An electric dipole of moment $\vec{p}$ is placed at the origin along the X-axis. The electric field at a point $P$, whose position vector makess an angle $\theta$ with the X-axis, will make an angle $\varphi$ with the $X$-axis, then $\varphi$ is ( \(\alpha=\tan ^{-1}\left(\frac{1}{2}…

Question

An electric dipole of moment $\vec{p}$ is placed at the origin along the X-axis. The electric field at a point $P$, whose position vector makess an angle $\theta$ with the X-axis, will make an angle $\varphi$ with the $X$-axis, then $\varphi$ is ( $\alpha=\tan ^{-1}\left(\frac{1}{2} \tan \theta\right)$)

Aicharya · Accepted Answer

An electric dipole of moment $=\bar{p}$
electric field $\mathrm{x}$-axis at a point $=\mathrm{p}$
angle $=\theta$ with $x-$ axis
$\tan a=\frac{1}{2} \tan \theta$
$(\theta+\alpha)=$ the value of the position vector makes an angle $\theta$
$\theta=60^{\circ}+\alpha$
now resolving $\mathrm{E}$ into its components
$E \cos \alpha=\frac{2 p \cos 60^{\circ}}{4 \pi \epsilon_0 r^3} \rightarrow(1)$
$E \sin \alpha=\frac{p \sin 60^{\circ}}{4 \pi \epsilon_0 r^3} \quad \rightarrow(2)$
Dividing 2 by 1
$\tan \alpha=\frac{1}{2} \tan \theta$
$\tan \alpha=\tan (\theta+\alpha)$
$\alpha=\theta+\alpha$

An electric dipole of moment \(\vec{p}\) is placed at the origin along the X-axis. The electric field at a…

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