(D) When a dielectric slab of dielectric constant $\epsilon_r$ is inserted into a capacitor while the cell remains connected,the potential difference

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Charge stored will be increased

(D) When a dielectric slab of dielectric constant $\epsilon_r$ is inserted into a capacitor while the cell remains connected,the potential difference $V$ across the plates remains constant because it is fixed by the cell.The new capacitance $C'$ becomes $C' = \epsilon_r C$. Since $\epsilon_r > 1$,the capacitance increases.The charge stored on the capacitor is given by $Q = CV$. Since $V$ is constant and $C$ increases,the charge $Q$ stored on the plates will increase $(Q' = \epsilon_r Q)$.

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

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$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

A
Capacitance will be decreased
B
Potential difference between the plates will be decreased
C
Charge stored will be decreased
D
Charge stored will be increased

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Electrostatic Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

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The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

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