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Question

A charge $$q$$ is placed at the centre of the line joining two equal charges $$Q$$. Show that the system of three charges will be in equilibrium if $$q = -\dfrac {Q}{4}$$.

Solution
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Charge $$q$$ is in equilibrium since charges $$A$$ and $$B$$ exert equal and opposite forces on it.

For equilibrium of charge $$Q$$ at $$B$$;

$$F_{BC} + F_{AB} = 0$$

$$\Rightarrow \dfrac {1}{4\pi \epsilon_{0}} \dfrac {qQ}{(l/2)^{2}} + \dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q . Q}{l^{2}} = 0$$

$$\Rightarrow \dfrac {1}{4\pi \epsilon_{0}} \dfrac {Q}{l^{2}} (4q + Q) = 0 \Rightarrow q = -\dfrac {Q}{4}$$.

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