Question

The given figure shows a set of equipotential surfaces. Find the magnitude and direction of electric field that exists in the region.

Answer 1

The electric field $E$ will be perpendicular to the equipotential surfaces as shown (i.e. at an angle ${45}^{\circ }$ with the positive x-axis).

We know,

$E=\left|\frac{-dv}{dL}\right|=\frac{v_2-v_1}{dcos{45}^0}$

Putting all the values

$E=\frac{40-30}{1\times cos{45}^0}=10\sqrt{2}$ V/m

Hence the magnitude of the electric field is $10\sqrt{2}$ V/m