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)|=(DeltaV)/(Deltad)`
Where, `DeltaV=` potential difference between two equipotential surfaces.
`Deltad=` perpendicular distance between two equipotential surfaces.
So `|vec(E)|=60/((10 sin 37^(º))xx10^(-2))=1000 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 kinetic field, electric decreases, so the correct direction is direction 2.
Hence `E=1000 V//m`, making an angle `127^(@)` with the x-axis