Question

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

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

Answer 1

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

electric field $=100V/m$

The potential difference $V_B-V_A=$

$OA=2m$

$OB=4m$

Electric field in vector form can be written as,

$\overline{E}=(100\cos 30\hat{i}+100\sin {30}^0\hat{j})$

$\overline{E}=50\sqrt{3}\hat{i}+50\hat{j}$

$A=(-2m,0)$

$B=(0,4m)$

$V_{BA}=V_B-V_A=-\int \frac{B}{E}.\overline{dr}$

$={\int }_{(-2,0)}^{(0,4)}(50\sqrt{3}dx+50dy)$

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

$=-(100\sqrt{3}+200)$

$=-100(2+\sqrt{3})$