Some equipotential surfaces are shown in figure(29.E3) 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