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Test: Potential due to a Point Charge & electric Dipole(4 Nov) - JEE MCQ


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10 Questions MCQ Test Daily Test for JEE Preparation - Test: Potential due to a Point Charge & electric Dipole(4 Nov)

Test: Potential due to a Point Charge & electric Dipole(4 Nov) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Test: Potential due to a Point Charge & electric Dipole(4 Nov) questions and answers have been prepared according to the JEE exam syllabus.The Test: Potential due to a Point Charge & electric Dipole(4 Nov) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Potential due to a Point Charge & electric Dipole(4 Nov) below.
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Test: Potential due to a Point Charge & electric Dipole(4 Nov) - Question 1
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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(4 Nov) - 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(4 Nov) - Question 2
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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(4 Nov) - 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(4 Nov) - Question 3
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What is the electric potential of the system?

Detailed Solution for Test: Potential due to a Point Charge & electric Dipole(4 Nov) - 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(4 Nov) - Question 4
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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(4 Nov) - 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(4 Nov) - Question 5
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Electric field intensity is equal to
Detailed Solution for Test: Potential due to a Point Charge & electric Dipole(4 Nov) - Question 5

Definition based

Test: Potential due to a Point Charge & electric Dipole(4 Nov) - Question 6
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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(4 Nov) - 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(4 Nov) - Question 7
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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(4 Nov) - 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(4 Nov) - Question 8
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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(4 Nov) - 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(4 Nov) - Question 9
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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(4 Nov) - 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(4 Nov) - Question 10
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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(4 Nov) - 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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