Q 12-01-068JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
A point charge $q_1=4q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x=12\ \text{cm}$. Charge of proton is $q_0$. The proton is placed on $x$-axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ______ cm.
Numerical value type. Enter your answer.
Answer: 24
The charges are unlike, so the null point lies outside them, on the side of the smaller charge ($x>12\ \text{cm}$):
$$\frac{4q_0}{x^2}=\frac{q_0}{(x-12)^2}\ \Rightarrow\ 2(x-12)=x\ \Rightarrow\ x=24\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics