 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-4/36d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

P

ρ_{b} = - \nabla \cdot P, σ_{b} = P \cdot n .

Indeed, summing the scalar potential of its dipoles and integrating by parts gives

ϕ (x) = \frac{1}{4 π ϵ _{0}} \int_{V} P (x^{'}) \cdot \nabla^{'} \frac{1}{∣ x - x ^{'} ∣} d^{3} x^{'}

= \frac{1}{4 π ϵ _{0}} [\int_{V} \frac{- \nabla ^{'} \cdot P ( x ^{'} )}{∣ x - x ^{'} ∣} d^{3} x^{'} + \int_{S} \frac{P ( x ^{'} ) \cdot n}{∣ x - x ^{'} ∣} d S^{'}] .

This is exactly the potential of

ρ_{b}

and

σ_{b}

D (r) = \frac{q}{4 π r ^{2}} \overset{r}{^}, E = \frac{D}{ϵ}, P = D - ϵ_{0} E = (1 - \frac{ϵ _{0}}{ϵ}) \frac{q}{4 π r ^{2}} \overset{r}{^} .

Thus on the sphere,

σ_{b} = (1 - \frac{ϵ _{0}}{ϵ}) \frac{q}{4 π R ^{2}} .

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