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(B) The voltmeter is connected across the two capacitors on the left,which are in parallel. Since the capacitors are in parallel,the potential differe

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$\pm 5 	imes 10^{-3} \, C$

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(B) The voltmeter is connected across the two capacitors on the left,which are in parallel. Since the capacitors are in parallel,the potential difference across each of them is the same as the voltmeter reading,$V = 200 \, V$. The capacitance of each capacitor is $C = 25 \, \mu F = 25 	imes 10^{-6} \, F$. The charge $Q$ on each plate of a capacitor is given by the formula $Q = CV$. Substituting the values: $Q = (25 	imes 10^{-6} \, F) 	imes (200 \, V)$ $Q = 5000 	imes 10^{-6} \, C$ $Q = 5 	imes 10^{-3} \, C$. Thus,the charge on each plate is $\pm 5 	imes 10^{-3} \, C$.

As shown in the figure,four capacitors each of capacitance $25 \, \mu F$ are connected. The voltmeter reads $200 \, V$. Find the charge on each plate of the capacitors.

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