A capacitor P with capacitance 10 times 10^{- | vedclass

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(D) Initial charge on capacitor $P$ is $Q_1 = C_1 V_1 = (10 	imes 10^{-6} 	ext{ F}) 	imes (6.0 	ext{ V}) = 60 	imes 10^{-6} 	ext{ C} = 6 	imes

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$4$

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(D) Initial charge on capacitor $P$ is $Q_1 = C_1 V_1 = (10 	imes 10^{-6} 	ext{ F}) 	imes (6.0 	ext{ V}) = 60 	imes 10^{-6} 	ext{ C} = 6 	imes 10^{-5} 	ext{ C}$.When the two capacitors are connected in parallel, the charge is redistributed until they reach a common potential $V$.The common potential $V$ is given by $V = \frac{Q_{total}}{C_{total}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.Given $C_1 = 10 	imes 10^{-6} 	ext{ F}$, $V_1 = 6.0 	ext{ V}$, $C_2 = 20 	imes 10^{-6} 	ext{ F}$, and $V_2 = 0 	ext{ V}$.$V = \frac{(10 	imes 10^{-6} 	imes 6) + (20 	imes 10^{-6} 	imes 0)}{10 	imes 10^{-6} + 20 	imes 10^{-6}} = \frac{60 	imes 10^{-6}}{30 	imes 10^{-6}} = 2 	ext{ V}$.The charge on capacitor $Q$ at equilibrium is $Q_2 = C_2 V = (20 	imes 10^{-6} 	ext{ F}) 	imes (2 	ext{ V}) = 40 	imes 10^{-6} 	ext{ C} = 4 	imes 10^{-5} 	ext{ C}$.Comparing this with $\alpha 	imes 10^{-5} 	ext{ C}$, we get $\alpha = 4$.

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