A uniform electric field of magnitude 325 V/m is directed in the negative y-direction in figure. The coordinates of point A are (-0.2 m, -0.3 m) and those of point B are (0.4 m, 0.5 m). Calculate the potential difference ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}$ (in volts), along the path shown in the figure.

The potential difference between two points A and B in electric field as, ${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=-{\int }_{\mathrm{A}}^{\mathrm{B}} \overrightarrow{\mathrm{E}}\cdot \overrightarrow{\mathrm{d}}\mathrm{r}$

Here the electric field is constant, here we can write the potential difference between points A and B as

${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=-\left[{\mathrm{E}}_{\mathrm{x}}\left({\mathrm{x}}_{\mathrm{B}}-{\mathrm{x}}_{\mathrm{A}}\right)+{\mathrm{E}}_{\mathrm{y}}\left({\mathrm{y}}_{\mathrm{B}}-{\mathrm{y}}_{\mathrm{A}}\right)\right]$

$=-\left[\right(-325\left)\right\{(0.5)-(-0.3)\} +0] \left[\because {\mathrm{E}}_{\mathrm{x}}=0\right]$

$=325\times 0.8=260 \mathrm{V}$

$\Rightarrow {\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=+260 \mathrm{V}$