Talk:Coulomb's law

The equation show V=e^2/( 4 pi r). Check me here, but V is proportional to e, not e^2??? Olness (talk) 16:01, 27 September 2023 (UTC) Reply

@Fgnievinski In this edit you removed

• Two charges can be approximated as point charges, if their sizes are small compared to the distance between them.

which was sourced to

• Srinivasan, M. V. (2025). Physics Part - I. National Council for Education Research and Training (NCERT). p.20. ISBN 978-81-7450-631-3.

Your edit summary was

• move definition of point charge to respective article and follow NIST Guide rule 7.13 (Distinction between an object and its attribute)

What is "NIST Guide rule 7.13"? Is this content invalid? I'm fine if you think this sentence is not relevant to the paragraph, but you left the source. Why? Johnjbarton (talk) 04:09, 18 December 2025 (UTC) Reply

The definition of "point charge" was moved to the corresponding article, Point charge, so as to avoid cluttering the lead of this separate article (an explanatory footnote might be a more acceptable way of repeating the information in the present article).

NIST Guide rule 7.13 advices against referring to a point charge, a charge carrier, or a charged body as a "charge".

fgnievinski (talk) 04:18, 18 December 2025 (UTC) Reply

You left the source: Why? That was my main question all along. Johnjbarton (talk) 04:24, 18 December 2025 (UTC) Reply

I was editing on a cellphone. I removed the source now, thanks! fgnievinski (talk) 04:32, 18 December 2025 (UTC) Reply

I didn't find any distinction between a experimental law and simply a law. Maybe there is a distinction but from my current evidence, i would think that the term "experimental law" could create the feeling that there is a specific class of laws which Coulomb's law belongs to despite not being the case. Before editing the introductory definition, i'm open to any opinion on the matter. Tucopu (talk) 11:38, 10 May 2026 (UTC) Reply

Law is ambiguous, so "scientific law" or "physical law". Johnjbarton (talk) 18:56, 10 May 2026 (UTC) Reply

I agree that "law" is ambiguous. I will edit it for "scientific law". Tucopu (talk) 20:54, 12 May 2026 (UTC) Reply

On the current CODATA web page, the value of ${\displaystyle \epsilon _{0}}${\displaystyle \epsilon _{0}} is given as ${\displaystyle \epsilon _{0}=8.8541878188\times 10^{-12},円{\textrm {F}},円{\textrm {m}}^{-1}}${\displaystyle \epsilon _{0}=8.8541878188\times 10^{-12},円{\textrm {F}},円{\textrm {m}}^{-1}}. If I use the full precision calculator (https://www.mathsisfun.com/calculator-precision.html) to derive the Coulomb constant with the expression 1/(8.854 187 8188e-12)/4/pi, I get 8987551786.170798670... which does not agree with that is there now. How was this calculated? Mgolden (talk) 21:31, 22 July 2026 (UTC) Reply

I corrected the content to match the given source from 2006. This is not satisfactory but at least it can be verified.

Unfortunately the full story post 2019 is surprisingly complicated. It will take some time to sort out. This source

• Haug, E. G. (2022). Potential simplification of charge and the Coulomb force without affecting predictions. Journal of Applied Mathematics and Physics, 10, 3003. https://doi.org/10.4236/jamp.2022.1010201

points out that Coulomb constant is just (exact) ${\displaystyle k_{e}=c^{2}\times 10^{-7}}${\displaystyle k_{e}=c^{2}\times 10^{-7}} if you just use formula, but because of issues of precision in the 2019 SI works out not quite that clean. We need a source that clarifies this directly. Johnjbarton (talk) 00:12, 23 July 2026 (UTC) Reply

You can't use sources from before 2019 for this. The definition of the units was changed in 2019 and ${\displaystyle \epsilon _{0}}${\displaystyle \epsilon _{0}} and hence ${\displaystyle k}${\displaystyle k} is no longer an exact value, stated in many sources. I haven't found a direct source for ${\displaystyle k}${\displaystyle k}, but I haven't spent a long time looking. If you don't believe it's within policy to simply compute ${\displaystyle 1/4\pi \epsilon _{0}}${\displaystyle 1/4\pi \epsilon _{0}} from a reliable source, then the value shouldn't appear at all. It definitely shouldn't be an out of date value. Mgolden (talk) 13:02, 23 July 2026 (UTC) Reply

NIST provides this table: https://physics.nist.gov/cuu/pdf/wall_2022.pdf with ${\displaystyle \mu _{0}/(4\pi \cdot 10^{-7})=0.999,999円,999円,87円(16)}${\displaystyle \mu _{0}/(4\pi \cdot 10^{-7})=0.999,999円,999円,87円(16)}. From this, the Coulomb constant is ${\displaystyle k_{e}=c^{2}\cdot 0.999,999円,999円,87円(16)\cdot 10^{-7}{\text{N}}{\text{A}}^{-2}}${\displaystyle k_{e}=c^{2}\cdot 0.999,999円,999円,87円(16)\cdot 10^{-7}{\text{N}}{\text{A}}^{-2}}. Jähmefyysikko (talk) 14:17, 23 July 2026 (UTC) Reply

Ok cool, this solves the precision issue: the NIST source gives the precision and c^2 is exact. So I'd be ok with this source and basic math.

