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Difference between Electrical Currents and Electrostatics
Posted Mar 5, 2015, 4:08 a.m. EST Low-Frequency Electromagnetics, Modeling Tools & Definitions, Parameters, Variables, & Functions Version 5.0 1 Reply
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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
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
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Posted:
10 years ago
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
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