Question

An electric dipole of moment $\overrightarrow{p}=\left(-\hat{i}-3\hat{j}+2\hat{k}\right)\times 10^{-29}$ C.m is at the origin $\left(0,0,0\right)$. The electric field due to this dipole at $\overrightarrow{r}=+\hat{i}+3\hat{j}+5\hat{k}$
(note that $\overrightarrow{r}\cdot \overrightarrow{p}=0$ ) is parallel to :

• A. $\left(+\hat{i}-3\hat{j}-2\hat{k}\right)$
• B. $\left(-\hat{i}-3\hat{j}+2\hat{k}\right)$
• C. $\left(+\hat{i}+3\hat{j}-2\hat{k}\right)$
• D. $\left(-\hat{i}+3\hat{j}-2\hat{k}\right)$

Answer 1

Answer: (C)

$\overrightarrow{P}=(-\hat{1}-3\hat{j}+2\hat{k})\times 10^{-29}cm$
position vector $\vec{r}=\hat{i}+3\hat{ȷ}+5\hat{k}$
$\vec{p}\cdot \vec{r}=(-\hat{1}-3\hat{j}+2\hat{k})\cdot (\hat{1}+3\hat{j}+5\hat{k})$
$=-1-9+10=0$
$\therefore$ Vectors $\overrightarrow{P}$ and $\overrightarrow{r}$
are perpendicular to each other.
From the figure, electric field at $\vec{r}$ will be anti-parallel to the direction of $\overrightarrow{p}$ (The given point is an axial point and electric field at an axial point is anti-parallel to the direction of $\vec{p}$ ) $\therefore \overrightarrow{E}=-\lambda \overrightarrow{p}$
where $\lambda >0$ $=-\lambda (-\hat{1}-3\hat{ȷ}+2\hat{k})\times 10^{-29}$
$=\lambda (\hat{i}+3\hat{j}-2\hat{k})\times 10^{-29}$
$\therefore \vec{E}$ is parallel to $\hat{i}+3\hat{j}-2\hat{k}$