With no free surface charge, tangential $E$ and normal $D$ are continuous. The image ansatz gives
$$
\frac{q'}q=\frac{1-\epsilon_r}{1+\epsilon_r},\qquad \frac{q''}q=\frac{2\epsilon_r}{1+\epsilon_r}.
$$
At $z=0^-$, with $R=(r^2+d^2)^{1/2}$,
$$
E_r=\frac{2q}{1+\epsilon_r}\frac{r}{4\pi\epsilon_0R^3},\qquad E_z=\frac{2\epsilon_rq}{1+\epsilon_r}\frac{d}{4\pi\epsilon_0R^3}.
$$

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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