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A charge q is placed at middle point of the line joining two equal charges +Q. The value of q such that system is in equilibrium is :

$(A)\;\frac{-Q}{2}\\ (B)\;\frac{-Q}{4} \\ (C)\; \frac{+Q}{2}\\ (D)\;\frac{+Q}{4} $

1 Answer

system will be in equilibrium only when force between Q and q + force between Q and Q is zero
$\large\frac{1}{4 \pi \in _0} \frac{Qq}{(r/2)^2}$$+\frac{1}{4 \pi \in _0} \frac{ Q.Q }{r^2}$$=0$
=> $q= \large\frac{-Q}{4}$
Hence B is the correct answer.


answered Jan 3, 2014 by meena.p

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