A 600; pF capacitor is charged by a 200; V su | vedclass

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(A) Initial capacitance $C_1 = 600\; pF = 600 	imes 10^{-12}\; F$ and potential $V_1 = 200\; V$.Initial energy $E_i = \frac{1}{2} C_1 V_1^2 = \frac{1

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

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(A) Initial capacitance $C_1 = 600\; pF = 600 	imes 10^{-12}\; F$ and potential $V_1 = 200\; V$.Initial energy $E_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} 	imes 600 	imes 10^{-12} 	imes (200)^2 = 1.2 	imes 10^{-5}\; J$.When connected to an identical uncharged capacitor $C_2 = 600\; pF$, the charge $Q = C_1 V_1 = 600 	imes 10^{-12} 	imes 200 = 1.2 	imes 10^{-7}\; C$ is shared.The common potential $V'$ is given by $V' = \frac{Q}{C_1 + C_2} = \frac{1.2 	imes 10^{-7}}{1200 	imes 10^{-12}} = 100\; V$.Final energy $E_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} 	imes 1200 	imes 10^{-12} 	imes (100)^2 = 0.6 	imes 10^{-5}\; J$.Energy loss $\Delta E = E_i - E_f = 1.2 	imes 10^{-5} - 0.6 	imes 10^{-5} = 0.6 	imes 10^{-5}\; J = 6 	imes 10^{-6}\; J$.

$A$ $600\; pF$ capacitor is charged by a $200\; V$ supply. It is then disconnected from the supply and is connected to another uncharged $600\; pF$ capacitor. How much electrostatic energy is lost in the process?

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