Why is the electric field perpendicular to the equipotential surface?

Hint: To define the reason for the given question we should that it is asked in contradiction of the definition of equipotential surface as we know that equipotential surfaces are always perpendicular to the fields.

Complete answer:
Let us first explain the equipotential surfaces, their properties and the relation between electric field and electrostatic potential.
An equipotential surface is a surface with a constant value of the potential at all the points on the surface.
Since to move a charge $q$ from a region of one potential to a region of different potential we know the work done is given by
$W = q({V_B} - {V_A}){\text{J}}$
Where $V$ denotes the potential in two points respectively.
Now in equipotential, the potential is equal in every point therefore the work done to move a point charge from one to another is zero.
The equipotential surfaces are always perpendicular to the direction of the field. This is because if the electric field is not perpendicular to the equipotential surface there would have existed some nonzero component along the surface. Hence to move a charge along the direction of the field some amount of work had to be done. This contradicts the above definition of equipotential that we just derived. Hence the electric field is perpendicular to the equipotential surface.

Note: We should note that for an electric dipole, the equipotential is perpendicular to the axis of the electric dipole. Since the distance of each point on a plane passing through the centre of the electric dipole and perpendicular to the axis of the electric dipole is equal, the potential throughout the plane is zero.