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(B) Let $l = 10 \ cm = 0.1 \ m$ be the length of the string. The angle between the strings is $60^{\circ}$,so the angle with the vertical is $	heta =

Vedclass · Accepted Answer

$0.6235$

Vedclass · Answer

(B) Let $l = 10 \ cm = 0.1 \ m$ be the length of the string. The angle between the strings is $60^{\circ}$,so the angle with the vertical is $	heta = 30^{\circ}$.In equilibrium,the forces acting on each sphere are tension $T$,gravitational force $mg$,and electrostatic force $F_e = \frac{kq^2}{x^2}$,where $x$ is the distance between the centers of the spheres.From the geometry,$x = 2l \sin 	heta = 2(0.1) \sin 30^{\circ} = 0.2 	imes 0.5 = 0.1 \ m$.Resolving forces:$T \sin 	heta = F_e$$T \cos 	heta = mg$Dividing the two equations: $	an 	heta = \frac{F_e}{mg} = \frac{kq^2}{x^2 mg}$.Rearranging for mass $m$: $m = \frac{kq^2}{x^2 g 	an 	heta}$.Substituting the values:$m = \frac{(9 	imes 10^9) 	imes (2 	imes 10^{-6})^2}{(0.1)^2 	imes 10 	imes 	an 30^{\circ}}$$m = \frac{9 	imes 10^9 	imes 4 	imes 10^{-12}}{0.01 	imes 10 	imes (1/\sqrt{3})}$$m = \frac{36 	imes 10^{-3}}{0.1 	imes (1/\sqrt{3})} = 36 	imes 10^{-2} 	imes \sqrt{3} \approx 0.36 	imes 1.732 = 0.6235 \ kg$.

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