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Coulomb's law is strict:

$F=k\dfrac{q_1*q_2}{r^2}$

that means that between two charges occurs force. I.e. occurs force on $q_1$ and on $q_2$.

If there are two electrons in vacuum with 300 00 000 meters distance between them, one electron feels force due to another, and another should instantly feel force due to first electron.

How does it possible? You probably will say something about energy propagation, speed of light, but it is not actually a light. Does it occurs instantly or not, and why?

The general idea of this question is the fact that there are limits only for propagation of changable processes, as far as I know(like em waves). Coulomb's force is about two fixed charges in vacuum, there are no moving.

And the second idea -, even if, let's say, there is a need in time to one electron affect another - when it happens, first electron, according to law above, should instantly feel force.

Qmechanic
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    Possible duplicate of The propagation of electric field – Jasper Oct 21 '18 at 21:10
  • "Does it occurs instantly or not, and why?" - I honestly don't understand what it is. In your second paragraph, you stipulate that there are two electrons 300 00 000 meters apart. What do you mean by "and another should instantly feel force to to first electron"? If the distance between the electrons is constant, the force between the electrons is constant and so what does it mean to 'instantly' feel a constant force? You have not made any mention of the distance changing so what is it that you're asking about? – Alfred Centauri Oct 21 '18 at 22:55

3 Answers3

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Coulomb's Law is an electrostatic approximation (in which we assume the charges have always been where they are, and don't move) of a broader concept: the force a charge feels is the direct result of the electric and magnetic fields produced by other charges.

One of the most important facts in electromagnetism is the following statement: changes in the electromagnetic field travel at the speed of light. This makes sense, because light itself is a propagating change in the electromagnetic field.* This also means that the electromagnetic field produced by a charge encodes its history: for example, if it was moving in the past and is stationary now, then there is a region closer to the charge where the field is consistent with a stationary charge, a region further away from the charge in which the field is consistent with a moving charge, and a region of electromagnetic waves in between that were produced during the necessary acceleration; this boundary between the two regions expands outward at the speed of light.

So, to answer your question, you have to look at the history of the charges. Coulomb's Law is only valid if the charges' electric fields are pointing radially outward from their current position across all of space; this implies that both charges have been sitting in their current positions for all time (because if they weren't, then there would be a certain distance away from them where this wouldn't be true). If you were to quickly move both charges into their current positions at some point in time, it would take a nonzero amount of time for the resulting change in the electric field (from "no field" to "the electric field from a stationary charge") to propagate outward, and as such it would take some time for the two charges to feel force on each other.

So, long story short, if Coulomb's Law is assumed valid, then the charges have always been where they currently are, and the question of whether one feels the force of the other "instantly" is a moot point.

*Note that, while all changes in the electromagnetic field propagate at the speed of light, not all changes in the electromagnetic field correspond to the emission of electromagnetic radiation (i.e. "light"). In particular, only changes that propagate significantly to infinity are classified as such (in other words, only accelerating charges produce electromagnetic radiation, not charges moving at constant velocity).

probably_someone
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  • Charges moving at constant speed do not generate a propagating change in the electric field, due to the principle of relativity. Is it the charge that is moving or is it the observer? –  Oct 21 '18 at 22:26
  • @Wolphramjonny Charges that are moving at constant speed in a particular frame do produce propagating changes in the electric field, but these changes do not propagate to infinity (they "move along with the charge" in the near field). As such, they do not carry away energy and are not electromagnetic radiation. This is the difference I tried to make clear in the * section. – probably_someone Oct 21 '18 at 22:33
  • While this answer is correct, I feel like I answered somewhat the same question foe example about where you are that Coulomb's law is valid, then charges (fields) have always been there and it is assumed instant. I too answered that changes in the field propagate at speed of light. And I specifically answered his question about the charges being 300 000 km away, and how long it takes to effect. Now I do not understand the downvotes. And I corrected as per your comments. – Árpád Szendrei Oct 21 '18 at 23:20
  • Energy can be carried away without electromagnetic radiation by virtual photons. They don't exist, they are not radiation, but sure enough they do carry energy and momentum. For example, if you hold two repulsing charges at a constant distance and then release them, they would start moving the instant you let them go without waiting for any radiation from the other. What do you say? – safesphere Oct 22 '18 at 06:12
  • About the constant moving charge fields. When I was playing with moving charge simulators like this https://phet.colorado.edu/sims/radiating-charge/radiating-charge_en.html, I actually was wondering about why when charge is moving, although with constant speed, there are no "ripple" in the field, that should be. –  Oct 22 '18 at 06:27
  • You write, that these changes do not propagate to infinity, they "move along with the charge"(as in simulator) and it is a "near field". Can You detail? –  Oct 22 '18 at 06:28
  • @ÁrpádSzendrei Those are likely because of your unconventional identification of "electric field" with "near field." – probably_someone Oct 22 '18 at 09:00
  • @safesphere You don't need to bring virtual photons into this at all. They only make things more confusing, in a question which can be easily answered within the realm of classical electromagnetism. If you hold two repulsing charges at a constant distance, they are repulsing because they are each being acted upon by the electric field of the other charge. When you let them go, this repulsion continues, because the electric field is already there (if it wasn't, then the charges wouldn't be repulsing when you held them). There is nothing that needs to travel in that situation. – probably_someone Oct 22 '18 at 09:05
  • @safesphere In fact, this situation illustrates the basic principle I've been trying to get across in my answer this whole time: forces come from fields, not from charges. – probably_someone Oct 22 '18 at 09:06
  • you are right, I missinterpreted your writting –  Oct 22 '18 at 12:41
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You actually do, the classical-classical model studied in electrostatics doesn't take it into account. In reality it's not instant as Quantum Field Theory (QFT) tells us. For electrostatic potential, it travels usually at the speed of light through the medium.

