A uniformly charged ring of radius $3a$ and total charge $q$ is placed in $x$-$y$ plane centred at origin. A point charge $q$ is moving towards the ring along the $z$-axis and has speed $v$ at $z = 4a$. The minimum value of $v$ such that it crosses the origin is:
Answer: (D) $\sqrt{\dfrac2m}\left(\dfrac2{15}\cdot\dfrac{q^2}{4\pi\epsilon_0a}\right)^{1/2}$
Potential of the ring on its axis: $V(z) = \dfrac{kq}{\sqrt{9a^2+z^2}}$, so $V(4a) = \dfrac{kq}{5a}$ and $V(0) = \dfrac{kq}{3a}$ (the maximum).
The charge just reaches the origin if
$$\frac12mv^2 = q\left[V(0) - V(4a)\right] = \frac{kq^2}{a}\left(\frac13-\frac15\right) = \frac{2}{15}\cdot\frac{q^2}{4\pi\epsilon_0a}$$
$$v = \sqrt{\frac2m}\left(\frac2{15}\cdot\frac{q^2}{4\pi\epsilon_0a}\right)^{1/2}$$
Solution by Sreeraj P, M.Sc Physics