# Electric Force and Time

Atran
Hi all,

[This is not a homework, I post this for my own calculations]
Let's say that there are two different charged copper cubes with the same mass.
Mass = 2 kg
Q(1) = +10 C
Q(2) = -4 C

If the distance between them is 10 m, then how can I calculate the remaining time for the distance to be 5 m or 0 m?

I see that cubes' acceleration changes over time when I use Coloumb' law and F=m*a.
I only want to know what (t) is equal and velocity if possible.

Thanks for help.

## Answers and Replies

Hi all,

[This is not a homework, I post this for my own calculations]
Let's say that there are two different charged copper cubes with the same mass.
Mass = 2 kg
Q(1) = +10 C
Q(2) = -4 C

If the distance between them is 10 m, then how can I calculate the remaining time for the distance to be 5 m or 0 m?

I see that cubes' acceleration changes over time when I use Coloumb' law and F=m*a.
I only want to know what (t) is equal and velocity if possible.

Thanks for help.

As a minor aside, I have asked this thread to be moved in the Homework forums. These forums are the right place for all questions like this, whether they are for private study or assigned homework. You'll get a better response there as well, I think.

There are some additional complexities in this case because accelerated charges radiate energy. I'll have to think about that. Apart from that, the problem you pose is similar to that of deriving motions of two particles attracted by their mutual gravitational attraction, and should be solved in the same way; by integrating. You are effectively calculating an orbit.

I presume you have the two particles initially at rest in empty space.

Cheers -- sylas

Let's say that there are two different charged copper cubes with the same mass.
Mass = 2 kg
Q(1) = +10 C
Q(2) = -4 C

If the distance between them is 10 m, then how can I calculate the remaining time for the distance to be 5 m or 0 m?

Followup. A coulomb is an enormous amount of charge. The force between these two copper cubes would be 3.6*109 newtons. So the immediate acceleration of the cubes would be 1.8*109 m/s2, each.

I think they'll be shredded. You can't get charges like that onto copper cubes. There will be all kinds of relativistic effects and you won't be able to ignore the special effects of accelerating charges, which will lead to radiation and additional forces.

A more realistic example would be to calculate the motions of two charged particles (a proton and anti-proton, for example) which approach each other from a large distance. I am not sure of the proper way to calculate that. Rather than guess, I'll sit back and wait for comment from some more competent advisors.

Cheers -- sylas

Atran
accelerated charges radiate energy.
How that works?

Followup. A coulomb is an enormous amount of charge. The force between these two copper cubes would be 3.6*109 newtons. So the immediate acceleration of the cubes would be 1.8*109 m/s2, each.

I think they'll be shredded. You can't get charges like that onto copper cubes. There will be all kinds of relativistic effects and you won't be able to ignore the special effects of accelerating charges, which will lead to radiation and additional forces.

Sorry, I didn't consider the size of the force. Speed of the cubes (since a = 1.8*109 m/s2) goes over the speed of light.
If we say that the force between the cubes is 2.7 N, then their charges must be too small.
Fr2/Ke = Q1 * Q2 = 2.7*102/(9*109) = 3*10-8

A more realistic example would be to calculate the motions of two charged particles (a proton and anti-proton, for example) which approach each other from a large distance.

That can help me too, but it may arise another question.
Thanks for the reply.

ashokanand_n
Atran,

The question you have formulated is not simple as it looks. And that is because there are lot of complexities associated with the situation.

The distribution of the charge (since they are free to move on a conductor) itself is dependent on the shape, relative orientation of the cubes. Plus the distribution is affected by the charge on the other cube. The force of interaction is in turn dependent on this charge distribution.

The accelerating charges will radiate energy as well.

All these things really complicate the situation and make it difficult to calculate the time evolution of the system. But the problem is in principle solvable.

What exactly is it that you want to know??

Is it the method to find out the dynamics of the system or is it some other fundamental doubts related to the problem which bothers you??

Atran
The question you have formulated is not simple as it looks. And that is because there are lot of complexities associated with the situation.

The distribution of the charge (since they are free to move on a conductor) itself is dependent on the shape, relative orientation of the cubes. Plus the distribution is affected by the charge on the other cube. The force of interaction is in turn dependent on this charge distribution..
If so, is it better to consider an electron and a proton instead of cubes? But it may be harder since the mass of the proton is different form the electron's.

What if I have an electron and a positron?
They have the same mass, and their shape (if they have any) would be neglected.

The accelerating charges will radiate energy as well.
I don't know how this works.

What exactly is it that you want to know??

Is it the method to find out the dynamics of the system or is it some other fundamental doubts related to the problem which bothers you??
I want to know how long does it take for charged objects/particles to move a certain distance [t = ?]? And if possible, knowing velocity in a given time would be helpful.
I want to learn the method, because I think the method can be also used for the gravitational force between two masses.

It's the first year of my gymnasiet (high school), and my physics book is very superficial, sometimes I read some definitions in other books or internet.
Therefore, I haven't worked with a changing acceleration before. I tried to make a graph of the changing acceleration today, but it didn't make much sense... I didn't thought about the graph for long because I drew it quickly and afterwards I went to another lesson/class. Tommorow, I have a test, therefore I don't have much time for physics today.

Thanks for the reply.