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What is the significance of writing it as $\frac{1}{4\pi \epsilon _0}$? Why not just name the whole thing $\epsilon _0$? And if there is a significance, why not do the same thing for gravitation?

I suspect that writing it in terms of $\pi$ has something to do with the interpretation of the inverse square laws fields as something spreading out uniformly in 3D space (in the form of an expanding sphere), but I don't know much about how that interpretation works. But then again, why not do it for gravitation as it also obeys an inverse square law?

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In electrostatics we do deal with spherically symmetric charge distributions with inverse square law fields, but we often meet uniform planar charge distributions producing uniform fields. If we didn't put $4\pi$ in the formula for field strength due to a spherically symmetric distribution, we'd have to have one in the planar case (as we can show using Gauss's Law). In setting up our so-called 'rationalised' system of units and constants the decision was made to choose constants such that the $4\pi$ appears in the spherical formula (where it doesn't seem too out of place), so it doesn't appear in the planar case – where we might think: What's that $4\pi$ doing here?

Most of the gravitational fields we meet are from spherically symmetric, or approximately spherically symmetric, bodies. We rarely if ever encounter gravitational fields due to planar mass distributions. So we don't have to worry that a $4\pi$ would appear in planar formulae if we didn't put one in the spherical formulae. So we don't put one in the spherical formulae!

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  • $\begingroup$ Maybe say that it's $4π$ because of the surface area formula (as OP suspected) and that the factor of $4π$ does show up in the field equation of general relativity. $\endgroup$
    – benrg
    Commented Aug 27, 2020 at 15:57

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