Q. Three charges $+Q, q, + Q$ are placed respectively, at distance, $d/2$ and $d$ from the origin, on the $x$-axis. If the net force experienced by $+Q$, placed at $x = 0$, $Ls$ zero, then value of $q$ is :
Solution:
For equilibrium,
$\vec{F}_{a} + \vec{F}_{B} = 0 $
$\vec{F}_{a} = - \vec{F_{B} } $
$\frac{kQQ}{d^{2}} = - \frac{kQq}{\left(d/2\right)^{2}} $
$ \Rightarrow q = - \frac{Q}{4} $
