Question

In the uniform electric field shown in figure, magnitude of $V_C-V_D$ is :

• A. 10 V
• B. 40 V
• C. 30 V
• D. 20 V

Answer 1

The correct option is D 20 V

$|V_C-V_D|=E\times 1=20V$
As $\overrightarrow{E}$ is directed from point A to B, point A is at a higher potential than point B.
planes perpendicular to an uniform electric field are equipotential planes. As B and C lie on the same plane perpendicular to $\overrightarrow{E}$ , they are at same potential.
As A and D lie on the same plane perpendicular to $\overrightarrow{E}$ , they are at same potential.
$|V_C-V_D|=+20V$