(C) The potential energy stored in a capacitor is given by $U = \frac{1}{2} CV^2$.Since the battery remains connected,the potential difference $V$ acr

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Assertion is true but the reason is false.

(C) The potential energy stored in a capacitor is given by $U = \frac{1}{2} CV^2$.Since the battery remains connected,the potential difference $V$ across the plates remains constant.Therefore,$U \propto C$.When a dielectric material is inserted,the capacitance $C$ increases $(C = K C_0)$.As $C$ increases,the potential energy $U$ also increases.Thus,the assertion is true.However,when the battery remains connected,the charge $Q = CV$ changes because $C$ changes while $V$ remains constant.Therefore,the reason is false.

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

Medium

Assertion: When a battery remains connected,the electric potential energy increases if a dielectric material is inserted between the plates of a capacitor.
Reason: When a battery remains connected,the charge on the plates of the capacitor remains the same.

A
Both assertion & reason are true & the reason is a correct explanation of the assertion.
B
Both assertion & reason are true but reason is not a correct explanation of the assertion.
C
Assertion is true but the reason is false.
D
Both assertion & reason are false.

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Class 12

Physics Questions

Chapter

Electrostatic Potential and Capacitance

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Effect of Dielectric Inside Capacitor

Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

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$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)

MediumMHT CET 2025

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Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?

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$A$ parallel plate capacitor of capacitance $12.5 \ pF$ is charged by a battery connected between its plates to a potential difference of $12.0 \ V$. The battery is now disconnected and a dielectric slab $(\epsilon_{r}=6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is . . . . . . . $\times 10^{-12} \ J$.

DifficultJEE MAIN 2024

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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 $kV$?

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Assertion: When a battery remains connected,the electric potential energy increases if a dielectric material is inserted between the plates of a capacitor.Reason: When a battery remains connected,the charge on the plates of the capacitor remains the same.

Similar Questions

Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

$A$ parallel plate air capacitor has capacity $C$ and distance of separation between plates is $d$. If a conducting sheet of thickness $\frac{2d}{3}$ is inserted between the plates,the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is (in $:1$)

Two parallel plate capacitors are connected in series and then connected to a $100 \ V$ battery. $A$ dielectric slab of dielectric constant $K = 4.0$ is inserted between the plates of the second capacitor. What will be the potential difference across each capacitor respectively?

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Assertion: When a battery remains connected,the electric potential energy increases if a dielectric material is inserted between the plates of a capacitor.
Reason: When a battery remains connected,the charge on the plates of the capacitor remains the same.