Please wait...

Electrostatic Potential Test - 7
Menu grid icon
Result Result point icon
Electrostatic Potential Test - 7
  • Goals icon

    /

    Score
  • Trophy icon

    -

    Rank
White alarm icon Time Taken: -
Result frame illustration
  • Question 1/10
    1 / -0.25

    For any charge configuration, equipotential surface through a point is _____to the electric field at that point.

    Solutions

    Definition based

  • Question 2/10
    1 / -0.25

    Two equipotential surfaces have a potential of -10V and 90V respectively, what is the difference in potential between these surfaces?

    Solutions

    Potential = high potential - low potential
    so our high potential is 90V and low potential is 10V,
    potential = 90-(-10)=100V

  • Question 3/10
    1 / -0.25

    Optical analogue of an equipotential surface is

    Solutions

    Definition based

  • Question 4/10
    1 / -0.25

    Electric field is steepest in the direction in which the potential

    Solutions

    In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of the electric field.

  • Question 5/10
    1 / -0.25

    Equipotential surfaces cannot

    Solutions

    The electric field lines are perpendicular to the equipotential surface. The field lines can not intersect each other because the electric force can not have two directions at a point.

  • Question 6/10
    1 / -0.25

    The shape of equipotential surface for an infinite line charge is:​

    Solutions

    The shape of equipotential surface for an infinite line charge is coaxial cylindrical because A curved surface on which potential is constant is equipotential curve . If we consider the line charge then the focus of the point should have the same potential hence it is a coaxial cylinder.

  • Question 7/10
    1 / -0.25

    When a positive charge is moved in an electrostatic field from a point at high potential to a low potential, its kinetic energy

    Solutions

    We can generalise it by assuming a positive charge moving away from another positive charge. now both of them are repelling each other with some force. so that positive charge will accelerate which results in the increase in K.E.

  • Question 8/10
    1 / -0.25

    Shape of equipotential surfaces for a uniform electric field along x-axis are

    Solutions


    Shape of equipotential surfaces for a uniform electric field along x-axis are Planes normal to the field.

  • Question 9/10
    1 / -0.25

    The value of electric field vector  along the surface of constant potential of 100 V

    Solutions

    As we know that electric field is in the reverse sense of the directional gradient of electric potential, so if V is constant, E should be zero. So, for E to be zero, either V has to be zero, or constant, or the 3 directional derivative components of V must cancel out each other in space. So we can not claim that V always has to be zero, if E is zero in a region, since E is a vector &V is a scalar quantity. But if V is zero, then surely E has to be zero; the reverse case is not always true.
     

  • Question 10/10
    1 / -0.25

    The amount of work done in moving a unit positive charge through distance of 10 cm on an equipotential surface is

    Solutions

    Definition based

Close button icon
User Profile
-

Correct (-)

Wrong (-)

Skipped (-)

  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10
Mockers logo Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Arrow pointer icon
Click on Allow to receive notifications
Notification bell icon ×
Open Now