Consider a fixed uniformly charged insulating sphere with radius $R$ and total charge $+Q$. A point charge $-q$ ($q << Q$) with mass $m$ is released from rest at a distance of $3R$ from the centre of the charged sphere. When the point charge reaches the surface of the sphere, its speed is :
($\epsilon_0$ is the permittivity of vacuum, neglect gravitational forces).
Answer: (C) $\sqrt{\dfrac{Qq}{3\pi\epsilon_0 mR}}$
Loss in potential energy = gain in kinetic energy:
$$\frac{1}{2}mv^2 = \frac{Qq}{4\pi\epsilon_0}\left(\frac{1}{R} - \frac{1}{3R}\right) = \frac{Qq}{4\pi\epsilon_0}\cdot\frac{2}{3R}$$
$$v^2 = \frac{4Qq}{3 \cdot 4\pi\epsilon_0 mR} = \frac{Qq}{3\pi\epsilon_0 mR}$$
$$v = \sqrt{\frac{Qq}{3\pi\epsilon_0 mR}}$$
Solution by Sreeraj P, M.Sc Physics