 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-3/36d/c/solution

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

Since

B = 0

E (r) = \frac{ϵ _{0} E ( r ) ^{2}}{2} = ⎩ ⎨ ⎧ \frac{q ^{2} r ^{2}}{32 π ^{2} ϵ _{0} R ^{6}}, \frac{q ^{2}}{32 π ^{2} ϵ _{0} r ^{4}}, r \leq R, r > R .

Radial integration gives

W_{in} = \frac{q ^{2}}{40 π ϵ _{0} R}, W_{out} = \frac{q ^{2}}{8 π ϵ _{0} R},

and therefore

W_{em} = \frac{3 q ^{2}}{20 π ϵ _{0} R} .

To first order in

v / c

E^{'} = E

and

B^{'} = - v \times E / c^{2}

. Thus

g_{z}^{'} = \frac{ϵ _{0} v}{c ^{2}} E^{2} sin^{2} θ .

Using the angular average

⟨ sin^{2} θ ⟩ = 2/3

and

\int ϵ_{0} E^{2} d V = 2 W_{em}

P_{z} = \frac{4}{3} \frac{W _{em}}{c ^{2}} v = \frac{q ^{2} v}{5 π ϵ _{0} R c ^{2}} .

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