As shown in the figure, a configuration of two equal point charges ($q_0=+2\ \mu\text{C}$) is placed on an inclined plane. Mass of each point charge is $20\ \text{g}$. Assume that there is no friction between charge and plane. For the system of two point charges to be in equilibrium (at rest) the height $h=x\times10^{-3}$ m. The value of $x$ is ______. (Take $\dfrac1{4\pi\epsilon_0}=9\times10^9\ \text{N m}^2\,\text{C}^{-2}$, $g=10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 300
Along the smooth incline the upper charge is held by the repulsion of the lower one. Their separation is $r=\dfrac{h}{\sin30^\circ}=2h$.
$$\frac{kq_0^2}{(2h)^2}=mg\sin30^\circ\ \Rightarrow\ h^2=\frac{kq_0^2}{2mg}=\frac{9\times10^9\times4\times10^{-12}}{2\times0.02\times10}=0.09$$
$h=0.3\ \text{m}=300\times10^{-3}\ \text{m}$.
Solution by Sreeraj P, M.Sc Physics