Three charged particles of each mass 0.1 ,g a | vedclass

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(C) Let $m = 0.1 \,g = 10^{-4} \,kg$, $L = 1 \,m$, $a = 3 \,cm = 0.03 \,m$. The distance from the centroid of the equilateral triangle to a corner is

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(C) Let $m = 0.1 \,g = 10^{-4} \,kg$, $L = 1 \,m$, $a = 3 \,cm = 0.03 \,m$. The distance from the centroid of the equilateral triangle to a corner is $r = \frac{a}{\sqrt{3}} = \frac{0.03}{\sqrt{3}} = 0.01\sqrt{3} \,m$. The angle $	heta$ with the vertical is $\sin 	heta \approx 	an 	heta = \frac{r}{L} = 0.01\sqrt{3}$. The forces on one particle are gravity $mg$, tension $T$, and electrostatic repulsion from the other two particles. The resultant electrostatic force $F_e$ from two charges $q$ at distance $a$ is $F_e = 2 \cdot \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{a^2} \cos(30^\circ) = 2 \cdot (9 	imes 10^9) \cdot \frac{q^2}{(0.03)^2} \cdot \frac{\sqrt{3}}{2} = \sqrt{3} 	imes 10^{13} q^2$. In equilibrium, $	an 	heta = \frac{F_e}{mg}$. Thus, $0.01\sqrt{3} = \frac{\sqrt{3} 	imes 10^{13} q^2}{10^{-4} 	imes 10}$. Solving for $q^2$: $q^2 = 0.01 	imes 10^{-3} / 10^{13} = 10^{-18} \,C^2$. Therefore, $q = 10^{-9} \,C = 1 \,nC$.

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