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(A) The initial capacitance of the parallel plate capacitor is $C = 0.025 \mu F$. When a metal plate of thickness $t = d/3$ is introduced,the new capa

Vedclass · Accepted Answer

$62.5$

Vedclass · Answer

(A) The initial capacitance of the parallel plate capacitor is $C = 0.025 \mu F$. When a metal plate of thickness $t = d/3$ is introduced,the new capacitance $C'$ becomes $C' = \frac{\varepsilon_0 A}{d - t} = \frac{\varepsilon_0 A}{d - d/3} = \frac{3}{2} \frac{\varepsilon_0 A}{d} = \frac{3}{2} C$. Since the capacitor remains connected to the $100 \ V$ source,the potential difference $V$ remains constant at $100 \ V$. The energy stored in the capacitor initially is $U_i = \frac{1}{2} C V^2$. The energy stored in the capacitor with the metal plate is $U_f = \frac{1}{2} C' V^2 = \frac{1}{2} (\frac{3}{2} C) V^2 = \frac{3}{4} C V^2$. The work done by the external agent to remove the plate is $W = U_i - U_f$ (since the battery does work to maintain potential). $W = \frac{1}{2} C V^2 - \frac{3}{4} C V^2 = -\frac{1}{4} C V^2$. The magnitude of work done by the external agent is $|W| = \frac{1}{4} C V^2$. $|W| = \frac{1}{4} 	imes (0.025 	imes 10^{-6} \ F) 	imes (100 \ V)^2 = \frac{0.025 	imes 10^{-6} 	imes 10^4}{4} = \frac{0.025 	imes 10^{-2}}{4} = 6.25 	imes 10^{-5} \ J = 62.5 \mu J$.

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