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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 $\vec{r_{A}} = \hat{i} + 2 \hat{j}$ and $\vec{r_{B}} = 2 \hat{i} + \hat{j} + 3 \hat{k} .$

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$- 1 V$

$1 V$

$2 V$

$3 V$

answer is A.

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Detailed Solution

$V_{B} - V_{A} = - [\int_{1}^{2} 2 d x + \int_{2}^{1} 3 d y] = - [2 (2 - 1) + 3 (1 - 2)] = - [2 - 3] = 1 V$

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