(A) Since the battery remains connected,the potential difference across the plates remains constant. Therefore,$V = V_0$.The electric field is given b

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(A) Since the battery remains connected,the potential difference across the plates remains constant. Therefore,$V = V_0$.The electric field is given by $E = V/d$. Since $V$ and the separation $d$ remain unchanged,$E = E_0$.Let the initial capacitance be $C_0$. When a dielectric slab of constant $K$ is inserted,the new capacitance becomes $C = K C_0$,where $K > 1$.The charge on the capacitor is $Q = CV = (K C_0) V_0 = K Q_0$. Since $K > 1$,it follows that $Q > Q_0$.The energy stored in the capacitor is $U = (1/2) CV^2 = (1/2) (K C_0) V_0^2 = K U_0$. Since $K > 1$,it follows that $U > U_0$.Thus,the correct relation is $Q > Q_0$.

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

Medium

$A$ parallel plate capacitor is connected to a battery. The quantities charge,voltage,electric field,and energy associated with the capacitor are given by $Q_0, V_0, E_0$,and $U_0$ respectively. $A$ dielectric slab is introduced between the plates of the capacitor,but the battery remains connected. The corresponding quantities now given by $Q, V, E$,and $U$ related to the previous ones are:

A
$Q > Q_0$
B
$V > V_0$
C
$E > E_0$
D
$U < U_0$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ capacitor of $10 \mu F$ capacitance,whose plates are separated by $10 \text{ mm}$ through air and each plate has an area of $4 \text{ cm}^2$,is now filled equally with two dielectric media of $K_1=2$ and $K_2=3$ respectively,as shown in the figure. If the new force between the plates is $8 \text{ N}$,the supply voltage is . . . . . . $V$.

DifficultJEE MAIN 2024

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The figure given below shows two identical parallel plate capacitors connected to a battery with switch $S$ closed. The switch is now opened and the free space between the plates of both capacitors is filled with a dielectric of dielectric constant $K = 3$. What will be the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric?

DifficultIIT 1983

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If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

EasyAP EAMCET 2025

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$A$ parallel plate capacitor has a plate area of $40\,cm^2$ and a plate separation of $2\,mm$. The space between the plates is filled with a dielectric medium of thickness $1\,mm$ and dielectric constant $5$. The capacitance of the system is:

MediumJEE MAIN 2023

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When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

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If a dielectric slab of dielectric constant $3$ is introduced between the plates of a capacitor having electric field $1.5 \times 10^{-9} \pi \ N C^{-1}$,then the electric displacement is

$A$ parallel plate capacitor has a plate area of $40\,cm^2$ and a plate separation of $2\,mm$. The space between the plates is filled with a dielectric medium of thickness $1\,mm$ and dielectric constant $5$. The capacitance of the system is:

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