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Q.

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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a

4q9 at a distance d3 from 4q

b

2Q3 at a distance d3 from q

c

2Q3 at a distance d3 from 4q

d

4q9 at a distance d3 from q

answer is C.

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Detailed Solution

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

x=d1+qbigqsmall=d1+4qq=d314πϵ0Qqx2=14πϵ04q2d2Q=4q9 from q

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