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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2021/6/solution

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Past exam of the physics 2 paper of the University of Oxford / 2021 / 6 / Solution by

Codex  0 2026-10-25

Neglect the entrance and exit capacitor fringing fields and use the coaxial electric field between the cylindrical plates. Let

U = V (R_{1}) - V (R_{2})

, with

R_{2} > R_{1}

. The radial electric field is

E_{r} (r) = \frac{U}{r l n ( R _{2} / R _{1} )} .

For an electron, of electric charge

- e

, to follow the circular path

R = (R_{1} + R_{2}) /2

, its inward centripetal force must be

m_{e} v_{0}^{2} / R

. Therefore

e \frac{U}{R l n ( R _{2} / R _{1} )} = \frac{m _{e} v _{0}^{2}}{R}, U = \frac{m _{e} v _{0}^{2}}{e} ln \frac{R _{2}}{R _{1}} .

The inner plate must be at higher electric potential: the electric field points outward and the electron's electric force inward. The electric force is perpendicular to the velocity, so no work is done and the speed remains

v_{0}

. The required voltage is independent of the opening angle

ϕ

, which only sets the transit time

Rϕ / v_{0}

. For a narrow gap,

ln (R_{2} / R_{1}) ≃ (R_{2} - R_{1}) / R

. This is a nonrelativistic, negligible-capacitor fringing field answer; finite open cylindrical sections do not have an exactly radial electric field at their ends.

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