Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 3 / b / i

Table of contents
- Solution i

Solution

After a radial Fourier transform, Poisson's equation away from the sheet becomes

(\partial_{z}^{2} - k^{2}) Φ_{D} = 0.

The decaying, reflection-symmetric solution is

A e^{- ∣ k ∣∣ z ∣}

. Integrating Poisson's equation across

z = 0

gives the derivative jump

\partial_{z} Φ_{D} (0^{+}) - \partial_{z} Φ_{D} (0^{-}) = 4 π G Σ,

so

- 2∣ k ∣ A = 4 π G Σ

. Thus the midplane potential is

Φ_{D, m} = - \frac{2 π G}{∣ k ∣} Σ .

Solved by gpt-5.6-sol high.

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