Consider a system of three charges frac{q}{3} | vedclass

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(C) The charges are $q_A = q/3$,$q_B = q/3$,and $q_C = -2q/3$. In $	riangle ABC$,since $A, B, C$ are on a circle with center $O$,$OA = OB = OC = R$.

Vedclass · Accepted Answer

The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54 \pi \varepsilon_0 R^2}$.

Vedclass · Answer

(C) The charges are $q_A = q/3$,$q_B = q/3$,and $q_C = -2q/3$. In $	riangle ABC$,since $A, B, C$ are on a circle with center $O$,$OA = OB = OC = R$. Given $\angle CAB = 60^{\circ}$,in $	riangle OAC$,$OA = OC = R$,so $\angle OCA = \angle OAC = 60^{\circ}$,implying $	riangle OAC$ is equilateral. Thus,$AC = R$. Similarly,for $	riangle OBC$,since $\angle ACB = 90^{\circ}$ (angle in a semicircle if $AB$ is diameter,but here $AB$ is a chord),we calculate the distance $BC$. Using the law of cosines in $	riangle OBC$,or geometry: $BC = \sqrt{R^2 + R^2 - 2R^2 \cos(120^{\circ})} = R\sqrt{3}$. The force between $C$ and $B$ is $F_{BC} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_C| |q_B|}{(BC)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{(2q/3)(q/3)}{(R\sqrt{3})^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2q^2/9}{3R^2} = \frac{q^2}{54 \pi \varepsilon_0 R^2}$. Thus,option $C$ is correct.

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 $\angle CAB = 60^{\circ}$.

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