(D) When the battery is disconnected,the charge $q$ on the capacitor remains constant because there is no path for the charge to flow.The capacitance

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The electrostatic energy stored in the capacitor increases

(D) When the battery is disconnected,the charge $q$ on the capacitor remains constant because there is no path for the charge to flow.The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.When the distance $d$ between the plates is increased,the capacitance $C$ decreases.Since $q = CV$,the voltage $V = \frac{q}{C}$ must increase as $C$ decreases.The electrostatic energy stored in the capacitor is given by $U = \frac{q^2}{2C}$.Since $q$ is constant and $C$ decreases,the energy $U$ increases.

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

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$A$ parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles,then

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A
The charge on the capacitor increases
B
The voltage across the plates decreases
C
The capacitance increases
D
The electrostatic energy stored in the capacitor increases

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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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$A$ capacitor is charged by a battery and the energy stored is $U$. Now,the battery is removed and the distance between the plates is increased to four times its original value. The new energy stored becomes:

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$A$ parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles,then

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$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. The graph between electric field $(E)$ and distance $(x)$ from the left plate will be:

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