Question

An 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 $\frac{\pi }{3}$ with the x-axis. If the electric field at P makes an angle θ with the x-axis, the value of $\theta$ would be:

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{3}+ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• C. $\frac{2\pi }{3}$
• D. $ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

The correct option is:(B)

Given Data:

Point P is at a distance of $\text{=20cm}$

OP makes an angle, $=\frac{\pi }{3}$

Fro fig, $\theta$ is an exterior angle of triangle,which is sum of two opposite interior angle.

$\theta =\frac{\pi }{3}+\alpha$

At point P, electric field will be divided into horizontal and vertical component,

So, Electric feild along X-axis; $E_H=Ecos\alpha =\frac{2P_{H}cos\alpha }{r^3}....\left(1\right)$

Electric field on vertical axis, $E_V=Esin\alpha =\frac{P_{V}cos\alpha }{r^3}....\left(2\right)$

Dividing eq(2) by eq(1)

$\frac{Esin\alpha }{Ecos\alpha }=\frac{\frac{P_V}{r^3}}{\frac{2P_H}{r^3}}=\frac{P_V}{2P_H}......\left(3\right)$

Here, $P_H=$ dipole momentum along OP-axis.

$P_V=$ dipole momentum along perpendicular to OP -axis

Dividing, this diploe momentm into horizontal and vertical component,

$P_H=Pcos\frac{\pi }{3}=\frac{P}{2}$

$P_V=Psin\frac{\pi }{3}=P\frac{\sqrt{3}}{2}$

putting value of $P_H=Pcos\frac{\pi }{3}=\frac{P}{2}$ and $P_V=Psin\frac{\pi }{3}=P\frac{\sqrt{3}}{2}$ in eq(3)

$\frac{Esin\alpha }{Ecos\alpha }=\frac{\frac{P_V}{r^3}}{\frac{2P_H}{r^3}}=\frac{P_V}{2P_H}.$

$tan\alpha ==\frac{\frac{\sqrt{3}}{2}}{2\frac{1}{2}}=\frac{\sqrt{3}}{2}$

$\alpha =ta{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)$

So,

$\theta =\frac{\pi }{3}+\alpha ⟹\theta +\frac{\pi }{3}+ta{n}^{-1}\frac{\sqrt{3}}{2}$