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A particle having a charge of $\;2 \mu C\;$ is placed directly below and at a separation of 10 cm from the bob of a simple pendulum at rest . What charge should the bob be given so that the string becomes loose ? mass of the bob is 1 kg .

$(a)\;5.5\times10^{-6} C\qquad(b)\;55\times10^{-6} C\qquad(c)\;0.55\times10^{-6} C\qquad(d)\;5.5\times10^{-8} C$

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Answer : (a) $\;5.5\times10^{-6} C$
Explanation :
$F + T =mg$
$T=0$
$F=mg$
$\large\frac{k\;q_{1}\;q_{2}}{r^2} = mg$
$\large\frac{9\times10^{9}\times2\times10^{-6}\times q}{(0.1)^2} = 10\times1$
$q=\large\frac{10\times10^{-2}}{18\times10^{3}}$
$q=5.5\times10^{-6}\;.$
answered Feb 5, 2014 by yamini.v
 

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