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the given situation is:
F2q <———q———>FQ----------Q------------------------FQ<———2q———>Fq
d-x x
now if net force on 2q has to be 0 then Q has to be negative and hence 2q will experience a force towards Q(attraction).But 2q will also experience a force away from q(repulsion).Hence these forces can be calculated using coulombs law and equated and get x.Same thing can be done with q to get another equation and using the two equation you can eliminate Q.
Hence,
FQ=Fq and F2q=FQ
PLZ APPROVE!
q and 2q are d distance apart.
let charge Q be kept between them on the same line joining those two ..at a distance of x from q.
now a repulsive force will be acting on it by q as well as by 2q but in opp. directions.
( if q at left and 2q at right), then
forcE on Q by q= k (qQ/x^2) towards right
" " Q by 2q = k((2qQ/(d-x)^2) " left
since these 2 forces should be equal and opp. for net force on Q to be zero,
so by equating tjose two,
we get
x=[d(q+q^1/2)]/(q-2) or [d(q-q^1/2)]/(q-2)
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