(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Initially,the capacitance is $C = \frac{\epsilon_0 A}{d}$.When the battery is di

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(A) The energy stored in a capacitor is given by $U = \frac{Q^2}{2C}$.Initially,the capacitance is $C = \frac{\epsilon_0 A}{d}$.When the battery is disconnected,the charge $Q$ on the plates remains constant.When a dielectric slab of dielectric constant $K$ and thickness $d$ is introduced,the new capacitance becomes $C' = KC$.The new energy stored is $U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(KC)} = \frac{1}{K} \left( \frac{Q^2}{2C} \right) = \frac{U}{K}$.Therefore,the ratio of the energy stored before and after the slab is introduced is $\frac{U}{U'} = K$.

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

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$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced,is

A
$K$
B
$1/K$
C
$\frac{A}{d^2 K}$
D
$\frac{d^2 K}{A}$

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Class 12 Questions

Class 12

Physics Questions

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

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$A$ parallel plate capacitor with a dielectric slab completely occupying the space between the plates is charged by a battery and then disconnected. The slab is pulled out with a constant speed. Which of the following curves represent qualitatively the variation of the capacitance $C$ of the system with time?

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The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is

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$A$ parallel plate capacitor has plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?

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$A$ capacitor of capacitance $2 \mu F$ is charged to $50 V$ and then disconnected from the source. Later,the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by $25 \%$ of its initial value,then the dielectric constant of the dielectric material is

MediumAP EAMCET 2025

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The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax is:

The capacity of a parallel plate capacitor is $5 \mu F$. When a glass plate is placed between the plates of the capacitor,its potential difference reduces to $1/8$ of the original value. The magnitude of the relative dielectric constant of the glass is

$A$ parallel plate capacitor has plates of area $A$ separated by a distance $d$. If a dielectric slab of thickness $t$ and dielectric constant $K$ is inserted between the plates,what is the new capacitance?

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