 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/section-a/4/solution

Spherical symmetry makes the electric field radial and constant in magnitude on each concentric Gaussian surface. The enclosed charge is

Q (r) = 4 π \int_{0}^{r} (a s) s^{2} d s = πa r^{4} .

Gauss's law therefore gives

E (r) = \frac{a r ^{2}}{4 ϵ _{0}} r, ∣ E ∣ = \frac{∣ a ∣ r ^{2}}{4 ϵ _{0}} .

The direction is outward for

a > 0

and inward for

a < 0

; the field vanishes at the centre. No exterior boundary condition changes this spherically symmetric interior result.

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