The four capacitors,each of 25,mu F,are conne | vedclass

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(B) From the circuit diagram,the voltmeter is connected across the parallel combination of the two capacitors on the left. Since the capacitors are co

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

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(B) From the circuit diagram,the voltmeter is connected across the parallel combination of the two capacitors on the left. Since the capacitors are connected in parallel,the potential difference across each capacitor is equal to the reading of the voltmeter,which is $V = 200\,V$. 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 given 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 $\pm 5 	imes 10^{-3}\,C$.

The four capacitors,each of $25\,\mu F$,are connected as shown in the figure. The $dc$ voltmeter reads $200\,V$. The charge on each plate of the capacitor is

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