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

  • Thread starter JordanGo
  • Start date
  • #1
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Homework Statement



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)

Homework Equations



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

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?
 

Answers and Replies

  • #2
115
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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]
 
  • #3
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So, for water:
E=[itex]\lambda[/itex]KT/80???

And for a vacuum:
[itex]\lambda[/itex]KT???
 
  • #4
115
2
No, we have:
[tex]\lambda_B k_BT=\frac{e^2}{4\pi\varepsilon}[/tex]
so put it into formula for energy
 
  • #5
73
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I did do that, D=80 for water and D=1 for a vacuum
 
  • #6
115
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In my formula "D" denotes distance
 
  • #7
73
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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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