After a radial <Fourier transform>, Poisson's equation away from the sheet becomes
$$
(\partial_z^2-k^2)\widetilde\Phi_D=0.
$$
The decaying, reflection-symmetric solution is $A e^{-|k||z|}$. Integrating Poisson's equation across $z=0$ gives the derivative jump
$$
\partial_z\widetilde\Phi_D(0^+)
-\partial_z\widetilde\Phi_D(0^-)
=4\pi G\widetilde\Sigma,
$$
so $-2|k|A=4\pi G\widetilde\Sigma$. Thus the midplane potential is
$$
\boxed{\widetilde\Phi_{D,m}
=-\frac{2\pi G}{|k|}\widetilde\Sigma}.
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-321/3/b/i/solution