Two capacitors of 6 , pF are connected in ser | vedclass

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(B) In series connection, the equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \, pF^{-

Vedclass · Accepted Answer

$2222 \, V$

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(B) In series connection, the equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \, pF^{-1}$, so $C_s = 3 \, pF$.The total charge $Q$ stored in the series combination is $Q = C_s 	imes V = 3 	imes 10^{-12} \, F 	imes 5000 \, V = 1.5 	imes 10^{-8} \, C$.When the capacitors are disconnected and reconnected in parallel, the total charge $Q$ is redistributed across the new equivalent capacitance $C_p = 6 \, pF + 6 \, pF = 12 \, pF$.The new potential difference $V'$ is $V' = \frac{Q}{C_p} = \frac{1.5 	imes 10^{-8} \, C}{12 	imes 10^{-12} \, F} = 0.125 	imes 10^4 \, V = 1250 \, V$.Note: Based on the provided options, the calculation $V = \frac{Q_{total}}{C_{parallel}}$ yields $1250 \, V$. However, if the question implies the charge on each capacitor is conserved and then shared, the result is $1250 \, V$. Given the options provided, there may be a discrepancy in the original problem statement's values. Based on standard calculation, $1250 \, V$ is the correct physical result.

Two capacitors of $6 \, pF$ are connected in series and a potential difference of $5000 \, V$ is applied across the combination. If the combination is disconnected and then reconnected in parallel, what will be the potential difference between the plates?

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