Discussion Forum

Posted: 1 decade ago Aug 30, 2011, 5:59 p.m. EDT

The only idea that I have in my mind is that, may be when we increase relative permittivity, the material will show some characteristics of the metal, and as we know, we do not have electric field in metals. is this correct? However, in this case, I can not understand why in this forum people say we should use 1 for the relative permittivity of metals. Regards Bahar

Posted: 1 decade ago Aug 31, 2011, 1:53 a.m. EDT

Hi which electric field ? the electric field [V/m] or the Displacement field [C/m^2] ? if the latter check your formulas (or the underlying equations in COMSOL), it is proportional to epsilon0*epsilonr hence by changing the permittivity you change the displacement, but NOT the electric field in [V/m] -- Good luck Ivar

Posted: 1 decade ago Aug 31, 2011, 5:57 a.m. EDT

Bahar, I think it also depends on the boundary conditions. If the electrodes are at fixed potential, the electric field will essentially remain constant after introducing a dielectric. The predominant property of a metal is its high conductivity, so permittivity gets insignificant, it doesn't matter which value you apply. Cheers Edgar

Posted: 1 decade ago Sep 1, 2011, 11:54 a.m. EDT

Dear Ivar I think I should correct my description of problem. You are right, if I have a constant Voltage on electrode then the electric field on top of that will not chnage by changing the permittivity. However, Now I have to layers of material on top of the electrodes. like this: ------------- Top material : with permittivity of (FIRST TRY=1 SECOND TRY=100) ------------ First layer of material: with permittivity of lets say 10 ------------ Electrodes ------------ Now my problem is that the electric field in the top layer in the first try(Er=1) is higher that the second try (Er=100) and to me, I can not find any logic behind this :( Regards Bahar

Posted: 1 decade ago Sep 1, 2011, 11:58 a.m. EDT

Dear Edgar I would be grateful if you could take a look at description that I post for Ivar now. It would be great if you could tell me your idea too. Regards Bahar

Posted: 1 decade ago Sep 2, 2011, 3:28 a.m. EDT

Dear Bahar, your observation is exactly what should be expected. In your inhomogeneous model the electric field is expected to drop in the material with higher permittivity. You can take the surface of that highly permittive layer as some kind of virtual floating electrode. by introducing the dielectric the voltage on the virtual electrode will drop and thus the electric field drops. And maybe I should add, that in that case the field in the less permittive layer must rise because your metal electrodes are at fixed potential. In a lumped component analogy it is like a series circuit of capacitors. Voltage across capacitor is inversely proportional to capacitance. Hope this will help. Cheers Edgar

Posted: 1 decade ago Sep 2, 2011, 11:06 a.m. EDT

Dear Edgar Thanks a lot for your reply. your description is really helpful. Now I can understand the logic behind the results I have seen. Regards Bahar

Posted: 8 years ago Jun 9, 2016, 2:54 p.m. EDT

[QUOTE] Dear Ivar I think I should correct my description of problem. You are right, if I have a constant Voltage on electrode then the electric field on top of that will not chnage by changing the permittivity. However, Now I have to layers of material on top of the electrodes. like this: ------------- Top material : with permittivity of (FIRST TRY=1 SECOND TRY=100) ------------ First layer of material: with permittivity of lets say 10 ------------ Electrodes ------------ Now my problem is that the electric field in the top layer in the first try(Er=1) is higher that the second try (Er=100) and to me, I can not find any logic behind this :( Regards Bahar [/QUOTE] _______________________________ Dear Bahar, When, in your first try, you use the permitivity of 1, it means that your material is highly resistive and it may take very long for it to become eventually polarised (~unless the the electric potential is very high). So, the electric field can built up more and more.... However, when you increase the permitivity to 100, it is like as if you have indirectly told the material to become 100 times less resistive (or 100 times more conductive). And as you correctly mentioned, highly conductive materials (like metals) can not resist much the current and will soon discharge the potential through themselves; therefore, they have not a chance to build up such high electric field - compare to what you might expect form the lower permitivity materials (like polymers). Regards. Alireza