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(B) The energy stored in a capacitor is given by $U = \frac{1}{2} \frac{Q^2}{C}$.Initially,the capacitor is charged by a $100 \ V$ battery. The charge

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$2.25 	imes 10^{-6} \ J$

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(B) The energy stored in a capacitor is given by $U = \frac{1}{2} \frac{Q^2}{C}$.Initially,the capacitor is charged by a $100 \ V$ battery. The charge $Q$ on the capacitor is $Q = CV = 900 	imes 10^{-12} \ F 	imes 100 \ V = 9 	imes 10^{-8} \ C$.The initial energy is $U_i = \frac{1}{2} \frac{Q^2}{C} = 4.5 	imes 10^{-6} \ J$.When the charged capacitor is connected to an identical uncharged capacitor of $900 \ pF$ in parallel,the total charge $Q$ is redistributed between them.The equivalent capacitance of the system becomes $C' = C + C = 2C = 1800 \ pF$.Since the charge $Q$ is conserved,the final energy of the system is $U_f = \frac{1}{2} \frac{Q^2}{C'} = \frac{1}{2} \frac{Q^2}{2C} = \frac{1}{2} U_i$.Therefore,$U_f = \frac{4.5 	imes 10^{-6} \ J}{2} = 2.25 	imes 10^{-6} \ J$.

The energy stored in the capacitor shown in figure $(a)$ is $4.5 \times 10^{-6} \ J$. If the battery is disconnected and the capacitor is connected to another uncharged capacitor of $900 \ pF$ as shown in figure $(b)$,what will be the total energy of the system?

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