Two identical beads each have a mass m and charge q. When placed in a hemispherical bowl of radius R with frictionless, nonconducting walls, the beads move, and at equilibrium, they are a distance d apart (see figure below). Determine the charge q on each bead.
Fe = (Ke*q^2)/d^2)
The Attempt at a Solution
I drew a free body diagram first. Since the beads are at equilibrium, the net force must be zero.
So, considering the electric force, the gravitational force, and the normal force of the hemisphere (where x is the angle between R and d) I got..
Fx = Ncos(x) - Fe = 0
Fy = Nsin(x) - mg = 0
Ncos(x) = Fe
Nsin(x) = mg
N = mg/sin(x)
(mg/sin(x))*cos(x) = (Ke*q^2)/d^2
d^2*(mg/sin(x))*cos(x) = Ke * q^2
q = sqrt((d^2 * (mg/sin(x)) * cos(x))/ Ke)
Now, I've seen problems similar to this where d is equal to R, giving an equilateral triangle, however that is not the case in this problem. I don't know how to eradicate the sin and cos of x or how to put x into terms of R and D.