(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is

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(C) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\epsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is the distance between them.For circular plates of radius $r$, the area $A = \pi r^2$.Thus, $C_1 = \frac{\epsilon_0 \pi r^2}{d}$.In the second case, the new radius $r' = \sqrt{3}r$ and the new distance $d' = \frac{d}{2}$.The new area $A' = \pi (r')^2 = \pi (\sqrt{3}r)^2 = 3\pi r^2$.The new capacitance $C_2 = \frac{\epsilon_0 A'}{d'} = \frac{\epsilon_0 (3\pi r^2)}{d/2} = 6 \left( \frac{\epsilon_0 \pi r^2}{d} \right) = 6C_1$.Therefore, the ratio $C_1 : C_2 = C_1 : 6C_1 = 1 : 6$.

Class 12 Physics Electrostatic Potential and Capacitance Parallel Plate Capacitor

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Two circular plates each of radius 'r' are kept parallel to each other distance 'd' apart. The capacitance of the capacitor formed is '$C_1$'. If the radius of each of the plates is increased to $\sqrt{3}$ times the earlier radius and their distance of separation decreased to half the initial value, the capacitance now becomes '$C_2$'. The ratio $C_1 : C_2$ is

MHT CET 2026 All MHT CET

A
$1 : 2$
B
$1 : 4$
C
$1 : 6$
D
$6 : 1$

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Class 12

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Electrostatic Potential and Capacitance

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Parallel Plate Capacitor

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Five identical metal plates, each of area $A$, are arranged parallel to each other at a constant distance $d$ apart. Plates $1$ and $4$ are connected by a wire, and plates $3$ and $5$ are connected by a wire as shown in the figure. Find the equivalent capacitance of the system between points $A$ and $B$.

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In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \, V$ supply,what is the charge on each plate of the capacitor?

Medium

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In an insulated parallel-plate capacitor of capacitance $C$,the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plates is

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Four identical plates each of area $a$ are separated by a distance $d$. The connection is shown below. What is the capacitance between $P$ and $Q$?

DifficultWBJEE 2013

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In the diagram,the area of each plate is $A = 2 \,m^2$ and $d = 2 \times 10^{-3} \,m$. An electric charge of $q = 8.85 \times 10^{-8} \,C$ is given to the plate '$Q$'. Plates '$P$' and '$R$' are connected to the ground. Then the potential of plate '$Q$' is:

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In an insulated parallel-plate capacitor of capacitance $C$,the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plates is

Four identical plates each of area $a$ are separated by a distance $d$. The connection is shown below. What is the capacitance between $P$ and $Q$?

In the diagram,the area of each plate is $A = 2 \,m^2$ and $d = 2 \times 10^{-3} \,m$. An electric charge of $q = 8.85 \times 10^{-8} \,C$ is given to the plate '$Q$'. Plates '$P$' and '$R$' are connected to the ground. Then the potential of plate '$Q$' is:

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