A parallel-plate capacitor of plate area 10 t | vedclass

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(C) Initial capacitance $C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 	imes 10 	imes 10^{-4}}{3 	imes 10^{-3}} = \frac{\varepsilon_0}{3} 	e

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) Initial capacitance $C = \frac{\varepsilon_0 A}{d} = \frac{\varepsilon_0 	imes 10 	imes 10^{-4}}{3 	imes 10^{-3}} = \frac{\varepsilon_0}{3} 	ext{ F}$.Initial charge $Q = CV = \frac{\varepsilon_0}{3} 	imes 12 = 4 \varepsilon_0 	ext{ C}$.Initial energy $U_i = \frac{Q^2}{2C} = \frac{(4 \varepsilon_0)^2}{2(\varepsilon_0/3)} = \frac{16 \varepsilon_0^2}{2 \varepsilon_0 / 3} = 24 \varepsilon_0 	ext{ J}$.After inserting the dielectric,the new capacitance $C' = KC = 3 	imes \frac{\varepsilon_0}{3} = \varepsilon_0 	ext{ F}$.Since the battery is disconnected,the charge $Q$ remains constant.Final energy $U_f = \frac{Q^2}{2C'} = \frac{(4 \varepsilon_0)^2}{2 \varepsilon_0} = \frac{16 \varepsilon_0^2}{2 \varepsilon_0} = 8 \varepsilon_0 	ext{ J}$.Work done $W = U_f - U_i = 8 \varepsilon_0 - 24 \varepsilon_0 = -16 \varepsilon_0 	ext{ J}$.The magnitude of work done on the system is $16 \varepsilon_0$. Thus,$\alpha = 16$.

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