electricity/gravity?

Question

2 beads of mass m and charge q.
when placed in a hemispherical bowl of radius R with a frictionless, nonconducting walls, the beads move, and at equilibrium they are a distance R apart. Determine the charge on each bead.


____________/\....
___________/__\...
__________/____\
_________/______\
________/________\
_______/__________\
______/____________\
m [bead]=======[bead]m

the diagram is in an equilateral triangle ignore the "_"
so it will be R on each side of the triangle and the two beads m , m on the left and right side on bottom.
and ofcourse this triangle is inscribed in a semi sircle.. where the beads are the only parts of the triangle touching the semi sircle and the top of the semicircle is empty space.


The answer in the book is R(mg/[(k)(sqrt(3])^1/2 i tried forever and cant figure it out
Thanks

Answer 1

Pick a bead (either one, but let's choose the one on the left), and consider the all of the forces acting on it. It might help to draw a force diagram.

There are 3 forces acting on the left-hand bead:

1. The bead's weight, "mg", acting straight down.

2. The normal force (Fn) of the bowl pushing against the bead. This acts diagonally up and to the right, at an angle of 60 degrees from the horizontal.

3. the electrical repulsion due to the other bead. This pushes toward the left, with a force of kq²/R² (where "k" is shorthand for "1/(4πε)")

Now, since the left-hand bead is in equlibrium, all the forces acting on it are in balance. In particular, the vertical forces are in balance and (separately) the horizontal forces are in balance.

The vertical forces on the left-hand bead are:
Down: mg
Up: Fn(sin60) [vertical component of Fn]
Since they're balanced, we have:
(1): mg = Fn(sin60)

The horizontal forces on the left-hand bead are:
Left: kq²/R²
Right: Fn(cos60) [horizontal component of Fn]
Since they're balanced, we have:
(2): kq²/R² = Fn(cos60)

Now combine equations (1) and (2). There are a number of ways to do that, but maybe the simplest is to divide (1) by (2). That gives:
(3): mg / (kq²/R²) = tan60

Or, since tan60 = sqrt(3):

mg / (kq²/R²) = sqrt(3)

Now solve for "q", and you should get the same answer that's in the book