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Finding the potential different across an un-parallel plate capacitor

  1. Apr 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Two plates of equal area: length b, width w
    On one side, the plates are a distance d away from each other, on the other side, a distance d+a

    2. Relevant equations

    No equations given. Using
    E= kq/r^2, V=-∫E*dl


    3. The attempt at a solution

    E = kq ∫∫dxdz/(x^2 + z^2 + y(x)^2)

    y(x) = a + ax/b

    Integrating x from 0 to b, and z from 0 to w.

    Doing this gives a very very messy solution, give me the impression that this is the incorrect path.
     

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    Last edited: Apr 28, 2013
  2. jcsd
  3. Apr 28, 2013 #2

    ehild

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    Hi Phy00, welcome to PF.
    The plates are of metal, aren't they? What do you know about the potential along a metal surface?

    Is the charge on the plates given?

    ehild
     
  4. Apr 28, 2013 #3
    The only thing given is the information on the figure. Yes, two metal plates.
     
  5. Apr 28, 2013 #4

    ehild

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    Presumably the net charge on a plate is given. What do you know, can the potential change along a metal plate?

    ehild
     
  6. Apr 28, 2013 #5
    Do you mean that since the potential is same throughout a surface, I only need to find out the field at a certain point?
     
  7. Apr 28, 2013 #6

    ehild

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    No, the field changes along the plates while the potential is constant. You need to guess the form of the electric field lines and find the surface charge density along the plates. What do you know about the direction of the electric field lines at an equipotential surface?

    ehild
     
  8. Apr 28, 2013 #7
    The field lines are ⊥ to the surface... so they would curve.
     
  9. Apr 28, 2013 #8

    ehild

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    Yes. I think they can be considered circular arcs. Make a drawing. The potential difference at a distance from one edge of the capacitor can be obtained by integrating the electric field along the arc. You can consider the field constant along an arc. The electric field is connected to the surface charge density. You can get it in terms of the potential difference from the integral. Integrating the surface charge density for the plate, you get the relation between the charge and potential difference.

    ehild
     
  10. Apr 28, 2013 #9
    I'm a little lost when you say "The electric field is connected to the surface charge density."
    Also, I can see how the field lines can curve, but near the edges, they wouldn't be circular.
     
  11. Apr 28, 2013 #10

    ehild

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    Do not mind the edges. You need some assumption about the electric field lines, and circular shape fullfills the requirement that they are perpendicular to the plates.

    From Gauss' Law you know that q/ε0 field lines emerge from a charge q. In case of σ charge per unit area, the electric field inside the capacitor is σ/ε0 near a plate.

    ehild
     
  12. Apr 29, 2013 #11

    ehild

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