• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • #1
1
0

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!
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,737
5,033

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.
 

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

  • Last Post
Replies
10
Views
6K
Replies
28
Views
3K
  • Last Post
Replies
8
Views
1K
Replies
2
Views
8K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
2
Views
3K
Replies
7
Views
3K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
4
Views
3K
Replies
16
Views
770
Top