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(B) The electric field $\vec{E}$ at a point $P$ due to an electric dipole is given by its radial component $E_r$ and tangential component $E_t$. Let $

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

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(B) The electric field $\vec{E}$ at a point $P$ due to an electric dipole is given by its radial component $E_r$ and tangential component $E_t$. Let $\phi$ be the angle $OP$ makes with the dipole axis (the $x$-axis). Here,$\phi = \frac{\pi}{3}$. The radial component is $E_r = \frac{2kp \cos \phi}{r^3}$ and the tangential component is $E_t = \frac{kp \sin \phi}{r^3}$. 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 $\phi = \frac{\pi}{3}$,we get $	an \alpha = \frac{1}{2} 	an \frac{\pi}{3} = \frac{\sqrt{3}}{2}$. Thus,$\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 $OP$ with the $x$-axis and the angle $\alpha$. Therefore,$	heta = \phi + \alpha = \frac{\pi}{3} + 	an^{-1}\left(\frac{\sqrt{3}}{2}ight)$.

An electric dipole is placed along the $x$-axis at the origin $O$. $A$ point $P$ is at a distance of $20 \ cm$ from this origin such that $OP$ makes an angle $\frac{\pi}{3}$ with the $x$-axis. If the electric field at $P$ makes an angle $\theta$ with the $x$-axis,the value of $\theta$ would be

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