In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C . The switch S 1 is pressed first to fully charge the capacitor C 1 and the released. The switch S 2 is then pressed to charge the capacitor C 2 . After some time, S 2 is released and then is S 3 pressed. After some time,

the charge on the upper plate of C 2 is - CV 0 .

the charge on the upper plate of C 1 is 2CV 0 .

the charge on the upper plate of C 2 is 0.

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Capacitance of a parallel plate air capacitor depends on

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Important Questions on Electrostatics

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor is of area

6 c m^{2}

and a separation

3 m m .

The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant

K_{1} = 10, K_{2} = 12

and

K_{3} = 14 .

The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor of capacitance

C

has spacing

d

between two plates having area

A .

The region between the plates is filled with

N

dielectric layers, parallel to its plates, each with thickness

δ = \frac{d}{N} .

The dielectric constant of the

m^{t h}

layer is

K_{m} = K (1 + \frac{m}{N}) .

For a very large

N (> 10^{3}),

the capacitance

C

α (\frac{K \in_{0} A}{d l n 2}) .

The value of

α

will be____.

[\in_{0}

is the permittivity of free space]

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A capacitor is made of two square plates each of side

a

making a very small angle

α

between them, as shown in figure. The capacitance will be close to:
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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor with plates of area

1 m^{2}

each, are at a separation of

0.1 m .

If the electric field between the plates is

100 N C^{- 1},

the magnitude of charge on each plate is:

(Take ε_{0} = 8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2})

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

Two identical parallel plate capacitors of capacitance each are connected in series with a battery of emf, $E$ as shown. If one of the capacitors is now filled with a dielectric of dielectric constant $k$ , the amount of charge which will flow through the battery is (neglect internal resistance of the battery)

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor has uniform electric field $E$ in the space between the plates.If the distance between the plates is $d$ and area of each plate is $A$ , the energy stored in the capacitor is ;

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor has

1 μ F

capacitance. One of its two plates is given

+ 2 μ C

charge and the other plate,

+ 4 μ C

charge. The potential difference developed across the capacitor is:

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

How will the voltage

(V)

between the two plates of a parallel plate capacitor depend on the distance

(d)

between the plates, if the charge on the capacitor remains the same?

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3 \times 1 0^{4} V/m$ , the charge density of the positive plate will be close to :

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor having plates of area

S

and plate separation

d

, has capacitance

C_{1}

in air. When two dielectrics of different relative permittivities (

ε_{1} = 2

and

ε_{2} = 4

) are introduced between the two plates as shown in the figure, the capacitance becomes

C_{2}

. The ratio

\frac{C_{2}}{C_{1}}

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor having cross-sectional area $A$ and separation $d$ has air in between the plates. Now an insulating slab of the same area but the thickness, $\frac{d}{2}$ , is inserted between the plates as shown in figure having dielectric constant $K (= 4) .$ The ratio of new capacitance to its original capacitance will be,

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate air capacitor of capacitance

C

is connected to a cell of emf

V

and then disconnected from it. A dielectric slab of dielectric constant

K

, which can just fill the air gap of the capacitor, is now inserted in it. Which of the following is incorrect?

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

In a parallel plate capacitor, the capacity can be increased by decreasing

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor has plates of area A separated by distance

d

between them. It is filled with a dielectric which has a dielectric constant that varies as

K (x) = K_{0} (1 + α x)

where

x

is the distance measured from one of the plates. If

(α d) < < 1,

the total capacitance of the system is best given by the expression:

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor of two plates, each with area $90 π {cm}^{2}$ , is separated by $5 mm$ air-gap. The capacitor is initially connected to $100 V$ source. When an unknown liquid is filled between the plates so as to fill the air-gap, an additional charge of $95 nC$ is found to be flowing onto the capacitor from the $100 V$ source. The dielectric constant of the liquid is
(Assume $\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}$ )

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor has plate area

100 m^{2}

and plate separation of

10 m .

The space between the plates is filled up to a thickness

5 m

with a material of dielectric constant of

10 .

The resultant capacitance of the system is

x pF .

The value of

ε_{0} = 8.85 \times 10^{- 12} F m^{- 1}

. The value of

x

to the nearest integer is ______.

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant

K

is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is

\frac{3}{4} d

, where

d

is the separation between the plates of parallel plate capacitor. The new capacitance

C^{'}

in terms of original capacitance

C_{0}

is given by the following relation :

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. The total capacitance of the capacitor is C while that of the portion with dielectric in between is $C_{1}$ . When the capacitor is charged, the plate area covered by the dielectric gets charge $Q_{1}$ and the rest of the area gets charge $Q_{2}$ . The electric field in the dielectric is $E_{1}$ and that in the other portion is $E_{2}$ . Choose the correct option/options, ignoring edge effects.

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Physics>Electricity and Magnetism>Electrostatics>Capacitor

In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch S₁ is pressed first to fully charge the capacitor C₁ and the released. The switch S₂ is then pressed to charge the capacitor C₂. After some time, S₂ is released and then is S₃ pressed. After some time,
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Physics>Electricity and Magnetism>Electrostatics>Capacitor

In the figure is shown a system of four capacitors connected across a $10 V$ battery. The charge that will flow from switch S when it is closed is:
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Capacitance of a parallel plate air capacitor depends on

Important Questions on Electrostatics

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