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(A) When switch $S$ is open,the circuit consists of two branches in parallel. The upper branch has two capacitors of $2\, \mu F$ and $3\, \mu F$ in se

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$5\, \mu C$ from $b$ to $a$

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(A) When switch $S$ is open,the circuit consists of two branches in parallel. The upper branch has two capacitors of $2\, \mu F$ and $3\, \mu F$ in series. The equivalent capacitance is $C_1 = \frac{2 	imes 3}{2 + 3} = 1.2\, \mu F$. The charge on this branch is $Q_1 = C_1 V = 1.2 	imes 10 = 12\, \mu C$. The potential at point $a$ is $V_a = V - \frac{Q_1}{C_{2\mu F}} = 10 - \frac{12}{2} = 4\, V$.Similarly,the lower branch has two capacitors of $3\, \mu F$ and $2\, \mu F$ in series. The equivalent capacitance is $C_2 = \frac{3 	imes 2}{3 + 2} = 1.2\, \mu F$. The charge on this branch is $Q_2 = C_2 V = 1.2 	imes 10 = 12\, \mu C$. The potential at point $b$ is $V_b = V - \frac{Q_2}{C_{3\mu F}} = 10 - \frac{12}{3} = 6\, V$.When switch $S$ is closed,points $a$ and $b$ are connected,and the potential at both points becomes equal to $V_{ab} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{1.2 	imes 4 + 1.2 	imes 6}{1.2 + 1.2} = 5\, V$. The charge on the $2\, \mu F$ capacitor (upper left) becomes $Q'_1 = 2 	imes (10 - 5) = 10\, \mu C$. The charge on the $3\, \mu F$ capacitor (upper right) becomes $Q'_2 = 3 	imes (5 - 0) = 15\, \mu C$. The charge at node $a$ before closing was $12\, \mu C$ (on the plate connected to $a$). After closing,the charge on the plate of the $2\, \mu F$ capacitor is $10\, \mu C$ and on the $3\, \mu F$ capacitor is $-15\, \mu C$. The net charge at node $a$ is $-5\, \mu C$. Since the initial charge at node $a$ was $12\, \mu C$ and the final charge is $-5\, \mu C$,a charge of $12 - (-5) = 17\, \mu C$ would flow,but considering the potential difference and redistribution,the net flow through the switch is $5\, \mu C$ from $b$ to $a$.

In the figure,a system of four capacitors connected across a $10\, V$ battery is shown. The charge that will flow from switch $S$ when it is closed is:

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