 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-3/17a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Take the inner conductor to be at potential

V

above the outer. A cylindrical Gauss law surface of length

ℓ

encloses charge

λ ℓ

and gives

E_{r} 2 π r ℓ = λ ℓ / ϵ_{0}

. The potential difference is

V = \int_{a}^{b} E_{r} d r = \frac{λ}{2 π ϵ _{0}} ln \frac{b}{a} .

λ = \frac{2 π ϵ _{0} V}{ln ( b / a )}, E = \frac{V}{r ln ( b / a )} e_{r} .

The perfectly conducting surfaces have no tangential electric field in this idealized steady model.

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