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ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/section-a/a8/solution

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Past exam of the physics course of the year 4 of the natural sciences course of the University of Cambridge / 2015 / Paper 1 / Section A / A8 / Solution by

Codex  0 2026-10-20

An element at

a (cos ϕ, sin ϕ, 0)

carries charge

d q = λ_{0} a sin ϕ d ϕ

. Its field at the origin points opposite its radial position. By Coulomb's law,

E (0) = - \frac{λ _{0}}{4 π ϵ _{0} a} \int_{0}^{2 π} sin ϕ (cos ϕ \hat{x} + sin ϕ \hat{y}) d ϕ = - \frac{λ _{0}}{4 ϵ _{0} a} \hat{y} .

Consequently

a (0) = - \frac{q λ _{0}}{4 ϵ _{0} ma} \hat{y} .

For

q λ_{0} > 0

the particle starts along

- y

; for the opposite sign it starts along

+ y

. Reflection symmetries make the

y

axis invariant for motion released there from rest. The ring has zero net charge but a nonzero dipole moment, so zero total charge does not imply zero field at its centre.

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