(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the dis

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.When the distance $d$ is increased,the capacitance $C$ decreases.Since the battery is disconnected,the charge $Q$ on the plates remains constant.The relationship between charge,capacitance,and potential difference is $V = \frac{Q}{C}$.Since $Q$ is constant and $C$ decreases,the potential difference $V$ must increase.

Class 12 Physics Electric Potential and Capacitance Parallel Plate Capacitor

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$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

AIPMT 2006 All AIPMT

A
decreases
B
increases
C
does not change
D
becomes zero

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Class 12 Questions

Class 12

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Electric Potential and Capacitance

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Parallel Plate Capacitor

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Four metallic plates are arranged as shown in the figure. If the distance between each plate is $d$ and the area of each plate is $A$, then the capacitance of the given system between points $A$ and $B$ is (Given $d << \sqrt{A}$):

Medium

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Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

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Four identical metal plates are located in air at equal distance $d$ from each other. The area of each plate is $S$. If the outer most plates are connected by a conducting wire as shown in the figure, the capacitance between points $A$ and $B$ will be

EasyTS EAMCET 2019

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Between the plates of a parallel plate capacitor,there is a $1\,mm$ thick paper of dielectric constant $4$. It is charged to $100\,V$. The electric field in $V/m$ between the plates of the capacitor is:

Easy

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$A$ parallel plate air capacitor is charged to a potential difference of $V \text{ volts}$. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates

Medium

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$A$ parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates:

Similar Questions

Four metallic plates are arranged as shown in the figure. If the distance between each plate is $d$ and the area of each plate is $A$, then the capacitance of the given system between points $A$ and $B$ is (Given $d << \sqrt{A}$):

Five identical capacitor plates,each of area $A$,are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of $emf$ $V$ as shown in the figure. Then the charges on plates $1$ and $4$ are,respectively:

Four identical metal plates are located in air at equal distance $d$ from each other. The area of each plate is $S$. If the outer most plates are connected by a conducting wire as shown in the figure, the capacitance between points $A$ and $B$ will be

Between the plates of a parallel plate capacitor,there is a $1\,mm$ thick paper of dielectric constant $4$. It is charged to $100\,V$. The electric field in $V/m$ between the plates of the capacitor is:

$A$ parallel plate air capacitor is charged to a potential difference of $V \text{ volts}$. After disconnecting the charging battery,the distance between the plates of the capacitor is increased using an insulating handle. As a result,the potential difference between the plates

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