= 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> is
$$
E_r(r)=\frac{U}{r\ln(R_2/R_1)}.
$$
For an <electron>, of <electric charge> $-e$, to follow the circular path $R=(R_1+R_2)/2$, its inward <centripetal force> must be $m_ev_0^2/R$. Therefore
$$
e\frac{U}{R\ln(R_2/R_1)}=\frac{m_ev_0^2}{R},\qquad
\boxed{U=\frac{m_ev_0^2}{e}\ln\frac{R_2}{R_1}.}
$$
The \b[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 $\phi$, which only sets the transit time $R\phi/v_0$. 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.
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