Question

Three charged particles $A,B$ and $C$ with charges $-4q,2q$ and $-2q$ are present on the circumference of a circle of radius $d$. The charged particle $A,C$ and centre $O$ of the circle formed an equilateral triangle as shown in figure. Electric field at $O$ along $x-$ direction is:

• A. $\frac{\sqrt{3}q}{\pi {ϵ}_0{d}^2}$
• B. $\frac{2\sqrt{3}q}{\pi {ϵ}_0{d}^2}$
• C. $\frac{\sqrt{3}q}{4\pi {ϵ}_0{d}^2}$
• D. $\frac{3\sqrt{3}q}{4\pi {ϵ}_0{d}^2}$

Answer 1

The correct option is A $\frac{\sqrt{3}q}{\pi {ϵ}_0{d}^2}$


Electric field due to charge $+2q$ at centre $O$ points along $OC$ and makes an angle ${30}^{\circ }$ with $x-$ axis.

${\overrightarrow{E}}_1=\frac{1}{4\pi {\epsilon }_0}\times \frac{2q}{d^2}\left[\frac{\sqrt{3}\hat{i}-\hat{j}}{2}\right]$

Electric field due to charge $-2q$ at centre $O$ also points along $OC$ and makes an angle ${30}^{\circ }$ with $x-$ axis.

${\overrightarrow{E}}_2=\frac{1}{4\pi {\epsilon }_0}\times \frac{2q}{d^2}\left[\frac{\sqrt{3}\hat{i}-\hat{j}}{2}\right]$

Electric field due to charge $-4q$ at centre $O$ points along $OA$ and makes an angle ${30}^{\circ }$ with $x-$ axis.

${\overrightarrow{E}}_3=\frac{1}{4\pi {\epsilon }_0}\times \frac{4q}{d^2}\left[\frac{\sqrt{3}\hat{i}+\hat{j}}{2}\right]$

$\therefore$ Net electric filed at point $O$ is,

${\overrightarrow{E}}_0={\overrightarrow{E}}_1+{\overrightarrow{E}}_2+{\overrightarrow{E}}_3=\frac{\sqrt{3}q}{\pi {\epsilon }_0{d}^2}\hat{i}$

Hence, option $\left(A\right)$ is correct.