If two charges +q and +4q are separated by a distance 'd' and a point charge Q is placed on the line joining the above two charges and in between them such that all charges are in equilibrium. Then the charge Q and it's position are

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$- \frac{4 q}{9} at a distance \frac{d}{3} from 4 q$

$\frac{- 2 Q}{3} at a distance \frac{d}{3} from q$

$\frac{- 2 Q}{3} at a distance \frac{d}{3} from 4 q$

$- \frac{4 q}{9} at a distance \frac{d}{3} from q$

answer is C.

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Detailed Solution

At equilibrium, distance of Q from q is given by,

$\begin{array}{l} x = \frac{d}{1 + \sqrt{\frac{q_{big}}{q_{small}}}} = \frac{d}{1 + \sqrt{\frac{4 q}{q}}} = \frac{d}{3} \\ \frac{1}{4 π ϵ_{0}} \frac{Q q}{x^{2}} = \frac{1}{4 π ϵ_{0}} \frac{4 q^{2}}{d^{2}} \Rightarrow Q = - \frac{4 q}{9} from q \end{array}$