(D) When a dielectric slab is inserted into a parallel plate capacitor while the battery remains connected,the potential difference $V$ across the pla

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(D) When a dielectric slab is inserted into a parallel plate capacitor while the battery remains connected,the potential difference $V$ across the plates remains constant,i.e.,$V = {V_0}$.Since the capacitance $C$ increases by a factor of $K$ (dielectric constant),where $C = KC_0$,the charge $Q$ on the plates increases according to $Q = CV = K C_0 V_0 = K Q_0$. Thus,$Q > {Q_0}$.The electric field $E$ between the plates is given by $E = V/d$. Since $V$ and $d$ remain constant,$E = {E_0}$.The energy stored in the capacitor is $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ remains constant,$U$ increases,i.e.,$U > {U_0}$.Therefore,both statements $(a)$ and $(b)$ are correct.

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

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Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as:

IIT 1985 All IIT

A
$Q > {Q_0}$
B
$U > {U_0}$
C
$E > {E_0}$
D
Both $(a)$ and $(b)$

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

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The plates of a parallel plate capacitor are charged up to $200 \ V$. $A$ dielectric slab of thickness $4 \ mm$ is inserted between its plates. Then,to maintain the same potential difference between the plates of the capacitor,the distance between the plates is increased by $3.2 \ mm$. The dielectric constant of the dielectric slab is

DifficultAP EAMCET 2004

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An air-filled parallel plate capacitor has a capacity of $2 \ pF$. If the separation between the plates is doubled and the interspace between the plates is filled with a dielectric material,the capacity increases to $6 \ pF$. The dielectric constant of the material is:

MediumMHT CET 2021

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Half of the space between the plates of a parallel plate capacitor is filled with a dielectric material of dielectric constant $K$ parallel to the plates. If the initial capacitance is $C$,what will be the new (final) capacitance?

$A$ parallel plate capacitor has capacitance $C$, when there is vacuum within the parallel plates. $A$ sheet having thickness $t = d/3$ (where $d$ is the separation between the plates) and relative permittivity $K$ is introduced between the plates. The new capacitance of the system is:

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