Some equipotential surfaces are shown in figure. What can you say about the magnitude and the direction of the electric field ?

Explanation

Here we can say that the electric will be perpendicular to equipotential surfaces.
Also `|vec(E)| = (Delta V)/(Delta d)`
where `Delta V` = potential difference between two equipotential surfaces.
`Delta d` = perpendicular distance between two equipotential surfaces.
So `|vec(E)| = (10)/((10 sin 30^(@)) xx 10^(-2)) = 200 V//m`
Now there are two perpendicular directions either direction 1 or direction 2 as shown in figure, but since we know that in the direction of electric field electric potential decreases so the correct direction is direction 2.
Hence E = 200 V/m, making an angle `120^(@)` with the x-axis.