Q. Three charges +Q, q, + Q are placed respectively, at distance, 0, 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} $

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