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If a charge q is placed at the centre of the line joining two equal charges $+ Q$ and $+ Q,$ the system of the three charges will be in equilibrium, if q is

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$4 Q$

$- \frac{Q}{4}$

$- \frac{Q}{2 \sqrt{2}}$

$- \frac{Q}{2}$

answer is B.

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Detailed Solution

The net force on either of the charges, Q, should be zero.

$\Rightarrow F_{Q} = - F_{q}$

$\Rightarrow \frac{1}{4 {πε}_{0}} \frac{Q^{2}}{4 d^{2}} = - \frac{1}{4 {πε}_{0}} \frac{Qq}{d^{2}}$

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

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