Two capacitors of 3,pF and 6,pF are connected | vedclass

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(B) Step $1$: Calculate the equivalent capacitance in series.$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6

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$2222\,V$

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(B) Step $1$: Calculate the equivalent capacitance in series.$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}\,pF^{-1}$.Thus,$C_s = 2\,pF$.Step $2$: Calculate the total charge stored in the series combination.$Q = C_s 	imes V = 2 	imes 10^{-12}\,F 	imes 5000\,V = 10^{-8}\,C$.Step $3$: When disconnected and reconnected in parallel,the total charge $Q$ is redistributed across the new equivalent capacitance $C_p = C_1 + C_2 = 3\,pF + 6\,pF = 9\,pF$.Step $4$: Calculate the new potential $V'$ across the parallel combination.$V' = \frac{Q}{C_p} = \frac{10^{-8}\,C}{9 	imes 10^{-12}\,F} = \frac{10000}{9}\,V \approx 1111.11\,V$.Wait,re-evaluating the charge distribution: In series,each capacitor holds the same charge $Q = 10^{-8}\,C$. When reconnected in parallel,the total charge is $Q_{total} = Q_1 + Q_2 = 10^{-8} + 10^{-8} = 2 	imes 10^{-8}\,C$.$V' = \frac{2 	imes 10^{-8}}{9 	imes 10^{-12}} = \frac{20000}{9} \approx 2222.22\,V$.

Two capacitors of $3\,pF$ and $6\,pF$ are connected in series and a potential difference of $5000\,V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is

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