Two indentical beads w/ mass m and charge q in hemispherical

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The problem involves two identical beads with mass m and charge q in a hemispherical bowl, where they achieve equilibrium at a distance d apart. The forces acting on the beads include electric force, gravitational force, and the normal force from the bowl's surface. The equations derived indicate that the net force is zero, leading to relationships involving the angle x, radius R, and distance d. To solve for the charge q, it is suggested to use geometry to express sin(x) and cos(x) in terms of d and R. This approach will help eliminate the angle x from the equations and facilitate finding the charge on each bead.
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Homework Statement


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.

Homework Equations


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.
Help!
 
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ortiz123 said:

Homework Statement


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.

Homework Equations


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.
Help!
Use a bit of geometry to express sin(x), cos(x) in terms of d and R.
 
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