I've been reading through (and listening to) a collection of lectures form Richard Feynman's Lectures on Physics. In lecture #2, titled "Basic Physics", he makes the following statement:

Although the forces between two charged objects should go inversely as the square of the distance, it is found, when we shake a charge, that the influence extends very much farther out than we would guess at first sight. That is, the effect falls off more slowly than the inverse square.

This statement is in the context of a discussion of the attraction between objects with opposite [static] electrical charges. I am aware of the inverse square law as it applies to static electrical fields, but I was under the impression that the inverse square law also describes oscillating electrical fields, that is, electromagnetic radiation.

Assuming a simple point-source omnidirectional EM radiator, the amplitude of the outwardly propagating EM field (a.k.a. EM "wave") should also fall off according to inverse square law in free three-dimensional space, correct?

Thus, I do not understand his statement "the effect falls off more slowly than the inverse square". Did Fenyman simply misspeak or am I missing something (perhaps embarrassingly obvious) here?

The full text of the lecture in question can be found here. You can search for the phrase "more slowly than the inverse square" if you'd like to see the immediate context.

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    $\begingroup$ See Dipole radiation $\endgroup$ Commented Dec 13, 2016 at 19:29
  • $\begingroup$ @AccidentalFourierTransform I am not sure that "dipole radiation" completely addresses my question, where I am considering an [ideal] omnidirectional point-source radiator. But it is interesting. :) $\endgroup$ Commented Dec 13, 2016 at 21:27

3 Answers 3


I am aware of the inverse square law as it applies to static electrical fields, but I was under the impression that the inverse square law also describes oscillating electrical fields, that is, electromagnetic radiation.

The inverse-square law applies to the intensity of electromagnetic waves that are propagating outwards. You can see this by drawing an imaginary sphere of radius $r$ around a point source of radiation. Since the power crossing that sphere is the same, regardless of the size of the sphere, it must be the case that the intensity of the waves (power per area) is proportional to $1/r^2$.

But the intensity $I$ of an electromagnetic wave is proportional to the square of its electric field amplitude $E$. Since $I \propto E^2$ and $I \propto 1/r^2$, we conclude that $E \propto 1/r$ for a point source of EM radiation.

(This result can also be proven by much more rigorous means, but this is a quick-and-dirty heuristic to see why this must be so.)

EDIT: In response to a couple of comments in the extended discussion (now moved to chat): It is true that Feynman never actually mentions the electric field in the above paragraph. However, he is talking about the force between two charged objects, and the force on the stationary charge will be equal to the amount of charge it carries times the electric field created by the oscillating charge. Since the electric field falls off as $1/r$, the force felt by the stationary charge will be proportional to $1/r$ as well, i.e., "the effect will fall off more slowly than the inverse square."

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented Dec 13, 2016 at 23:46
  • $\begingroup$ @ACuriousMind some of the comments were specifically in regard to clarifying Michael Seifert's answer. Those comments should not have been moved to chat. $\endgroup$ Commented Dec 14, 2016 at 14:21
  • $\begingroup$ @KevinH.Patterson I recognized that the comments actually pertained to the answer, which is why I moved them to chat instead of deleting them outright. However, there's no use in keeping more than 15 comments around in plain view for others to sift through unless they really want to (in which case they can still read them in chat). $\endgroup$
    – ACuriousMind
    Commented Dec 14, 2016 at 15:27
  • $\begingroup$ Does the argument provided in this answer rely on drawing an imaginary sphere? Would the same argument still work if we chose to draw an imaginary cylinder, for example? Is the idea to draw any consistent shape that is large enough for the full wave to intercept the surface during a designated time Δt? $\endgroup$
    – jdphys
    Commented Apr 22, 2020 at 17:09
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    $\begingroup$ @RohitShekhawat: That's where the "much more rigorous means" come in. Showing that the $1/r$ piece only depends on the acceleration of the charge (and not on its velocity) can be done; see, for example, Griffiths or Jackson for the details. But the question wasn't asking for that; it was asking "does the electric field always obey an inverse square law?" and the answer to that is "no, not for accelerating charges", as I've explained above. $\endgroup$ Commented Dec 20, 2022 at 18:20

The inverse square law applies to the case of electrostatics, i.e., when the charges are stationary. What Feynman is trying to say is that, when you 'shake' a charge, the inverse square law does not apply, since the system isn't static anymore. If you shake a charge in a particular way, you find that the force will fall off as $$\frac{1}{r}$$ which is slower than inverse square.

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    $\begingroup$ Welcome on Physics SE :) You can find a help section on the TeX implementation of this website here =) $\endgroup$
    – Sanya
    Commented Dec 14, 2016 at 21:13
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    $\begingroup$ The start of a very nice explanation... maybe a few pointers or references as to the difference between electrostatics and the dynamic system you are describing, especially what is meant by "shake a charge in a particular way", and why that would be. $\endgroup$ Commented Dec 4, 2017 at 0:35

This is taught by Feynman in Chapter 28:


All the equations are there too.

  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Commented Oct 29, 2023 at 18:13

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