Question

Find out the magnitude of electric field intensity, $E$ at point $(2,0,0)$ due to a dipole moment, $\overrightarrow{p}=\hat{i}+\sqrt{3}\hat{j}$ kept at origin as shown in the figure and also find the potential, $V$ at that point.

$[k=\frac{1}{4\pi {\epsilon }_0}]$

• A. $E=\frac{k}{4},V=\frac{k}{8}$
• B. $E=\frac{\sqrt{7}k}{8},V=\frac{k}{4}$
• C. $E=\frac{8k}{3},V=\frac{5k}{2}$
• D. $E=\frac{k}{3},V=\frac{3k}{2}$

Answer 1

The correct option is B $E=\frac{\sqrt{7}k}{8},V=\frac{k}{4}$

Diagram for the problem:


Magnitude of the dipole moment, $p=\sqrt{1^2+(\sqrt{3}{)}^2}=2\text{Cm}$

Distance of point from origin , $r=2\text{m}$

Angle of $\overrightarrow{r}$ from the axis of dipole :

$\tan \theta =\frac{\sqrt{3}}{1}$

$\Rightarrow \theta ={60}^{\circ }$

Formula for net electric field $$E=\frac{kp}{r^3}\sqrt{1+3{\cos }^2\theta }$$Substituting the values,

$E=\frac{k\times 2}{2^3}\sqrt{1+3{\cos }^2{60}^{\circ }}$

$\Rightarrow E=\frac{k}{4}\sqrt{1+\frac{3}{4}}=\frac{\sqrt{7}k}{8}$

Formula for potential $(r,\theta )$
$$V=\frac{kp\cos \theta }{r^2}$$
$\Rightarrow V=\frac{k\times 2}{2^2}\cos {60}^{\circ }$

$\therefore V=\frac{k}{4}$

Hence, option (b) is the correct answer.