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 , is zero, then value of q is

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+Q/4

-Q/2

-Q/4

+Q/2

answer is A.

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Detailed Solution

Force due to charge +Q, $F_{a} = \frac{KQQ}{d^{2}}$ Force due to charge q, $F_{b} = \frac{KQq}{{(\frac{d}{2})}^{2}}$
For equilibrium ${\vec{F}}_{a} + {\vec{F}}_{b} = 0 \Rightarrow \frac{κQQ}{d^{2}} + \frac{kQq}{(d / 2)^{2}} = 0$

$\Rightarrow q = - \frac{Q}{4}$

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