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Electrostatic Constant Written Funny

  1. May 13, 2009 #1
    So the electrostatic constant k = 9*10^9, but I've also seen it written in my Fundamentals of Phys book (Resnick) and Wikipedia as k = 1/(4*pi*e0). Why is this? Thanks.
  2. jcsd
  3. May 13, 2009 #2
    No one knows?
  4. May 13, 2009 #3

    Doc Al

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    Staff: Mentor

    The ε0 is a constant called the electric permittivity of free space. Read about it here: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elefie.html#c3"
    Last edited by a moderator: Apr 24, 2017
  5. May 13, 2009 #4
    The capacitance of a parallel plate air capacitor of area A and plate separation d is

    C = e0 A/d

    where the permittivity of free space e0 = 1/(u0 c2) = 8.85 x 10-12 Farads per meter,
    and the permeability of free space is
    u0 = 4 pi x 10-7 Henrys per meter
  6. May 13, 2009 #5


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    Staff: Mentor

    It's partly for historical reasons, and partly a matter of minimizing the number of equations that have factors of [itex]4 \pi[/itex].

    Using [itex]\epsilon_0[/itex] Coulomb's Law is more complicated:

    [tex]F_{elec} = \frac {1} {4 \pi \epsilon_0} \frac {q_1 q_2} {r^2}[/tex]

    but other equations like Gauss's Law and the parallel-plate capacitor equation are simple:

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

    [tex]C = \frac {\epsilon_0 A}{d}[/tex]

    Whereas using k, Coulomb's Law is simpler:

    [tex]F_{elec} = k \frac {q_1 q_2} {r^2}[/tex]

    but you have to insert factors of [itex]4 \pi[/itex] into other equations:

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

    [tex]C = \frac {A}{4 \pi k d}[/tex]
  7. May 14, 2009 #6


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    Gold Member

    Well 1/4pi*e0 is where 9x10^9 came from.
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