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Boundary conditions on a conductor?

  1. Feb 12, 2013 #1
    I've been trying to get my head arround this problem for several days now, and while I deemed it relatively simple at first it turns out that I can't figure out the BCs on a conductor, to which we apply a potential U.

    In the simplified version of the problem, there is a rectangular conductor with resistance ρ to two opposing sites of which (call them A and C) we impose a given voltage 0 to U₀.

    Question remains: What are the boundary conditions on sides B and D?

    I don't want to influence the topic at this point so I won't really state what I thought could be done, I just say that no matter what BC I imagined, I could not ultimately justify it without requiring further assumptions than the given data.

    Addition: I should add that the conductor is interfaced into a circuit at sides A and C, so a current flows through it.
     
    Last edited: Feb 12, 2013
  2. jcsd
  3. Feb 12, 2013 #2

    Simon Bridge

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    Sketch the situation - how is the field being applied? Draw the field lines.
    What do you expect the "boundary conditions" to tell you?
     
  4. Feb 12, 2013 #3
    Dear Simon, I appreciate your intentions to "lead me onto the right track". As I said, thought, I have already made ample effort to figure this out. With the desire to spare us unncessary discussion I ask you, please try to answer the question directly and let us discuss it afterwards.

    Thank you!

    PS: We can't argue based upon symmetry or some other sort of "intuition". The real problem is slightly more complex and I'm seeking to determine which is the generic logic for the boundary condition on the insulating edges - which, hopefully, I can infer from how the BCs are modelled for the simplified version of the problem.

    (I take it you're probably getting at the insulating part, requiring that E is parallel to the edges - however, this logic only seems to hold at a first glance, nor is it a sufficient Neumann BC for obtaining a unique solution).
     
    Last edited: Feb 12, 2013
  5. Feb 12, 2013 #4

    Simon Bridge

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    OK, in general. Without he information asked for I cannot help you.
     
  6. Feb 12, 2013 #5
    Thank you. In the meantime: I did not understand that you were actually asking for information. The problem as stated does not give any further information than:

    U = 0 on A, U = U₀ on C

    ΔU = 0 ⇔ ∇×E = 0 ; ∇E = 0 in the conductor (time-stationary process)

    ( E = ρj )

    B and D are insulating boundaries.
    A and C are interfaces to the circuiting.
     
    Last edited: Feb 12, 2013
  7. Feb 12, 2013 #6

    Simon Bridge

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    OK - that kinda suggests that A and C are equi-potential surfaces doesn't it? What you have is a resistor-material completely filling the space between two conducting plates.

    Do I understand correctly that there is a current through the thing, and a voltage source maintaining a steady PD between A and C? Are you given the dimensions of the slab?

    (you are using Δ = ∇2?)
    Questions like I asked before serve several functions - one of which is to supply me with more information on the problem, another is to guide you to realizing what you need to do, and another is to illustrate your understanding so I know how to talk to you.

    I'm still in the dark about what you want D and B boundary conditions for.
    Presumably the slab also has front and back surfaces? Charges are moving along those boundaries.

    What are you supposed to be computing?
     
  8. Feb 12, 2013 #7
    Yes.
    Yes. Call them L between A and C and W between B and D. The Problem is supposed to be two-dimensional (I meant to imply this by talking of a rectangle).
    Yes.
    I'm supposed to compute U, thus E from which again follows j. In order to get a unique solution for U I need to know BCs. The slab has no front nor back because the problem is assumed to be 2-dimensional.

    FYI: I've tried finding a Finite-Element formulation of the problem, assuming only the direction of E at the insulating boundaries (so that j is parallel to those boundaries), but I concluded that this is not enough information (for a Neumann BC I obviously need the exact value of E). Also, I cannot consistently argue why E would have to cause j to be in parallel to the insulating surface, given that the insulating surface may aswell be modelled by infinite resistance. But that just as an aside.

    I may find a Finite-Differences formulation based only upon the direction of E on B and D, but that doesn't solve the underlying problem that when solved, I'd get no unique solution. So I ask myself which physical constraits there are, that I'm missing.
     
  9. Feb 12, 2013 #8

    Simon Bridge

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    What's wrong with dividing the PD by the length?
     
  10. Feb 12, 2013 #9
    We musn't use any vague arguments based upon symmetry or alike. I seek a unique solution to a well-posed problem.

    To clarify: The original problem (which I'd prefer not to discuss in order to keep it simple) assumes ρ rotates j with respect to E.
     
  11. Feb 12, 2013 #10

    Simon Bridge

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    OK - so what are you allowed to use?
    Can you use concrete (non-vague) symmetry arguments?

    U(x,y) is what you need.
    Putting A to the left and D on the bottom, placing the origin on the corner between A and D so the width is in the y direction would have you doing something like:$$\left ( \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2} \right )U(x,y)=0$$ subject to boundary conditions ##U(0,y)=0\; ,\; U(L,y)=U_0## ... the other two boundaries are insulators so: ##U(x,0)=U(x,w)##

    You feel you need to further justify the last boundary?
     
  12. Feb 12, 2013 #11
    Yes. To be honest, I don't understand how U(x,0) = U(x,W) would be enough even if we had it (it leaves the solution non-unique as before). But I don't think we can assume that in the first place unless it can more rigorously be proven by other than arguments based upon symmetry (which I don't have in my original problem).
     
  13. Feb 12, 2013 #12

    Simon Bridge

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    I'm not convinced you need it.
    That condition follows from the PDE (implies the free charge density is zero in the region of interest) and the fact that A and C are equipotential surfaces.

    You know the general solution to the PDE right?
     
  14. Feb 13, 2013 #13
    Do you or anyone else know how to prove this? I.e. that the solution is unique or, equally that the condition for existence of a solution is violated by any given ## \vec{E} |_{ B,D } ## other than one?

    You mean the analytical one? I haven't given it any thought but I assume the seperation ansatz is suitable? Anyway, I would prefer not to consider the problem under any kind of ansatz unless we have established the uniqueness of the solution.
     
    Last edited: Feb 13, 2013
  15. Feb 13, 2013 #14

    Simon Bridge

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    Yes - solve the PDE.

    It does not really matter what you prefer - this is the way to the kind of solution you need. It it quite common that you have to start out working on a problem without knowing what the solution will be like. You are over-thinking the problem! Stop it!

    But consider:
    Why should I bother trying to help you if you insist on being so obtuse?

    Good luck.
     
  16. Feb 13, 2013 #15
    I wouldn't know how to solve the PDE analytically by any other means than using an ansatz. If I find a solution to the problem - even if I can prove the solution is unique within the space allowed for by the ansatz - I still don't know whether it's globally unique - which is what I need to know.

    To my knowledge, separated functions X(x)Y(y) are a complete basis to the solution space of Laplace' Equation, so I could find that the solution is unique if I can show that

    ## \sum_i^\infty a_i X_i(x) Y_i(y) ##

    has only one solution for a given E (less the magnitude) on the boundaries B and D, this would suffice. Given that series, it sounds difficult though - thus I hoped someone knows a more generic proof which doesn't rely on an ansatz and the completeness thereof.

    PS: You're being rude.
     
    Last edited: Feb 13, 2013
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