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(B) The circuit is a balanced Wheatstone bridge. The ratio of capacitors in the arms is $10\,\mu F / 10\,\mu F = 10\,\mu F / 10\,\mu F = 1$. Since the

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$500$

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(B) The circuit is a balanced Wheatstone bridge. The ratio of capacitors in the arms is $10\,\mu F / 10\,\mu F = 10\,\mu F / 10\,\mu F = 1$. Since the bridge is balanced,the potential difference across the central $8\,\mu F$ capacitor is zero,and no charge flows through it. We can remove the central capacitor from the circuit. The circuit now consists of two parallel branches,each containing two $10\,\mu F$ capacitors in series. The equivalent capacitance of each branch is $C_{eq} = (10\,\mu F 	imes 10\,\mu F) / (10\,\mu F + 10\,\mu F) = 5\,\mu F$. The total equivalent capacitance of the circuit is $C_{total} = 5\,\mu F + 5\,\mu F = 10\,\mu F$. The potential difference across each branch is $100\,V$. For the top branch,the potential difference across the $10\,\mu F$ capacitor is $V' = (100\,V / 2) = 50\,V$. The charge on the $10\,\mu F$ capacitor is $Q = C 	imes V' = 10\,\mu F 	imes 50\,V = 500\,\mu C$.

Five capacitors are connected to a $DC$ potential of $100\,V$ as shown in the figure. Find the charge in the $10\,\mu F$ capacitor (top-left).

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