Calculating distance between charged Balls

[psy]

Hello! I have the following problem and attempt for solution.

Two equal, small conductive balls hang on two long non-conductive threads attached to a hook. The balls are charged with the same charges and are located 5 cm apart. One of the balls will now be decharged. What is the new distance between the balls?
So the Force between the balls : F= k ⋅ q1 ⋅ q2 / r^2 .

Also there is the gravitational force F= m⋅g.

sinα = (r/2)/l ; tanα= Fe/Fg.

l- length of the thread

sinα = tanα ; (r/2)/l = Fe/Fg → Fe= r⋅m⋅g / 2⋅l . Putting this in the above equation :

Fe= k ⋅ q1 ⋅ q2 / r^2 → r^2 = k ⋅ q1 ⋅ q2 / Fe → r^2 = (k ⋅ q1 ⋅ q2) / (r⋅m⋅g / 2⋅l ).

Now if i don't have any given length of the thread and mass of the balls, how is it possible to calculate the distance?
Is there another way to calculate it?

Kind regards

 

[tnich]

If there is zero charge on one ball, what is the electrostatic force between the two balls?

 

[psy]

0?

 

[tnich]

0?

Right. So how far would the balls be forced apart?

 

[haruspex]

This is a bit tricky. The balls are conductors. If one is charged and the other not charged then the force between them is not zero. And it will change subsequently, so I assume they are asking for the new equilibrium.
I note that the question says the ball is "decharged"; it does not say "earthed", which would create another complication.

 

[psy]

Well in case there is no Electrostatic force there is no electrostatic "push", so the only force left should be the gravitational.

Is this correct?

 

[haruspex]

Well in case there is no Electrostatic force

But there is.

 

[psy]

So one ball gets decharged, so its neutral after that. Yet the positive Charge of the other ball "reorganizes" the charges in the neutral one, it attracts the unlike charges to the nearer side of the ball and "pushes" away the like charges. As the unlike charges are closer, distance is smaller, the attracting force is stronger, so the balls are going to be attracted?

 

[haruspex]

So one ball gets decharged, so its neutral after that. Yet the positive Charge of the other ball "reorganizes" the charges in the neutral one, it attracts the unlike charges to the nearer side of the ball and "pushes" away the like charges. As the unlike charges are closer, distance is smaller, the attracting force is stronger, so the balls are going to be attracted?

Yes. The induced charge distribution results in attraction.
So what will happen subsequently?

Btw, sometimes it is asked whether there is a force between a charged object and an uncharged insulator. Surprisingly, there is. Even though charges cannot redistribute freely over the insulator, there are induced dipoles at the atomic level.

 

[psy]

So the half of the electrons from the neutral one are transferring to the charged ball. A new Equilibrium takes place, where the charges are +q/2 on both balls. So again they are pushing each other away, yet with smaller distance as the charges are smaller.

As the Coulombs law F = k * q1 * q2 / r^2 says the square of the distance and the electrostatic force are indirect proportional , the new distance will be :

F = (k * q/2 * q/2)/ r^2 => r^2 = k * q / F * 4 .

In first case it was r^2 = k * q^2 / F

Yet the force and the distance in the new Equilibrium are unknown. How can i get to calculate it?

 

[haruspex]

half of the electrons from the neutral one are transferring to the charged ball.

That isn't really meaningful. You could say a charge of -q/2 is transferred. But you got the right result.

F = (k * q/2 * q/2)/ r^2 => r^2 = k * q / F * 4 .

Go back to the last equation you had in post #1. You had r on both sides. You need to rearrange into the form r=[an expression not involving r] and apply it to this later condition too.