For any charge configuration, equipotential surface through a point is normal to the electric field. Justify.

Hint: There are different properties of an equipotential surface. The electric potential of all the points in an equipotential surface are the same. The electric force does not hinder or help the motion of an electric charge across this equipotential surface.

Complete answer:
We have learnt that an equipotential surface is a surface across which the potential is a constant. That means, the work done to move a charge from one point to another point in an equipotential surface is zero. The potential difference across any different points in an equipotential surface is also zero.
We already understood that the work done in moving a charge across an equipotential surface is zero.
That is, the Work done can be given as,
$W = Fs\cos \theta = 0$
Here, $F$ is the electric force and $s$ is the magnitude of displacement
Now we know that, for a non-zero displacement, the $\cos \theta $ should be equal to zero
That is, $\cos \theta = 0$
$\therefore \theta = 90^\circ $
$90^\circ $ means perpendicular.
From this we can understand that in an equipotential surface, the force acting on point charge is perpendicular.
It is known that the direction of electric force on a charge is given by the electric field lines. So we can conclude that, for any charge configuration, equipotential surface through a point is normal to the electric field.
Note: The statement, for any charge configuration, equipotential surface through a point is normal to the electric field can also be justified by using the contradiction method. In that method it is assumed that in an equipotential surface there can be a component of electric field and the work done is taken as non-zero. But it is known that the work done is zero and hence it is wrong. And it is evident that the electric field cannot have a component.