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Simplify the expression -- Electric field in a capacitor

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Between two metal plates a metal strip is inserted as shown on the figure. Captureh.PNG
    a) Calculate the electric potential anywhere inside the capacitor.
    b) Simplifly ##U(x,y)## for ##x>>a##.
    c) Calculate the electric field in the middle of the capactior.

    2. Relevant equations


    3. The attempt at a solution

    a) Ok, since we have no other charges in the capacitor, than ##\nabla ^2 U=0##. In kartezian coordinates this means that ##U(x,y)=\sum _m (A_me^{-k_mx}+B_me^{k_mx})(Csin(k_my)+Dcos(k_my))##.

    Using boundary conditions
    ##U(0,y)=U_0## and
    ##U(x,0)=0## and
    ##U(x,a)=0## we find out that

    ##U(x,y)=\sum _{n=1} ^{\infty}\frac{2U_0}{n\pi }(1-(-1)^n)e^{-\frac{n\pi}{a}x}\sin(\frac{n\pi }{a}y)##

    b) The result is that out of the sum only ##n=1## survives, therefore ##U(x,y)=\frac{4U_0}{\pi }e^{-\frac{\pi}{a}x}\sin(\frac{\pi }{a}y)##.

    I am having some troubles understanding this. Could anybody please try to explain how do I get this result? o_O

    c)

    ##U(x,a/2)=\sum _{n=1} ^{\infty}\frac{2U_0}{n\pi }(1-(-1)^n)e^{-\frac{n\pi}{a}x}\sin(\frac{n\pi }{2})##

    ##\vec E=-\nabla U(x,a/2)=\sum _{n=1} ^{\infty}\frac{2U_0}{a}(1-(-1)^n)\sin(\frac{n\pi }{2})e^{-\frac{n\pi}{a}x}##

    I think I should do something more with that last equation but I really don't know what or how.. :/
     
    Last edited: Nov 11, 2014
  2. jcsd
  3. Nov 11, 2014 #2

    SteamKing

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    Consider the expression (1 - (-1)n). What happens when n is even? When n is odd?
     
  4. Nov 11, 2014 #3
    How is this related to ##x>>a##?

    ps: SteamKing, I edited the post and added some more questions.
     
  5. Nov 11, 2014 #4

    SteamKing

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    Well, if you had examined the expression, you would have seen that about half the terms drop out, due to the expression being 0 for n even.
     
  6. Nov 12, 2014 #5
    Yes, I agree. But this still doesn't explain why only ##n=1## is ok? What happens with ##n=3##?
    I apologize but I still can't see how is this related to ##x>>a##.
     
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