A 400, pF capacitor is charged with a 100, V | vedclass

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Question

(A) Initial energy stored in the capacitor: $U_i = \frac{1}{2} C V^2 = \frac{1}{2} 	imes (400 	imes 10^{-12} \, F) 	imes (100 \, V)^2 = 2 	imes 10

Vedclass · Accepted Answer

$1	imes10^{-6}\, J$

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(A) Initial energy stored in the capacitor: $U_i = \frac{1}{2} C V^2 = \frac{1}{2} 	imes (400 	imes 10^{-12} \, F) 	imes (100 \, V)^2 = 2 	imes 10^{-6} \, J$. When the charged capacitor is connected to an identical uncharged capacitor,the charge redistributes until the potential difference across both is the same. The common potential is $V' = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{400 	imes 100 + 400 	imes 0}{400 + 400} = 50 \, V$. The final energy stored in the system is $U_f = \frac{1}{2} (C_1 + C_2) (V')^2 = \frac{1}{2} 	imes (800 	imes 10^{-12} \, F) 	imes (50 \, V)^2 = 1 	imes 10^{-6} \, J$. The energy loss is $\Delta U = U_i - U_f = 2 	imes 10^{-6} \, J - 1 	imes 10^{-6} \, J = 1 	imes 10^{-6} \, J$.

$A$ $400\, pF$ capacitor is charged with a $100\, V$ battery. After disconnecting the battery,this capacitor is connected to another uncharged $400\, pF$ capacitor. Find the energy loss.

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