A 4 mu F capacitor is charged by a 200 V bat | vedclass

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(B) Initial charge on the capacitor: $q = C_1 V_1 = 4 	imes 10^{-6} 	imes 200 = 800 	imes 10^{-6} \ C$.Initial energy stored: $U_i = \frac{1}{2} C_

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$2.67 	imes 10^{-2}$

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(B) Initial charge on the capacitor: $q = C_1 V_1 = 4 	imes 10^{-6} 	imes 200 = 800 	imes 10^{-6} \ C$.Initial energy stored: $U_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} 	imes 4 	imes 10^{-6} 	imes (200)^2 = 8 	imes 10^{-2} \ J$.When connected to an uncharged $2 \mu F$ capacitor,the common potential $V$ is given by $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{800 	imes 10^{-6} + 0}{4 	imes 10^{-6} + 2 	imes 10^{-6}} = \frac{800}{6} \ V$.Final energy stored: $U_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} 	imes (6 	imes 10^{-6}) 	imes (\frac{800}{6})^2 = 3 	imes 10^{-6} 	imes \frac{640000}{36} = \frac{64}{12} 	imes 10^{-2} \approx 5.33 	imes 10^{-2} \ J$.Loss in energy: $\Delta U = U_i - U_f = 8 	imes 10^{-2} - 5.33 	imes 10^{-2} = 2.67 	imes 10^{-2} \ J$.

$A$ $4 \mu F$ capacitor is charged by a $200 \ V$ battery. It is then disconnected from the supply and connected to another uncharged $2 \mu F$ capacitor. During the process,the loss of energy (in $J$) is:

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