Question

Consider a system of three charges $q/3,q/3,-2q/3$ placed at points A, B and C respectively as shown in the figure. If the radius of the circle is $R$ and $\angle$CAB is ${60}^o$, then the electric field at the centre $O$ is

• A. $\frac{q}{8\pi {ϵ}_0{R}^2}$
• B. $\frac{q^2}{54\pi {ϵ}_0{R}^2}$
• C. $\frac{q}{6\pi {ϵ}_0{R}^2}$
• D. $0$

Answer 1

The correct option is C $\frac{q}{6\pi {ϵ}_0{R}^2}$

Given a circle of radius $R$, three charges $\frac{q}{3}$, $\frac{q}{3}$ and $\frac{-2q}{3}$ plpaced at poits $A$, $B$ and $C$ respectively as shown in the figure. $\angle CAB={60}^{\circ }$.

We have to find the electric field at the center i.e, $O$.

So, the electric field at $O$ due to two charges $\frac{q}{3}$ will get canceled.

Electric field will be only due to $\frac{-2q}{3}$ and will be directed along the negative x-axis and is given as:

$E=\frac{1}{4\pi {\epsilon }_0}\frac{\left(2q/3\right)}{R^2}=\frac{q}{6\pi {\epsilon }_0{R}^2}$

where $R$ is distance of point $C$ to the center $\left(O\right)$.