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

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 isFor an electron, of electric charge $-e$, to follow the circular path $R=(R_1+R_2)/2$, its inward centripetal force must be $m_e{v}_0^2/R$. ThereforeThe 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 $\varphi$, which only sets the transit time $R\varphi /v_0$. For a narrow gap, $\ln (R_2/R_1)\simeq (R_2-R_1)/R$. This is a nonrelativistic, negligible-capacitor fringing field answer; finite open cylindrical sections do not have an exactly radial electric field at their ends.