Initially switch S was open and capacitors C, 2C were having charges Q and 2Q respectively as shown in figure. Find charge flown in the direction 1 shown in figure after switch S is closed.

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$(\frac{2 Q}{5})$

$(- \frac{4 Q}{5})$

$(\frac{4 Q}{5})$

$(- \frac{2 Q}{5})$

answer is C.

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Detailed Solution

$\begin{aligned} \frac{Q - Q_{1}}{C} - \frac{Q_{1}}{C} + \frac{(2 Q - Q_{1})}{2 C} = 0 \\ 2 Q - 4 Q_{1} + 2 Q - Q_{1} = 0 \\ Q_{1} = \frac{4 Q}{5} \end{aligned}$

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