Question

An electric field is expressed as $\overrightarrow{E}=2\hat{i}+3\hat{j}$. Find the potential difference $(V_A-V_B)$ between two points A and B whose position vectors are given by $r_A=\hat{i}+2\hat{j}$ and $r_B=2\hat{i}+\hat{j}+3\hat{k}.$

• A. $-1V$
• B. $+1V$
• C. $2V$
• D. $3V$

Answer 1

The correct option is A $-1V$

We know that $dV=-E.dr$
$dV=-E.(dx\hat{i}+dy\hat{j}+dz\hat{k})$
$dV=-(2\hat{i}+3\hat{j}$).($dx\hat{i}+dy\hat{j}+dz\hat{k})$
If we consider the intial position of $r_a$ and final position be $r_b$ then,
${\int }_{V_a}^{V_b}dV=-2{\int }_1^{2}dx+-3{\int }_2^{1}dy$
$⟹V_b-V_a$ = $1$
$\therefore V_a-V_b=-1$