Two point charges +Q and +4Q are kept at a separation l. A third negative point charge is kept at point such that the system of three charges is in equilibrium. Then the third charge is

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$- \frac{2 Q}{9}$

$- \frac{2 Q}{3}$

$- \frac{Q}{3}$

$- \frac{4 Q}{9}$

answer is C.

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Detailed Solution

Considering equilibrium of the third charge -2,

$K \frac{Q q}{x^{2}} = K \frac{4 Q q}{{(l - x)}^{2}} \Rightarrow l - x = 2 x \Rightarrow x = \frac{l}{3}$

Considering equilibrium of the charge $+ Q_{1}$

$K \frac{Q q}{x^{2}} = K \frac{Q .4 Q}{l^{2}} \Rightarrow q = 4 Q . {(\frac{x}{l})}^{2} = \frac{4 Q}{9}$

$∴$ The third charge is $- \frac{4 Q}{9}$

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