A 3 mu F capacitor is charged to a potential | vedclass

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(C) Initial charges on the capacitors are:$Q_1 = C_1 V_1 = 3\ \mu F 	imes 300\ V = 900\ \mu C$$Q_2 = C_2 V_2 = 2\ \mu F 	imes 200\ V = 400\ \mu C$Si

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(C) Initial charges on the capacitors are:$Q_1 = C_1 V_1 = 3\ \mu F 	imes 300\ V = 900\ \mu C$$Q_2 = C_2 V_2 = 2\ \mu F 	imes 200\ V = 400\ \mu C$Since they are connected with opposite polarities,the net charge is $Q_{net} = Q_1 - Q_2 = 900\ \mu C - 400\ \mu C = 500\ \mu C$.The common potential $V$ after connection is given by:$V = \frac{Q_{net}}{C_1 + C_2} = \frac{500\ \mu C}{3\ \mu F + 2\ \mu F} = 100\ V$.The final charge on the $3\ \mu F$ capacitor is $Q_1' = C_1 V = 3\ \mu F 	imes 100\ V = 300\ \mu C$.The charge that flows through the wire is the change in charge on one of the capacitors:$\Delta Q = Q_1 - Q_1' = 900\ \mu C - 300\ \mu C = 600\ \mu C$.

$A$ $3\ \mu F$ capacitor is charged to a potential of $300\ V$ and a $2\ \mu F$ capacitor is charged to $200\ V$. They are then connected in parallel with plates of opposite polarity joined together. What is the amount of charge (in $\mu C$) that flows through the connecting wires?

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