A particle of mass m and charge q is kept at | vedclass

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(D) At the point where the particle loses contact,the normal force $N = 0$. The radial forces acting on the particle are the component of gravity $mg

Vedclass · Accepted Answer

$\frac{3 - 2\sqrt{2}}{3}$

Vedclass · Answer

(D) At the point where the particle loses contact,the normal force $N = 0$. The radial forces acting on the particle are the component of gravity $mg \cos 	heta$ and the component of the electric force $qE \sin 	heta$. The net radial force provides the centripetal acceleration:$mg \cos 	heta - qE \sin 	heta = \frac{mv^2}{R}$ .........$(1)$Applying the work-energy theorem from the top of the sphere to the point at angle $	heta$:Work done by gravity + Work done by electric field = Change in kinetic energy$mgR(1 - \cos 	heta) + qER \sin 	heta = \frac{1}{2}mv^2$$mv^2 = 2mgR(1 - \cos 	heta) + 2qER \sin 	heta$ .........$(2)$Substituting $(2)$ into $(1)$:$mg \cos 	heta - qE \sin 	heta = \frac{2mgR(1 - \cos 	heta) + 2qER \sin 	heta}{R}$$mg \cos 	heta - qE \sin 	heta = 2mg - 2mg \cos 	heta + 2qE \sin 	heta$$3mg \cos 	heta - 2mg = 3qE \sin 	heta$Given $	heta = 45^{\circ}$,so $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$:$3mg(\frac{1}{\sqrt{2}}) - 2mg = 3qE(\frac{1}{\sqrt{2}})$$mg(\frac{3 - 2\sqrt{2}}{\sqrt{2}}) = qE(\frac{3}{\sqrt{2}})$$\frac{qE}{mg} = \frac{3 - 2\sqrt{2}}{3}$

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