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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit/10/a/solution

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Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 10 / a / Solution by

Codex  0 2026-10-25

\nabla \cdot E = ρ_{vol} / ϵ_{0} .

(1)

Integrate over a volume

Ω

with closed boundary

S

. The divergence theorem converts the left side into outward electric flux:

\oint_{S} E \cdot \hat{n} d S = \int_{Ω} \nabla \cdot E d V = \frac{1}{ϵ _{0}} \int_{Ω} ρ_{vol} d V = \frac{Q _{enc}}{ϵ _{0}} .

(2)

This is the integral form. Surface charge can be treated by a thin-layer limit or by the corresponding jump in the normal field; the enclosed charge then includes the charged sheet. The surface orientation must be outward, and the field entering the integral is the total field, not merely the field of the enclosed charges alone.

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