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A uniform electric field of 20 N/C exists along the x-axis in space. The potential difference $(V_{B} - V_{A})$ for the points $A (4 m, 2 m)$ and $B (6 m, 5 m)$ is

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$20 \sqrt{13} v$

-40 V

+40V

0 V

answer is B.

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Detailed Solution

$V_{B} - V_{A} = - [\bar{E} . (\bar{r_{B}} - \bar{r_{A}})]$

$= - [20 \hat{i} . (2 \hat{i} + 3 \hat{j})]$

$= - 40 V$

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