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Potential in parallel plate system from charge density?

  1. May 13, 2015 #1
    1. The problem statement, all variables and given/known data

    16987125673_015172f5f0_b.jpg

    Note - I have used an image here. For the simple reason I don't want to confuse anyone by missing out subscripts. But if that is not acceptable let me know and I will attempt to write it out.

    2. Relevant equations

    I have considered using Poissons equation - but couldn't get that to work.

    I have also attempted to consider this as a parallel plate capacitor, so: u = 0.5e(o)E^2


    3. The attempt at a solution

    I am getting stuck with this. Annoyingly I think it's probably quite simple - but it's confusing me.

    I have drawn a diagram here to understand what is going on, but I don't think I am allowed to post too many photo's. And have no idea how to draw that on a computer.

    So I thought I could find the E field magnitude. To do this I could use the boundary conditions. But I am not going from air into a dialectic medium. So I got a bit confused there. I don't think I can say the E=V/d because of the dialectic material - however I do think I can say: E = eV/d

    Subsitituting this into the above equation I get:

    u = 0.5e(0)(eV/d)^2

    Now the change in V is going to be V(0) so:

    u = 0.5e(0)e^2V(o)^2/d^2

    And I know that charge density = charge/volume

    therefore: charge/volume = p(0)/d

    But here I get stuck... I am at a bit of a loss. As I've said I think it's probably simpler than I am making it. But some hints would be amazing.

    Thanks

    Sam
     
  2. jcsd
  3. May 13, 2015 #2

    Orodruin

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    What were your thoughts on this exactly?
     
  4. May 13, 2015 #3
    Well as the plates are infinitely long - you are only dealing with d^2x/dx^2 = -p/e

    So I tried to integrate this twice to solve the equation - and then applied boundary condition are V(x=0)=0 and V(x=d) = V(o)

    But I still have no idea where to go after this...
     
  5. May 13, 2015 #4

    Orodruin

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    If you have done things correctly, you then have the potential and there should not be anywhere else to go ...
     
  6. May 13, 2015 #5
    Here is my problem - I am not doing it properly, or not understanding the question then!!!

    I've posted my work below. I can't type that out, as I don't know how to put the differentials in etc... Or the integral signs. But I am not sure how to do get rid of the x. I am failing to understand this badly.

    17423993030_a83193ed1f_b.jpg

    Basically - I am doing something awfully wrong. I though x=d, but when I do that all I get is V = V(o) which is clearly not what they are after.
     
    Last edited: May 13, 2015
  7. May 13, 2015 #6

    Orodruin

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    Why do you want to put x = d? The coordinate x is perfectly fine as a parameter in your answer as it describes the potential at the point x. Since the problem asks for the potential at any point, the result will generally be a function of the point, in this case x. You have essentially solved the problem without realising that you already have the answer.
     
  8. May 13, 2015 #7

    Orodruin

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  9. May 13, 2015 #8
    Thanks on both making me see how stupid I am being - and pointing out how I can use latex. I think I am trying to make a lot of this stuff harder than it has to be... Either that or I'm just a bit thick. But I appreciate you helping me out. So all I need to do now is substitute in the expression for the charge density and I'm good to do.

    Thanks again.
     
  10. May 13, 2015 #9

    Orodruin

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    Some of the most difficult problems are those whose answers are staring us in the face ... :smile:
    Yes.
     
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