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Two positively charged plates : interaction force

  1. Nov 8, 2013 #1
    Hello,

    I am facing a paradox, well it seems like one, resolving the interaction force between two equally charged plates each bearing a positive charge. Let us assume this as Q, and plates having area A.

    On one hand , we can claim the force = Q^2/2Ae0 where e0 is the vacuum permittivity, as each plate bears a charge Q and generates a field Q/2Ae0 at points not too far off from itself or close to the edges.

    But, if we look at the energy stored, I am getting a very different result. There is no field in the region between the two plates so the energy field density 1/2e0(E^2) is zero. Outside the plates, the electric field will remain ~ Q/2Ae0 from each plate, totalling Q/Ae0 at first and then begin trailing off as we go farther and farther. Eventually it will trail off completely. So if we move a plate, this trailing off merely starts at a different point and the net energy stored outside each plate is conserved. Hence energy is a constant therefore F = -dE/dr = zero. I can't accept this logically but neglecting the fringe field, this seems to imply some sort of internal screening of charges.
     
  2. jcsd
  3. Nov 8, 2013 #2

    Simon Bridge

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    How do you figure that?
     
  4. Nov 8, 2013 #3
    Because at points not too far off to the edge, or corners, the field due to each plate can be approximated as charge density/2e0. They act in opposite directions.

    I now realize though, this probably implies the fringe field is non-negligible in this case..
     
  5. Nov 8, 2013 #4

    Simon Bridge

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    Oh I see, you wanted to make an approximation for the case that the dimensions of the sheets are very large compared with their separation - the infinite sheet approximation?

    To use Gausses Law and F=qE ideas, each distribution of charge moves in the potential due to the other charges.

    So each plate feels the force due to the field due to the other plate alone.
    You appear to have been combining the fields in your arguments.

    When in doubt though: return to Coulomb's law.
     
  6. Nov 9, 2013 #5
    I know that each plate feels the field due to the other plate alone. I just want to derive that via energy stored in the electric field, and this seems to be impossible without the fringe field which is in turn very hard to numerically figure out. Ignoring it, I get F = zero which is absurd.
     
  7. Nov 9, 2013 #6

    Simon Bridge

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    Well yes - the shortcut you tried (via energy density between the plates) (a) combines the fields, and (b) relies on the infinite sheet approximation ... which is not valid when the separation is comparable to or bigger than the dimensions of the sheets - which is what happens for the stored energy calculation since the sheets start at infinite separation.
     
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