Q. Two point electric charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that the net force acting on q and 2q is zero. Calculate the position of charge Q in terms of q and d...???plz explain me in brief...

the given situation is: 

         F_2q <———q———>F_Q----------Q------------------------F_Q<———2q———>F_q                  

                                          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,

F_Q=F_q          and             F_2q=F_Q

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)