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(B) Let the angle made by the thread with the vertical be $	heta$. From the geometry,$\sin 	heta = \frac{10 \ cm}{20 \ cm} = \frac{1}{2}$,so $	heta

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$3$

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(B) Let the angle made by the thread with the vertical be $	heta$. From the geometry,$\sin 	heta = \frac{10 \ cm}{20 \ cm} = \frac{1}{2}$,so $	heta = 30^{\circ}$.At equilibrium,the forces acting on the particle are: tension $T$ along the thread,weight $mg$ downwards,and electric force $qE$ horizontally away from the wall.Resolving the forces: $T \sin 	heta = qE$ and $T \cos 	heta = mg$.Dividing the two equations: $	an 	heta = \frac{qE}{mg}$.Given $m = 2 \ g = 2 	imes 10^{-3} \ kg$,$E = 2 	imes 10^4 \ N/C$,$g = 10 \ m/s^2$,and $	heta = 30^{\circ}$.$	an 30^{\circ} = \frac{q 	imes 2 	imes 10^4}{2 	imes 10^{-3} 	imes 10} = \frac{q 	imes 2 	imes 10^4}{2 	imes 10^{-2}} = q 	imes 10^6$.$\frac{1}{\sqrt{3}} = q 	imes 10^6 \implies q = \frac{1}{\sqrt{3}} 	imes 10^{-6} \ C = \frac{1}{\sqrt{3}} \ \mu C$.Comparing with $\frac{1}{\sqrt{x}} \ \mu C$,we get $x = 3$.

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