0
$\begingroup$

I currently have a 3D geometry which is made from an isotropic material. In my case this material is simply a highly conductive metal. We can think of this geometry almost as a thin film with slightly varying thickness throughout. I am told that the thickness at a point on the film correlates with its conductance there. Simply put the thicker the section the better conductance it has.

Here is what a 2D projection of my geometry looks like. For the moment ignore the grid lines inside. enter image description here

What I am trying to solve is a boundary valued problem for a 2D model of this 3D geometry in MATLAB. The top curved edge has a voltage of +V0 applied to it and the bottom curved edge has a voltage of -V0 applied to it. These are my Dirichlet boundary conditions. The left and right sides have a boundary condition of zero flux (I think this means electric field flux but it may be current flux) and these are my Von Neumann boundary conditions.

So far this makes sense to me. It's when I try solving the governing equation that I become a bit lost. It seems I need to solve 2 partial differential equations simultaneously.

The first equation is Poisson's electrostatic equation shown here, note V(x,y) represent the voltage as a function of x and y,:

$\nabla^2V(x,y)=-\frac{\rho}{\epsilon}$

The other equation I think I need to solve is the continuity equation shown here:

$\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t}$

Now I will explain what my misunderstandings are. First off, since we are dealing with a static case I believe that both $\frac{\partial \rho}{\partial t}$ and $\frac{\rho}{\epsilon}$ (might be wrong to assume this is zero) are zero meaning both of the right hand sides of the PDE go to zero.

Secondly for isotropic materials (which this should be since its all the same metal) we can relate the electric field and current density as $\vec{J} = \sigma \vec{E}$. Does the thickness not effect this equation at all?

Based on my equations and understandings of the problem I do not see how the thickness is coming into play. Is it a mistake to be treating this 3D geometry as a 2D one with different conductances at different regions?

One person mentioned that if the thickness is a function like $\delta(x,y)$ where it varies based on my 2D location then the actual equation I should be solving is

$\nabla \cdot \big(\sigma \delta(x,y) \nabla V(x,y)\big) = 0$

but $\sigma$ should be constant so that just falls out. If someone could provide some help I would greatly appreciate it. Thank you and sorry for this long winded post.

$\endgroup$
  • $\begingroup$ It should be clear to you that the current is not equal through all parts of your film. Consider the case of an array of resistors of linearly-varying resistances. Go from there. $\endgroup$ – Jon Custer Dec 14 '18 at 15:19
  • $\begingroup$ @nick2225 Why not use a 3D model? $\endgroup$ – Alex Trounev Dec 14 '18 at 16:13
  • $\begingroup$ @JonCuster I think I get your idea. I tried to derive a similar expression by representing the film as infinitesimal squares with various heights. So in a 2D projection it would look like squares with side lengths of dl, differential length, and height delta. However, getting an expression for the conductance of a single infinitesimal doesn't shed light for me on the governing PDEs above. I tried performing a current balance at the same scale but I am afraid it was not too informative to me as I could not get my derivation to match any of the PDEs above, I can include my notes later. $\endgroup$ – nick2225 Dec 14 '18 at 17:45
  • $\begingroup$ Why is $\rho=0$ in static case? $\endgroup$ – Kyle Kanos Dec 18 '18 at 11:11
0
$\begingroup$

Here is a simple answer but may not be what you are looking for. I agree, as you know the thickness at every point then you know the resistance at these points. So it is just a matter of solving for voltage using an array of points and interation. Use a suitable numerical approximation for voltage. Keep the boundaries where the voltage is constant as constant, the other edges are a 3 point average.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.