Electrostatics Question 695
Question: 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.
Options:
A) $ R{{( \frac{mg}{k _{e}\sqrt{3}} )}^{1/2}} $
B) $ R{{( \frac{mg}{k _{e}\sqrt{2}} )}^{1/2}} $
C) $ R{{( \frac{mg}{k _{e}2\sqrt{3}} )}^{1/2}} $
D) $ R{{( \frac{2mg}{k _{e}\sqrt{3}} )}^{1/2}} $
Show Answer
Answer:
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}Nm^{2}/C^{2}\therefore q=R{{( \frac{mg}{k _{e}\sqrt{3}} )}^{1/2}} $
