Electrostatics Test

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Electrostatics Test - 15

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Weekly Quiz Competition

Question 1
1 / -0

Electric potential due to a point charge q at a distance r from the point is _______ (in the air).

A
q/r

B
q*r

C
q/r²

D
-q/r

Solution

Force on a unit point charge kept at a distance r from the charge = q/r².
Therefore, work done to bring that point charge through a small distance dr = q/r² * (-dr). Therefore, the potential of that point is = .
Question 2
1 / -0

Calculate electric potential due to a point charge of 10C at a distance of 8cm away from the charge.

A
1.125*10¹³V

B
1.125*10¹²V

C
2.25*10¹³V

D
0.62*10¹³V

Solution

In the SI system, electric potential due to a point charge at a distance r is . Substituting the values, we get potential = V = 1.125*10¹²V. Though in practice, this huge value of electric potential is not present.
Question 3
1 / -0

What is the amount of work done to bring a charge of 4*10^-3C charge from infinity to a point whose electric potential is 2*10²V?

A
0.8J

B
-0.8J

C
1.6J

D
-0.4J

Solution

Work done = potential*charge by definition. We know that the potential of a point is the amount of work done to bring a unit charge from infinity to a certain point. Therefore, work done W=q*V=4*10^-3*200J=0.8J. The work done is positive in this case.
Question 4
1 / -0

Two point charges 10C and -10C are placed at a certain distance. What is the electric potential of their midpoint?

A
Some positive value

B
Some negative value

C
Zero

D
Depends on medium

Solution

Electric potential is a scalar quantity and its value is solely dependent on the charge near it and the distance from that charge. In this case, the point is equidistant from the two point charges and the point charges have the same value but opposite nature. Therefore equal but opposite potentials are generated due to the charges and hence the net potential at midpoint becomes zero.
Question 5
1 / -0

What is the electric potential of the system?

A

B

C
Zero

D

Solution

Electric potential between +q and +q isand electric potential between two charges +q and –q is . Therefore the electric potential of the system is Electric potential is a scalar quantity and thus the system becomes zero-potential-system.
Question 6
1 / -0

A small charge q is rotated in a complete circular path of radius r surrounding another charge Q. The work done in this process is _________

A
Zero

B

C

D

Solution

We know that electric potential does not depend on the path, i.e. it is a state function. Therefore if we rotate a charge around another in a complete circular path, the potential energy of its initial and final points is the same. Therefore, there is no change of potential energy of the charge and hence net work done on the charge is zero.
Question 7
1 / -0

Two plates are kept at a distance of 0.1m and their potential difference is 20V. An electron is kept at rest on the surface of the plate with lower potential. What will be the velocity of the electron when it strikes another plate?

A
1.87*10⁶ m/s

B
2.65*10⁶ m/s

C
7.02*10¹² m/s

D
32*10^-19 m/s

Solution

Increase in potential energy of the electron when it strikes another plate = Potential difference*charge of electron=20*1.6*10^-19 J=3.201*10^-18J and as the electron was at rest initially, this energy will be converted to kinetic energy completely. Therefore, 0.5*mass of electron*(velocity)²=3.201*10^-18J. Substituting m=9.11*10^-31kg, we get v=2.65*10⁶ m/s.
Question 8
1 / -0

Electric field intensity and electric potential at a certain distance from a point charge is 32 N/C and 16 J/C. What is the distance from the charge?

A
50 m

B
0.5 m

C
10 m

D
7 m

Solution

Electric field due to a charge q at a distance r is and electric potential at that point is Therefore, and Dividing these two equations, we get r=0.5m. If the medium is air, k=1. Thus we can get the value of q=0.89 C.
Question 9
1 / -0

Three charges –q, Q and –q are placed in a straight line maintaining equal distance from each other. What should be the ratio q/Q so that the net electric potential of the system is zero?

A
1

B
2

C
3

D
4

Solution

Let the distance between any two charges is d. Therefore the net potential energy of the system is But the total energy of the system is zero. So, -qQ-qQ+ q²/2
=0 that means q=4Q, i.e. q/Q=4.