(A) When a capacitor is charged and the battery is disconnected,the charge $q$ on the plates remains constant.The capacitance of a parallel plate capa

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The stored energy of the capacitor increases

(A) When a capacitor is charged and the battery is disconnected,the charge $q$ on the plates remains constant.The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$. When the plates are moved apart,the distance $d$ increases,so the capacitance $C$ decreases.Since the charge $q$ is constant and $C$ decreases,the potential difference $V = \frac{q}{C}$ must increase.The energy stored in the capacitor is given by $W = \frac{q^2}{2C}$. Since $C$ decreases and $q$ is constant,the stored energy $W$ increases.

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

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$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

A
The stored energy of the capacitor increases
B
Charge on the capacitor increases
C
Voltage of the capacitor decreases
D
The capacitance increases

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Class 12 Questions

Class 12

Physics Questions

Chapter

Electrostatic Potential and Capacitance

Subtopic

Effect of Dielectric Inside Capacitor

Assertion: The electrostatic force between the plates of a charged isolated capacitor decreases when a dielectric fills the whole space between the plates.
Reason: The electric field between the plates of a charged isolated capacitor increases when a dielectric fills the whole space between the plates.

EasyAIIMS 2009

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$A$ parallel plate capacitor with air between the plates has a capacitance of $15 \, pF$. The separation between the plates is doubled and the space between them is filled with a medium of dielectric constant $3.5$. Then the capacitance becomes $\frac{x}{4} \, pF$. The value of $x$ is $............$

MediumJEE MAIN 2023

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$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

EasyMHT CET 2007

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$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

Difficult

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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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$A$ parallel plate capacitor is charged to a certain potential and the charging battery is then disconnected. Now,if the plates of the capacitor are moved apart,then:

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$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

$A$ parallel plate capacitor has plates of area $A$ and a separation of $10 \, mm$. Two dielectric sheets are placed between the plates with dielectric constants $k_1 = 10$ and $k_2 = 5$,and thicknesses $t_1 = 6 \, mm$ and $t_2 = 4 \, mm$ respectively. Calculate the capacitance of the capacitor.

$A$ 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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