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Electric Charges and Fields Test - 18

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Electric Charges and Fields Test - 18
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Under the influence of the Coulomb field of charge +Q, a charge -q is moving around it in an elliptical orbit. Find out the correct statements(s)

    Solution

    For a system of two charges under the action of centripetal and centrifugal forces, angular momentum remains constant. This is the same as Kepler’s law of planetary motion.

     

  • Question 2
    1 / -0

    In nature, the electric charge of any system is always equal to

    Solution

    According to quantization of charge, the charge of any system is an integral multiple of the charge of electron which is the least amount of charge on any system.

     

  • Question 3
    1 / -0

    Dielectric constant for metal is

    Solution

    The dielectric constant of metal is infinite. Metals are those materials which have abundance of valance electrons.

     

  • Question 4
    1 / -0

    The electric field inside a spherical shell of uniform charge density is

    Solution

    We know that electric field of an ideal spherical shell with uniformly distributed charge is zero inside the shell and equal to EF of a point charge on its centre.

    When we calculate the EF for a point on the surface of shell, it is equal to half of EF of same point charge EF.

     

  • Question 5
    1 / -0

    The electric field (E) at a point just outside a perfect conductor is

    Solution
    • The electric field is zero inside a conductor. Just outside a conductor, the electric field lines are perpendicular to its surface, ending or beginning on charges on the surface.
    • Any excess charge resides entirely on the surface or surfaces of a conductor.

     

  • Question 6
    1 / -0

    If a unit positive charge is taken from one point to another over an equipotential surface, then

    Solution

    Since the potential at each point of an equipotential surface is the same, the potential does not change while we move a unit positive charge from one point to another. Therefore work done in the process is zero.

     

  • Question 7
    1 / -0

    The magnitude of electric field intensity E is such that, an electron placed in it would experience an electrical force equal to its weight is given by:

    Solution

    Force on electron

    |F| = qE = eE = mg

    E = m/ge

     

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