Question

An electric dipole of moment vector $p=\left(-\hat{i}-3\hat{j}+2\hat{k}\right)\times {10}^{-29}cm$ is at the origin $\left(0,0,0\right)$. The electric field due to this dipole at$\begin{array}{l}\overrightarrow{r}=(\hat{i}+3\hat{j}+5\hat{k})\end{array}$ is parallel to [Note that $\overrightarrow{r}.\overrightarrow{p}=0$]

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

Answer 1

The correct option is C

$\hat{i}+3\hat{j}-2\hat{k}$

Step 1. Given data

An electric dipole of the moment, $p=\left(-\hat{i}-3\hat{j}+2\hat{k}\right)\times {10}^{-29}cm$

Position vector, $\overrightarrow{r}=\hat{i}+3\hat{j}+5\hat{k}$

$\overrightarrow{r}.\overrightarrow{p}=0$

Step 2. Finding the electric field, $E$

$\overrightarrow{r}.\overrightarrow{p}=0$ This means the field is along the tangential direction and the dipole is perpendicular to the radius vector. Since the electric field and dipole are along the same line, we can write

$\overrightarrow{E}=\lambda \left(\overrightarrow{p}\right)$ [Where $\lambda$is an arbitrary constant]

$\overrightarrow{E}=\left(-\hat{i}-3\hat{j}+2\hat{k}\right)\times {10}^{-29}\times \left(-1\times {10}^{29}\right)$

$\overrightarrow{E}=\hat{i}+3\hat{j}-2\hat{k}$

Therefore, the correct option is $\left(C\right)$.