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(B) For the upper charge to be in equilibrium along the inclined plane,the component of gravitational force acting down the plane must be balanced by

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$300$

Vedclass · Answer

(B) For the upper charge to be in equilibrium along the inclined plane,the component of gravitational force acting down the plane must be balanced by the electrostatic repulsive force acting up the plane.Let $r$ be the distance between the two charges. From the geometry,$h = r \sin 30^{\circ}$,so $r = \frac{h}{\sin 30^{\circ}} = 2h$.The force balance equation is: $mg \sin 30^{\circ} = \frac{1}{4 \pi \varepsilon_0} \frac{q_0^2}{r^2}$.Substituting $r = 2h$ and the given values $(m = 0.02 \, kg, g = 10 \, ms^{-2}, q_0 = 2 	imes 10^{-6} \, C, \sin 30^{\circ} = 0.5)$:$0.02 	imes 10 	imes 0.5 = 9 	imes 10^9 	imes \frac{(2 	imes 10^{-6})^2}{(2h)^2}$$0.1 = 9 	imes 10^9 	imes \frac{4 	imes 10^{-12}}{4h^2}$$0.1 = \frac{9 	imes 10^{-3}}{h^2}$$h^2 = \frac{9 	imes 10^{-3}}{0.1} = 9 	imes 10^{-2} \, m^2$$h = 0.3 \, m = 300 	imes 10^{-3} \, m$.Comparing this with $h = x 	imes 10^{-3} \, m$,we get $x = 300$.

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