 Solution

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

Codex  0 Created 2026-09-24 Updated 2026-09-25

With no free surface charge, tangential

E

and normal

D

are continuous. The image ansatz gives

\frac{q ^{'}}{q} = \frac{1 - ϵ _{r}}{1 + ϵ _{r}}, \frac{q ^{''}}{q} = \frac{2 ϵ _{r}}{1 + ϵ _{r}} .

z = 0^{-}

, with

R = (r^{2} + d^{2})^{1/2}

E_{r} = \frac{2 q}{1 + ϵ _{r}} \frac{r}{4 π ϵ _{0} R ^{3}}, E_{z} = \frac{2 ϵ _{r} q}{1 + ϵ _{r}} \frac{d}{4 π ϵ _{0} R ^{3}} .

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