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Solving Laplace's equation for a rectangle

  1. May 20, 2014 #1
    Hello,
    1. The problem statement, all variables and given/known data
    I am trying to solve Laplace's equation for the setup shown in the attachment, where f(x)=9sin(2πx)+3x and g(x)=10sin(πy)+3y. I have managed to solve it for the setup without the rectangle (PEC), and am now trying to solve ∇2[itex]\phi[/itex]=0 for that inner rectangle in order to then apply superposition and sum up the solutions.


    2. Relevant equations



    3. The attempt at a solution
    Since the inner rectangle is a perfect conductor, the electric field inside must be zero. Hence the potential must be constant, right (as E=-∇[itex]\phi[/itex])? d1=1/4 and d2=1/3, hence the boundary conditions are: [itex]\phi[/itex](x,y=0)=?, [itex]\phi[/itex](x=1/4,y)=?, [itex]\phi[/itex] (x,y=1/3)=?, [itex]\phi[/itex](x=0,y)=?. Now how should I proceed? Should all these potentials indeed be equated to constants or ought I to use something linear, such as (Ax+B)(Cy+D)?
     

    Attached Files:

  2. jcsd
  3. May 21, 2014 #2

    BvU

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    If you solve for the setup without the PEC rectangle, what boundary conditions do you use for x=0 and y=0 ?
    Looks to me as if those are conductors as well, grounded to boot!

    PEC potential is definitely constant: any deviation would cause charge to move until it's constant again.
     
  4. May 21, 2014 #3
    Yes, without the PEC the bottom and left sides of the original rectangle are indeed grounded.
    The original boundary conditions are:
    ϕ(x=0,y)=0; ϕ(x,y=0)=0;ϕ(x=1,y)=g(y);ϕ(x,y=1)=f(x)
    and the solution is given as:
    ϕ(x,y)=3xy + 9sin(2pi*x)sinh(2pi*y)/sinh(2pi) + 10sinh(pi*x)sin(pi*y)/sinh(pi)
    How do I now solve Laplace's equation for the inner PEC?
     
  5. May 21, 2014 #4

    BvU

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    OK, looks good. I even played with a spreadsheet relaxation and got a nice 3D plot.
    As far as I can see you now have to find solutions that are equal and opposite to ϕ on the bounds of the PEC and 0 on what remains of the bounds of the original unit square.
    I don't see a way to deal with the irregularity.
    So, like you, I am stuck for the moment...
     
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