(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 Electric 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

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to:

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When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

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The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

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$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

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$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.

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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:

Similar Questions

$A$ parallel plate capacitor is made of two circular plates separated by a distance $5 \ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4 \ Vm^{-1}$,the charge density of the positive plate will be close to:

When air in a capacitor is replaced by a medium of dielectric constant $K$,the capacity

The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

$A$ capacitor is fully charged with a battery and then disconnected. $A$ dielectric is then inserted into the capacitor. How do the charges on the surface of the dielectric and the outer surface of the plates of the capacitor change, respectively?

$A$ parallel plate capacitor has a capacitance of $2\ \mu F$ with a separation of $0.4\ cm$ between the plates. If the distance between the plates is halved and the space is filled with a dielectric material of constant $K = 2.8$,the final capacitance of the capacitor will be .....$\mu F$.

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