If two charges +q and +9q 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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−4q9 at a distance d3 from 4q

−2Q3 at a distance d3 from q

−16Q9 at a distance d4 from q

−2Q3 at a distance d3 from 4q

answer is C.

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Detailed Solution

x=d9qq+1=d4 from qF1=F214π∈0Qqx2=14π∈09q2d2∴x=d4⇒q=−16Q9

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