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Is it correct to say that electro-static potential of a charge is the energy of a motionless charge?

I ask this to better understand this (great) answer;

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    $\begingroup$ Does electrostatic potential have the right dimensions to be energy? $\endgroup$ – G. Smith Sep 5 at 18:56
  • $\begingroup$ Are you speaking of electro-static potential or electrostatic potential energy? $\endgroup$ – Bob D Sep 5 at 18:58
  • $\begingroup$ @BobD I don't know myself; I quote from the answer in the linked session: practically we can't measure the (electrostatic potential) energy of a charge. $\endgroup$ – JohnDoea Sep 5 at 19:06
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    $\begingroup$ @JohnDoea I have added an answer which touches on this side remark at the end. Feel free to ask clarifications regarding any part(s) of the question. Hope it helps! :) $\endgroup$ – Dvij Mankad Sep 5 at 23:45
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    $\begingroup$ @DvijMankad thank you, $\endgroup$ – JohnDoea Sep 6 at 4:19
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Is it correct to say that electro-static potential of a charge is the energy of a motionless charge?

If you are speaking about electrostatic potential energy then a motionless charge, as well as a charge in motion, can have an electrostatic potential energy. Just like a motionless mass $m$ a height $h$ above and close enough to the surface of the earth that the acceleration due to gravity is constant, can have a gravitational potential energy of $mgh$. If you are talking about the electrostatic potential, then that would be the electrostatic potential energy per unit charge, or in the case of gravitational potential energy, the gravitational potential energy per unit mass.

The reason why your linked session says "practically we can't measure the (electrostatic potential) energy of a charge" is that they are talking about an absolute value for this energy. In general there is no absolute potential energy of a charge unless an absolute value of potential energy (or potential) is arbitrarily assigned a value of zero to some position.

Consider the gravitational potential energy analogy. If a motionless mass $m$ is a height $h$ above the floor in a room, it is said to have gravitational potential energy of $mgh$ with respect to the floor of the room. But if the floor of the room is also a height $h$ above the ground outside the building containing the room, the mass has a gravitational potential energy with respect to the ground outside of $2mgh$. If there is a surface of table in the room a height $1/2$ h from the floor, the mass has a gravitational potential energy of $(1/2)(mgh)$ with respect to the surface of the table and the floor of the room.

I think you get the idea.

Hope this helps.

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Before directly addressing your question, I would like to elaborate a bit on the concept of the electrostatic potential energy. There are two notions of electrostatic potential energy in electrostatics. The first is a "potential-based" definition and the other is a "field-based" definition.

So, according to the first definition, the electrostatic potential energy of an electrostatic configuration is given by $\int V(\vec{r})\rho(\vec{r})d^3{\vec{r}}$ where $\rho(\vec{r})$ is the charge density at position $\vec{r}$ and $V(\vec{r})$ is the electrostatic potential at position $\vec{r}$ and is given by $V(\vec{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|}d^3\vec{r'}$.

According to the second definition, the electrostatic potential energy of an electrostatic configuration is given by $\frac{1}{2\epsilon_0}\int |\vec{E}(\vec{r})|^2d^3\vec{r}$ where $\vec{E}(\vec{r})$ is the electric field at position $\vec{r}$ which is given by $\vec{E}(\vec{r})=\frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r'})}{|\vec{r}-\vec{r'}|^3}(\vec{r}-\vec{r'})d^3\vec{r'}$.

Obviously, the two definitions can be shown to be equivalent. In particular, it can be easily shown that $\frac{1}{2\epsilon_0}\int |\vec{E}(\vec{r})|^2d^3\vec{r}=\int V(\vec{r})\rho(\vec{r})d^3{\vec{r}}$. However, the two definitions are philosophically quite different. The first suggests that this electrostatic potential energy is "located" where ever the charge density $\rho(\vec{r})$ is non-zero. On the other hand, the second definition suggests that the electrostatic potential energy is "located" where ever the electric field $\vec{E}(\vec{r})$ is non-zero. This distinction is important in talking about the electrostatic potential energy of a charged particle because the former view tells us that such potential energy is really located at the position of the particle whereas the second view tells us that such potential energy is really just the potential energy associated with this charged particle by virtue of its electric field (and that this energy is actually located throughout the space with an energy density proportional to $|\vec{E}(\vec{r})|^2$). Keep in mind that no matter which way one calculates it, the answer would be the same, but, "pictures" associated with the two approaches are distinct.

So, is the electrostatic potential energy of a charged particle just the energy of a charged particle at rest? Well, in the first picture, it unambiguously is! In the second picture, it is the total energy of the configuration which is implied by a charged particle at rest. So, it is the total energy associated with the charged particle at rest, but, it doesn't really reside within the charged particle.

Now, there is no way to distinguish between these two pictures in electrostatics. However, from the full theory of classical electrodynamics, we know that the second approach, i.e., the field approach is the correct one. In fact, there is some ambiguity about the uniqueness of the field-based definition of energy density but everyone has a good amount of confidence that it is correct, for good reasons. See, Section 27.4, Volume II, Feynman Lectures on Physics.

So, we finally conclude that the electrostatic potential energy of a charged particle is the total energy of the configuration which is implied by a charged particle at rest. In other words, it is the total energy associated with the charged particle at rest, but, it doesn't really reside within the charged particle.


The notion of the "electrostatic potential energy of a charged particle" is a ridiculously irritating concept than you might imagine. If you treat a charged particle as a truly point-like particle, its electrostatic potential energy turns out to be infinity. So, one imagines a charged particle to be actually a spherical shell or a spherical ball. If one makes this move, one is forced to address the question of the structure of this sphere which opens up a whole different can of worms. There is also another very attractive but failed idea associated with the electrostatic potential energy of a charged particle, the idea of the electromagnetic mass. It was thought that if a charged particle at rest has some energy associated with it then, according to $E_0=mc^2$, this energy should show up as the mass of the charged particle. This thought led people to propose that the mass of the electron must be coming from this potential energy associated with it. Apart from being wrong (in modern understanding, the mass of the electron, which is a pointlike particle, comes from the Higgs mechanism), the idea is riddled with internal paradoxes such as the famous and the notorious $4/3$ problem.

All these issues, with the exception of the $4/3$ problem, are solved by replacing classical electrodynamics with quantum electrodynamics as the underlying theory of electrodynamics. The solutions to the $4/3$ problem usually appeal to a careful understanding of the concepts of relativity itself rather than recoursing to the quantum electrodynamics.

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