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.