A capacitor of capacitance 50 ; pF is charged | vedclass

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Question

(D) The initial energy stored in the first capacitor is $U_i = \frac{1}{2} C V^2 = \frac{1}{2} 	imes 50 	imes 10^{-12} 	imes (100)^2 = 250 	imes 1

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$125$

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(D) The initial energy stored in the first capacitor is $U_i = \frac{1}{2} C V^2 = \frac{1}{2} 	imes 50 	imes 10^{-12} 	imes (100)^2 = 250 	imes 10^{-9} \; J = 250 \; nJ$.When connected to an identical uncharged capacitor,the charge redistributes until the potential across both is $V' = \frac{V}{2} = 50 \; V$.The final energy stored in the system is $U_f = 2 	imes (\frac{1}{2} C V'^2) = C 	imes (\frac{V}{2})^2 = 50 	imes 10^{-12} 	imes 2500 = 125 	imes 10^{-9} \; J = 125 \; nJ$.The energy loss is $\Delta U = U_i - U_f = 250 \; nJ - 125 \; nJ = 125 \; nJ$.Alternatively,using the formula for energy loss: $\Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 = \frac{1}{2} \frac{50 	imes 50}{50 + 50} 	imes 10^{-12} 	imes (100 - 0)^2 = 125 	imes 10^{-9} \; J = 125 \; nJ$.

$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.

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