 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-1/15b/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

Using

ρ = ε_{0} \nabla \cdot E

E = - \nabla ϕ

ϕ = 0

on the boundary gives

U = \frac{ε _{0}}{2} \int_{V} ∣ E ∣^{2} d^{3} x .

Put

C = q / (4 π ε_{0})

. Gauss's law gives the radial field

E_{r} = ⎩ ⎨ ⎧ 0, C / r^{2}, - C / r^{2}, 0, r < R, R < r < 2 R, 2 R < r < 3 R, r > 3 R .

Taking zero potential outside,

ϕ = ⎩ ⎨ ⎧ C / (3 R), C (1/ r - 2/ (3 R)), C (1/ (3 R) - 1/ r), 0, r < R, R < r < 2 R, 2 R < r < 3 R, r > 3 R .

The charge formula gives

U = \frac{1}{2} \sum_{i} Q_{i} ϕ (r_{i}) = \frac{q ^{2}}{12 π ε _{0} R} .

The field formula gives the same result after integrating

4 π r^{2} E_{r}^{2}

over the two nonzero annuli.

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