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(A) From the circuit diagram, the capacitors $9\, \mu F, 5\, \mu F$, and $10\, \mu F$ are connected in parallel between the two nodes. Their equivalen

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$4\, \mu F, 50\, \mu C$

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(A) From the circuit diagram, the capacitors $9\, \mu F, 5\, \mu F$, and $10\, \mu F$ are connected in parallel between the two nodes. Their equivalent capacitance is $C_p = 9 + 5 + 10 = 24\, \mu F$. Now, the circuit simplifies to three capacitors in series: $8\, \mu F, 24\, \mu F$, and $12\, \mu F$. The equivalent capacitance $C_{eq}$ is given by $\frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{24} + \frac{1}{12} = \frac{3+1+2}{24} = \frac{6}{24} = \frac{1}{4}$. Thus, $C_{eq} = 4\, \mu F$. The total charge $q$ flowing from the source is $q = C_{eq} 	imes V = 4\, \mu F 	imes 60\, V = 240\, \mu C$. Since the $24\, \mu F$ equivalent capacitor is in series with the others, the charge on the parallel combination is $240\, \mu C$. The potential difference across the parallel combination is $V_p = \frac{q}{C_p} = \frac{240\, \mu C}{24\, \mu F} = 10\, V$. The charge on the $5\, \mu F$ capacitor is $q_5 = C_5 	imes V_p = 5\, \mu F 	imes 10\, V = 50\, \mu C$.

Five capacitors together with their capacitances are shown in the adjoining figure. The potential difference between the points $A$ and $B$ is $60\, V$. The equivalent capacitance between the points $A$ and $B$ and the charge on the $5\, \mu F$ capacitor will be respectively:

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