Question

Consider a system of three charges $\frac{q}{3}$ , $\frac{q}{3}$ and $-\frac{2q}{3}$ placed at points $A,B$ and $C$, respectively, as shown in the figure. Take $O$ to be the centre of the circle of radius $R$ and angle $CAB={60}^o$

• A. The electric field at point $O$ is $\frac{q}{8\pi {ϵ}_0{R}^2}$ directed along the negative x-axis
• B. The potential energy of the system is zero
• C. The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54\pi {ϵ}_0{R}^2}$
• D. The potential at point $O$ is $\frac{q}{12\pi {ϵ}_{0}R}$

Answer 1

The correct option is C The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54\pi {ϵ}_0{R}^2}$

The electric field at O due to charges at A and B are cancel out because both charges are equal and same sign at A and B. Thus only charge at C will contribute the field at O.

The filed at O is $E=\frac{1}{4\pi {ϵ}_0}\frac{(-2q/3)}{R^2}=-\frac{q}{6\pi {ϵ}_0{R}^2}$ along negative x axis.

Potential energy $U=\frac{1}{4\pi {ϵ}_0}[\frac{\left(q^2/9\right)}{AB}+\frac{(-2q^2/9)}{BC}+\frac{(-2q^2/9)}{AC}]=\frac{1}{4\pi {ϵ}_0}[\frac{\left(q^2/9\right)}{2R}+\frac{(-2q^2/9)}{2R\left(\sqrt{3}/2\right)}+\frac{(-2q^2/9)}{2R\left(1/2\right)}]\ne 0$

The force between charges at B and C is $F=\frac{1}{4\pi {ϵ}_0}\frac{\left(q/3\right)(-2q/3)}{B{C}^2}=-\frac{1}{4\pi {ϵ}_0}\frac{2q^2}{9(2R\sin 60{)}^2}=-\frac{1}{4\pi {ϵ}_0}\frac{2q^2}{9(4R^2\times 3/4)}$

$|F|=\frac{q^2}{54\pi {ϵ}_0{R}^2}$

Potential at O is $V=\frac{1}{4\pi {ϵ}_0}[\frac{\left(q/3\right)}{OA}+\frac{\left(q/3\right)}{OB}+\frac{(-2q/3)}{OC}]=\frac{1}{4\pi {ϵ}_0}[\frac{\left(q/3\right)}{R}+\frac{\left(q/3\right)}{R}+\frac{(-2q/3)}{R}]=0$