Some equipotential surfaces are shown in figure(29.E3) What can you say about the magnitude and the direction of the electric field?

Explanation

First we will find the direction of the `vec(E )` noting that the `vec(E )` lines are always perpendicular to the local equipotential surfaces and points in that direction in which the potential is decreasing . So it will be as shown below.

Note that we have drawn the `vec(E )` lines perpendicular to all equipotential surface and it si pointing in that direction in which the potential is decreasing. so the total angle `vec(E ) ` is making with the positve x-axis is `120^(@)` as shown . .

Now, how to calculate the magnitude of the electric field for this we need to jump from one equipotential surface to another through shortest (perpendicular ) distance
then calculate, E = `("Change in potential")/("distance travelled")`
Let us suppose we are jumping from 20 V line to 10V line from B to A. note that the BA line is the shortest length between the two equipotential surface . the potential difference between the two lines is (20 V-10 V) = 10 volt now, calculate the length AB = 10sin `30^(@)` cm = 5 cm = `5xx 10^(-2)` m
Therefore, E `= (10V)/(5 xx 10^(-2)m)` = 200 [ V/m]
Hence, the complete answer is that the `vec(E )` has the value 200 V/m and makes `120^(@)` with positve x-axis.