First I'd like to explain why I don't think this question is a duplicate of the many other questions about the electric potential and potential energy. Despite there being many questions and answers on the forum about this topic, many just further the confusion for me and did not seem to help (I will mention some of them throughout this). In addition, I also have questions about some formulas regarding the electric potential, potential energy, and energy/charge density which I have not seen answered at all.

It is also important to note that these questions deal purely with electrostatics.

From what I understand, the relationship between the electric potential and potential energy is the same as that of force and the electric field: one is for a specific charge, and one is for a standard unit positive charge. The formula connecting them is given in this answer, namely $U = q \cdot V$. But something else that I've heard a lot is that "potential energy is a property of the system". This appears in the same answer mentioned above, and also this one. This seems completely contradictory to me. The electric potential is a function of position. And so if $U = q \cdot V$, it must also be a function of position, not some constant of the system. My confusion about this is only furthered by the formulas discussed below.

At the start of the course, we saw this formula for potential energy:

$$U = \frac{1}{2} \sum_{i \neq j} k\frac{q_i q_j}{r_{ij}}$$

And that the potential energy for a particle is

$$U = \sum_{other-particles} k\frac{q_i}{r_{i}}$$

If I had to guess despite my confusion, the former is for the energy of the entire system and the latter is for the potential energy of a particle (which somehow both exist).

We later saw this formula:

$$u = \frac{1}{2}\epsilon_0\left|E\right|^2$$

Where $u$ is the potential energy density.

This raises more questions for me. Firstly, triply integrating this yields the energy not of the entire system nor at a single point in space. In addition, I struggle to find meaning in the fact that you can have non-zero energy in a region in space with no particles, only a field. If someone has a good explanation for this, it would help. I also imagine that by triply integrating over all of space, you would get the formerly mentioned "energy of the system". What is the relationship between this formula and the $\sum$ sum discussed above? If you were to compute both, would they be equal? And is there a way to use this energy density formula to find the energy for a specific particle? The 3 last questions also apply to the following formula:

$$U = \int_{\Bbb R^3} \rho(\vec r)V(\vec r)d\vec r$$

  • $\begingroup$ Too much text, could you trim it out to the essential part? And also editing the title suitably would help. $\endgroup$
    – Babu
    Commented May 13, 2023 at 19:42
  • $\begingroup$ Your $U = \sum_\text{other-particles} k\frac{q_i}{r_{i}}$ does not have units of potential energy. Do you mean electric potential? The electric potential (volt) is not potential energy (joule). Also, you are using the same symbol $U$ for many different things. Please edit your question and be clear with your notation. $\endgroup$
    – Themis
    Commented May 13, 2023 at 19:45
  • $\begingroup$ I had typed up a very long answer in the earlier version you deleted, right before I could post it. $\endgroup$ Commented May 14, 2023 at 14:09
  • $\begingroup$ What earlier version? I haven't changed the question. $\endgroup$
    – Remeraze
    Commented May 14, 2023 at 17:27

1 Answer 1


It is important not to conflate $U = qV$ and $$U = \frac{1}{2} \sum_{i \neq j} k\frac{q_i q_j}{r_{ij}}.$$In general they don't mean the same thing. The potential energy as we normally use it is a property of the entire configuration of charges. It is the work that must be done to assemble the configuration, by bringing each of the constituent charges in from infinity.

$\Delta U=qV$ (where I've added $\Delta$ to avoid confusion) should be understood as the work that must be done to bring a charge $q$ in from infinity to a location at which the electrostatic potential is $V$, with the positions of the all the other charges kept fixed. We sometimes eliminate the requirement that the other charges stay fixed and just say this works when $q$ is a small "test charge", small enough that it doesn't cause the other charges to rearrange. With this understanding, $\Delta U$ is just the increase in the system's potential energy when you add the charge $q$.

In electrostatics, the energy can be viewed as being stored in the charges or in the field, both are valid: see this answer. The integral of $u=\frac12\epsilon_0E^2$ gives the potential energy of the system. It can give result that differs from energy calculated with $u=\frac12\rho V$ (or the sum over discrete charges) by a constant value, but remember that potential energy is only really uniquely defined up to a constant, and all we can measure is differences in potential energy (by the work-energy theorem). Poynting's theorem strongly motivates thinking of energy as being stored in the field with $u_E=\frac12\epsilon_0 E^2$: this applies in time-varying situations as well. In particular, keep in mind that energy transfer can occur via radiation, even though no charges may be directly involved in this process.

You can't really use $u = \frac{1}{2}\epsilon_0 E^2$ to calculate the "self-energy" of a single point charge: this will give you infinity. This is of little concern in practice. You can conveniently sweep the problem under the rug by considering the charge as a tiny, uniformly charged sphere; this allows you to calculate a finite self-energy. This energy doesn't change as you move charges around, so it's just a constant. To more concretely illustrate how the "potential energy in the field" $U_\text{field}$ differs from the "potential energy in the charges", if you have two charges $q_1$ and $q_2$ separated by $r$, we can write $$U_\text{field}=U_\text{1,self} + U_\text{2,self}+k\frac{q_1 q_2}{r}$$ where $U_{i\text{,self}}$ is the self-energy of charge $i$.

  • $\begingroup$ Thank you very much, this answer is very helpful. I just want to confirm some conclusions I got from it. The very last formula you wrote seems to suggest that the sigma sum does not accurately describe the potential energy. But this does not matter since the separation is a constant for some system of particles, and only the change in potential matters, yes? And also, the $U=qV$ formula is simply a description of how much energy it would take to move a charge to a certain position in the already existing (and non-changing) system? $\endgroup$
    – Remeraze
    Commented May 13, 2023 at 20:52
  • $\begingroup$ @Remeraze Your descriptions are correct, just one clarification. There isn't really one correct potential energy of the system; we have two different, correct ways of expressing it. They just differ by a constant, but like you said this doesn't matter. The sum over charges is also a valid formulation of potential energy. The one in terms of the field is sometimes preferred because it also applies to problems with time variation. $\endgroup$
    – Puk
    Commented May 13, 2023 at 20:58
  • 1
    $\begingroup$ Got it. Thanks a bunch!! $\endgroup$
    – Remeraze
    Commented May 13, 2023 at 21:08

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