A parallel plate capacitor has plate area A a | vedclass

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Question

(D) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d}$. Initial energy $U_i = \frac{1}{2} C_i V_0^2 = \frac{1}{2} \frac{\varepsilon_0 A}{d} V_0^2$.

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$\frac{\varepsilon_0 AV_0^2}{d}$

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(D) Initial capacitance $C_i = \frac{\varepsilon_0 A}{d}$. Initial energy $U_i = \frac{1}{2} C_i V_0^2 = \frac{1}{2} \frac{\varepsilon_0 A}{d} V_0^2$.Since the battery is disconnected,the charge $Q$ on the plates remains constant. $Q = C_i V_0 = \frac{\varepsilon_0 A V_0}{d}$.New separation $d' = 3d$. New capacitance $C_f = \frac{\varepsilon_0 A}{3d} = \frac{C_i}{3}$.Final energy $U_f = \frac{Q^2}{2 C_f} = \frac{(C_i V_0)^2}{2 (C_i/3)} = \frac{3}{2} C_i V_0^2 = 3 U_i$.Work done $W = U_f - U_i = 3 U_i - U_i = 2 U_i$.$W = 2 	imes (\frac{1}{2} \frac{\varepsilon_0 A}{d} V_0^2) = \frac{\varepsilon_0 A V_0^2}{d}$.

$A$ parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

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