 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/4d/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

A capacitor consists of two conductors carrying equal and opposite charges. Its capacitance is

C = \frac{Q}{V},

where

Q

is the magnitude of the charge on either conductor and

V

is their potential difference.

For

a < r < b

ℓ

encloses charge

λ ℓ

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

Taking

V

to mean the inner potential minus the outer potential,

V = \int_{a}^{b} E (r) d r = \frac{λ}{2 π ϵ _{0}} lo g \frac{b}{a} .

Since

Q = λ L

C = \frac{2 π ϵ _{0} L}{lo g ( b / a )} .

U = \frac{ϵ _{0}}{2} \int_{a}^{b} E (r)^{2} (2 π r L) d r = \frac{λ ^{2} L}{4 π ϵ _{0}} lo g \frac{b}{a} = \frac{1}{2} (λ L) (\frac{λ}{2 π ϵ _{0}} lo g \frac{b}{a}) = \frac{1}{2} Q V .

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