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The solutions to many problems involving electric fields are simplified by making use of equipotential surfaces. An equipotential surface is a surface: 1. On which the potential is the same everywhere 2. The movement of charge over such a surface would require no work 3. The tangential electric field is zero 4. The normal electric field is zero Which of the above statements are correct?

The solutions to many problems involving electric fields are simplified by making use of equipotential surfaces. An equipotential surface is a surface: 1. On which the potential is the same everywhere 2. The movement of charge over such a surface would require no work 3. The tangential electric field is zero 4. The normal electric field is zero Which of the above statements are correct? Correct Answer 1, 2 and 3 only

Explanation:

  • The surface which is the locus of all points which are at the same potential is known as the equipotential surface.
  • No work is required to move a charge from one point to another on the equipotential surface.

Example of the equipotential surface is shown below.

EQUIPOTENTIAL DIAGRAM

Work done

The work done in moving a charge between two points in an equipotential surface is zero.

If a point charge is moved from point VA to VB, in an equipotential surface, then the work done in moving the charge is given by

W = q0(VA –VB)

As VA – VB is equal to zero, the total work done is W = 0.

Properties

  1. The electric field is always perpendicular to an equipotential surface.
  2. Two equipotential surfaces can never intersect.
  3. For a point charge, the equipotential surfaces are concentric spherical shells.
  4. For a uniform electric field, the equipotential surfaces are planes normal to the x-axis
  5. The direction of the equipotential surface is from high potential to low potential.
  6. In a uniform electric field, any plane normal to the field direction is an equipotential surface.
  7. The spacing between equipotential surfaces enables us to identify regions of a strong and weak field i.e. E = −dV/dr ⇒ E ∝ 1/dr

 

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