 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-4/7c/solution

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

\nabla \cdot E = ρ / ε_{0}

gives

0 = μ_{0} \nabla \cdot J + μ_{0} \frac{\partial ρ}{\partial t},

ρ_{t} + \nabla \cdot J = 0

J = σ E

, so

\nabla \cdot J = σ ρ / ε_{0}

. Therefore

ρ (t) = ρ (0) e^{- σ t / ε_{0}},

with charge-relaxation rate

σ / ε_{0}

ε_{0} / σ

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