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

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(C) Given charges are $q_A = \frac{q}{3}$,$q_B = \frac{q}{3}$,and $q_C = \frac{-2q}{3}$.In $	riangle ABC$,since $A, B, C$ lie on a circle of radius $

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) Given charges are $q_A = \frac{q}{3}$,$q_B = \frac{q}{3}$,and $q_C = \frac{-2q}{3}$.In $	riangle ABC$,since $A, B, C$ lie on a circle of radius $R$ and $\angle CAB = 60^{\circ}$,the chord $BC$ subtends an angle at the circumference. The distance $BC$ can be calculated using the law of sines or geometry. Since $O$ is the center,$\angle COB = 2 \angle CAB = 120^{\circ}$.The distance $BC = 2R \sin(60^{\circ}) = 2R \frac{\sqrt{3}}{2} = R\sqrt{3}$.The magnitude of the electrostatic force between charges at $C$ and $B$ is given by Coulomb's Law:$F = \frac{1}{4\pi \varepsilon_0} \frac{|q_C \cdot q_B|}{(BC)^2} = \frac{1}{4\pi \varepsilon_0} \frac{|(\frac{-2q}{3}) \cdot (\frac{q}{3})|}{(R\sqrt{3})^2}$$F = \frac{1}{4\pi \varepsilon_0} \frac{2q^2/9}{3R^2} = \frac{2q^2}{108\pi \varepsilon_0 R^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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