Two identical small beads each have mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, non-conductive walls, the beads move and at equilibrium the line joining the beads is horizontal and the distance between them is R (see figure). Neglect any induced charge on the hemispherical bowl. Then the charge on each bead is $(here K = \frac{1}{4 {πε}_{0}})$

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$q = R {(\frac{\sqrt{3} mg}{K})}^{1 / 2}$

$q = {(R \frac{\sqrt{3} mg}{K})}^{1 / 2}$

$q = R {(\frac{mg}{K \sqrt{3}})}^{1 / 2}$

$q = {(R \frac{mg}{K \sqrt{3}})}^{1 / 2}$

answer is A.

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Detailed Solution

The bowl exerts a normal force N on each bead, directed along the radius line or at 60° above the horizontal. Consider the free-body diagram of the bead on the left with the electric force F_e applied.

$\sum F_{y} = Nsin 60^{\circ} - mg = 0, \Rightarrow N = mg / \sin 60^{\circ} \sum F_{x} = - F_{e} + Ncos 60^{\circ} = 0 \Rightarrow \frac{{Kq}^{2}}{R^{2}} = Ncos 60^{\circ} = \frac{mg}{\tan 60^{\circ}} = \frac{mg}{\sqrt{3}} Thus, q = R {(\frac{mg}{K \sqrt{3}})}^{1 / 2}$