A uniform electric field of 100 V/m is directed at 30º with the positive x-axis as shown in figure. Find the potential difference V_B – V_A if OA = 2 m and OB = 4 m.

Draw equipotential surface through ‘B’ using geometry  $\perp$ distance from ‘A’ to equipotential surface through B is

$\mathrm{MN}=\left(\frac{4}{\sqrt{3}}+2\right)\times \frac{\sqrt{3}}{2}=(2+\sqrt{3})$

Potential decreases in the direction of the field.

$$\begin{gathered} \Rightarrow {\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}=(2+\sqrt{3})100 \mathrm{V} \\ \Rightarrow {\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=-(2+\sqrt{3})100 \mathrm{V} \end{gathered}$$