 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-1/15d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

0 = μ_{0} \nabla \cdot J + μ_{0} ϵ_{0} \frac{\partial}{\partial t} (\nabla \cdot E) = μ_{0} (\nabla \cdot J + \frac{\partial ρ}{\partial t}) .

Thus

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

\frac{d Q}{d t} = - \int_{\partial V} J \cdot n d S .

Charge in

V

is conserved provided no current crosses

\partial V

. For total charge in all space, the corresponding assumption is sufficient decay of

J

so that the flux at infinity vanishes.

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