 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/4d/solution

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

\oint_{\partial V} E \cdot d S = \frac{Q _{enclosed}}{ε _{0}} .

Cylindrical symmetry and a coaxial Gaussian cylinder give

E (r) = ⎩ ⎨ ⎧ 0, \frac{Q}{2 π ε _{0} r} e_{r}, 0, 0 < r < a, a < r < b, r > b .

The field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.

Choose the outer conductor's potential to be zero. Since

E = - \nabla V

V (r) = ⎩ ⎨ ⎧ \frac{Q}{2 π ε _{0}} lo g (b / a), \frac{Q}{2 π ε _{0}} lo g (b / r), 0, 0 < r \leq a, a < r < b, r \geq b .

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

U = \frac{ε _{0}}{2} \int_{a}^{b} ∣ E ∣^{2} 2 π r d r = \frac{Q ^{2}}{4 π ε _{0}} lo g \frac{b}{a} .

Substitution of

C

verifies

U = Q^{2} / (2 C)

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