However, when I look at most sources I see Coulomb's law written with ${\displaystyle 1/4\pi \epsilon _{0}}${\displaystyle 1/4\pi \epsilon _{0}}, not ${\displaystyle k_{e}}${\displaystyle k_{e}}. And we don't find "Coulomb's constant" in the NIST table or pretty much anywhere else. I think we should down play the term in the article to match this. More like "sometimes called " with the engineering ref. Johnjbarton (talk) 17:23, 23 July 2026 (UTC) Reply

Ok I rewrote the whole section. I used 9 figures since I assume (16) means plus or minus 16 in the last positions but needs to be propagated.

I'd prefer to start the section with the value in SI units. Maybe just swap the two paragraphs.

I chose to write out the logic and source it because I constantly find these types of conversions questioned on Wikipedia, basic math or not. Johnjbarton (talk) 23:38, 23 July 2026 (UTC) Reply

I am refraining from editing this, but it's currently misleading. The term "MKSA system" is an old term for what is currently called the SI. (Note that the linked article distinguishes between MKS and MKSA.)

The linked article states "The MKS units with the ampere as a fourth base unit is sometimes referred to as the MKSA system. This system was extended by adding the kelvin and candela as base units in 1960, thus forming the International System of Units. The mole was added as a seventh base unit in 1971."

This implies that the term "MKSA" means the subset of the SI without the kelvin, candela, and mole. This conforms to my experience. I don't think anyone would regard the current use of the term MKSA to imply that the definitions of M K S and A are frozen in amber.

If it's felt to be necessary to discuss the old value of ${\displaystyle k}${\displaystyle k} it would be better to simply discuss the pre-2019 and post-2019 definitions of ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}}, and their corresponding values of $k$. I think this is a reasonable idea.

Lastly, we need the error on the final value of ${\displaystyle k}${\displaystyle k}. Mgolden (talk) 14:10, 24 July 2026 (UTC) Reply

I swapped the two paragraphs to highlight the post 2019 values.

The sources use "MKS" and I think its better to use that term than meander through the various dated SI versions and then not match the sources anyway.

NIST has a documented procedure to propagate errors but my patience has been exhausted here. Johnjbarton (talk) 18:20, 24 July 2026 (UTC) Reply

My concern is that you are drawing an incorrect distinction. Regarding what is being discussed here, since 1960 the term "SI units" is the same as "MKS units" is the same as "MKSA units". These words have always meant the same system of units (and hence the same value of ${\displaystyle k}${\displaystyle k}) but that system of units has changed over time. In particular, there were several changes between 1970 (the Feynman source you quote) and the present (see https://www.nist.gov/system/files/documents/pml/div683/museum-timeline.pdf)

My view is that the article should either discuss the 2018 definitions and the 2019 definition (which is where ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}} because measurable) or only the current definition. I don't think we need to talk about anything older than that. Mgolden (talk) 22:54, 24 July 2026 (UTC) Reply

I edited the section. The idea that SI is a different unit system from MKS is not correct.

I tried to leave things as they were as much as I could. I left the main points of the presentation intact. I left in the discussion of the old SI (which was called MKS), and clarified the timeframe in which it applied.

There was a sentence that mentioned that the constant depended on the units, but the article didn't discuss any units other than SI, so I added a couple of sentences about Gaussian CGS and Heaviside-Lorentz units. Logically these had to go at the beginning of the article.

I removed the equation ${\displaystyle \epsilon _{0}=1/\mu _{0}c^{2}}${\displaystyle \epsilon _{0}=1/\mu _{0}c^{2}} since that duplicated what was said in the first paragraph.

There were a couple of minor corrections:

1) The first equation for ${\displaystyle k_{e}}${\displaystyle k_{e}} had ${\displaystyle 10^{7}}${\displaystyle 10^{7}} instead of ${\displaystyle 10^{-7}}${\displaystyle 10^{-7}}

2) The equation was missing the units of ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}}, which are needed to make it dimensionally correct.

3) I added the error to the value. Mgolden (talk) 01:42, 26 July 2026 (UTC) Reply

Thanks I fixed a few things and a ref. Johnjbarton (talk) 03:16, 26 July 2026 (UTC) Reply

I made these changes:

1) I fixed the first "citation needed". Everything in the first paragraph is discussed in the Jackson reference, so I moved the footnote to the end of the paragraph. In Jackson this whole matter is discussed in an entire section, so I gave the page numbers in the reference itself.

2) The link [ [ MKS ] ] points to a disambiguation page, I meant to point it to MKS_units.

3) The sentence "Prior to 1983, the constant would be exactly ${\displaystyle k_{e}=10^{-7}c^{2}}${\displaystyle k_{e}=10^{-7}c^{2}}, but numerical value depended on a measured value of the speed of light." runs the risk of leaving people with the impression that ${\displaystyle k_{e}}${\displaystyle k_{e}} was an exact value prior to 1983. The current wording makes it clear that ${\displaystyle k_{e}}${\displaystyle k_{e}} was a measured value.

