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

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(C) Given charges: $q_A = q/3, q_B = q/3, q_C = -2q/3$. Since $\angle CAB = 60^\circ$ and $OA=OB=OC=R$,$	riangle OAC$ and $	riangle OAB$ are equilat

Vedclass · Accepted Answer

The magnitude of the force between the charges at $C$ and $B$ is $\frac{q^2}{54\pi \epsilon_0 R^2}$.

Vedclass · Answer

(C) Given charges: $q_A = q/3, q_B = q/3, q_C = -2q/3$. Since $\angle CAB = 60^\circ$ and $OA=OB=OC=R$,$	riangle OAC$ and $	riangle OAB$ are equilateral triangles. Thus,$A, B, C$ are on the circle.Electric field at $O$: $\vec{E}_A = \frac{1}{4\pi\epsilon_0} \frac{q/3}{R^2}$ (away from $A$),$\vec{E}_B = \frac{1}{4\pi\epsilon_0} \frac{q/3}{R^2}$ (away from $B$),$\vec{E}_C = \frac{1}{4\pi\epsilon_0} \frac{2q/3}{R^2}$ (towards $C$).Resultant field $\vec{E}_O = \vec{E}_A + \vec{E}_B + \vec{E}_C$. Resolving components,the net field is $\frac{q}{6\pi\epsilon_0 R^2}$ along the negative $x$-axis.Potential at $O$: $V_O = \frac{1}{4\pi\epsilon_0 R} (q/3 + q/3 - 2q/3) = 0$.Force between $C$ and $B$: Distance $CB = \sqrt{R^2 + R^2 - 2R^2 \cos(120^\circ)} = \sqrt{3}R$. Force $F = \frac{1}{4\pi\epsilon_0} \frac{|q_C q_B|}{(CB)^2} = \frac{1}{4\pi\epsilon_0} \frac{(2q/3)(q/3)}{3R^2} = \frac{2q^2}{108\pi\epsilon_0 R^2} = \frac{q^2}{54\pi\epsilon_0 R^2}$.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 $\angle CAB = 60^\circ$.

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