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Interatomic coulombic potential

  1. Jun 28, 2012 #1
    I am trying to calculate coulombic potentials. The equation given in wikipedia is

    [tex]F=k_e\frac{q_{1}q_{2}}{r^2}[/tex]

    where k_e is in units of N m^2 C^-2, q are the partial charges, and r is the distance.

    However, I look here and they give k_e=2.31*10^19 and remove the square of the distance. I assume that the difference is related to the units, which is in J*nm in this other resource, but if k_e is a constant then how can an equation of the form 1/r^2 give the same answer as one with the form 1/r?

    Also, is the given value of k_e (2.31*10^19 J*nm) very accurate? What is the more accurate value?
     
  2. jcsd
  3. Jun 28, 2012 #2

    tiny-tim

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    Hi Hypatio! :smile:

    (try using the X2 button just above the Reply box :wink:)
    the force is ~ 1/r2

    the potential is ~ 1/r :wink:
    i don't recognise that figure

    the usual constant in Coulomb's law is 8.988 x 109 N m2/C2

    (or 8.987 551 787 368 176 4, see http://en.wikipedia.org/wiki/Coulomb's_law#The_law)
     
  4. Jun 28, 2012 #3
    The formula in the paper is for energy and not for force:
    [tex]E=k_e\frac{q_{1}q_{2}}{r}[/tex]
    The numerical constant comes from the units: they use the numerical charge (number of electron charges) and not the charge in coulombs.
    They also use the distance in nm.
    So the number you see is

    [tex]k_e\times e^2\times 10^9[/tex]
     
  5. Jun 28, 2012 #4
    Thanks!

    What is the difference between the partial charge and the 'numerical charge'? Also, what is the parameter e?
     
  6. Jun 28, 2012 #5
    Numerical charge is the charge in units of elementary (electron charge).
    e is the value of the elementary charge (1.6x10^(-19) C).
     
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