Figure shows equipotential lines in  $x-y$ plane. Distances  $AB,BC$ and  $CD$ are all equal to  $4cm.$ Find the magnitude of electric field (in V/cm) in the region.

Electric field lines are directed perpendicular to the given equipotential lines. Their direction is as indicated in the figure given below.
Perpendicular distance between two consecutive equipotentials given in the figure is
 $\Delta r=4\sin 30°=2cm$
This implies that when we move 2 cm in the direction of the field, the potential drops by 10V.
 $\left|E\right|=\left|\frac{\Delta V}{\Delta r}\right|=\frac{10V}{2cm}=5Vc{m}^{-1}$

Obviously, direction of electric field makes 60° angle with negative   direction as shown.