# Two Charges in Hemisphere

1. Sep 2, 2009

### Hov

1. The problem statement, all variables and given/known data

"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 R apart. Determine the charge on each bead. (Use k_e for ke, g for the acceleration due to gravity, m, and R as necessary.)"

Here's a picture:

http://img3.imageshack.us/i/p2368.gif

2. Relevant equations

Relevant equations include Coulomb's Law, and other elementary physics equations.

F = Ke * (q1) * (q2) / (distance)2

Where Ke = 8.9 x 109

3. The attempt at a solution

So what I've gotten so far is this:

Using the left-most ball as a model, there's a force in the negative x direction from the ball on the right. That force is exactly

F = -Ke * (qe)2/R2

Now there's a y component of the normal force from the hemisphere given by:

mgsin(x) = N

So, I also figured that the normal and the gravitational forces should each have an x component and that it should be equal and opposite of the force from the charge on the right:

Ncos(x) = Force from the charge

So here's what I did from there:

Ncos(x) = Ke * (qe)2/R2
Nsin(x) = mg

tan(x) = mg / ( Ke * (qe)2/R2 )

R2 / tan(x)mg = Ke * (qe)2

R2 / tan(x)mg * Ke = (qe)2

And ended up with:

qe = sqrt ( R2 / tan(x)mg * Ke )

Which is incorrect.

Last edited: Sep 2, 2009
2. Sep 2, 2009

### rl.bhat

One component of weight of the bead balances the normal reaction and other component pulls it down.
One component of force of the bead balances the normal reaction and other component pulls it up along the hemisphere.

3. Sep 2, 2009

### Hov

Thank you for your reply.

So just to be clear, the charge force has a y component, even in equilibrium? I was under the assumption that it only had an x component.

4. Sep 2, 2009

### rl.bhat

Yes. It has both the components.

5. Sep 5, 2009

### Hov

I'm still stuck on this problem. :(

6. Sep 6, 2009

### rl.bhat

If θ is the angle between electrostatic force and normal, then
F*cosθ = mg*sinθ
F*sinθ + mg*cos = N

7. Sep 22, 2009

### Hov

This was a homework question on Webassign. I got it wrong.

But I need to know how to do this for my test on Thursday. Some classmates said that there is no reference to theta in the final answer because it's an equilateral triangle.

Can anyone help me further? I'm very lost on this.

Here is another image since the first one was apparently taken down.

http://img5.imageshack.us/img5/9582/p2368.gif [Broken]

Last edited by a moderator: May 4, 2017
8. Sep 22, 2009

### Hov

Never mind, got it. I wasn't too far off in my original attempt.

Nsinθ = mg

is the correct equation. It makes perfect sense now. But I still don't understand why this equation:

N = mgcosθ

Doesn't work. Does anyone know why? According to my diagram, the two forces should be equal.

9. Sep 22, 2009

### rl.bhat

Sorry.In my post #6, the sin and cos are interchanged. It should be
F*sinθ = mg*cosθ. Here θ = 60 degrees. Try it.

10. Sep 23, 2009

### Hov

I don't think that's correct.

The system of equations should be:

Y component: Nsinθ = mg
X component: Ncosθ = Fe

I understand how to do this problem now, and got the correct answer (finally). My question now is why couldn't the first equation be written as:

N = mgcosθ

Because according to the diagram, wouldn't these two be equal?

Thanks again for your help rl.bhat.

11. Sep 23, 2009

### rl.bhat

In equilibrium condition your method is correct.
You will get the same result if you use the equations in my post 9#
Here N consist of two components.If you take the angle between the vertical and normal as θ, then
N = mg*cosθ + F*sinθ

Last edited: Sep 23, 2009
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