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A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges
will 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
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..:)
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