Question

Find out the magnitude of electric field intensity at point $(2,0,0)$ due to a dipole of dipole moment $\overrightarrow{P}=\hat{i}+\sqrt{3}\hat{j}$ kept at origin. Also find the potential at that point.

• A. $\left|E\right|=\frac{K}{8},V=\frac{K}{2}$
• B. $\left|E\right|=\frac{K}{2\sqrt{2}},V=\frac{7K}{8}$
• C. $\left|E\right|=\frac{\sqrt{7}K}{8},V=\frac{K}{4}$
• D. $\left|E\right|=\frac{K}{2},V=\frac{K}{4}$

Answer 1

The correct option is C $\left|E\right|=\frac{\sqrt{7}K}{8},V=\frac{K}{4}$

Given,
the dipole moment $\overrightarrow{P}=\hat{i}+\sqrt{3}\hat{j}$

So, the dipole moment along $\text{x}-$ axis,
${\overrightarrow{P}}_x=\hat{i}$
So, the dipole moment along $\text{y}-$ axis,
${\overrightarrow{P}}_y=\sqrt{3}\hat{j}$


Position vector, $\overrightarrow{r}=2\hat{i}+0\hat{j}+0\hat{k}$

$\therefore r=\sqrt{2^2+0^2+0^2}=2$

As electric field intentsity at any axial point due a dipole,
${\overrightarrow{E}}_x=\frac{2k{P}_x}{r^3}[k=\frac{1}{4\pi {\epsilon }_0}]$

Substituting the values,
$\Rightarrow {\overrightarrow{E}}_x=\frac{2k\times \hat{i}}{2^3}=\frac{k\hat{i}}{4}$

similarly, electric field along equitorial line is,

${\overrightarrow{E}}_y=\frac{k{P}_y}{r^3}=\frac{k\sqrt{3}}{8}\hat{j}$

So, the magnitude of net electric field due the given dipole,

$\left|E\right|=\sqrt{E_x^2+E_y^2}$

$\Rightarrow \left|E\right|=\sqrt{\frac{k^2}{16}+\frac{3k^2}{64}}$ $=\frac{k\sqrt{7}}{8}\text{units}$

Now potential at point $\text{A}$ is given by,
$V_A=\frac{k\overrightarrow{P}\cdot \hat{r}}{r^2}$

$\therefore V_A=\frac{k{P}_x}{r^2}$

Substituting the values,
$\Rightarrow V_A=\frac{k\times 1}{4}=\frac{k}{4}\text{units}$

Hence, option (c) is the correct answer.

Why this Question ? Tip : At point $\text{A}$ the net potential is due to dipole component $P_x$ and $P_y$. The orientation of $\text{A}$ is equitorial for $P_y$ hence $V=Ed=0$. Thus $V_A$ is only due to axial position of $\text{A}$ with respect to $P_x$.