Test: Potential due to a Point Charge & Electric Dipole (November 17) - NEET MCQ

Test: Potential due to a Point Charge & Electric Dipole (November 17) - NEET MCQ

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Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 1

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

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 1

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 = .

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 2

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?

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 2

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.

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Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 3

What is the electric potential of the system?

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 3

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.

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 4

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?

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 4

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.

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 5

Electric field intensity is equal to

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 5

Definition based

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 6

A charge of 6 mC is located at the origin. The work done in taking a small charge of -2 x 10-9 C from a point P (0, 3 cm, 0) to a Q (0,4 cm, 0) is​

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 6

q=6x10-10 c
Q=-2x10-9c
r1=3x10-2m
r2=4x10-2m
△vQ=w/q
(1/4πεo)-2x10-9/10-2[(1/4)-(1/3)]=W/6x10-3
9x109x2x10-9x6x10-3/12x10-2=W
W=9x10-3x102
W=0.9J

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 7

On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 7

Potential difference between two points is given by

Va - Vb = W/q0

Work, W = 2 J

Charge, q0 = 20 C

Potential difference = 2/20 = 0.1 V

The correct option is C.

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 8

The amount of work done in moving a charge from one point to another along an equipotential line or surface charge is

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 8

Since Potential difference between two points in equipotential surfaces is zero, the work done between two points in equipotential surface is also zero.

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 9

Electric field intensity at point ‘B’ due to a point charge ‘Q’ kept at a point ‘A’ is 12 NC-1 and the electric potential at a point ‘B’ due to same charge is 6 JC-1. The distance between AB is​

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 9

E.l = V   where,
E = electric field intensity = 12 N/C
V = electric potential = 6 J/C
=> distance between A and B,
l = (6 / 12) m or (1 / 2) m = 0.5 m

Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 10

Work done in carrying 2C charge in a circular path of radius 2m around a charge of 10C is​

Detailed Solution for Test: Potential due to a Point Charge & Electric Dipole (November 17) - Question 10

The overall work performed in carrying a 2coulomb charge in a circular orbit of radius 3 m around a charge of 10 coulomb is calculated below.
It is a well-known fact that W=qdv.
Here dV is the change in overall potential. In the circular orbit of r potential at each point is similar.
Most significantly, the value of r is 3.
The value of dv=0 and hence W=q0=0.

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