# Potential energy stored in a system of point charges

1. Sep 12, 2014

### davidbenari

The basic equation is as follows:

U=∑kqiqj/Rj with (i≠j)

My question is how is this definition useful? What makes you want to say that is the potential energy of the system?

For example, I thought that it is useful because if all particles change in position then you could evaluate that equation to find the total work done. However, this wouldn't be true because all of the charges would move in that hypothetical situation.

So why is this useful? How is it derived, or proven?

Thanks.

2. Sep 12, 2014

### Khashishi

If you start with a collection of charges at infinite distance apart and start with one charge in the center, this is the work it takes to assemble the composite object by moving charges in one by one.

3. Sep 12, 2014

### davidbenari

Nice. What I don't like is the reference to a static charge (one charge at the center). Is this a problem? Or is there a way to avoid it? Thanks.

4. Sep 12, 2014

### Staff: Mentor

Start with all the charges at (practically) infinity, so there's (practically) no electric field. Now bring in one charge to its final position. How much work does that require?

5. Sep 12, 2014

### davidbenari

Nice. But of what use is this? I mean, sure it takes that amount of work to get the final configuration if I'm assembling the system piece-by-piece, but I don't see how this is "potential energy" because I don't see how this energy is going to be released:

If all charges are separated from each other simultaneously, I think the calculation is going to be too complicated and my "potential energy" equation will be of no use.

So how is this energy stored? And how can it be released?

Thanks.

6. Sep 13, 2014

### DivergentSpectrum

the work energy theorem states that potential energy= -∫Force dx can be converted to kinetic energy= m(dx/dt)2/2, and back again.
with this in mind, you can add potential energy to the kinetic energy and it will always be the same number.

in other words,
m(dx/dt)2/2-∫Force dx=constant.
coulombs law says that force=kQ1Q2/x^2 where x is distance and k is coulombs constant
then ∫Force dx =-kQ1Q2/x
and energy= m(dx/dt)2/2+kQ1Q2/x
so, you can plug in the initial speed and velocity to find the energy, then if you know the position at any time you plug that into the energy equation and find the velocity, and vice versa.

really these quantities work and energy and such are more of a mathematical shortcut than any type of concrete thing.

7. Sep 14, 2014

### voko

This sounds as if "energy" were less concrete than "force". This is false. Force is also a mathematical abstraction, and it is possible to formulate electrodynamics without using this concept.

8. Sep 14, 2014

### Staff: Mentor

If you add this quantity to the total kinetic energy (sum of $(mv^2)/2$ for each particle) you get the total energy of the system. That's pretty much the definition of "potential energy", so that's what we call it.

9. Sep 15, 2014

### DivergentSpectrum

i was just referring to the general notion of energy in the vernacular- ie if the energy is negative you have a closed orbit and if its positive you get a deflection. im sure its as "real" as any other mathematical quantity

10. Sep 20, 2014

### davidbenari

voko: Voko what you said has been very interesting to me the last few days (about formulating electrodynamics without using the concept of force), I've done a cursory search on google for this and can't find any info. Do you think you can help me find anything about this? Thanks!

11. Sep 21, 2014

### voko

The keywords here are "Lagrangian" and "Least Action". An example of that can be found in the Course of Theoretical Physics by Landau & Lifschitz, vol. 2, even though it is not completely "pure" in that respect. This is advanced stuff, however, introductory EM texts always follow a force-centric approach.