Photons, the elemental particles of "light" are the "middle men" of the electromagnetic force. Light is way more general than just "visible light", When you have two charges, the only way they know about each other is through photons.

zivo
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    Classical electrodynamics also tells us that changes in the electromagnetic field do not propagate instantly. That's the whole point of Maxwell's equations defining a finite speed of light, and this is clearly evidenced in any problem involving retarded potentials. No need to bring QFT into this. – probably_someone Oct 21 '18 at 21:37
  • @probably_someone, I thought he wanted a physical explanation. I for one never got the physical side of the finite propagation in Maxwell's equations. QFT's physical explanation does make sense, though. – zivo Oct 21 '18 at 21:45
  • Your second paragraph is incorrect. Light or photons do not mediate electromagnetic interactions. Virtual photons do. The difference between real and virtual photons is that real photons exist, but virtual do not. Also the electrostatic potential does not travel. Otherwise it would not be static. – safesphere Oct 22 '18 at 05:45
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First, you are asking about the electron's electric field, and we should distinguish between the near and far field.

The electrons' electric field's (near field) force is mediated by virtual photons, they are just a mathematical way to describe this force. These are not real photons and are not real particles.

Though, the electric field (near field) extends with the speed of light, so it is not instantaneous.

In your case, since the electrons are stationary, and are already there, they already have a near field extended (around a certain position in space), and if the two electrons are close enough, they will feel measurable repulsion as you suggest.

Now the reason they repel each other instantly is because the near field they create is already there, always, and cannot be turned off. This way you cannot test the speed of these interactions.

The way to test what you are asking for, is with electric charges (artificial, man-made), that can be turned off/turned on. Once you put two of these charges close enough, you can test, what you are asking, that is how fast will these near fields interact?

So your question is basically, after you turn on the charges, how soon will they interact and repel?

The answer is, after you turn on the charges, the near field (electric) extends in spherical shape in space with the speed of light. So depending on their distance, it might take a while until they feel the other one's near field.

Let's disregard that in your question you are talking about distances of 300 000 km, which is too far away for man-made electric charges to create a near field strong enough to feel measurable repulsion. Let's say that these charges are 300 000 km apart. Let's say that we have strong enough electric fields, that will create a near field strong enough so that it will have a measurable effect 300 000 km away on the other charge.

Now let's say that we can time the charges to turn on simultaneously at t0. The near fields of the charges will extend with speed c. The two fields will not meet half way, at 150 000 km distance, because of the correct comments and because two near fields will not effect each other, but they will obey the principle of superposition, they will add without interacting.

So the field of one electromagnet will reach the other electromagnet at 300 000 km, in 1 second, and will start repelling it.

Árpád Szendrei
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  • The "near fields" don't "effect each other." The electric fields from two charges obey the principle of superposition, i.e. they add without interacting. – probably_someone Oct 21 '18 at 21:43
  • @probably_someone oh I should have written electromagnets that repel and can be turned on/off. Is the reasoning in the case of electromagnets correct at the speed of light and 0.5seconds? – Árpád Szendrei Oct 21 '18 at 21:45
  • No, the fields still add without interacting. This is a basic property of classical electromagnetic fields, which (irrelevant to this question) carries over into QFT in that there are no photon-photon interactions at tree level. – probably_someone Oct 21 '18 at 21:48
  • @probably_someone thank you, but then how can we explain to him in everyday language, that two electromagnets will repel, after 0.5seconds? – Árpád Szendrei Oct 21 '18 at 21:49
  • It doesn't take 0.5 seconds. For 300,000 km, it takes 1 second for one electromagnet to feel force from the other electromagnet. When the electromagnet is turned on, the current induces a change in the electromagnetic field, which propagates outward at the speed of light. The other electromagnet feels force when its moving charges encounter the magnetic field of the other electromagnet. As such, the field of the other electromagnet must propagate 300,000 km, to reach the charges of the other electromagnet, before any force is felt. – probably_someone Oct 21 '18 at 21:55
  • +1 for the edit. – safesphere Oct 22 '18 at 05:52
  • @probably_someone, what You are talking about is em radiation, not a static electric field. –  Oct 22 '18 at 06:30