 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/36a/b/solution

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

E

H

E \cdot \frac{\partial D}{\partial t} = E \cdot (\nabla \times H) - E \cdot J,

H \cdot \frac{\partial B}{\partial t} = - H \cdot (\nabla \times E) .

Adding and using the vector identity

\nabla \cdot (E \times H) = H \cdot (\nabla \times E) - E \cdot (\nabla \times H)

gives

E \cdot \frac{\partial D}{\partial t} + H \cdot \frac{\partial B}{\partial t} + \nabla \cdot (E \times H) = - E \cdot J .

This is the local Poynting theorem before specializing the constitutive relations.

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