I also have comments:

1) I don't understand what the Feynman reference is doing. The only point it seems to support is that ${\displaystyle \mu _{0}\epsilon _{0}=c^{-2}}${\displaystyle \mu _{0}\epsilon _{0}=c^{-2}}, which was already referenced above. I believe it should be removed.

2) The remaining "citation needed" I assume refers to the "formerly called". As I stated, I beleive that there has been an increasing tendency to call the units SI, the term MKS or MKSA are also sometimes still used. (For example, Jackson's second edition referred to them in the referenced section as MKSA, but in the third edition, they're called SI.) In fact, it is used in this very article above. If you revert to "also called" then there is no citation needed.

I am not going to make either of these changes. Mgolden (talk) 15:13, 26 July 2026 (UTC) Reply

1) Feynman does not say ${\displaystyle \mu _{0}\epsilon _{0}=c^{-2}}${\displaystyle \mu _{0}\epsilon _{0}=c^{-2}}, he says ${\displaystyle k_{e}=10^{-7}c^{2}}${\displaystyle k_{e}=10^{-7}c^{2}}

2) You claim that SI == MKS, but in my understanding the systems diverged when SI was invented. SI began with MKS but it was not just a name change. Anyway I added a ref and a content tweek to match. Johnjbarton (talk) 17:25, 26 July 2026 (UTC) Reply

1) OK, I take your point.

The conventional presentation for this is that in the pre-2019 system, the Ampere was defined by the sentence "The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed one metre apart in vacuum, would produce between these conductors a force equal to ×ばつ10−7 newtons per metre of length." (See https://en.wikipedia.org/wiki/Ampere#Former_definition_in_the_SI) this is equivalent to the statement that ${\displaystyle \mu _{0}=4\pi \times 10^{-7}}${\displaystyle \mu _{0}=4\pi \times 10^{-7}} N/A${\displaystyle ^{2}}${\displaystyle ^{2}} exactly.

Feynman, for pedagogical reasons, chose to start from the ${\displaystyle k_{e}=10^{-7}}${\displaystyle k_{e}=10^{-7}}N A${\displaystyle ^{-2}c^{2}}${\displaystyle ^{-2}c^{2}} expression, which one derives exactly as we show on this page. He didn't give all of this because it requires introducing ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}}, hence magnetism, which comes later in his presentation.

My one remaining concern is that the insertion of the words "the defined value of ${\displaystyle \epsilon _{0}}${\displaystyle \epsilon _{0}}..." makes it sound as though ${\displaystyle \epsilon _{o}}${\displaystyle \epsilon _{o}} has a defined value. It does not. The definition is of the value of ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}}, and ${\displaystyle \epsilon _{0}}${\displaystyle \epsilon _{0}} follows from that and is not exact.

Can you suggest some other wording? Or revert the change?

2) I don't say SI == MKSA. My statement is that MKSA is the subset of SI not including the three additional base units mole, candela, and kelvin. If you are only using the units based on M K S and A, referring to the system as MKSA is fine, though increasingly out of fashion. What is not true is that now or any time in the past did the use of the term MKSA rather than SI imply that some old SI or non-SI definition of the meter, kilogram, second, or ampere was intended. At any given moment there is only one definition of each of these units, as established by an international treaty.

(Interestingly, I asked a guy at the Particle Data Group about whether he thought that the redefinition of the kilogram in 2019 implies that the definition of the gram in the CGS system changed as well. He said that that was the view that they were taking. This means that, at least in their view, the definition of the statcoulomb also changed in 2019. Of course there is no international body in charge of Gausian CGS.)

Feynman's lectures are from just after 1960, so he would have been referring to the system where the meter was defined they wavelength of radiation from krypton, as would Jackson 2nd ed, but not Jackson 3rd ed. Mgolden (talk) 01:02, 27 July 2026 (UTC) Reply

One other point which I had forgotten about above. Feynman is a bit idiosyncratic in his presentation in that he never uses ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}}. You can see this on this page https://www.feynmanlectures.caltech.edu/II_13.html looking at Eqn 13.13. Even when talking about a purely magnetic situation, he writes ${\displaystyle 1/c^{2}\epsilon _{0}}${\displaystyle 1/c^{2}\epsilon _{0}} instead of ${\displaystyle \mu _{0}}${\displaystyle \mu _{0}}. This continues throughout the book. Mgolden (talk) 03:20, 27 July 2026 (UTC) Reply

Oh, one other thing... What is the point of saying that 8.988 is close to 9? I left it in before because it came from Feynman as a mnemonic hint, but as it stands it just seems weird. Mgolden (talk) 15:16, 26 July 2026 (UTC) Reply

The point is that 9E9 is a good approximation that won't get you into trouble unless you're doing precision work. It's very rare that rounding to 1 digit of precision gets you within 0.14% of the actual value. Headbomb {t · c · p · b} 21:36, 27 July 2026 (UTC) Reply