Question 258 of 311

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

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Answer: (C) $\left(+\hat{i}+3\hat{j}-2\hat{k}\right)$

As $\vec{r}\cdot \vec{p}=0$, this means that $\vec{r}$ is perpendicular to $\vec{p}$.
For an electric dipole, electric field at equatorial position is antiparallel to $\vec{p}$.
$\Rightarrow \vec{E}=-\lambda \vec{p}$,
where, $\lambda$ is an arbitrary positive constant.
$\Rightarrow \vec{E}=-\lambda \left(-\hat{i}-3\hat{j}+2\hat{k}\right)\times {10}^{-29}C{m}^{-2}$
$\Rightarrow \vec{E}=\lambda \left(\hat{i}+3\hat{j}-2\hat{k}\right)\times {10}^{-29}C{m}^{-2}$
For $\lambda =\dfrac{1}{{10}^{-29}}$
$\Rightarrow \vec{E}=\left(\hat{i}+3\hat{j}-2\hat{k}\right)C{m}^{-2}$