A parallel-plate capacitor of capacitance 40 | vedclass

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Question

(C) Initial capacitance $C = 40 \mu F$,Voltage $V = 100 V$,Dielectric constant $K = 2$. New capacitance $C' = KC = 2 	imes 40 \mu F = 80 \mu F$. Extr

Vedclass · Accepted Answer

$4 mC$ and $0.2 J$

Vedclass · Answer

(C) Initial capacitance $C = 40 \mu F$,Voltage $V = 100 V$,Dielectric constant $K = 2$. New capacitance $C' = KC = 2 	imes 40 \mu F = 80 \mu F$. Extra charge $\Delta q = q' - q = (C' - C)V = (80 - 40) 	imes 10^{-6} 	imes 100 = 40 	imes 10^{-6} 	imes 100 = 4 	imes 10^{-3} C = 4 mC$. Change in electrostatic energy $\Delta U = U' - U = \frac{1}{2}C'V^2 - \frac{1}{2}CV^2 = \frac{1}{2}(K-1)CV^2$. $\Delta U = \frac{1}{2} 	imes (2-1) 	imes 40 	imes 10^{-6} 	imes (100)^2 = \frac{1}{2} 	imes 1 	imes 40 	imes 10^{-6} 	imes 10000 = 20 	imes 10^{-2} J = 0.2 J$. Thus,the extra charge is $4 mC$ and the change in energy is $0.2 J$.

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