Question

A uniform electric field of $100\text{V/m}$ is directed at ${30}^{\circ }$ with the positive $\text{x-}$axis as shown in figure. Find the potential difference $V_{BA}$ if $\text{OA}=2\text{m}$ and $\text{OB}=4\text{m}$.

• A. $50(2+\sqrt{3})\text{V}$
• B. $-200(2+\sqrt{3})\text{V}$
• C. $-100(2+\sqrt{3})\text{V}$
• D. $-300(2+\sqrt{3})\text{V}$

Answer 1

The correct option is C $-100(2+\sqrt{3})\text{V}$

$\text{Method 1:}$
The given electric field in vector form can be written as
$\overrightarrow{E}=(100\cos {30}^{\circ }\hat{i}+100\sin {30}^{\circ }\hat{j})\text{V/m}$

$\therefore \overrightarrow{E}=(50\sqrt{3}\hat{i}+50\hat{j})\text{V/m}$

Given points, $A=(-2,0,0)\text{m}$
and $B=(0,4,0)\text{m}$

We know that,
$V_{BA}=V_B-V_A=-{\int }_A^{B}E.dr$

Now, substituting the values,

$\Rightarrow V_{BA}=-{\int }_{(-2,0,0)}^{(0,4,0)}(50\sqrt{3}\hat{i}+50\hat{j}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$

$\Rightarrow V_{BA}=-{[50\sqrt{3}x+50y]}_{(-2,0,0)}^{(0,4,0)}$

$\Rightarrow V_{BA}=-\left[50\sqrt{3}\right(0-(-2))+50(4-0\left)\right]\text{V}$

$\therefore V_{BA}=-100(\sqrt{3}+2)\text{V}$

$\text{Method 2:}$
We can also use, $V_B-V_A=-E.d_{AB}$

with the view that $V_A>V_B$ or $V_B-V_A$ will be negative.

From figure,

$d_{AB}=OA\cos {30}^{\circ }+OB\sin {30}^{\circ }$

$\therefore d_{AB}=2\times \frac{\sqrt{3}}{2}+4\times \frac{1}{2}=(\sqrt{3}+2)$

$\therefore V_B-V_A=-E{d}_{AB}=-100(2+\sqrt{3})$

Hence, option (c) is correct answer.