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If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of q is

$(a)\;\large\frac{Q}{2}\qquad(b)\;-\large\frac{Q}{2}\qquad(c)\;\large\frac{Q}{4}\qquad(d)\;- \large\frac{Q}{4}$

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Answer : (d) $\;-\large\frac{Q}{4}$
Explanation :
Let's see the force on one Q
$F= \large\frac{k Q^2}{4 a^2} + \large\frac{k\; q\; Q}{a^2}$
For the system to be in equilibrium F =0
$ \large\frac{k Q^2}{4 a^2} + \large\frac{k\; q\; Q}{a^2}=0$
answered Jan 30, 2014 by yamini.v
edited Feb 13, 2014 by yamini.v

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