Question

The electric dipole is placed along the $x$ - axis at the origin $O$. A point $P$ is at a distance of $20cm$ from this origin such that $OP$ makes an angle ${60}^o$ with the $x$ - axis. If the electric field at $P$ makes an angle $\theta$ with the $x$ - axis, the value of $\theta$ would be

• A. $\frac{\pi }{3}$
• B. ${\tan }^{-1}\frac{\sqrt{3}}{2}$
• C. $\frac{2\pi }{3}$
• D. $\frac{\pi }{3}+{\tan }^{-1}\frac{\sqrt{3}}{2}$

Answer 1

The correct option is D $\frac{\pi }{3}+{\tan }^{-1}\frac{\sqrt{3}}{2}$

An electric dipole is placed along the $x-$axis at the origin $P.A$ point $P$ is a distance of $20cm$ from this origin such that $OP$ makes an angle ${60}^0$ with $x-$axis. If electric field at $P$ makes on angle $\theta$ with $x-$axis, then

By the fig.

$\theta =60+\alpha$

Now, resolving $E$ into its components

$E\cos \alpha =\frac{2P\cos {60}^0}{4\pi {ϵ}_0{r}^3}⟶\left(1\right)$

$E\sin \alpha =\frac{P\sin {60}^0}{4\pi {ϵ}_0{r}^3}⟶\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$

$\tan \alpha =\frac{1}{2}\tan {60}^0$

$\tan \alpha =\frac{\sqrt{3}}{2}$

or, $\alpha ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Now, $\theta =60+\alpha$

$=\frac{\pi }{3}+{\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$