In free space, a particle $A$ of charge $1\ \mu\text{C}$ is held fixed at point $P$. Another particle $B$ of the same charge and mass $4\ \mu\text{g}$ is kept at a distance of $1\ \text{mm}$ from $P$. If $B$ is released, then its velocity at a distance of $9\ \text{mm}$ from $P$ is: $\left[\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}\right]$
Answer: (C) $2.0\times10^3\ \text{m s}^{-1}$
Loss of potential energy $=$ gain of kinetic energy:
$$\frac12mv^2 = kq^2\left(\frac1{r_1} - \frac1{r_2}\right) = 9\times10^9\times10^{-12}\left(1000 - \frac{1000}{9}\right) = 8\ \text{J}$$
With $m = 4\ \mu\text{g} = 4\times10^{-9}\ \text{kg}$:
$$v = \sqrt{\frac{2\times8}{4\times10^{-9}}} = 2.0\times10^3\ \text{m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics