The electric field at point A due to an elect | vedclass

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(D) Let $\alpha$ be the angle that the resultant electric field vector makes with the position vector $r$ of point $A$.The angle between the dipole mo

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$	an^{-1}(\sqrt{2})$

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(D) Let $\alpha$ be the angle that the resultant electric field vector makes with the position vector $r$ of point $A$.The angle between the dipole moment $p$ and the position vector $r$ is $	heta$.The formula for the angle $\alpha$ between the resultant electric field and the position vector is given by $	an \alpha = \frac{	an 	heta}{2}$.Given that the electric field is perpendicular to the dipole moment $p$,the angle between the electric field and the dipole moment is $90^{\circ}$.From the geometry,the angle between the electric field and the position vector $\alpha$ and the angle $	heta$ are related as $\alpha + 	heta = 90^{\circ}$,so $\alpha = 90^{\circ} - 	heta$.Substituting this into the tangent formula: $	an(90^{\circ} - 	heta) = \frac{	an 	heta}{2}$.Since $	an(90^{\circ} - 	heta) = \cot 	heta$,we have $\cot 	heta = \frac{	an 	heta}{2}$.This simplifies to $\frac{1}{	an 	heta} = \frac{	an 	heta}{2}$,which gives $	an^2 	heta = 2$.Therefore,$	an 	heta = \sqrt{2}$,or $	heta = 	an^{-1}(\sqrt{2})$.

The electric field at point $A$ due to an electric dipole with dipole moment $p$ is perpendicular to the dipole moment vector $p$. The angle $\theta$ is:

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