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(B) The electric field of a dipole at a point $(r, \phi)$ in polar coordinates is given by the radial component $E_r = \frac{2kp \cos \phi}{r^3}$ and

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The electric field of a dipole at a point $(r, \phi)$ in polar coordinates is given by the radial component $E_r = \frac{2kp \cos \phi}{r^3}$ and the tangential component $E_t = \frac{kp \sin \phi}{r^3}$.Here,$\phi = \pi/3$ is the angle the position vector $OP$ makes with the dipole axis (the $x$-axis).The angle $\alpha$ that the resultant electric field makes with the radial vector $OP$ is given by $	an \alpha = \frac{E_t}{E_r} = \frac{kp \sin \phi / r^3}{2kp \cos \phi / r^3} = \frac{1}{2} 	an \phi$.Substituting $\phi = \pi/3$,we get $	an \alpha = \frac{1}{2} 	an(\pi/3) = \frac{\sqrt{3}}{2}$,so $\alpha = 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$.The angle $	heta$ that the electric field makes with the $x$-axis is the sum of the angle of the position vector $\phi$ and the angle $\alpha$,so $	heta = \phi + \alpha = \frac{\pi}{3} + 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$.

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