Capacitors Test

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Capacitors Test - 1

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Weekly Quiz Competition

Question 1
4 / -1

Increasing the charge on the plates of a capacitor means

A
increasing the capacitance

B
increasing the potential difference between the plates

C
both

D
none

Solution

Q = CV, where C is decided only by the geometrical factors
Question 2
4 / -1

In a parallel plate capacitor the potential difference of 150 V is maintained between the plates. If distance between the plates is 6 mm, what will be the electric field at points A and B?

A
2.5×10³

B
2.5×10⁴

C
2.5×10⁵

D
3.5×10⁴

Solution
Question 3
4 / -1

Which of the following is a non polar dielectric?

A
Water

B
Alcohol

C
HCl

D
Benzene

Solution

Alcohol and HCl are polar molecules since they have a net dipole moment towards a particular direction. Both water and benzene are non-polar molecules. But water is a conductor of electricity, whereas benzene is a dielectric (insulator).
Question 4
4 / -1

The distance between the plates of a capacitor is d. What will be the new capacitance if a metal plate of thickness d/2 is introduced between the plates without touching them

A
it will be thrice of its initial value

B
it will be double of its initial value

C
remains the same

D
it will be half of its initial value

Solution

Definition based
Question 5
4 / -1

What will be the effect on capacitance, if the distance between the parallel plates reduced to one-third of its original value?

A
decrease by a factor 3

B
increase by a factor 3

C
increase by a factor 3/2

D
increase by a factor ½

Solution

C=(Kε0A/d)∝K/d
Hence,C₁/C₂=d2/d1=(d/3)d=3
C₂=3C₁
Question 6
4 / -1

What should be the radius of an isolated spherical conductor so that it has a capacity of 2μF?

A
1.8X10⁵ m

B
1.8X10⁴ m

C
2.5X10⁴ m

D
2.5X10⁵ m

Solution

Capacitance of an isolated spherical conductor =4π€R , where €=permittivity of vacuum
C=2×10^-6F
C=4π€R
4×3.14×8.85×10^-12×R=2×10^-6
R=2×10^-6/4×3.14×8.85×10^-12
=1.8×10⁴ m
Question 7
4 / -1

Work done in placing a charge of 8 x 10^-18 C on a condenser of capacity 100 microfarad is

A
4 x 10^-10 J

B
32 x 10^-32 J

C
16 x 10^-32 J

D
3.1 x 10^-26 J

Solution

Given, charge =q=8×10−18C and Capacitance =C=100μF
Here the work done is the energy stored in the capacitor. i.e,
W=q²/C= (8×10−18)²/100×10⁻⁶=32×10⁻³²J
Question 8
4 / -1

Capacitor preferred when there is a high frequency in the circuits is ______

A
Electrolyte capacitor

B
Air capacitor

C
Mica capacitor

D
Glass capacitor

Solution

Mica capacitors are preferred for high frequency circuits because they have low ohmic losses and less reactance.
Question 9
4 / -1

A capacitor stores 0.24 coulombs at 10 volts. Its capacitance is

A
0.024 F

B
0.12 F

C
0.6 F

D
0.8 F

Solution

Q = CV
Question 10
4 / -1

Work done in charging a capacitor is stored in it in the form of electrostatic energy, given by

A

B

C

D

Solution

W = Energy stored in the capacitor

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Capacitors Test - 1

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Capacitors Test - 1

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SHARING IS CARING

Question 1

increasing the capacitance

increasing the potential difference between the plates

both

none

Solution

Question 2

2.5×103

2.5×104

2.5×105

3.5×104

Solution

Question 3

Water

Alcohol

HCl

Benzene

Solution

Question 4

it will be thrice of its initial value

it will be double of its initial value

remains the same

it will be half of its initial value

Solution

Question 5

decrease by a factor 3

increase by a factor 3

increase by a factor 3/2

increase by a factor ½

Solution

Question 6

1.8X105 m

1.8X104 m

2.5X104 m

2.5X105 m

Solution

Question 7

4 x 10-10 J

32 x 10-32 J

16 x 10-32 J

3.1 x 10-26 J

Solution

Question 8

Electrolyte capacitor

Air capacitor

Mica capacitor

Glass capacitor

Solution

Question 9

0.024 F

0.12 F

0.6 F

0.8 F

Solution

Question 10

Solution

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2.5×10³

2.5×10⁴

2.5×10⁵

3.5×10⁴

1.8X10⁵ m

1.8X10⁴ m

2.5X10⁴ m

2.5X10⁵ m

4 x 10^-10 J

32 x 10^-32 J

16 x 10^-32 J

3.1 x 10^-26 J