(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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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

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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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$A$ frictionless dielectric plate $S$ is placed on a frictionless table $T$. $A$ charged parallel plate capacitor $C$ (whose plates are also frictionless) is kept near it. The plate $S$ is partially between the plates of the capacitor. When the plate $S$ is released, what happens?

Easy

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$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

MediumAIIMS 2019

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An air capacitor has capacitance $C_1$. The space between the two plates of the capacitor is filled with two dielectrics as shown in the figure. The new capacitance of the capacitor is $C_2$. The ratio $\frac{C_1}{C_2}$ is ($d=$ distance between two plates of the capacitor,$K_1$ and $K_2$ are dielectric constants of the two dielectrics respectively).

MediumMHT CET 2024

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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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The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

$A$ frictionless dielectric plate $S$ is placed on a frictionless table $T$. $A$ charged parallel plate capacitor $C$ (whose plates are also frictionless) is kept near it. The plate $S$ is partially between the plates of the capacitor. When the plate $S$ is released, what happens?

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

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