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

• kevin_tee
In summary, the conversation discusses using Coulomb's law and equations for capacitance to find the force between two charged spheres given their radius, permittivity, distance, and voltage. The calculation involves replacing the variables in Coulomb's law with those from the equations for voltage-dependent capacitors and capacitance of a sphere. However, it is important to carefully consider the geometry and how the equations may be affected by the non-uniform charge distribution when another charged sphere is introduced. Using energy considerations may be a more accurate approach.
kevin_tee

## Homework Statement

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.

## Homework 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

## 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.

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

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

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.

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.

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Your approach is correct, but there are a few things to consider in your calculation. First, make sure that the units are consistent throughout your equation. In Coulomb's law, the unit for charge (Q) is Coulombs (C) and the unit for distance (r) is meters (m). In the voltage-dependent capacitors equation, the unit for voltage (V) is volts (V) and the unit for capacitance (C) is Farads (F). So when you substitute these values into Coulomb's law, you will need to convert the units accordingly.

Secondly, when substituting the capacitance of a sphere equation into Coulomb's law, you should use the capacitance of one sphere (C) and the voltage of the other sphere (V), rather than multiplying both capacitances and both voltages together.

Finally, remember that Coulomb's law gives the magnitude of the force between two charged objects, but not the direction. The direction of the force will depend on the signs of the charges on the two spheres, which will affect the overall sign of the force.

Overall, your approach is a valid way to find the force between two charged spheres, but make sure to double check your units and consider the direction of the force.

## What is the formula for finding the force between two charged spheres?

The formula for finding the force between two charged spheres is F = (q1 * q2) / (4πε0 * r2), where q1 and q2 are the charges on the spheres, ε0 is the permittivity of free space, and r is the distance between the spheres.

## How does the distance between the spheres affect the force?

The force between the charged spheres is inversely proportional to the square of the distance between them. This means that as the distance between the spheres decreases, the force between them increases, and vice versa.

## How does the magnitude of the charges on the spheres affect the force?

The force between the charged spheres is directly proportional to the product of the charges on the spheres. This means that as the magnitude of the charges increases, the force between them also increases, and vice versa.

## What is the role of the permittivity of free space in the calculation of force?

The permittivity of free space, represented by ε0, is a constant that determines the strength of the electric force between two charged objects. It is a measure of how easily electric fields can pass through a vacuum. Without this constant, the force between two charged spheres would be infinite.

## How does the relative permittivity of the medium between the spheres affect the force?

The relative permittivity, also known as the dielectric constant, of the medium between the spheres affects the force by altering the strength of the electric field between the spheres. This is due to the fact that the relative permittivity changes the value of ε0. As the relative permittivity increases, the force between the spheres decreases, and vice versa.

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