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Coulomb's constant, electric force constant

  1. Sep 30, 2012 #1
    Coulomb's Law states F = q**2/ r**2

    q= statCoulombs

    In SI units, F = Ke (q**2/r**2)

    q= Coulombs

    I read that 1 C = 2.997 x 10E+09 statC

    Then Ke = 8.987 x 10E+09

    (1) Does anyone know where the conversion for C to statC came from?
    (2) and how you get the Ke from those values?

    Thanks!
     
  2. jcsd
  3. Oct 1, 2012 #2

    vanhees71

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    As said earlier in this forum, "Coulomb's Constant" is just a conversion factor from the SI unit of charge to more natural units of the same quantity. There's no physics involved in this but just the definition of the unit of charge.

    In the official definitions this is however not that straight forward, because the SI units were chosen not on physical or didactical grounds but to have a continuity in earlier definitions of electromagnetic quantities and handy numbers in everyday practice of electrical engineering.

    That's why the SI chooses to work in a four-unit system for mechanics+electromagnetism. Besides the mechanical base units for mass (kilogram still defined as the mass of a platinum-iridium alloy cylinder kept save in Paris), time (second defined by the frequency of a hyperfine transition in Cesium), length (meter defined by fixing the speed of light in the vacuum to a certain value) one introduces a new unit for the electric current, i.e., for the amount of electric charge running through a cross sectional area per unit time. The definition is quite abstract but simple

    "The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to [itex]2 \cdot 10^{-7}[/itex] newton per metre of length."

    Then the unit of charge is derived from this as that amount of charge running through a surface within one second when the current is one Ampere.

    As a consequence, in Maxwell's fundamental equations there appear two conversion factors instead of the one physical fundamental constant, which is the speed of light:

    [tex]\vec{\nabla} \times \vec{B}-\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}=\mu_0 \vec{j}, \quad \epsilon_0 \vec{\nabla} \cdot \vec{E}=\rho.[/tex]
    [tex]\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{E} + \frac{\partial \vec{B}}{\partial t}=0.[/tex]
    Finally the force on a point charge is given by
    [tex]\vec{F}=Q (\vec{E} + \vec{v} \times \vec{B}).[/tex]
    From this it immediately follows that in the SI the electric and magnetic components of the electromagnetic fields have not the same unit (very unnatural by the way).

    The connection between the equations, written in the SI and the more natural older units (Gaussian or Heaviside-Lorentz units) is then given by deriving from the equations in the SI the equations, containing only really physical constants of nature. The most fundamental one is the speed of light in the vacuum. It's not only a constant concerning electromagnetic phenomena but a universal constant of nature. It comes out not to be just 1, as in natural units, because we choose to measure time intervals and spatial distances with different units (in the SI in seconds and meters respectively, in the Gauss or Heaviside-Lorentz systems in seconds and centimeters respectively; since also the mass is measured in kg or g, for pure mechanical quantities there are only powers of 10 as conversion factors between the SI and the Gauss or Heaviside-Lorentz units, and the latter two systems coincide completely). That's why the speed of light is just fixed to an exact value:
    [tex]c:=299792458 \cdot \mathrm{m/s} \simeq 3 \cdot 10^8 \cdot \mathrm{m/s}. \qquad (*)[/tex]
    Now using Maxwell's equations written in the above SI version, leads after some vector calculus to the relation
    [tex]\mu_0 \epsilon_0=\frac{1}{c^2}.[/tex]
    Next we have to use the definition of the Ampere ot settle one of the two unnatural constants. For steady currents (DC), as used in the definition, the Maxwell-Ampere Law simplifies to the Ampere Law,
    [tex]\vec{\nabla} \times \vec{B}=\mu_0 \vec{j}.[/tex]
    Now take an infinite straight cylindrical wire with a DC [itex]I_1[/itex] running through. Together with [itex]\vec{\nabla} \cdot \vec{B}[/itex] you find the magnetic field components to be
    [tex]\vec{B}=\frac{I_1}{2 \pi \mu_0 r} \vec{e}_{\varphi}.[/tex]
    Here I have used standard cylinder coordinates [itex](r,\varphi,z)[/itex] and assumed that the wire is along the [itex]z[/itex] axis with the current counted as positive when it's running in positive [itex]z[/itex] direction.
    Using then the force law one obtains for the force on another very long wire along the [itex]z[/itex] axis with a DC [itex]I_2[/itex] running through:
    [tex]\vec{F}=-\frac{I_1 I_2 L}{2 \pi \mu_0 r},[/tex]
    where [itex]L[/itex] is the length of this wire. Here the condition for the validity of this approximate calculation is that [itex]L \gg r[/itex]. Plugging in the values in the definition of the Ampere, i.e., [itex]r=1 \; \mathrm{m}[/itex] and [itex]I_1=I_2=1 \; \mathrm{A}[/itex], you find
    [tex]\mu_0=4 \pi 10^{-7} \mathrm{N}/\mathrm{A}^2=4 \pi 10^{-7} \frac{\mathrm{kg} \; \mathrm{m}}{\mathrm{s}^2 \mathrm{A}^2}=4 \pi 10^{-7} \frac{\mathrm{kg} \; \mathrm{m}}{\mathrm{C}^2}.[/tex]
    From (*) then also the conversion factor [itex]\epsilon_0[/itex] is fixed.

