question_answer

Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, nonconductive walls, the beads move, and at equilibrium the distance between them is R (Fig.). Determine the charge on each bead.

A) \[R{{\left( \frac{mg}{{{k}_{e}}\sqrt{3}} \right)}^{1/2}}\]

B)        \[R{{\left( \frac{mg}{{{k}_{e}}\sqrt{2}} \right)}^{1/2}}\]

C) \[R{{\left( \frac{mg}{{{k}_{e}}2\sqrt{3}} \right)}^{1/2}}\]

D)        \[R{{\left( \frac{2\,mg}{{{k}_{e}}\sqrt{3}} \right)}^{1/2}}\]

Correct Answer: A

Solution :

[a] The bowl exerts a normal force N on each bead, directed along the radial line or at \[60.0{}^\circ \]above the horizontal. Consider the free-body diagram of the bead on the left with electric force \[{{F}_{e}}\] applied: \[\Sigma {{F}_{y}}=N\sin 60{}^\circ -mg=0\] \[\Rightarrow N=\frac{mg}{\sin 60{}^\circ }\] \[\Sigma {{F}_{x}}=-{{F}_{e}}+N\cos 60{}^\circ =0\] \[\Rightarrow \frac{{{k}_{e}}{{q}^{2}}}{{{R}^{2}}}=N\cos 60{}^\circ =\frac{mg}{\tan 60{}^\circ }=\frac{mg}{\sqrt{3}}\] \[{{k}_{e}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\approx 9.0\times {{10}^{9}}N{{m}^{2}}/{{C}^{2}}\therefore q=R{{\left( \frac{mg}{{{k}_{e}}\sqrt{3}} \right)}^{1/2}}\]