Question

A circular ring of radius $R$ with uniform positive charge $q$ is located in the $y-z$ plane with its centre at the origin $O$. A particle of mass $m$ and positive charge $q$ is projected from the point $P[\sqrt{3}R,0,0]$ on the positive x-axis directly towards $O$, with initial speed $V$. Find the smallest (non zero) value of the speed such that the particle does not return to $P$.

Answer 1

$V_x=\frac{kq}{\sqrt{R^2+x^2}}=\frac{kq}{\sqrt{R^2+(\sqrt{3R}{)}^2}}=\frac{kq}{2R}$
P.E. of change $q$ at $P(\surd 3,0,0)$ is $U_p=V_q=\frac{k{q}^2}{2R}$
P.E. of charge $q$ at the centre of ring
$U_0=V_{0}q=\frac{k{q}^2}{R}$
In order that the charged particle does not return to $P$, it must just across the centre $O$ and there after it will be repelled on the other side.
$(KE{)}_p+U_p=U_0$
$⟹\frac{1}{2}m{v}^2+\frac{k{q}^2}{2R}=\frac{k{q}^2}{R}⟹v=\sqrt{\frac{k{q}^2}{mR}}=\sqrt{\frac{q^2}{4\pi {\in }_{0}mR}}$