    The force between two point charges [itex]Q[/itex] at rest at a distance [itex]r[/itex] is according to the electrostatic Maxwell Laws and the Lorentz-force Law
    [tex]|\vec{F}|=\frac{Q^2}{4 \pi \epsilon_0 r^2}. \qquad (\text{SI})[/tex]
    In the original nonrationalized Gauß units the same force reads
    [tex]|\vec{F}|=\frac{q^2}{r^2}. \qquad (\text{Gauß}).[/tex]
    Of course both forces are the same, and we finally get
    [tex]Q=\sqrt{4 \pi \epsilon_0} q.[/tex]
    The conversion factor now this leads to
    [tex]\sqrt{4 \pi \epsilon_0}=\sqrt{\frac{4 \pi}{c^2 \mu_0}}=\frac{1}{c} \sqrt{\frac{10^{7} \mathrm{C}^2}{\mathrm{kg} \; \mathrm{m}}}=\frac{10}{c} \frac{\mathrm{C}}{\sqrt{\mathrm{g} \; \mathrm{cm}}}.[/tex]
    Now the Gaussian charge unit is
    [itex]1 \; \text{esu}=1 \sqrt{\frac{\mathrm{g} \; \mathrm{cm}^3}{\mathrm{s}^2}}.[/itex]
    Finally we have for [itex]Q=1 \; \mathrm{C}[/itex]
    [tex]q_{1 \text{C}}=\frac{1 \; \mathrm{C}}{\sqrt{4 \pi \epsilon_0}}=\frac{c}{10} \sqrt{\mathrm{g} \; \mathrm{cm}} \approx 3 \cdot 10^9 \; \mathrm{esu},[/tex]
    where in the final step I've set [itex]c \approx 3 \cdot 10^8 \mathrm{m}/\mathrm{s}=3 \cdot 10^{10} \mathrm{cm}/\mathrm{s}[/itex].
     
  4. Oct 1, 2012 #3
    Thank you vorhees71!

    That is an incredible reply to my question.

    I owe you a beer!!

    Thanks!
     
  5. Oct 2, 2012 #4

    vanhees71

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    Well, you are welcome, but you are right. Doing electromagnetism in the SI of units is only possible with some beer :smile: This desease is, however, increasing even in the theoretical textbook literature. Even Jackson, who wrote THE bible on the subject, changed to the SI with the 3rd (and also the 4th German) edition.:cry:
     
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