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(A) The circuit consists of two batteries of $18\,V$ and $13\,V$ connected in series with two capacitors of $3\,\mu F$ and $2\,\mu F$.The net potentia

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$6$

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(A) The circuit consists of two batteries of $18\,V$ and $13\,V$ connected in series with two capacitors of $3\,\mu F$ and $2\,\mu F$.The net potential difference across the series combination of capacitors is $V_{net} = 18\,V - 13\,V = 5\,V$.Since the capacitors are in series,the same charge $Q$ flows through both. The equivalent capacitance $C_{eq}$ is given by:$C_{eq} = \frac{C_1 	imes C_2}{C_1 + C_2} = \frac{3 	imes 2}{3 + 2} = \frac{6}{5}\,\mu F$.The charge $Q$ on each capacitor is:$Q = C_{eq} 	imes V_{net} = \left(\frac{6}{5}\,\mu Fight) 	imes 5\,V = 6\,\mu C$.Therefore,the charge on the $2\,\mu F$ capacitor is $6\,\mu C$.

Two batteries and two capacitors are connected as shown in the figure. The charge on the $2\,\mu F$ capacitor is.....$\mu C$.

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