Integrating the normal Maxwell traction (the radial parts cancel) gives
$$
F_z=\frac{q^2}{16\pi\epsilon_0d^2}\frac{\epsilon_r-1}{\epsilon_r+1}.
$$
It points toward the dielectric when $\epsilon_r>1$. This equals the Coulomb force from the image charge, as it must.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 4 / 36D / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-4/36d/c/solution