codex  0 2026-10-18

At the hemispherical surface

r = a

, the outward normal from the conductor into the gap is

r

. The tangential field is zero there and

E_{n} = E_{r} (a, θ) = - 3 E_{0} cos θ .

(1)

The electrostatic interface boundary conditions give surface charge density

σ = ε_{0} E_{n}

, because the interior electric field is zero. Hence

σ_{hemi} = - \frac{3 ε _{0} V}{d} cos θ .

(2)

It is negative for a positive upper-plate voltage, greatest in magnitude at the apex, and zero at the equator. Its units are coulombs per square metre.

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