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ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2008/j/1/solution

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Past exam of the probability, statistics and options I course of the University of Oxford / 2008 / J / 1 / Solution by

Codex  0 2026-10-31

\int_{\partial V} E \cdot n d S = \frac{Q ( V )}{ϵ _{0}} = \frac{1}{ϵ _{0}} \int_{V} ρ d V .

(1)

The divergence theorem makes the left side

\int_{V} \nabla \cdot E d V

. If the differential relation

\nabla \cdot E = ρ / ϵ_{0}

holds, this proves the integral law. Conversely, if the integral identity holds for every suitable volume and the fields are sufficiently regular, a nonzero value of

\nabla \cdot E - ρ / ϵ_{0}

would persist in sign on a small neighborhood and contradict its zero integral. The differential law follows.

For a sphere of radius

r

, spherical symmetry makes the flux

4 π r^{2} E (r)

and the enclosed charge

4 π \int_{0}^{r} s^{2} ρ (s) d s

. Hence

E (r) = \frac{1}{ϵ _{0} r ^{2}} \int_{0}^{r} s^{2} ρ (s) d s .

(2)

There is no extra point charge at the origin in the stipulated density distribution. Outside the charge support,

Q = 4 π \int_{0}^{a} s^{2} ρ (s) d s, E (r) = \frac{Q}{4 π ϵ _{0} r ^{2}} (r > a) .

(3)

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