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Attraction or Repulsion?

  1. Jul 20, 2005 #1

    somasimple

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    Hi All,

    In the attached figure, I have drawn some "virtual" ions of two kind: blue are sodium, and violet for potassium (normally solvated with water).
    There is more ions outside of my "vat" (bilipidic layer) than inside.

    All ions are supposed as positive.

    Since there is a difference in number of ions between the two sides, is the internal ions attracted (because outside is more positive than inside) or repulsed (because ions are positive on the two sides), or anything else?
     

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  2. jcsd
  3. Jul 20, 2005 #2

    Gokul43201

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    I can see what's holding the internal ions in place - the mutual repulsion is forcing them to the walls of the container. But what is holding the external ions in place ? What is stopping them from flying off ?

    Assuming there is some mechanism that holds the external ions in place, let's proceed. The external ions have absolutely no effect on the internal ions. By Gauss' Law, the electric field inside the container, due to the external ions is zero. So, the only force on the internal ions comes from their mutual repulsion, which basically pushes them as far out as possible (till they are up against the wall and can go no farther outwards).
     
  4. Jul 20, 2005 #3

    somasimple

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    Thanks for this fast response,

    It is assumed that there is other water/ions associations. It may be a cell that is represented with the extracellular milieu.
     
  5. Jul 20, 2005 #4

    somasimple

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    Hi all,

    Another question about this hypothesis:

    1/ If there is m1 ions outside and m2 ions inside, is it possible to compute a possible voltage difference between the two sides?

    2/ If there is also some negative ions on the two sides, does it change a lot the previous response?
     
  6. Jul 21, 2005 #5

    Gokul43201

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    The voltage difference is virtually independent of the ions outside the vat (thick green cylinder). Assuming the vat is indeed a cylinder of radius R, wall thickness T, and height H, and that n1 is low enough that there is not more than a single monolayer of charge on the inside (as shown in the picture), the voltage difference across the wall (just my best guess) is of the order of
    [tex]\Delta V \approx \frac {n_1 eT}{4 \pi \epsilon_0 \epsilon_r RH} [/tex]

    If there are uniformly distributed negative charges on the inside, they can be treated as reducing the total positive cherges by their number, so [itex]n_1(eff) = n_1(pos) - n_1(neg) [/itex]
     
    Last edited: Jul 21, 2005
  7. Jul 21, 2005 #6

    somasimple

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    Thanks,
    What is the meanings of the employed e?
    [tex]e \epsilon_0 \epsilon_r[/tex]
     
    Last edited: Jul 21, 2005
  8. Jul 21, 2005 #7

    Gokul43201

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    e = charge on electron = 1.6*10^-19 C
    [itex]\epsilon_0[/itex] = permittivity of free space = 8.85*10^-12 C^2/Nm^2
    [itex]\epsilon_r[/itex] = dielectric constant of the material of the wall
     
  9. Oct 11, 2005 #8

    somasimple

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    Hello,

    I have a question concerning the picture/hypothesis below.

    it is a cynlinder with another one in it:
    the external one Out has negative fixed charges on it.
    the internal one has a negative core.

    there is free positive charges allowed to move freely.
    it is the start of hypothese:

    1/ are positive charges moving/attracted to the inner blue cylinder?
     

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  10. Oct 11, 2005 #9

    Gokul43201

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    Yes. The outer charges have no effect on them (Gauss' Law).
     
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