Solving Laplace's equation for a rectangle

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Homework Help Overview

The discussion revolves around solving Laplace's equation within a rectangular domain, particularly focusing on a setup involving a perfect electric conductor (PEC) and boundary conditions derived from a previous solution without the PEC. The original poster is exploring the implications of the PEC on the potential within the rectangle.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply superposition to solve Laplace's equation for the inner rectangle, questioning the nature of the boundary conditions and whether they should be constants or linear functions. Other participants inquire about the boundary conditions used in the absence of the PEC and discuss the implications of grounding on the potential.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the boundary conditions and the behavior of the potential in relation to the PEC. Some guidance has been offered regarding the nature of the potential at the PEC boundaries, but no consensus has been reached on the next steps for solving the equation.

Contextual Notes

Participants note that the original boundary conditions involve grounded sides of the rectangle, which may influence the approach to solving for the PEC. The presence of an irregularity in the setup is also acknowledged, contributing to the complexity of the problem.

peripatein
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Hello,

Homework Statement


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\phi=0 for that inner rectangle in order to then apply superposition and sum up the solutions.


Homework Equations





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=-∇\phi)? d1=1/4 and d2=1/3, hence the boundary conditions are: \phi(x,y=0)=?, \phi(x=1/4,y)=?, \phi (x,y=1/3)=?, \phi(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)?
 

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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.
 
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?
 
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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