(D) Initially,both capacitors have capacity $C$ and are connected in parallel. The initial equivalent capacity is $C_{eq,i} = C + C = 2C$.After one ca

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(D) Initially,both capacitors have capacity $C$ and are connected in parallel. The initial equivalent capacity is $C_{eq,i} = C + C = 2C$.After one capacitor is filled with a dielectric of constant $K$,its new capacity becomes $C' = KC$. The other capacitor remains $C$.The new equivalent capacity of the parallel combination is $C_{eq,f} = KC + C = C(K+1)$.The change in the effective capacity is $\Delta C = C_{eq,f} - C_{eq,i} = C(K+1) - 2C = CK + C - 2C = C(K-1)$.

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

Medium

Two parallel plate air capacitors of same capacity $C$ are connected in parallel to a battery of e.m.f. $E$. Then,one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacity of the parallel combination is:

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A
$\frac{C}{(K-1)}$
B
$\frac{KC}{K-1}$
C
$KC+1$
D
$C(K-1)$

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Class 12 Questions

Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.

MediumAIIMS 2008

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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.

Easy

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The force between two point charges kept with a separation of $9 \ cm$ in air is $98 \ N$. If a dielectric slab of constant $4$,thickness $6 \ cm$ and another dielectric slab of constant $9$,thickness $3 \ cm$ are introduced between the two charges,then the new force becomes (in $N$)

MediumAP EAMCET 2025

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$A$ parallel plate air capacitor has capacity $C$ farad,potential $V$ volt,and energy $E$ joule. When the gap between the plates is completely filled with a dielectric material of dielectric constant $K > 1$,what happens to the potential $V$ and energy $E$?

EasyMHT CET 2017

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In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

Medium

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Two parallel plate air capacitors of same capacity $C$ are connected in parallel to a battery of e.m.f. $E$. Then,one of the capacitors is completely filled with a dielectric material of constant $K$. The change in the effective capacity of the parallel combination is:

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Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.Reason : The surface density of charge on the plate remains constant or unchanged.

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

The force between two point charges kept with a separation of $9 \ cm$ in air is $98 \ N$. If a dielectric slab of constant $4$,thickness $6 \ cm$ and another dielectric slab of constant $9$,thickness $3 \ cm$ are introduced between the two charges,then the new force becomes (in $N$)

$A$ parallel plate air capacitor has capacity $C$ farad,potential $V$ volt,and energy $E$ joule. When the gap between the plates is completely filled with a dielectric material of dielectric constant $K > 1$,what happens to the potential $V$ and energy $E$?

In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

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Assertion : $A$ parallel plate capacitor is connected across a battery through a key. $A$ dielectric slab of dielectric constant $K$ is introduced between the plates. The energy stored becomes $K$ times.
Reason : The surface density of charge on the plate remains constant or unchanged.