$- 10 μC$ , $40 μC$ and q are the charges on three identical spherical conductors P, Q and R respectively. Now P and Q attract each other with a force F when they are separated by a distance d. Now P and Q are made in contact with each other and then separated. Again Q and R are touched and they are separated by a distance d. The repulsive force between Q and R is 4F. Then the charge q is

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$10 μC$

$40 μC$

$30 μC$

$65 μC$

answer is D.

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Detailed Solution

The force between P & Q

$F = 9 \times 10^{9} \frac{10 μC \times 40 μC}{d^{2}} = \frac{3 .6}{d^{2}}$

when P & Q are in contact with each other the charge on each =

$q' = \frac{q_{1} + q_{2}}{2} = \frac{- 10 + 40}{2} = 15 μC [∵ identical spheres]$

when Q and R are touched and separated, the charge on each, $Q = (\frac{15 + q}{2}) μC [q is in μC]$

$\begin{array}{l} \Rightarrow 9 \times 10^{9} \frac{{(\frac{15}{2} + q)}^{2} \times (μC)^{2}}{d^{2}} = 4 \times \frac{3 .6}{d^{2}} \\ \Rightarrow q = 65 μC . \end{array}$

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