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(B) The angle $\phi$ that the electric field vector $\vec{E}$ makes with the radial vector $\vec{r}$ (the line $OP$) is given by $	an \phi = \frac{E_

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

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(B) The angle $\phi$ that the electric field vector $\vec{E}$ makes with the radial vector $\vec{r}$ (the line $OP$) is given by $	an \phi = \frac{E_{\perp}}{E_{\|}} = \frac{1}{2} 	an \alpha$,where $\alpha$ is the angle the position vector makes with the dipole axis.Here,$\alpha = \frac{\pi}{3}$.Therefore,$	an \phi = \frac{1}{2} 	an \frac{\pi}{3} = \frac{1}{2} \cdot \sqrt{3} = \frac{\sqrt{3}}{2}$.So,$\phi = 	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 $\phi$.Thus,$	heta = \alpha + \phi = \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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