A capacitor of capacitance C = 900 , pF is ch | vedclass

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(B) $1$. Initial charge on the first capacitor: $Q = C V = 900 	imes 10^{-12} \, F 	imes 100 \, V = 9 	imes 10^{-8} \, C$.$2$. When connected to an

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$2.25$

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(B) $1$. Initial charge on the first capacitor: $Q = C V = 900 	imes 10^{-12} \, F 	imes 100 \, V = 9 	imes 10^{-8} \, C$.$2$. When connected to an uncharged capacitor of the same capacitance $C$,the charge $Q$ is shared equally between the two capacitors because they are in parallel.$3$. The common potential $V'$ is given by: $V' = \frac{Q_{total}}{C_{total}} = \frac{Q}{C + C} = \frac{9 	imes 10^{-8} \, C}{1800 	imes 10^{-12} \, F} = 50 \, V$.$4$. The total electrostatic energy stored in the system is: $U = \frac{1}{2} (C + C) (V')^2 = \frac{1}{2} (1800 	imes 10^{-12} \, F) (50 \, V)^2$.$5$. $U = 900 	imes 10^{-12} 	imes 2500 = 225 	imes 10^{-8} \, J = 2.25 	imes 10^{-6} \, J$.

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