Question

An electric dipole is placed at the origin O such that its equator is y-axis. At a point P far away from the dipole, the electric field direction is along the y-direction. OP makes an angle $\alpha$ with the x-axis such that:
A. $\tan \alpha =\sqrt{3}$
B. $\tan \alpha =\sqrt{2}$
C. $\tan \alpha =1$
D. $\tan \alpha =\dfrac{1}{\sqrt{2}}$

Answer 1

Hint : Electric field, at a point making $\theta$ angle with the centre of dipole if make angle $\phi$ with the position vector OP
Then $\tan \phi =\dfrac{1}{2}\tan \phi$

Complete Step By Step Answer:
Let assume the resulting electric field at point P makes angle $\theta$ with $\overrightarrow{OP}$ .

In question we are given that ${{E}_{P}}$ is parallel to y-axis which mean from figure we can state that angle $\phi +\alpha ={{90}^{\circ }}$ because angle between $\overrightarrow{OP}$ and x-axis is $\alpha$ we also know that if A resultant electric field makes angle $\theta$ with the position vector and if positive vector makes angle $\alpha$ with the dipole’s direction then $\tan \theta =\dfrac{1}{2}\tan \alpha$
because $\theta +\alpha ={{90}^{\circ }}\Rightarrow \theta =\left( {{90}^{\circ }}-\alpha \right)$
$\tan \left( 90-\alpha \right)=\dfrac{1}{2}\tan \alpha$
$\Rightarrow \cot \alpha =\dfrac{1}{2}\tan \alpha$
$\Rightarrow \dfrac{01}{\tan \alpha }=\dfrac{1}{2}\tan \alpha$
$\Rightarrow {{\tan }^{2}}\left( \alpha \right)=2$
$\Rightarrow \tan \alpha =\sqrt{2}$ .

Note :
Magnitude of resultant electric field at a point P whose position vector with centre of dipole if makes angle $\theta$ with direction dipole moment. then
$\left| \overrightarrow{{{E}_{P}}} \right|=\dfrac{2kp\sqrt{3{{\cos }^{2}}\theta H}}{{{r}^{3}}}$
angle $\theta$ is not the angle of resultant electric field. Students should take care of it because the resultant electric field makes a new angle with the position vector of point P and centre of dipole.