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  4. An electric dipole of moment p = (- hati-3 hatj+2 hatk)×10-29 C.m is at the origin (0,0,0). The electric field due to this dipole at r = + hati + 3 hatj + 5 hatk (note that r ⋅ p = 0 ) is parallel to :

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Q. An electric dipole of moment $\overrightarrow{p} = \left(-\hat{i}-3\hat{j}+2\hat{k}\right)\times10^{-29}$ 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 :

JEE MainJEE Main 2020Electric Charges and Fields

Solution:

$\overrightarrow{ P }=(-\hat{1}-3 \hat{ j }+2 \hat{ k }) \times 10^{-29} cm$
position vector $\vec{r}=\hat{i}+3 \hat{\jmath}+5 \hat{k}$
$\vec{p} \cdot \vec{r}=(-\hat{1}-3 \hat{j}+2 \hat{k}) \cdot(\hat{1}+3 \hat{j}+5 \hat{k})$
$=-1-9+10=0$
$\therefore$ 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{\jmath}+2 \hat{k}) \times 10^{-29} $
$ =\lambda(\hat{i}+3 \hat{j}-2 \hat{k}) \times 10^{-29} $
$\therefore \vec{E}$ is parallel to $\hat{i}+3 \hat{j}-2 \hat{k}$