Two capacitors of capacitance 2 mu F and 3 mu | vedclass

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(D) The equivalent capacitance $C_{eq}$ of two capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \fra

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$400$

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(D) The equivalent capacitance $C_{eq}$ of two capacitors in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.Thus,$C_{eq} = \frac{6}{5}\ \mu F$.The total charge $Q$ supplied to the series combination is $Q = C_{eq} 	imes V_{total} = \frac{6}{5} 	imes 1000 = 1200\ \mu C$.Since the capacitors are in series,the charge on each capacitor is the same,$Q = 1200\ \mu C$.The potential difference across the $2\ \mu F$ capacitor is $V_1 = \frac{Q}{C_1} = \frac{1200}{2} = 600\ V$.If the outer plate of the first capacitor is at $1000\ V$,the potential at the inner plate (which is shared with the second capacitor) is $V_{inner} = 1000\ V - V_1 = 1000 - 600 = 400\ V$.

Two capacitors of capacitance $2\ \mu F$ and $3\ \mu F$ are joined in series. The outer plate of the first capacitor is at $1000\ V$ and the outer plate of the second capacitor is earthed (grounded). The potential on the inner plates of each capacitor will be: (in $V$)

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