Some equipotentialo surfaces are shown in (Fig. 3.31.) What can you say about the magnitude and the direction of the electric field ?
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Explanation

Here we can say that the electric field will be perpendicular to equipotential surfaces. Also
`|vec E| = (Delta V)/(Delta d)`
where `Delta V` = potential diference between two euipotential surfaces.
`Delta d` = perpendicular distance between two equipotential surfaces.
`|vec E | = (10)/((10 sin 30^(@)) xx 10^-2 ) = 200 Vm^-1`
Now there are two perpendicular directions (1 or 2) as shown in (Fig . 3.32), but since we know that but since we know that electric potential dereases in the direction of electric field, so the correct direction is 2.
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Hence, `E = 200 Vm^-1`, making an angle `120^(@)` with the x - axis.