Question

What is the direction of electric field E at point P in the following diagram:

Answer 1

Hint: Use the fact that: Electric field always moves from higher to the lower potential and perpendicular to the equipotential surface.

Complete step by step answer:
Here, in the given diagram all the lines shown are equipotential surfaces.
Equipotential surface: An equipotential surface is a surface over which the potential is constant everywhere in its plane. Or we can also say that the potential difference between any two points on an equipotential surface is equal to zero.
Below are some important points related to equipotential surfaces:
1. The work done in moving a charge over an equipotential surface is always zero.
2. The direction of the electric field is always perpendicular to an equipotential surface.
3. Any two equipotential surfaces can never intersect, if they could intersect, then at their point of intersection there would have been two values of electric potential which is not possible in real.
So, as we know electric fields move from higher potential to the lower potential and perpendicular to the equipotential surface.
Therefore, at point P, the electric field moves towards left direction and perpendicular to the −10 V surface.

Note:
 The potential difference between any two points can be expressed in terms of electric field as:
\[{V_a} - {V_b} = \int {E.dl} \]
Therefore, Electric field (E) as a function of potential can be expressed as:
\[{\vec E_r} = - \dfrac{{dV}}{{dr}}\]
where \[{\vec E_r}\] is the component of electric field along the direction of \[\dfrac{{dV}}{{dr}}\] which is known as the potential gradient and the negative sign here, infers that electric field acts in a direction of decrease of potential.
 The expression above also signifies that E is not certainly zero if V is zero it will be constant.