(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 Electric 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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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ capacitor stores $60 \ \mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,a charge of $120 \ \mu C$ flows through the battery. If the initial capacitance of the capacitor was $2 \ \mu F$,the amount of heat produced when the dielectric is inserted is ... $\mu J$.

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In a parallel plate capacitor setup,the plate area of the capacitor is $2 \, m^{2}$ and the plates are separated by $1 \, m$. If the space between the plates is filled with a dielectric material of thickness $0.5 \, m$ and area $2 \, m^{2}$ (see figure),the capacitance of the setup will be $......... \, \varepsilon_{0}$. (Dielectric constant of the material $= 3.2$) (Round off to the nearest integer).

DifficultJEE MAIN 2021

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$A$ capacitor has a capacitance of $5 \mu F$ when its parallel plates are separated by an air medium of thickness $d$. $A$ slab of material with a dielectric constant of $1.5$,having an area equal to that of the plates but a thickness of $\frac{d}{2}$,is inserted between the plates. The capacitance of the capacitor in the presence of the slab will be $..........\mu F$.

MediumJEE MAIN 2023

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In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

Medium

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In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

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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.

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In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

In a parallel plate capacitor, the radius of each circular plate is $12\,cm$ and the distance between the plates is $5\,mm$. A glass slab of $3\,mm$ thickness and radius $12\,cm$ with a dielectric constant of $6$ is placed between the plates. The capacitance of the capacitor will be:

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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.