(C) $1$. When the battery is disconnected,the charge $Q$ on the capacitor plates remains constant because there is no path for the charge to flow.$2$.

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Charge remains constant,and both voltage and electrostatic potential energy decrease.

(C) $1$. When the battery is disconnected,the charge $Q$ on the capacitor plates remains constant because there is no path for the charge to flow.$2$. When a dielectric slab of constant $K > 1$ is inserted,the capacitance $C$ increases according to the formula $C' = KC$.$3$. Since $Q = CV$,the new voltage $V' = Q / C'$. Since $C' > C$,the voltage $V'$ decreases $(V' = V/K)$.$4$. The electrostatic potential energy $U$ is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the potential energy $U$ decreases $(U' = U/K)$.

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

Easy

$A$ capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates,choose the correct statement.

A
Charge increases,voltage decreases,and electrostatic potential energy increases.
B
Charge remains constant,voltage increases,and electrostatic potential energy decreases.
C
Charge remains constant,and both voltage and electrostatic potential energy decrease.
D
None of the above.

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Class 12

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Chapter

Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

Easy

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Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$,then the amount of charge which will flow through the battery is (neglect internal resistance of the battery).

DifficultKVPY 2014

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$A$ slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$,where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be (Given $C_{0} =$ capacitance of capacitor with air as medium between plates):

DifficultJEE MAIN 2022

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Between the plates of a parallel plate capacitor, a dielectric plate is introduced to completely fill the space between the plates. The capacitor is charged and then disconnected from the battery. The dielectric plate is slowly drawn out of the capacitor parallel to the plates. The plot of the potential difference $V$ across the plates versus the length $x$ of the dielectric plate drawn out is:

Medium

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$A$ simple pendulum of mass $m$, length $l$, and charge $+q$ is suspended in the electric field produced by two conducting parallel plates as shown. The value of the deflection of the pendulum in the equilibrium position will be -

DifficultJEE MAIN 2021

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$A$ capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates,choose the correct statement.

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When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

$A$ simple pendulum of mass $m$, length $l$, and charge $+q$ is suspended in the electric field produced by two conducting parallel plates as shown. The value of the deflection of the pendulum in the equilibrium position will be -

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