Electrostatics Test

Result

Electrostatics Test - 150

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Weekly Quiz Competition

Question 1
4 / -1

Four capacitors are connected in a circuit as shown in the following figure. Calculate the effective capacitance between the points A and B.

A

B

C

D

Solution
Question 2
4 / -1

A parallel plate capacitor with air as the dielectric has capacitance C. A slab of dielectric constant K and having the same thickness as the separation between the plates is introduced so as to fill one-fourth of the capacitor as shown in the figure. The new capacitance will be

A

B

C

D

Solution
Question 3
4 / -1

A

B

C

D

Solution
Question 4
4 / -1

What is the potential difference between points A and B in the circuit shown?

A
2 V

B
4 V

C
3 V

D
12 V

Solution
Question 5
4 / -1

Three capacitors of capacitances 1µF, 2 µF and 4µF are connected first in a series combination and then in a parallel combination. The ratio of their equivalent capacitances will be

A
2 : 49

B
49 : 2

C
4 : 49

D
49 : 4

Solution
Question 6
4 / -1

The equivalent capacitance of the combination of the capacitors is

A
3.20 µF

B
7.80 µF

C
3.90 µF

D
2.16 µF

Solution
Question 7
4 / -1

Each capacitor shown in figure is 2 µF. Then, the equivalent capacitance between points A and B is

A
2 µF

B
4 µF

C
6 µF

D
8 µF

Solution
Question 8
4 / -1

Two plates are 20 cm apart and the potential difference between them is 10V. The electric field between the plates is

A
50 Vm^-1

B
500 Vm^-1

C
0.5 Vm^-1

D
20 Vm^-1

Solution
Question 9
4 / -1

An electron enters the space between the plates of a charged capacitor as shown. The charge density on the plate is σ, electric intensity in the space between the plates is E. A uniform magnetic field B also exists in the space perpendicular to the direction of E. The electron moves perpendicular to both E and B without any change in direction. The time taken by the electron to travel a distance l in the space is

A

B

C

D

Solution
Question 10
4 / -1

A parallel plate capacitor having air as dielectric medium is charged by a potential difference of V volt. After disconnecting the battery, the distance between the plates of the capacitor is increased using an insulated handle. As a result, potential difference between the plates

A
increases

B
does not change

C
becomes zero

D
decreases

Solution