Solving Laplace's equation for a rectangle

[peripatein]

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[itex]\phi[/itex]=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=-∇[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)?

 

[BvU]

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.

 

[peripatein]

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?

 

[BvU]

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...

 

[paranormal]

Hello,

It seems like you are on the right track in solving Laplace's equation for this setup. Since the inner rectangle is a perfect conductor, the electric field inside must be zero, and therefore the potential must be constant. This means that the potential at each of the boundaries of the rectangle must be equal to a constant value. So, your boundary conditions should indeed be equated to constants.

To solve for the potential at each boundary, you can use the method of separation of variables. This involves assuming a solution of the form φ(x,y)=X(x)Y(y) and plugging it into Laplace's equation. This will result in two separate ordinary differential equations, one for X(x) and one for Y(y), which can then be solved using the given boundary conditions. The solution to the overall problem will then be the product of the solutions for X(x) and Y(y).

Alternatively, you can also use the method of images to solve for the potential in this setup. This involves introducing a fictitious charge at the same location as the inner rectangle, and then solving for the potential using the method you have already used for the setup without the rectangle. This will provide a solution for the potential inside the rectangle, which can then be used in the overall solution using superposition.

I hope this helps. Good luck with your problem!