Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.

Difference between Electrical Currents and Electrostatics

Please login with a confirmed email address before reporting spam

Hi everybody,
just a quick question.
using the electrostatic package I can easily put a dielectric constant that depends on the electric field using :

epsilon(es.normE)

yet with the electrical currents package I cannot use a conductivity which depends on the current, i.e. :

sigma(ec.normJ)

Does anyone has a simple explanation for that ? It's probably very simple..

Thanks
Alex

1 Reply Last Post Mar 9, 2015, 5:04 a.m. EDT
Sven Friedel COMSOL Employee

Please login with a confirmed email address before reporting spam

Posted: 9 years ago Mar 9, 2015, 5:04 a.m. EDT
Dear Alex,

the reason is the following:

Electrostatics:

div D = 0
D= epsilon_0*epsilon_r*E
E=-grad V
--> D = epsilon_0*epslion_r(E)*E

That means, even by introducing the nonlinearity, all terms containing E remain on the right side.
The system can be solved.

Electric currents:
div J = 0
J = sigma*E
E=-grad V
--> J = sigma(E)*E would be possible (in analogy to electrostatics) but
--> J = sigma(J)*E would create a circular dependency, as J appears on both sides of the eq.

There are two workarounds:
1) express your conductivity as a function of E instead of J (this is the better option) or
2) introduce an auxillary variable (creates some numerical overhead and will typically converge slower but is possible)

Best regards,

Sven


Dear Alex, the reason is the following: Electrostatics: div D = 0 D= epsilon_0*epsilon_r*E E=-grad V --> D = epsilon_0*epslion_r(E)*E That means, even by introducing the nonlinearity, all terms containing E remain on the right side. The system can be solved. Electric currents: div J = 0 J = sigma*E E=-grad V --> J = sigma(E)*E would be possible (in analogy to electrostatics) but --> J = sigma(J)*E would create a circular dependency, as J appears on both sides of the eq. There are two workarounds: 1) express your conductivity as a function of E instead of J (this is the better option) or 2) introduce an auxillary variable (creates some numerical overhead and will typically converge slower but is possible) Best regards, Sven

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.