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An electric field is expressed as $\vec{E} = 2 \hat{i} + 3 \hat{j} .$ Find the potential difference $(V_{A} - V_{B})$ between two points A and B whose position vectors are given by $r_{A} = \hat{i} + 2 \hat{j} a n d r_{B} = 2 \hat{i} + \hat{j} + 3 \hat{k} .$

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3 V

-1 V

2 V

1 V

answer is A.

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Detailed Solution

$V_{B} - V_{A} = - \int_{1}^{2} 2 d x + \int_{2}^{1} 3 d y$

$= - [2 (2 - 1) + 3 (1 - 2)]$

$= - [2 - 3] = 1 V$

Hence, $V_{A} - V_{B} = - 1 V$

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