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  1. B

    Find potential within a pipe using Laplace's equation

    If I follow the steps illustrated in post 3 but instead solve for d, I obtain -Csin(ak)=Csin(ak), so two possibility is C=0 or sin(ak)=0, so k=n*pi/a for this instance
  2. B

    Find potential within a pipe using Laplace's equation

    C need not be zero as well, so there must be another way to determine coefficients
  3. B

    Find potential within a pipe using Laplace's equation

    How does this fact help me in determining which (C or Do) is 0? Sorry for all the questions!
  4. B

    Find potential within a pipe using Laplace's equation

    I think I see what your point is. The other possibility would be that cos(ak) = 0, meaning k = n*pi/(2a)
  5. B

    Find potential within a pipe using Laplace's equation

    Either x = 0 or y = 0 Edit. I think I see what your point is. The other possibility would be that cos(ak) = 0, meaning k = n*pi/(2a)
  6. B

    Find potential within a pipe using Laplace's equation

    Another possibility is D=0? I'm unsure of what else it could be
  7. B

    Find potential within a pipe using Laplace's equation

    If I do the same but instead solve for C, I can find that C= -C.
  8. B

    Find potential within a pipe using Laplace's equation

    My reason is as follows - If I plug y = a into (Csin(ky)+Dcos(ky)), I can solve for C. Then if I plug in y = -a and my value for C, I can get -Dcos(ak) = Dcos(ak) (because sin is odd function and cos is even function) I'm not sure if my reasoning is correct so feedback is appreciated. I'm not...
  9. B

    Find potential within a pipe using Laplace's equation

    Homework Statement Hello, I'm trying to solve laplaces equation to find a solution for the potential in a pipe with the given boundary conditions: at x=b, V=V_0 at x= -b, V = -V_0 at y=a, V=0 at y=-a, V=0 (Assume this configuration is centered on the origin, pipe as dimensions -b<=x<=b...
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