An electric dipole is placed along the x-axis | vedclass

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(B) Let the angle made by the position vector $OP$ with the dipole moment vector $\vec{p}$ (along the $x$-axis) be $\phi = \frac{\pi}{3}$.The angle $\

Vedclass · Accepted Answer

$\frac{\pi}{3} + 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$

Vedclass · Answer

(B) Let the angle made by the position vector $OP$ with the dipole moment vector $\vec{p}$ (along the $x$-axis) be $\phi = \frac{\pi}{3}$.The angle $\alpha$ between the electric field vector $\vec{E}$ and the position vector $OP$ is given by the relation $	an \alpha = \frac{1}{2} 	an \phi$.Substituting $\phi = \frac{\pi}{3}$,we get $	an \alpha = \frac{1}{2} 	an \frac{\pi}{3} = \frac{1}{2} 	imes \sqrt{3} = \frac{\sqrt{3}}{2}$.Therefore,$\alpha = 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$.The angle $	heta$ that the electric field makes with the $x$-axis is the sum of the angle of the position vector with the $x$-axis and the angle $\alpha$ between the electric field and the position vector.Thus,$	heta = \phi + \alpha = \frac{\pi}{3} + 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$.

Column-$I$ (Angle between $\vec{p}$ and $\vec{E}$)	Column-$II$ (Potential energy of the dipole)
$a. 180^{\circ}$	$i. -pE$
$b. 120^{\circ}$	$ii. pE$
$c. 90^{\circ}$	$iii. \frac{1}{2} pE$
	$iv. 0$

An electric dipole is placed along the $x$-axis with its center at the origin. The line $OP$ makes an angle of $\frac{\pi}{3}$ with the $x$-axis. If the electric field at point $P$ makes an angle $\theta$ with the $x$-axis,then $\theta$ is equal to:

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