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(B) Initially,the energy stored in the capacitor of capacitance $C = 2\; F$ at potential $V$ is given by:$E_1 = \frac{1}{2} CV^2 = \frac{1}{2} (2) V^2

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$1: 2$

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(B) Initially,the energy stored in the capacitor of capacitance $C = 2\; F$ at potential $V$ is given by:$E_1 = \frac{1}{2} CV^2 = \frac{1}{2} (2) V^2 = V^2$When this charged capacitor is connected in parallel to an identical uncharged capacitor,the total charge $Q = CV = 2V$ is redistributed equally between the two capacitors because they are identical.The new potential $V'$ across each capacitor is:$V' = \frac{Q_{total}}{C_{total}} = \frac{2V}{2C} = \frac{2V}{4} = \frac{V}{2}$The total energy $E_2$ stored in the combination is:$E_2 = 2 	imes \left( \frac{1}{2} C (V')^2 ight) = C \left( \frac{V}{2} ight)^2 = 2 	imes \frac{V^2}{4} = \frac{V^2}{2}$Therefore,the ratio $E_2 / E_1$ is:$\frac{E_2}{E_1} = \frac{V^2 / 2}{V^2} = \frac{1}{2}$

$A$ parallel plate capacitor of capacitance $2\; F$ is charged to a potential $V$. The energy stored in the capacitor is $E_1$. The capacitor is now connected to another uncharged identical capacitor in parallel combination. The energy stored in the combination is $E_2$. The ratio $E_2 / E_1$ is

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