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Coulomb's Constant

  1. Dec 27, 2007 #1
    Can anyone tell me where the 4*pi comes from in coulomb's law? I know the law is often compared to netwon's law of gravity, but there is no 4*pi in that one. What is the meaning of 4*pi in the equation of coulomb's law?
  2. jcsd
  3. Dec 27, 2007 #2
    Basically the [tex] 4\pi[/tex] is for convenance.

    Coulombs constant is, by experiment:K= 8.988...E9,Nm^2C^-2 an equivalent way of expressing this is [tex] \frac{1}{4\pi \epsilon_o}[/tex]. Where [tex]\epsilon_o[/tex] is the permitivity of free space.
    Nothin too special, just fo convenance.
  4. Dec 27, 2007 #3
    4*pi is hardly convenient if you got to choose a value by convenience =/

    where is the equivalence that you say exists?
  5. Dec 27, 2007 #4


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    it's more than that. why didn't they pull [itex]\pi[/itex] or e or Euler's [itex]\gamma[/itex] or some number like that outa their butt and scale it with one of those numbers?

    the [itex]4 \pi[/itex] comes from the fact that the surface area of a sphere is [itex]4 \pi r^2[/tex]. the relevant topics to study is inverse-square law, Gauss's Law, and the concept of "flux" and "flux density". try Wikipedia to look that up. that will explain where the [itex]1/(4 \pi \epsilon_0)[/itex] comes from in Coulomb's Law and why you don't see it in Maxwell's equations.
    Last edited: Dec 27, 2007
  6. Dec 27, 2007 #5

    Ben Niehoff

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    In fact, the entire [itex]1/4\pi r^2[/itex] factor comes from the area of a sphere. The [itex]\epsilon_0[/itex] is the actual relevant physical quantity (i.e., the permittivity of free space).

    The electrostatic force is ultimately due to the exchange of virtual photons between the particles. If you imagine these photons traveling outward along rays from a particle (much like from a light bulb), you will see that the density of virtual photons falls off in proportion to the area of a sphere of increasing radius. The density of virtual photons is proportional to the rate of particle interactions (which in turn leads to a macroscopic change in momentum; i.e., force); therefore, we must have that the force on a charge, F, is inversely proportional to the area of a sphere of radius r:

    [tex]F \propto \frac{1}{4\pi r^2}[/tex]

    where r is the distance between the test charge and the source charge.
  7. Dec 27, 2007 #6


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    The factor of 4*Pi allows the relevant Mawell's equation to be expressed as:
    [tex]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex]

    Rather than something like

    [tex]\nabla \cdot \vec{E} = \frac{\rho}{4\pi\alpha}[/tex]

    This all comes from the fact that [tex]\nabla \cdot \frac{\vec{r}}{r^2} = 4\pi\delta^3(\vec{r})[/tex]

    Thus if we take [tex]\rho = q\delta^3(\vec{r})[/tex] (ie a point charge), we require [tex]\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\vec{r}[/tex], for the Mawell equation to hold, in the convenient form.
  8. Dec 28, 2007 #7
    ya last night when I got a chance to look in my text book it said in a footnote that it would be explained in a later chapter when discussing gauss's law so you guys are right and thank you for these posts, I will read them carefully because I am trying to get an intuition for electromagnetics :)
  9. Dec 28, 2007 #8


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    first get an intuition about inverse-square laws. why are they 1/r2 instead of 1/r or 1/r3 or 1/r3/2 ?
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