Three charges $+ Q_{1}, + Q_{2}$ and $q$ are placed on a straight line such that $q$ is somewhere in between $+ Q_{1}$ and $+ Q_{2}$ . If this system of charges is in equilibrium, what should be the magnitude and sign of charge $q$ ?

\frac{Q_{1} Q_{2}}{(\sqrt{Q_{1}} + \sqrt{Q_{2}})^{2}}, + v e

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\frac{Q_{1} + Q_{2}}{2}, + v e

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\frac{Q_{1} Q_{2}}{(\sqrt{Q_{1}} + \sqrt{Q_{2}})^{2}}, - v e

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\frac{Q_{1} + Q_{2}}{2}, - v e

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Solution

The correct option is C $\frac{Q_{1} Q_{2}}{(\sqrt{Q_{1}} + \sqrt{Q_{2}})^{2}}, - v e$
Force on $Q_{2}$ is zero ( $q$ should be -ve)
Let the distance between the charge $q$ and $Q_{2}$ is $x$

For $Q_{2}$ to be in equilibrium,
$\Rightarrow \frac{k Q_{1} Q_{2}}{R^{2}} = \frac{k q Q_{2}}{x^{2}}$
$\Rightarrow \frac{x}{R} = \sqrt{\frac{q}{Q_{1}}}$ ...(i)

Net force on $q$ is zero:
$\frac{k Q_{1} q}{(R - x)^{2}} = \frac{k q Q_{2}}{x^{2}}$
$\Rightarrow \frac{R - x}{x} = \frac{\sqrt{Q_{1}}}{\sqrt{Q_{2}}}$ ...(ii)

From equation (i) and (ii)
$\frac{R}{x} = \frac{\sqrt{Q_{1}} + \sqrt{Q_{2}}}{\sqrt{Q_{2}}}$
$\Rightarrow \frac{\sqrt{Q_{1}}}{\sqrt{q}} = \frac{\sqrt{Q_{1}} + \sqrt{Q_{2}}}{\sqrt{Q_{2}}}$
$\Rightarrow q = \frac{Q_{1} Q_{2}}{(\sqrt{Q_{1}} + \sqrt{Q_{2}})^{2}}$

Hence, option (c) is correct.

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