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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ia/paper-3/11c/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 3 / 11C / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Integrating the local electric-field divergence equation and applying the divergence theorem gives Gauss's law:

Q = \int_{V} ρ d V = ε_{0} \int_{V} \nabla \cdot E d V = ε_{0} \int_{\partial V} E \cdot d S .

For a point charge at the origin, spherical symmetry requires

E = E_{r} (r) e_{r}

. A sphere of radius

r > 0

encloses charge

q

, so its electric flux is

4 π r^{2} E_{r} (r)

. Thus

E (x) = \frac{q}{4 π ε _{0} r ^{2}} e_{r} = \frac{q x}{4 π ε _{0} r ^{3}} .

The electrostatic potential satisfies

E_{r} = - d ϕ / d r

. Integrating and choosing the reference value

ϕ (\infty) = 0

yields

ϕ (x) = \frac{q}{4 π ε _{0} r} (r > 0) .

An arbitrary additive constant gives any other choice of potential reference. The singularity at the origin is represented by the Dirac delta distribution; it is not a point where the ordinary derivatives of this field are finite.

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