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Hint: Equipotential surfaces are surfaces with equal potential throughout their surface. For a point charge these equipotential surfaces appear around as concentric spherical shells.
Complete step by step answer:
An equipotential surface is a surface over which potential is the same at every point.
The figure given below depicts the equipotential surfaces formed due to an isolated point charge +q and Electric field coming out of it.
Properties of equipotential surfaces are:
1. The direction of electric field is always perpendicular to the equipotential surface coming from a region of high potential to a region of low potential.
2. No two equipotential surfaces can intersect each other because at the point of intersection there would be two different values of potential which is not possible.
The equipotential surfaces are close together as the distance between the surface and source charge increases because of a strong electric field near the source charge.
We know that the relation between the electric field and the potential is given by:
$E=\dfrac{-dV}{dr}$
Since for equipotential surfaces
$\text{dV=constant}$
We have
$E\propto \dfrac{1}{dr}$
Which depicts that as the distance r decreases the Electric field strength increases and therefore the equipotential surfaces come close together near the source charge.
Note: Students should note that the work done in moving a unit charge over an equipotential surface is zero.
Complete step by step answer:
An equipotential surface is a surface over which potential is the same at every point.
The figure given below depicts the equipotential surfaces formed due to an isolated point charge +q and Electric field coming out of it.
Properties of equipotential surfaces are:
1. The direction of electric field is always perpendicular to the equipotential surface coming from a region of high potential to a region of low potential.
2. No two equipotential surfaces can intersect each other because at the point of intersection there would be two different values of potential which is not possible.
The equipotential surfaces are close together as the distance between the surface and source charge increases because of a strong electric field near the source charge.
We know that the relation between the electric field and the potential is given by:
$E=\dfrac{-dV}{dr}$
Since for equipotential surfaces
$\text{dV=constant}$
We have
$E\propto \dfrac{1}{dr}$
Which depicts that as the distance r decreases the Electric field strength increases and therefore the equipotential surfaces come close together near the source charge.
Note: Students should note that the work done in moving a unit charge over an equipotential surface is zero.
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