A parallel plate capacitor with air as dielec | vedclass

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Question

(D) Initial capacitance with air $C_0 = 4 \mu F$. After filling with dielectric $(K = 5)$,the new capacitance is $C = K C_0 = 5 	imes 4 \mu F = 20 \m

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(D) Initial capacitance with air $C_0 = 4 \mu F$. After filling with dielectric $(K = 5)$,the new capacitance is $C = K C_0 = 5 	imes 4 \mu F = 20 \mu F$. The charge on the capacitor is $Q = C V = 20 \mu F 	imes 100 \ V = 2000 \mu C = 2 	imes 10^{-3} \ C$. Since the capacitor is disconnected from the battery,the charge $Q$ remains constant. Initial energy stored with dielectric: $U_i = \frac{Q^2}{2C} = \frac{(2 	imes 10^{-3})^2}{2 	imes 20 	imes 10^{-6}} = \frac{4 	imes 10^{-6}}{40 	imes 10^{-6}} = 0.1 \ J$. Final capacitance after removing the dielectric: $C_f = C_0 = 4 \mu F$. Final energy stored: $U_f = \frac{Q^2}{2C_f} = \frac{(2 	imes 10^{-3})^2}{2 	imes 4 	imes 10^{-6}} = \frac{4 	imes 10^{-6}}{8 	imes 10^{-6}} = 0.5 \ J$. Work done by external agent $W = U_f - U_i = 0.5 \ J - 0.1 \ J = 0.4 \ J$.

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