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Q.

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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a

−4q9  at  a  distance  d3  from  4q

b

−2Q3  at  a  distance  d3  from  q

c

−16Q9  at  a  distance  d4  from  q

d

−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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