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(B) Let the angle made by the resultant electric field vector with the radial vector be $\alpha$. The relation between the angle of the position vecto

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$	an 	heta = \sqrt{2}$

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(B) Let the angle made by the resultant electric field vector with the radial vector be $\alpha$. The relation between the angle of the position vector $	heta$ and the angle of the electric field $\phi$ with the radial direction is given by $	an \alpha = \frac{1}{2} 	an 	heta$.The electric field is along the $y$-axis,and the position vector $OP$ makes an angle $	heta$ with the $x$-axis. The angle of the electric field with the $x$-axis is $90^{\circ}$.Therefore,the angle $\alpha$ between the electric field and the radial vector is $\alpha = 90^{\circ} - 	heta$.Substituting this into the formula: $	an(90^{\circ} - 	heta) = \frac{1}{2} 	an 	heta$.Since $	an(90^{\circ} - 	heta) = \cot 	heta$,we have $\cot 	heta = \frac{1}{2} 	an 	heta$.This simplifies to $	an^2 	heta = 2$,or $	an 	heta = \sqrt{2}$.

An electric dipole is placed on the $x$-axis with its center at the origin. The line $OP$ makes an angle $\theta$ with the $x$-axis. If the electric field at point $P$ is in the direction of the $y$-axis,what is the value of $\theta$?

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