(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.When the plates are pulled apart,the separation distance $d

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the potential difference increases

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.When the plates are pulled apart,the separation distance $d$ increases.Since $C \propto \frac{1}{d}$,the capacitance $C$ decreases.Because the battery is removed,the charge $Q$ on the plates remains constant.Using the relation $V = \frac{Q}{C}$,since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

Class 12 Physics Electrostatic Potential and Capacitance Parallel Plate Capacitor

Medium

$A$ parallel plate capacitor is charged. Then the battery is removed. Now,if the plates are pulled apart,

A
the capacitance increases
B
the potential difference increases
C
the total charge increases
D
the charge and potential difference remain the same

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Class 12

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Electrostatic Potential and Capacitance

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Parallel Plate Capacitor

Four metallic plates, each with a surface area of one side $A$, are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

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The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

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Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d$. The plates are connected to a potential difference of $V \text{ volts}$. The charge on plates $1$ and $4$ will be

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In a parallel plate capacitor,if $10^{12}$ electrons pass from one plate to another,a potential difference of $10 \,V$ is developed across the plates. The capacitance of the capacitor is

MediumAP EAMCET 2022

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The capacity of a parallel plate capacitor depends on:

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$A$ parallel plate capacitor is charged. Then the battery is removed. Now,if the plates are pulled apart,

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Four metallic plates, each with a surface area of one side $A$, are placed at a distance $d$ from each other. The plates are connected as shown in the circuit diagram. Then the capacitance of the system between $a$ and $b$ is

The distance between the plates of a parallel plate capacitor is $4 \ mm$ and the potential difference is $60 \ V$. If the distance between the plates is increased to $12 \ mm$,then:

Five identical plates each of area $A$ are joined as shown in the figure. The distance between the plates is $d$. The plates are connected to a potential difference of $V \text{ volts}$. The charge on plates $1$ and $4$ will be

In a parallel plate capacitor,if $10^{12}$ electrons pass from one plate to another,a potential difference of $10 \,V$ is developed across the plates. The capacitance of the capacitor is

The capacity of a parallel plate capacitor depends on:

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