(A) Initial capacitance with air is $C = 80 \times 10^{-6} \ F$. When the dielectric slab of constant $K = 20$ is inserted,the new capacitance becomes

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(A) Initial capacitance with air is $C = 80 \times 10^{-6} \ F$. When the dielectric slab of constant $K = 20$ is inserted,the new capacitance becomes $C' = K \times C = 20 \times 80 \times 10^{-6} = 1600 \times 10^{-6} \ F$. The charge on the capacitor with the dielectric is $q_1 = C' V = (1600 \times 10^{-6}) \times 30 = 48000 \times 10^{-6} \ C = 48 \times 10^{-3} \ C$. After removing the dielectric,the capacitance returns to $C = 80 \times 10^{-6} \ F$. The new charge on the capacitor is $q_2 = C V = (80 \times 10^{-6}) \times 30 = 2400 \times 10^{-6} \ C = 2.4 \times 10^{-3} \ C$. The charge that flows through the wire is $\Delta q = q_1 - q_2 = 48 \times 10^{-3} - 2.4 \times 10^{-3} = 45.6 \times 10^{-3} \ C$.

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

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$A$ parallel plate capacitor has a capacity of $80 \times 10^{-6} \ F$ when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant $K = 20$. The capacitor is connected to a battery of $30 \ V$. The dielectric slab is then removed while the capacitor remains connected to the battery. Calculate the charge that passes through the wire.

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A
$45.6 \times 10^{-3} \ C$
B
$25.3 \times 10^{-3} \ C$
C
$120 \times 10^{-3} \ C$
D
$125 \times 10^{-3} \ C$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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Two identical parallel plate capacitors are placed in series and connected to a constant voltage source of $V_0 \, V$. If one of the capacitors is completely immersed in a liquid with dielectric constant $K$,the potential difference between the plates of the other capacitor will change to:

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