An electric field is expressed as E→=2i^+3j^ V/m . The potential difference VA−VB between two points A and B is n×−0.25V whose position vectors are given by rA=i^+2j^ m and rB=2i^+j^+3k^ m. Find n.

dV= −E→.dr→ so, VA−VB=E→.(rB→−rA→)=2i^+3j^.i^-j^+3k^=-1VSo, n=4

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An electric field is expressed as $\vec{E} = 2 \hat{i} + 3 \hat{j} V / m$ . The potential difference $V_{A} - V_{B}$ between two points A and B is $n \times - 0.25 V$ whose position vectors are given by $r_{A} = \hat{i} + 2 \hat{j} m$ and $r_{B} = 2 \hat{i} + \hat{j} + 3 \hat{k} m$ . Find n.

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answer is 4.

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$d V = - \vec{E} . \vec{d r}$

so, $V_{A} - V_{B} = \vec{E} . (\vec{r_{B}} - \vec{r_{A}})$

$= (2 \hat{i} + 3 \hat{j}) . (\hat{i} - \hat{j} + 3 \hat{k}) = - 1 V$

So, n=4

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