# Finding force between charge sphere given r,ε,a,v

1. Feb 24, 2014

### kevin_tee

1. The problem statement, all variables and given/known data
This is not homework but it's kind of similar so may be I will post it here. I want to find the force between two charge sphere given radius, permittivity,distance, and voltage by diverge other equation to coulomb's law.

2. Relevant equations
Coulomb law F=k Q1 Q2 /r2
F = force
k = electric constant
Q = electrostatic quality
r= distance

Voltage-dependent capacitors Q=CV
Q = electrostatic quality
C = capacitance
V = voltage

Capacitance of sphere C=4πεa
C = capacitance
ε = permittivity

3. The attempt at a solution
2. F=k (CV)1 (CV)2/r2 replace Q with CV since Q=CV
3. F=k (4πεaV)1 (4πεaV)2/r2 replace C with 4πεa since C=4πεa

So F=k (4πεaV)1 (4πεaV)2/r2

Is this correct, I replace voltage-dependent capacitors and capacitance of sphere equation into coulomb law to find how much force is acting on it. So when I evaluate radius, permittivity,distance, and voltage it should gives me amount of force.

Last edited: Feb 24, 2014
2. Feb 24, 2014

### Simon Bridge

You are thinking that r>>a and V is measured wrt the edge of the Universe?

3. Feb 25, 2014

### kevin_tee

r is distance between two metal ball and a is radius of the ball. I don't really understand what you mean edge of the universe? V is voltage ex.30000 volts.

4. Feb 25, 2014

### Simon Bridge

The "voltage" is always a value at some place that is compared with a value at some other place.
So if you have 30000V on each sphere, where is the place that has 0V?

I'm trying to check the geometry you have in mind - I can make guesses from the equations you choose but that only takes me so far. If I am to answer your question properly I have to know what you intended rather than what I think you may have intended.

If r is the separation of the centers (you didn't say - maybe it's the separation between the surfaces?), then r>a means the balls are totally separate and r>>a means the distance between the balls is large compared with the radius of the balls. Initially I thought you were thinking of concentric shells - but the radii are the same so it's not that. I'm also guessing that by "electrostatic quality" you mean "electric charge".

Assuming something like - conducting balls equal radius a, equal charge Q, one centered at z=-r/2 and one at z=+r/2 where r > a, then:

The equation you use for capacitance is for a conducting sphere by itself (which is the same as for two concentric spheres, with the outer sphere radius infinity). This has a uniform charge distribution. Introduce another charged sphere and the charge distribution is no longer uniform so the equation may no longer be valid. How did you check?

How were you including the effect, if any, of the capacitance between the two balls?

I think you'd do better looking at energy considerations instead of forces.

Last edited: Feb 25, 2014