A charge q is placed at the centre of the line joining two equal charges Q. The system of the three chargeswill be in equilibrium of q is equal to(a) -(Q/2) (b) -(Q/4) (c) + (Q/4) (d) +(Q/2)

Here use this formula

U = 1/(4Πε₀)[2qQ/r + Q²/2r] = 0

q=-Q/4

Approved

well...you hv not mentioned what q is in ur ques....

,,solving ques assuming that q is the third charge that u mentioned

for the first Q charge to be in equilibrium...net force on it shud be zero..

i.e... force due to q + force due to Q =0

let the distance btw Q charge to Q charge be 2l.....so dat distance btw q and Q is l

kqQ/l²+kQ²/4l²=0

solving it we get..q=-Q/4...

..hope it helps..:)