 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-2/4b/solution

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

\nabla \cdot E = ρ / ε_{0}, \nabla \cdot B = 0, \nabla \times E = - \partial_{t} B, \nabla \times B = μ_{0} J + μ_{0} ε_{0} \partial_{t} E .

\partial_{t} ρ + \nabla \cdot J = 0.

Therefore

\dot{Q} = - \int_{\partial D} J \cdot n d S = 0

because

J

vanishes on the boundary. Componentwise integration by parts also gives

\frac{d}{d t} \int_{D} x ρ d^{3} x = - \int_{D} x \nabla \cdot J d^{3} x = \int_{D} J d^{3} x,

with the boundary term again zero.

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