Q 12-01-094JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
Two identical charged particles each having a mass $10\ \text{g}$ and charge $2.0\times10^{-7}\ \text{C}$ are placed on a horizontal table with a separation of $L$ between them such that they stay in limiting equilibrium. If the coefficient of friction between each particle and the table is $0.25$, find the value of $L$. [Use $g = 10\ \text{m s}^{-2}$]
Answer: (A) $12\ \text{cm}$
In limiting equilibrium the Coulomb repulsion equals the maximum friction:
$$\frac{kq^2}{L^2} = \mu mg\ \Rightarrow\ L^2 = \frac{9\times10^9\times(2\times10^{-7})^2}{0.25\times0.01\times10} = \frac{3.6\times10^{-4}}{0.025} = 0.0144$$
$$L = 0.12\ \text{m} = 12\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics