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

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

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

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$A$ parallel-plate capacitor of capacitance $40 \mu F$ is connected to a $100 V$ power supply. Now,the intermediate space between the plates is filled with a dielectric material of dielectric constant $K=2$. Due to the introduction of the dielectric material,the extra charge and the change in the electrostatic energy in the capacitor,respectively,are:

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A
$2 mC$ and $0.2 J$
B
$8 mC$ and $2.0 J$
C
$4 mC$ and $0.2 J$
D
$2 mC$ and $0.4 J$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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