 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-1/15d/a/solution

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

r

L

, symmetry makes the electric field radial and constant on the curved surface. Gauss's law gives

E (2 π r L) = \frac{λ L}{ε _{0}},

E (r) = \frac{λ}{2 π ε _{0} r} e_{r} .

Since

E = - \nabla ϕ

, a potential relative to an arbitrary reference radius

r_{0}

ϕ (r) = - \frac{λ}{2 π ε _{0}} lo g \frac{r}{r _{0}} .

An infinite line charge has no finite convention

ϕ (\infty) = 0

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