(B) Since the battery remains connected to the capacitor,the potential difference $V$ across the plates remains constant.When a dielectric slab of die

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charge on the plates becomes $K$ times

(B) Since the battery remains connected to the capacitor,the potential difference $V$ across the plates remains constant.When a dielectric slab of dielectric constant $K$ is inserted to fill the entire space between the plates,the new capacitance becomes $C' = KC$.Using the relation for charge $q = CV$,the new charge $q'$ on the plates is given by $q' = C'V = (KC)V = K(CV) = Kq$.Therefore,the charge on the plates becomes $K$ times the original charge.Thus,option $B$ is correct.

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

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$A$ parallel plate capacitor having capacitance $C$ farad is connected to a battery of emf $V$ volt. Keeping the capacitor connected to the battery,a dielectric slab of dielectric constant $K$ is inserted between the plates. The dimensions of the slab are such that it fills the whole space between the capacitor plates. Then $:-$

A
charge on the capacitor plates remains the same
B
charge on the plates becomes $K$ times
C
potential difference between the plates decreases to $V/K$
D
all of the above

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

Two dielectric slabs having dielectric constants $K_1$ and $K_2$ of thicknesses $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates of a capacitor as shown in the figure. The net capacitance between $A$ and $B$ is [where $\varepsilon_0$ is the permittivity of free space].

DifficultMHT CET 2023

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In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

MediumMHT CET 2019

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Two identical capacitors $1$ and $2$ are connected in series to a battery as shown in the figure. Capacitor $2$ contains a dielectric slab of dielectric constant $k$ as shown. $Q_1$ and $Q_2$ are the charges stored in the capacitors. Now the dielectric slab is removed and the corresponding charges are $Q'_1$ and $Q'_2$. Then:

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$A$ parallel plate capacitor is charged by connecting a $2 \ V$ battery across it. It is then disconnected from the battery and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

MediumKCET 2022

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Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be

MediumJEE MAIN 2021

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Two dielectric slabs having dielectric constants $K_1$ and $K_2$ of thicknesses $\frac{d}{4}$ and $\frac{3d}{4}$ are inserted between the plates of a capacitor as shown in the figure. The net capacitance between $A$ and $B$ is [where $\varepsilon_0$ is the permittivity of free space].

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

$A$ parallel plate capacitor is charged by connecting a $2 \ V$ battery across it. It is then disconnected from the battery and a glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

Two capacitors of capacities $2C$ and $C$ are joined in parallel and charged up to potential $V$. The battery is removed and the capacitor of capacity $C$ is filled completely with a medium of dielectric constant $K$. The potential difference across the capacitors will now be

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