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

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(C) Let the electric dipole be along the $x$-axis. The electric field $\vec{E}$ at a point $(r, 	heta)$ makes an angle $\alpha$ with the radial vecto

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the electric dipole be along the $x$-axis. The electric field $\vec{E}$ at a point $(r, 	heta)$ makes an angle $\alpha$ with the radial vector $\vec{r}$,given by $	an \alpha = \frac{1}{2} 	an 	heta$.Here,$	heta$ is the angle the position vector makes with the dipole axis.The angle that the electric field vector makes with the dipole axis is $\phi = 	heta + \alpha$.For the electric field to be perpendicular to the dipole axis,the angle $\phi$ must be $90^{\circ}$.Therefore,$	heta + \alpha = 90^{\circ}$,which implies $\alpha = 90^{\circ} - 	heta$.Substituting this into the relation $	an \alpha = \frac{1}{2} 	an 	heta$,we get:$	an(90^{\circ} - 	heta) = \frac{1}{2} 	an 	heta$$\cot 	heta = \frac{1}{2} 	an 	heta$$	an^2 	heta = 2$$	an 	heta = \sqrt{2}$$	heta = 	an^{-1}(\sqrt{2})$.

The electric field at a point due to an electric dipole,on an axis inclined at an angle $\theta$ $(< 90^{\circ})$ to the dipole axis,is perpendicular to the dipole axis,if the angle $\theta$ is

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