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

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$500$

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(B) The given circuit is a balanced Wheatstone bridge because the ratio of capacitors in the arms is equal $(10\,\mu F / 10\,\mu F = 10\,\mu F / 10\,\mu F = 1)$.Since the bridge is balanced,no charge flows through the central $8\,\mu F$ capacitor.Thus,the circuit simplifies to two parallel branches,each containing two $10\,\mu F$ capacitors in series.The equivalent capacitance of each branch is $C_{eq} = (10\,\mu F 	imes 10\,\mu F) / (10\,\mu F + 10\,\mu F) = 5\,\mu F$.Since there are two such branches in parallel,the total potential difference across each branch is $100\,V$.The potential difference across each $10\,\mu F$ capacitor in a branch is $V = 100\,V / 2 = 50\,V$.The charge on each $10\,\mu F$ 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 (in $\mu C$).

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