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Electric potential inside a pipe

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data

    EGDQm.jpg

    2. Relevant equations

    2V = 0

    3. The attempt at a solution


    I'm guessing I need to use Laplace's equation in 2 dimensions since the potential depends on x and y. I have no idea how I would do this.
     
  2. jcsd
  3. Oct 21, 2012 #2

    vela

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    This is a pretty standard type of problem. Consult your textbook for similar examples, and then post your work.
     
  4. Oct 21, 2012 #3
    The reason I'm having trouble is my textbook doesn't really have any examples for this. I figured I could at least find a similar example online but I'm not even sure what to search for.
     
  5. Oct 21, 2012 #4

    vela

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    That's surprising. You could check another E&M text. It's also covered in math methods texts, like Arfken, as well. Look up "separation of variables" and "Laplace's equation".
     
  6. Oct 21, 2012 #5
    There actually is at least one example in the separation of variable section, I didn't think I needed to go into that for this problem, thanks.
     
  7. Oct 23, 2012 #6
    Ok so I tried working out a solution but I went wrong somewhere.

    From separation of variables I got:

    V(x,y) = (Aekx+Be-kx)(Csin(ky)+Dcos(ky))

    Boundary conditions:

    V=0 when x=0
    V=0 when y=0
    V=0 when y=b
    V=v0(y) when x=a

    D=0 by the second condition, and I can then absorb C into A and B:

    (Aekx+Be-kx)sin(ky)

    -A = B by the first condition:

    A[ekx-e-kx]sin(ky) = Asinh(kx)sin(ky)

    k = n∏/b by the third condition:

    Asinh(n∏x/b)sin(n∏y/b)

    Now I'm pretty sure I'm supposed to do some sort of linear combination here to solve for A.

    I did this instead. I'm pretty sure it's wrong but I'm not sure why?

    By the 4th boundary condition:

    Vo(y) = Asinh(n∏a/b)sin(n∏y/b)

    solved for A, then plugged that into Asinh(n∏x/b)sin(n∏y/b)

    I guess I can't do that since Vo(y) isn't constant? Still though... why?
     
  8. Oct 24, 2012 #7

    Sure you can do that as long is A comes out to be a constant. It must be a constant because you made the assumption that it was a constant.

    The variation with y of Vo(y) is taken care of by something other than A in your solution.
     
  9. Oct 24, 2012 #8

    vela

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    Each value of n corresponds to a solution Vn, so you should write
    $$V_n(x,y) = A_n \sinh (k_nx) \sin (k_ny).$$ The general solution is a linear combination of the individual solutions
    $$V(x,y) = \sum_{n=1}^\infty A_n \sinh (k_nx) \sin (k_ny)$$ where ##k_n = n\pi/b##.

     
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