Original Question:

The potential energy stored in a system of $n$ charges is:

$$\frac{1}{2}∑_{i=1}^{n}q_i ∑_{j≠i}^{n}\frac{kq_j}{r_{ij}}=\frac{1}{2}∑_{i=1}^{n}q_i \phi(q_i)$$

In the above, $n$ is the number of charges, $r_{ij}$ is the distance between the $q_i$ and $q_j$, and $\phi(q_i)$ is the value of the potential function due to the other $n-1$ charges at the location of $q_i$.

I've seen the usual derivation, where this "total energy" is made equivalent to the energy that the system of charges would do if we let the charges fly away, one by one, from the original distribution. It would be given by:

$$\frac{1}{2}∑_{i=1}^{n}q_i ∑_{j≠i}^{n}\frac{kq_j}{r_{ij}}=∑_{i=1}^{n}q_i ∑_{j>i}^{n}\frac{kq_j}{r_{ij}}$$

The factor of $\frac{1}{2}$ is introduced to the left sum because of the $j>i$ requirement in the right sum, accounting for the fact that once the $i_{th}$ charge flies away, it won't be contributing to the work done to the charges still in the system, so we don't want to count it.

However, I'm confused about the meaning of the $\frac{1}{2}$. At first, I thought it was just a way to account for the "double-counting" in the left sum. But, then I read somewhere that it implies that each of the charges have half of the potential energy they would if the other charges were fixed in space.

I can't really see this intuitively.

For example, say we have two charges $q_1$ and $q_2$.

If $q_1$ were fixed in space, the work that it would do on $q_2$ as we let $q_2$ fly infinitely far away would be $\frac{kq_1q_2}{r_{12}}$.

Similarly, if $q_2$ were fixed in space, the work that it would do on $q_1$ as we let $q_1$ fly infinitely far away would be $\frac{kq_1q_2}{r_{12}}$.

However, why is it that if we let them both fly away at the same time, the work that would be done on each would be $\frac{1}{2}$ of $\frac{kq_1q_2}{r_{12}}$?

How can we show this from the fact that both particles fly away in opposite directions at the same speed? I can't see why this leads to half the work done on each...

Would it always lead to half the work done on each? What if the forces of the charges on one another weren't inverse square?

And does the same argument apply in the general case, with more than 2 charges? If we let them all fly away from one another at the same time, do they each get half the energy than they would if we let them fly away, while holding the rest in place?


Edit 1:

After reading over my question, I realized that more fundamentally, my question is the following:

Why is it that, when calculating the total energy stored in a system of charges (aka the total work done on the charges if we let them all fly away from the original distribution), we can let them fly away one by one and sum over the work done on each of them, or let them all fly away at the same time, and the total work done would be the same in either case?

Edit 2:

I understand the explanation that the energy must get symmetrically distributed once we accept conservation of energy...but, how can we derive the same result had we not accepted it from the get-go? Take a look at my comment under Aaron's response!

Thanks again.


1 Answer 1


I think looking at a system of just two point charges is sufficient.

Let's say we have two point charges each of charge $q>0$ separated a distance $r$. Then the potential energy relative to infinity is $$U=\frac{kq^2}{r}$$

Let's fix one charge and let the other charge fly away. Then by energy conservation, the change in potential energy of the system is equal to the negative change in kinetic energy of the moving charge $\Delta U=-\Delta K$. By the work-energy theorem this means that

$$\Delta K=-\Delta U=W$$

Where $W$ is the work done on the moving charge. No work is done on the other charge since it does not move.

Now let's let both charges fly away. We start off with the same amount of potential energy, but this now has to be distributed across both charges in terms of their kinetic energy. Therefore, for each charge $$\Delta K_i=-\frac12\Delta U=W_i$$

If you added up all of the $W_i$ then you get back $\sum W_i=-\Delta U$, so we get the same thing either way.

So, to summarize, we get the same work done either way due to energy conservation. The more charges that can move, the less work is done on each charge$^*$, but you now have more charges that work is being done on. So either way, you get the same total amount of work done.

With general distributions of more than two charges of various charge values you just apply the same reasoning to each pair of particles, making sure to not double count.

$^*$In thinking of work done being $W=\int\mathbf F\cdot\text d\mathbf x$ this makes sense. If both charges move away from each other then the force drops of faster than if only one charge moved, as the force weakens as the charges move father apart.

  • $\begingroup$ Thanks Aaron...yep, I can see that it would work, like you mentioned at the end. But, how can we mathematically show it so, without just accepting conservation of energy? Let's say that they start at a distance of $r$ from one another. How can we show that at as they push one another away, and do work on one another, the fact that they're both flying away at the same rate, makes them end up with a fourth of the velocity (since they have half the energy) than a single one would if it flew away, without simply assigning half the energy to each, but following from Newton's Laws? Thanks! $\endgroup$ Commented Mar 22, 2020 at 19:45
  • $\begingroup$ @JoshuaRonis I'll ping you when I've edited that in there. Might be a while though $\endgroup$ Commented Mar 22, 2020 at 20:23
  • $\begingroup$ Thanks Aaron! Looking forward to it! $\endgroup$ Commented Mar 22, 2020 at 20:27
  • $\begingroup$ Hey Aaron...any time recently? Don't mean to rush u...take your time, just, I'm still really curious. Thanks man! $\endgroup$ Commented Mar 27, 2020 at 22:05

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