Question

Find potential difference $V_{AB}$ between two points $\text{A}$ and $\text{B}$ whose position vectors are $r_A=(\hat{i}-2\hat{j}+\hat{k})\text{m}$ and $r_B=(2\hat{i}+\hat{j}-2\hat{k})\text{m}$, in an uniform electric field $E=(2\hat{i}+3\hat{j}+4\hat{k})\text{N/C}$.

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

Answer 1

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

Given,
$r_A=(\hat{i}-2\hat{j}+\hat{k})\text{m}$
$r_B=(2\hat{i}+\hat{j}-2\hat{k})\text{m}$
$E=(2\hat{i}+3\hat{j}+4\hat{k})\text{N/C}$.

Here, the given field is uniform So, we can write

$dV=-E.dr$

Now, the potential difference, between the points $\text{A}$ and $\text{B}$,

$V_{AB}=V_A-V_B$

$V_{AB}=-{\int }_B^{A}E.dr$

Substituting the values in the above equation,

$\Rightarrow V_{AB}=-{\int }_{(2,1,-2)}^{(1,-2,1)}(2\hat{i}+3\hat{j}+4\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$

$\Rightarrow V_{AB}=-{\int }_{(2,1,-2)}^{(1,-2,1)}(2dx+3dy+4dz)$

$\Rightarrow V_{AB}=-[2x+3y+4z{]}_{(2,1,-2)}^{(1,-2,1)}$

$\Rightarrow V_{AB}=-\left[2\right(1-2)+3(-2-1)+4(1-(-2)\left)\right]$

$\therefore V_{AB}=-1\text{V}$

Hence, option (d) is correct answer.