Question

A point charge $q_1=4 q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x=12 \,cm$. 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 ____ $cm$.

Answer 1

$\frac{q_0}{x^2}=\frac{4 q_0}{(x+12)^2} $
$ x+12=2 x$
$ x=12$
Distance from origin $=x+12=24\, cm$.