(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$.From this,we can see that $C \propto \frac{1}{d

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(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$.From this,we can see that $C \propto \frac{1}{d}$.Therefore,the ratio of the capacitances is $\frac{C_1}{C_2} = \frac{d_2}{d_1}$.Given: $C_1 = 15\ \mu F$,$d_1 = 6\, cm$,and $d_2 = 2\, cm$.Substituting these values into the ratio formula:$\frac{15}{C_2} = \frac{2}{6}$$\frac{15}{C_2} = \frac{1}{3}$$C_2 = 15 \times 3 = 45\ \mu F$.Thus,the new capacitance is $45\ \mu F$.

Class 12 Physics Electric Potential and Capacitance Parallel Plate Capacitor

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When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

A
$15$
B
$30$
C
$45$
D
$60$

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Class 12

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Electric Potential and Capacitance

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Parallel Plate Capacitor

The capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$,the dielectric constant of the paper is $2.5$,and the width of the paper is $400\,mm$,then the length of the foil will be.....$m$.

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$A$ capacitor of capacitance $C$ with a distance $d$ between its plates is connected to a battery of $V$ volts. What is the force acting between the two plates?

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Does the capacitance of a parallel plate capacitor depend on the potential difference $(V)$ between its plates?

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$A$ parallel plate capacitor has $1\,\mu F$ capacitance. One of its two plates is given $+2\,\mu C$ charge and the other plate,$+4\,\mu C$ charge. The potential difference developed across the capacitor is......$V$

MediumJEE MAIN 2019

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Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).

DifficultAP EAMCET 2023

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When the distance between the two plates of a parallel plate capacitor is $6\, cm$,its capacitance is $15\ \mu F$. If the distance between the plates is reduced to $2\, cm$,the capacitance of the parallel plate capacitor becomes ... $\mu F$.

Similar Questions

The capacitance of a capacitor made by a thin metal foil is $2\,\mu F$. If the foil is folded with paper of thickness $0.15\,mm$,the dielectric constant of the paper is $2.5$,and the width of the paper is $400\,mm$,then the length of the foil will be.....$m$.

$A$ capacitor of capacitance $C$ with a distance $d$ between its plates is connected to a battery of $V$ volts. What is the force acting between the two plates?

Does the capacitance of a parallel plate capacitor depend on the potential difference $(V)$ between its plates?

$A$ parallel plate capacitor has $1\,\mu F$ capacitance. One of its two plates is given $+2\,\mu C$ charge and the other plate,$+4\,\mu C$ charge. The potential difference developed across the capacitor is......$V$

Four metal plates,each with surface area $A$ on one side,are placed with separation '$d$' as shown in the figure. The capacitance between $a$ and $b$ is ($\varepsilon_0$ = permittivity of free space).

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