A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of three charges will be in equilibrium if $q$ is equal to:

- \frac{Q}{2}

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- \frac{Q}{4}

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+ \frac{Q}{4}

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+ \frac{Q}{2}

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Solution

The correct option is A $- \frac{Q}{4}$
For $Q$ to be in equilibrium, net force acting on $Q$ must be $0$ .
So, $F = \frac{k Q q}{(\frac{r}{2})^{2}} + \frac{k Q^{2}}{r^{2}} = 0$
$\Rightarrow \frac{4 q}{r^{2}} + \frac{Q}{r^{2}} = 0$
$\Rightarrow q = - \frac{Q}{4}$

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A charge q is placed at the center of the line joining two equal charges Q. The system of three charges will be in equilibrium if q is equal to

Q = \frac{- q}{4}

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