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(D) Let the length of each thread be $L = 3\, m$. The angle between the threads is $120^{\circ}$,so each thread makes an angle $	heta = 60^{\circ}$ w

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None of the above

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(D) Let the length of each thread be $L = 3\, m$. The angle between the threads is $120^{\circ}$,so each thread makes an angle $	heta = 60^{\circ}$ with the vertical.The distance $r$ between the two charges is given by $r = 2L \sin(60^{\circ}) = 2 	imes 3 	imes \frac{\sqrt{3}}{2} = 3\sqrt{3}\, m$.The electrostatic force $F_e$ between the charges is $F_e = \frac{1}{4\pi\varepsilon_0} \frac{Q^2}{r^2} = 9 	imes 10^9 	imes \frac{(10 	imes 10^{-6})^2}{(3\sqrt{3})^2} = 9 	imes 10^9 	imes \frac{10^{-8}}{27} = \frac{10}{27} \approx 0.37\, N$.In equilibrium,the forces acting on one ball are tension $T$ (along the thread),electrostatic force $F_e$ (horizontal),and weight $mg$ (vertical). However,the mass $m$ is not given. Assuming the question implies the tension is balanced by the horizontal component of the electrostatic force,we look at the force balance: $T \sin(60^{\circ}) = F_e$ and $T \cos(60^{\circ}) = mg$.Given the options provided,there seems to be a discrepancy in the provided solution logic. Re-evaluating: $T \sin(60^{\circ}) = F_e \implies T = \frac{F_e}{\sin(60^{\circ})} = \frac{10/27}{\sqrt{3}/2} = \frac{20}{27\sqrt{3}} \approx 0.427\, N$. None of the options match this result. Given the provided solution structure in the prompt,it appears the question is flawed or missing mass information.

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