 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-311/1/b/iv/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

- g = r^{3} sin θ cos θ

. On

e^{- iω t}

ω + q A_{t} = ω - \frac{qQ}{r ^{2}} .

\frac{1}{r ^{3}} \frac{d}{d r} (r^{3} f \frac{d R}{d r}) + \frac{( ω - qQ / r ^{2} ) ^{2}}{f} R + \frac{R}{r ^{2} Θ s i n θ c o s θ} \frac{d}{d θ} (sin θ cos θ \frac{d Θ}{d θ}) = 0.

λ

gives

\frac{1}{sin θ cos θ} \frac{d}{d θ} (sin θ cos θ \frac{d Θ}{d θ}) + λ Θ = 0

and

\frac{1}{r ^{3}} \frac{d}{d r} (r^{3} f \frac{d R}{d r}) + [\frac{( ω - qQ / r ^{2} ) ^{2}}{f} - \frac{λ}{r ^{2}}] R = 0 .

λ = ℓ (ℓ + 2)

with

ℓ = 0, 2, 4, \dots

, the even values selected by the absence of

ϕ

and

ψ

dependence.

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