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Definition of Eo (Coulomb's law)

  1. Jun 24, 2010 #1

    If epsilon 0 was defined as the proportionality constant for this equation, why was 4 pi not included in Eo? Why does there have to be a 4pi in the equation, instead of just having Eo equal its current value times 4pi?
  2. jcsd
  3. Jun 24, 2010 #2


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    Because in differential form, the equation is written as:

    [tex]\nabla \cdot E = \frac{\rho}{\epsilon_0}[/tex]

    If you apply Gauss Theorem to it, you get the 1/4pi term.

    More fundamentally, it's because epsilon works as polarization coefficient for vacuum.

    It also works neatly with the wave equation. [itex]\epsilon_0\mu_0 = \frac{1}{c^2}[/itex]
  4. Jun 25, 2010 #3


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    The [itex]4\pi[/itex] and the [itex]\epsilon_0[/itex] appear where they do because the equations assume SI units. In other systems of units, they appear in different places, or not at all. For example, in Gaussian cgs units, [itex]\epsilon_0 = 1[/itex] and we have

    [tex]F = \frac {q_1 q_2}{r^2}[/tex]

    [tex]\nabla \cdot E = 4 \pi \rho[/tex]

    See http://en.wikipedia.org/wiki/Gaussian_units
  5. Jun 29, 2010 #4
    Epsilon 0 has that silly form because it came to the party last and Mu 0 got in first.:smile:

    The Ampere was defined in terms of the force between two parallel conductors one meter apart. The force is caused by the magnetic field, which as you know is circular around the wire.
    That's where the 4 pi came from (actually it's 2 *2 pi because there are two wires)

    The clincher is the fact that Epsilon 0 * Mu 0 = 1/c2

    Once Mu has that pi in it, Epsilon has no choice.

    If Epsilon had got in first, Mu 0 would be stuck with the pi instead.
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