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Electrostatic Potential Test - 2
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Electrostatic Potential Test - 2
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  • Question 1/10
    1 / -0.25

    The minimum velocity v with which charge q should be projected so that it manages to reach the centre of the ring starting from the position shown in figure is

    Solutions

  • Question 2/10
    1 / -0.25

    The work done in moving a positive test charge qo from infinity to a point P at a distance r from the charge q is

    Solutions

    Definition based

  • Question 3/10
    1 / -0.25

    Electric field intensity is equal to

    Solutions

    Definition based

  • Question 4/10
    1 / -0.25

    Electric potential at a point located far away from the charge is taken to be

    Solutions

    Electric potential at a point located far away from the charge is taken to be zero Because the distance is much larger so in other words there is no effect of other charges on that charge.

  • Question 5/10
    1 / -0.25

    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​

    Solutions

    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

  • Question 6/10
    1 / -0.25

    A particle of mass 1 Kg and charge 1/3 μC is projected towards a non conducting fixed spherical shell having the same charge uniformly distributed on its surface. The minimum intial velocity V0 of projection of particle required if the particle just grazes the shell is

    Solutions

    From conservation of angular momentum, 
    mr0 = r/2 = mvr
    ⇒ v = v0/2
    From conservation of energy,

  • Question 7/10
    1 / -0.25

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

    Solutions

    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.

  • Question 8/10
    1 / -0.25

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

    Solutions

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

  • Question 9/10
    1 / -0.25

      A charge is uniformly distributed inside a spherical body of radius r1 = 2r0 having a concentric cavity of radius r2 = r0 (ρ is charge density inside the sphere). The potential of a point P or a distance 3r0/2 from the centre is

    Solutions

    Electric field at a distance r,
    (r0 < r < 2r0) from the center is given by

  • Question 10/10
    1 / -0.25

    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​

    Solutions

    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

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