$- 20 μ C, 40 μ C$ 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 2F. Then the charge q is

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$10 μ C$

$30 μ C$

$40 μ C$

$70 μ C$

answer is D.

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Detailed Solution

$\begin{matrix} D u r i n g c o n d u c t i o n, d u e t o c o n t a c t e a c h s p h e r e a c q u i r e s a v e r a g e o f b o t h t h e c h a r g e s . \\ F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{d^{2}} \\ q_{1} = - 20 μ C = - 20 \times 10^{- 6} C a n d q_{2} = 40 μ C = 40 \times 10^{- 6} C \\ F = 9 \times 10^{9} \times \frac{20 \times 40 \times 10^{- 12}}{d^{2}} - - - (1) \\ w h e n P, Q a r e t o u c h e d a n d s e p a r t e d t h e y g e t e a c h o f c h a r g e = \frac{- 20 + 40}{2} = 10 μ C \\ 2 F = \frac{9 \times 10^{9} \times {(\frac{10 + q}{2})}^{2}}{d^{2}} - - - - (2) \\ F o r m (1) a n d (2) w e g e t \\ \Rightarrow q = 70 μ c \end{matrix}$