5
$\begingroup$

I'm solving an electrostatic problem governed by Laplace equation $$-\nabla \cdot (\rho^{-1} \nabla u) = 0$$ in the following domain: a brick ($\Omega_1$) with a cylindrical inclusion ($\Omega_2$), where $\rho$ - electric resistivity, $u$ - potential. There are two plane electrodes attached to opposite faces of the brick.

On top and bottom faces Dirichlet boundary conditions are defined: $$u|_{\partial\Omega_{top}} = u_1$$ $$u|_{\partial\Omega_{bottom}} = u_2$$ On other faces of the brick homogeneous Neumann boundary conditions are held. $$\rho^{-1}\frac{\partial u}{\partial n}|_{\partial\Omega_{side}} = 0$$

If the inclusion is strictly inside the brick, everything is good. Here is a 2D view of potential distribution with isolines.

2D view of potential distribution for the case of cylinder lying inside brick

I am interested in such a situation when inclusion goes through the brick. In this case it crosses top and bottom faces, and they become consisting of two subdomains.

If I impose the same boundary conditions as in previous case on both subdomains, I get the following potential distribution:

2D view of potential distribution for the case of cylinder crossing brick

I think it's wrong. Therefore I would like to ask - what should I change to get right solution - boundary conditions on top/bottom faces, or something else?

Parameters: $\rho_{\Omega_1} = 10^6 (Ohm \cdot m)$, $\rho_{\Omega_2} = 10^{-3} (Ohm \cdot m)$, $u_1 = 1 (V)$, $u_2 = 0$

Edit

Why I think that the result I obtain in the second case is wrong. The task I'm solving is a task about conductive inclusion inside non-conductive (or low conductive) medium. I was taught that there is no electric field in conductive bodies, and therefore electrostatic potential is constant there. That can be seen from the result of the first simulation when inclusion is fully inside a brick. And I expect the same behaviour of electric field in the second case. But that doesn't happen. Potential equally changes in conductive and non-conductive media, and electric field is the same in the whole domain. Therefore I think something is wrong there. I would be happy if somebody could say that the model I use is correct, but I'm wrong only in interpretation of my results.

Edit 2

My thanks to all of those who took part in discussion of my problem. But the main part of the answers was about explaining that the solution of the problem in the second case (when inclusion goes through the brick) is the right solution of the boundary value problem I provided. I agree with all these answers. And as @StefanoM said "The second problem is trivial and does not need a FEM code to be solved". I agree 100% with that. But as @StefanoM wisely noticed "one should never forget that phenomenological equations are based on a number of simplifying assumptions and cannot be applied if those assumptions are not true". Therefore the main question that worries me is - what is the right model for the problem I'm trying to solve? Perhaps I need another equation, or another boundary conditions somewhere. Could anyone suggest the model that would correctly describe the electrostatic problem when the computational domain consists of two subdomains with contrasting electrical resistivity, especially in the case when condutive inclusion goes through non-conductive media. I started from cylindrical inclusion because it's the simplest case. In fact I will need to work with more complicated domains (inclusions), therefore I do need to use FEM code.

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ What is the "right solution?" In the case where there's no vertical change in the resistance, this is what one would expect from the potential. $\endgroup$
    – Peter Brune
    Sep 9, 2013 at 14:04
  • $\begingroup$ silly question: are you sure you're looking to an inner slice and not the outer boundary $ \partial\Omega_1 $ ? (which would give these parallel isolines for both scenarios) $\endgroup$
    – SAAD
    Sep 9, 2013 at 16:31
  • $\begingroup$ @SAAD :) yes, I'm sure it's an inner slice. I build a 2D view in a code, not in visualization program; and the code stays the same, I change only input mesh. $\endgroup$
    – martemyev
    Sep 9, 2013 at 23:48
  • $\begingroup$ @PeterBrune, it's a very good question indeed. I answered editing my question. $\endgroup$
    – martemyev
    Sep 10, 2013 at 1:37
  • $\begingroup$ About edit 2: I would be very happy to help you, but at the present stage it is unclear to me what type of problem you are trying to solve. May be you should formulate a new question, clearly stating the physical problem, its context, and the quantities you are looking for. May be the description of an experimental setup could be of help. This said I fear that such a question would be off topic on SciComp SE. $\endgroup$
    – Stefano M
    Sep 12, 2013 at 21:08

1 Answer 1

Reset to default
6
$\begingroup$

I don't have electrostatics training but your second picture looks OK to me. Your conductive inclusion is essentially shorting out the plates u1 and u2.

I was taught that there is no electric field in conductive bodies, and therefore electrostatic potential is constant there.

If the conductive body touches both plates u1 and u2 which are forced to have different potentials according to the boundary conditions, then the conductive body cannot have constant electrostatic potential. It has a potential gradient and has a high current going through it.

Improve this answer
$\endgroup$
7
  • $\begingroup$ I believe this answer is likely the correct one. There is no electric field inside a perfect conductor. Your inclusion is not perfectly conducting, therefore it is producing the gradients it must to solve the governing equations. $\endgroup$
    – Godric Seer
    Sep 10, 2013 at 2:32
  • $\begingroup$ @GodricSeer, so why there was no electric field in the first case (inclusion inside brick)? If I set electrical resistivity of inclusion to $10^{-16} (Ohm \cdot m)$ which corresponds to a perfect conductor (I believe), changes in potential distribution won't be visible. $\endgroup$
    – martemyev
    Sep 10, 2013 at 3:18
  • 2
    $\begingroup$ @martemiev Think about a circuit that looks like this: (neg terminal)--100ohm--1ohm--1ohm--1ohm--100ohm--(pos terminal). In this case, the high-conductance resistors in the middle will each not have much potential difference across them. Now consider (neg terminal)--1ohm--1ohm--1ohm--1ohm--1ohm--(pos terminal). In this case, the three resistors in the middle will each have relatively high potential difference across them. The first case is like your first picture, and the second case is like your second picture. $\endgroup$
    – post-as-guest
    Sep 10, 2013 at 3:30
  • $\begingroup$ @post-as-guest, well, you are right that the result I'm obtaining corresponds to the mathematical model I'm considering. But the result is not physical - in my belief. Therefore I think that the model itself describes the physical process incorrectly. $\endgroup$
    – martemyev
    Sep 10, 2013 at 3:35
  • 2
    $\begingroup$ @post-as-guest has obviously the correct answer and the right analogy with an electric circuit. The second problem is trivial and does not need a FEM code to be solved... (one can check by hand that the computed solution is the correct unique solution to the given BVP.) As what regards the question whether the solution to the linear Laplace equation is physical or not, one should never forget that phenomenological equations are based on a number of simplifying assumptions and cannot be applied if those assumptions are not true. $\endgroup$
    – Stefano M
    Sep 10, 2013 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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