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(D) Let $m = 10 \, mg = 10 	imes 10^{-6} \, kg$,$L = 0.5 \, m$,$r = 0.2 \, m$,and $g = 10 \, ms^{-2}$.At equilibrium,the forces acting on each sphere

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) Let $m = 10 \, mg = 10 	imes 10^{-6} \, kg$,$L = 0.5 \, m$,$r = 0.2 \, m$,and $g = 10 \, ms^{-2}$.At equilibrium,the forces acting on each sphere are tension $T$,weight $mg$,and electrostatic force $F_e = \frac{kq^2}{r^2}$.From the geometry,$\sin 	heta = \frac{r/2}{L} = \frac{0.1}{0.5} = 0.2$. Thus,$\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - 0.04} = \sqrt{0.96}$.Resolving forces: $T \cos 	heta = mg$ and $T \sin 	heta = F_e$.Dividing the two equations: $	an 	heta = \frac{F_e}{mg} = \frac{kq^2}{r^2 mg}$.$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{0.2}{\sqrt{0.96}} = \frac{0.2}{0.9798} \approx 0.204$.$q^2 = \frac{r^2 mg 	an 	heta}{k} = \frac{(0.2)^2 	imes (10^{-5}) 	imes (0.204)}{9 	imes 10^9} = \frac{0.04 	imes 10^{-5} 	imes 0.204}{9 	imes 10^9} \approx 9.06 	imes 10^{-17} \, C^2$.$q \approx 9.52 	imes 10^{-9} \, C = 0.952 	imes 10^{-8} \, C$.Given $q = \frac{a}{21} 	imes 10^{-8} \, C$,so $\frac{a}{21} = 0.952 \implies a = 0.952 	imes 21 \approx 20$.

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