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Electrostatic energy of two opposite charges in water and in a vacuum

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data

    Compare the electrostatic energy of two opposite charges e and -e, a distance 7 angstroms apart in water at room temperature and that in vacuum (express the energy in terms of Bjerrum length)

    2. Relevant equations

    E = 1/(4(p[itex]\pi[/itex][itex]\epsilon[/itex]D)*(-e^2)/r^2 ???

    3. The attempt at a solution

    First of all is this the right equation to use? If so, is the Bjerrum lenght the distance between the charges? And for a vacuum is the dialectic constant (D) just 1?
     
  2. jcsd
  3. Sep 29, 2012 #2
    Electrostatic energy is:[tex]E=-\frac{e^2}{4\pi \varepsilon D}[/tex]
    Bjerrum length:
    [tex]\lambda_B=\frac{e^2}{4\pi \varepsilon k_BT}[/tex]
     
  4. Sep 29, 2012 #3
    So, for water:
    E=[itex]\lambda[/itex]KT/80???

    And for a vacuum:
    [itex]\lambda[/itex]KT???
     
  5. Sep 30, 2012 #4
    No, we have:
    [tex]\lambda_B k_BT=\frac{e^2}{4\pi\varepsilon}[/tex]
    so put it into formula for energy
     
  6. Sep 30, 2012 #5
    I did do that, D=80 for water and D=1 for a vacuum
     
  7. Sep 30, 2012 #6
    In my formula "D" denotes distance
     
  8. Sep 30, 2012 #7
    Ok that makes sense, so D is dialectric constant and lets say r now is the distance, which is missing in the equations.
     
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