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Electrostatics MCQ Level - 2 - Physics MCQ


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10 Questions MCQ Test Topic wise Tests for IIT JAM Physics - Electrostatics MCQ Level - 2

Electrostatics MCQ Level - 2 for Physics 2024 is part of Topic wise Tests for IIT JAM Physics preparation. The Electrostatics MCQ Level - 2 questions and answers have been prepared according to the Physics exam syllabus.The Electrostatics MCQ Level - 2 MCQs are made for Physics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Electrostatics MCQ Level - 2 below.
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Electrostatics MCQ Level - 2 - Question 1

The electric field inside a sphere which carries a volume charge density proportional to the distance from the origin ρ = α r, (α is a constant) is:

Detailed Solution for Electrostatics MCQ Level - 2 - Question 1

We can consider all the charge inside the sphere to be concentrated on the center of sphere
Consider an elementary shell of radius x and thickness dx.

The correct answer is:  

Electrostatics MCQ Level - 2 - Question 2

Two concentric uniformly charged spheres of radius 10 cm and 20 cm are arranged as shown in the figure. Potential difference between the spheres is :

Detailed Solution for Electrostatics MCQ Level - 2 - Question 2

Potential difference due to inner 10C charge 

Potential difference due to outer charge 
∴ Potential difference = 

The correct answer is: 4.5 × 1011 V

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Electrostatics MCQ Level - 2 - Question 3

A neutral conducting spherical shell is kept near a charge q as shown. The potential at point P due to the induced charges is

Detailed Solution for Electrostatics MCQ Level - 2 - Question 3

Electrostatics MCQ Level - 2 - Question 4

An infinitely long plate has surface charge density ρ. As shown in the figure, a point charge q is moved from A to B. Net work done by electric field is:

Detailed Solution for Electrostatics MCQ Level - 2 - Question 4

Electrostatics MCQ Level - 2 - Question 5

A dipole of dipole moment p is kept at the centre of a ring of radius R and charge Q. The dipole moment has direction along the axis of the ring. The resultant force on the ring due to the dipole is:

Detailed Solution for Electrostatics MCQ Level - 2 - Question 5

Electric field due to ring at point P on its axis at a distance x from centre O of ring is

The correct answer is:  

Electrostatics MCQ Level - 2 - Question 6

In the figure shown. A is a fixed charge. B (of mass m) is given a velocity v perpendicular to line AB. At this moment the radius of curvature of the resultant path of B is

Detailed Solution for Electrostatics MCQ Level - 2 - Question 6



Electrostatics MCQ Level - 2 - Question 7

Figure shows three circular arcs, each of radius R and total charge as indicated. The net electric potential at the centre of curvature is :

Detailed Solution for Electrostatics MCQ Level - 2 - Question 7

V = V1 + V2 + V3
(where  V1, V2  and  V3  are potentials due to the three parts of ring)


The correct answer is: 

Electrostatics MCQ Level - 2 - Question 8

A conducting disc of radius R rotates about its axis with an angular velocity ω. Then the potential difference between the centre of the disc and its edge is (no magnetic field is present): 

Detailed Solution for Electrostatics MCQ Level - 2 - Question 8

 (Balancing of forces of e–)


Also, 

Electrostatics MCQ Level - 2 - Question 9

Two identical small balls each have a mass m and charge q. When placed in a hemispherical bowl of radius R with friction less, non-conducting walls, the beads move and at equilibrium, the line joining the balls is horizontal and the distance between them is R (figure). Neglect any induced charge on the hemispherical bowl. Then the charge on each bead is: (here, 

Detailed Solution for Electrostatics MCQ Level - 2 - Question 9

The bowl exerts a normal force N on each bead, directed along the radial line or at 60° above the horizontal. Consider the free-body diagram of the bead on the left with the electric force Fe  applied.


The correct answer is: 

Electrostatics MCQ Level - 2 - Question 10

Two positively charged particles of charges q1 and q2 have mass m each. A uniform electric field having magnitude E exists in positive x direction as shown in the figure. The given two charged particles are released from rest at t = 0 as shown in the figure. If position of q1 at t = 2sec is given by coordinate (+2a, 0) then the x-coordinate of q2 at t = 2sec is (neglect gravitational interaction between the particles)

Detailed Solution for Electrostatics MCQ Level - 2 - Question 10

The acceleration of centre of mass of system of particles is


∴   x-coordinate of centre of mass at t = 2 second is

Let the x-coordinates of q1 and q2 at t = 2 sec be x1 and x2 ; [x1 = 2a at t = 2 sec.]

The correct answer is: 

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