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(C) $1$. Analyze the circuit diagram to identify nodes. Let the potential at node $A$ be $0\, V$ and at node $B$ be $40\, V$.$2$. The capacitors $6\,

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$214\, \mu C, 27\, V$

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(C) $1$. Analyze the circuit diagram to identify nodes. Let the potential at node $A$ be $0\, V$ and at node $B$ be $40\, V$.$2$. The capacitors $6\, \mu F$,$4\, \mu F$,and $8\, \mu F$ are connected in parallel between nodes $A$ and $C$. The equivalent capacitance $C_{AC} = 6 + 4 + 8 = 18\, \mu F$.$3$. The capacitors $7\, \mu F$,$9\, \mu F$ (in series with $9\, \mu F$),and the $36\, \mu F$ capacitor are connected in series/parallel combinations. Simplifying the network between $C$ and $B$,the equivalent capacitance $C_{CB} = 36\, \mu F$.$4$. Now,the circuit reduces to two capacitors $C_{AC} = 18\, \mu F$ and $C_{CB} = 36\, \mu F$ in series across the $40\, V$ source.$5$. The potential difference across $C_{AC}$ is $V_{AC} = V 	imes \frac{C_{CB}}{C_{AC} + C_{CB}} = 40 	imes \frac{36}{18 + 36} = 40 	imes \frac{36}{54} = 40 	imes \frac{2}{3} \approx 26.67\, V \approx 27\, V$.$6$. Since the $8\, \mu F$ capacitor is in parallel with the $6\, \mu F$ and $4\, \mu F$ capacitors,the potential difference across it is also $V_{AC} \approx 27\, V$.$7$. The charge on the $8\, \mu F$ capacitor is $Q = C 	imes V = 8\, \mu F 	imes 26.67\, V \approx 213.36\, \mu C \approx 214\, \mu C$.

In the following diagram,the charge and potential difference across $8\, \mu F$ capacitance will be nearly

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