A 60; pF capacitor is fully charged by a 20; | vedclass

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(B) Initial charge on the capacitor: $Q = CV = 60 	imes 10^{-12} \; F 	imes 20 \; V = 1200 	imes 10^{-12} \; C = 1.2 	imes 10^{-9} \; C$.Initial e

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$6$

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(B) Initial charge on the capacitor: $Q = CV = 60 	imes 10^{-12} \; F 	imes 20 \; V = 1200 	imes 10^{-12} \; C = 1.2 	imes 10^{-9} \; C$.Initial electrostatic energy stored: $U_i = \frac{1}{2} CV^2 = \frac{1}{2} 	imes 60 	imes 10^{-12} 	imes (20)^2 = 30 	imes 10^{-12} 	imes 400 = 12000 	imes 10^{-12} \; J = 12 \; nJ$.When connected in parallel,the total capacitance becomes $C_{eq} = C + C = 2C = 120 \; pF$. The total charge $Q$ is conserved and redistributes equally as $Q/2$ on each capacitor.Final electrostatic energy stored: $U_f = \frac{(Q/2)^2}{2C} + \frac{(Q/2)^2}{2C} = 2 	imes \frac{Q^2/4}{2C} = \frac{Q^2}{4C} = \frac{1}{2} 	imes \frac{Q^2}{2C} = \frac{1}{2} U_i = \frac{1}{2} 	imes 12 \; nJ = 6 \; nJ$.Energy lost: $\Delta U = U_i - U_f = 12 \; nJ - 6 \; nJ = 6 \; nJ$.

$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

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