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(D) The initial energy stored in the capacitor is $U = \frac{Q^2}{2C}$,where $Q$ is the initial charge and $C$ is the capacitance.When the charged cap

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$\frac{U}{9}$

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(D) The initial energy stored in the capacitor is $U = \frac{Q^2}{2C}$,where $Q$ is the initial charge and $C$ is the capacitance.When the charged capacitor is disconnected from the battery and connected in parallel to two other identical uncharged capacitors,the total charge $Q$ is shared equally among the three capacitors because they are identical.Therefore,the charge on each capacitor becomes $Q' = \frac{Q}{3}$.The new energy $U'$ stored in each capacitor is given by $U' = \frac{(Q')^2}{2C}$.Substituting $Q' = \frac{Q}{3}$ into the equation:$U' = \frac{(\frac{Q}{3})^2}{2C} = \frac{Q^2}{9 	imes 2C} = \frac{1}{9} 	imes \frac{Q^2}{2C}$.Since $U = \frac{Q^2}{2C}$,we get $U' = \frac{U}{9}$.

Energy of a charged capacitor is $U$. Now it is removed from a battery and then connected to two other identical uncharged capacitors in parallel. What will be the energy of each capacitor?

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