Question

Calculate the potential difference between points $\text{A}$ and $\text{B}$ in an electric field $\overrightarrow{E}=(2\hat{i}+3\hat{j}+4\hat{k}){\text{NC}}^{-1}$, where the position vectors of points $A$ and $B$ are ${\overrightarrow{r}}_A=(\hat{i}-2\hat{j}+\hat{k})\text{m}$ and ${\overrightarrow{r}}_B=(2\hat{i}+\hat{j}-2\hat{k})\text{m}$ respectively.

• A. $-4\text{V}$
• B. $-3\text{V}$
• C. $-2\text{V}$
• D. $-1\text{V}$

Answer 1

The correct option is D $-1\text{V}$

Given,
Position vector of point $\text{A}$,
${\overrightarrow{r}}_A=(\hat{i}-2\hat{j}+\hat{k})\text{m}$
Position vector of point $\text{B}$,
${\overrightarrow{r}}_B=(2\hat{i}+\hat{j}-2\hat{k})\text{m}$
The relation between electric field ($\overrightarrow{E}$) and electric potential difference $\left(dV\right)$ is given by
$dV=-\overrightarrow{E}.\overrightarrow{dr}$
where, $\overrightarrow{dr}=dx\hat{i}+dy\hat{j}+dz\hat{k}$

The potential difference between point $\text{A}$ and $\text{B}$ is
$\Rightarrow \underset{B}{\overset{A}{\int }}dV=-\underset{B}{\overset{A}{\int }}\overrightarrow{E}.\overrightarrow{dr}$

$\Rightarrow \underset{B}{\overset{A}{\int }}dV=-\underset{(2,1,-2)}{\overset{(1,-2,1)}{\int }}(2\hat{i}+3\hat{j}+4\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$

$\Rightarrow \underset{B}{\overset{A}{\int }}dV=-\underset{(2,1,-2)}{\overset{(1,-2,1)}{\int }}(2dx+3dy+4dz)$

$\Rightarrow V_A-V_B=-\left[\right[2x{]}_2^1+[3y{]}_1^{-2}+[4z{]}_{-2}^1]$

$\Rightarrow V_A-V_B=-\left[2\right(1-2)+3(-2-1)+4(1+2\left)\right]$

$\Rightarrow V_A-V_B=-[-2-9+12]=-1\text{V}$

Hence, option (d) is correct answer.