A capacitor of capacitance 20 mu F is charged | vedclass

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(C) Initial charge on the first capacitor $C_1 = 20 \mu F$ is $Q_0 = C_1 V = 20 \mu F 	imes 24.3 \ V = 486 \mu C$.When $C_1$ is connected to an uncha

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$64$

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(C) Initial charge on the first capacitor $C_1 = 20 \mu F$ is $Q_0 = C_1 V = 20 \mu F 	imes 24.3 \ V = 486 \mu C$.When $C_1$ is connected to an uncharged capacitor $C_2 = 10 \mu F$,the charge redistributes. The common potential $V'$ is given by $V' = \frac{Q_{total}}{C_1 + C_2} = \frac{Q}{C_1 + C_2}$.The new charge on $C_1$ becomes $Q' = C_1 V' = Q \left( \frac{C_1}{C_1 + C_2} ight)$.Here,the ratio $\frac{C_1}{C_1 + C_2} = \frac{20}{20 + 10} = \frac{20}{30} = \frac{2}{3}$.After each process,the charge on $C_1$ is multiplied by a factor of $\frac{2}{3}$.After $n$ processes,the charge $Q_n = Q_0 \left( \frac{2}{3} ight)^n$.For the fifth process,$n = 5$,so $Q_5 = 486 	imes \left( \frac{2}{3} ight)^5$.$Q_5 = 486 	imes \frac{32}{243} = 2 	imes 32 = 64 \mu C$.

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