(B) When a dielectric is inserted into a capacitor while it remains connected to a battery,the potential difference $V$ remains constant. However,if t

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(B) When a dielectric is inserted into a capacitor while it remains connected to a battery,the potential difference $V$ remains constant. However,if the capacitor is disconnected from the battery,the charge $Q$ remains constant.Assuming the capacitor is disconnected from the battery: $1$. The new capacitance becomes $C' = KC$,where $K > 1$.$2$. Since the charge $Q$ is constant,the new potential $V' = Q/C' = Q/(KC) = V/K$. Since $K > 1$,$V' $3$. The new energy $E' = Q^2 / (2C') = Q^2 / (2KC) = E/K$. Since $K > 1$,$E' Therefore,both $V$ and $E$ decrease.

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

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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 $E$?

MHT CET 2017 All MHT CET

A
Both $V$ and $E$ increase
B
Both $V$ and $E$ decrease
C
$V$ decreases,$E$ increases
D
$V$ increases,$E$ decreases

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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MediumAIIMS 2008

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MediumMHT CET 2023

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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 $E$?

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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.

Three parallel plate capacitors each with area $A$ and separation $d$ are filled with two dielectrics ($k_1$ and $k_2$) in the following fashion. Which of the following is true? $(k_1 > k_2)$

The potential energy of a charged parallel plate capacitor is $U_0$. If a slab of dielectric constant $K$ is inserted between the plates,then the new potential energy will be

An air capacitor is connected to a battery. The effect of filling the space between the plates with a dielectric is to increase:

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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.