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 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.