(D) For a parallel plate capacitor,the initial potential difference $V$ is given by $V = E \times d$.Given $E = 3 \times 10^4\, V/m$ and $d = 0.05\, m

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(D) For a parallel plate capacitor,the initial potential difference $V$ is given by $V = E \times d$.Given $E = 3 \times 10^4\, V/m$ and $d = 0.05\, m$,we have $V = 3 \times 10^4 \times 0.05 = 1500\, V = 1.5\, kV$.Since the capacitor is disconnected from the battery,the charge $q$ remains constant.The potential difference $V'$ after inserting a slab of thickness $t$ and dielectric constant $K$ is given by $V' = V \times \frac{(d - t) + (t/K)}{d}$.For a dielectric slab with $K = 2$ and $t = 0.01\, m$:$V' = 1.5 \times \frac{(0.05 - 0.01) + (0.01/2)}{0.05}$$V' = 1.5 \times \frac{0.04 + 0.005}{0.05}$$V' = 1.5 \times \frac{0.045}{0.05}$$V' = 1.5 \times 0.9 = 1.35\, kV$.

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

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The distance between the plates of a parallel plate capacitor is $0.05\, m$. An electric field of $3 \times 10^4\, V/m$ is established between the plates. The capacitor is disconnected from the battery,and an uncharged metal plate of thickness $0.01\, m$ is inserted. If a dielectric slab of dielectric constant $K = 2$ is inserted instead of the metal plate,what will be the potential difference in $kV$?

A
$1.89$
B
$2.15$
C
$1.20$
D
$1.35$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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The function of a dielectric in a capacitor is

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An air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).

MediumMHT CET 2024

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