Question

An electric field $\overrightarrow{E}=(2\hat{i}+3\hat{j})N/C$ exists in space. The potential difference $(V_P-V_Q)$ between two points $P$ and $Q$ whose position vectors $\overrightarrow{r_P}=\hat{i}+2\hat{j}$ and $\overrightarrow{r_Q}=2\hat{i}+\hat{j}+\hat{k}$ is

• A. $-1V$
• B. $2V$
• C. $-3V$
• D. $4V$

Answer 1

The correct option is A $-1V$

Position vector $\overrightarrow{r_{PQ}}={\overrightarrow{r}}_Q-{\overrightarrow{r}}_P$
$\therefore$ ${\overrightarrow{r}}_{PQ}=(2\hat{i}+\hat{j}+\hat{k})-(\hat{i}+2\hat{j})=\hat{i}-\hat{j}+\hat{k}$
Potential difference $V_Q-V_P=-\overrightarrow{E}.{\overrightarrow{r}}_{PQ}$
$\therefore$ $V_Q-V_P=-(2\hat{i}+3\hat{j}).(\hat{i}-\hat{j}+\hat{k})=-(2-3)=1V$
$⟹V_P-V_Q=-1V$