(A) When the capacitor is disconnected from the source,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.I

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(A) When the capacitor is disconnected from the source,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.Inserting a dielectric slab increases the capacitance $C$ of the capacitor,as $C = K C_0$,where $K > 1$ is the dielectric constant.Since the charge $Q$ is constant and the capacitance $C$ increases,the potential difference $V$ across the plates changes according to the relation $V = Q / C$. As $C$ increases,$V$ decreases.Therefore,the potential difference $(i)$ and the capacitance $(ii)$ change,while the charge $(iii)$ remains constant.

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

Medium

$A$ parallel plate capacitor is charged to a potential difference of $100\,V$ and disconnected from the source of $emf$. $A$ slab of dielectric is then inserted between the plates. Which of the following three quantities change?
$(i)$ The potential difference
$(ii)$ The capacitance
$(iii)$ The charge on the plates

A
only $(i)$ and $(ii)$
B
only $(i)$ and $(iii)$
C
only $(ii)$ and $(iii)$
D
All $(i), (ii)$ and $(iii)$

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

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If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will

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The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used,is $C$. If the dielectric is replaced by another with dielectric constant $20$,the capacity will become

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$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

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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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$A$ parallel plate capacitor has a plate area of $50 \ cm^2$ and a plate separation of $3 \ mm$. The space between the plates is filled with a dielectric medium of thickness $1 \ mm$ and a dielectric constant of $4$. Calculate the capacitance. ($\epsilon_0$ is the permittivity of free space)

If the distance between the plates of a parallel plate capacitor is halved and the dielectric constant of the dielectric is doubled,then its capacity will

The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used,is $C$. If the dielectric is replaced by another with dielectric constant $20$,the capacity will become

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

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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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.