Consider two uncharged capacitors of equal ca | vedclass

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(C) Initial energy $U_i = \frac{1}{2} C V^2$. Given $C = 200 	ext{ pF} = 200 	imes 10^{-12} 	ext{ F}$ and $V = 100 	ext{ V}$. $U_i = \frac{1}{2} \

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$0.5 	imes 10^{-6} 	ext{ J}$

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(C) Initial energy $U_i = \frac{1}{2} C V^2$. Given $C = 200 	ext{ pF} = 200 	imes 10^{-12} 	ext{ F}$ and $V = 100 	ext{ V}$. $U_i = \frac{1}{2} 	imes 200 	imes 10^{-12} 	imes (100)^2 = 100 	imes 10^{-12} 	imes 10^4 = 10^{-6} 	ext{ J}$. When the charged capacitor is connected to an identical uncharged capacitor, the charge redistributes until the potential becomes common. The common potential $V' = \frac{CV + 0}{C+C} = \frac{V}{2} = \frac{100}{2} = 50 	ext{ V}$. Final energy $U_f = \frac{1}{2} (C+C) (V')^2 = C (\frac{V}{2})^2 = \frac{1}{4} CV^2 = \frac{U_i}{2}$. $U_f = \frac{10^{-6}}{2} = 0.5 	imes 10^{-6} 	ext{ J}$. The energy lost is $\Delta U = U_i - U_f = 10^{-6} - 0.5 	imes 10^{-6} = 0.5 	imes 10^{-6} 	ext{ J}$.

Consider two uncharged capacitors of equal capacitance $200 \text{ pF}$. One of them is charged by a $100 \text{ V}$ supply and disconnected. Now, this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

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