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Big difference between analytical solution and COMSOL simulation (line charge, Image charge method)

Posted 15 mar 2016, 16:02 GMT-4 5 Replies

Woojin Ahn
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I have an analytical formula for potential distribution (by image charge method) when the line charge is floating on grounded plane, which is,
V=P/(2*pi*eps)*ln (ri/r)

Also I did numerical simulation exactly same system with that of analytical one, the result is different, please find attached COMSOL simulation file.
If I scale the analytical function with 1.7, then the curve matches beautifully, see attached picture file

Red dots : simulation result.
Green line: analytical model
Black line: analytical model * 1.7

I manually found out the value 1.7, at this point, I have no idea where this 1.7 is coming from.
I think the simulation set is fine..

Is there anyone who had similar experience with me?
or know why the curve matches with COMSOL result with scaling factor 1.7?
or maybe my set up in the simulation wrong?

If simulation set up is fine, I think either Gauss Law or COMSOL simulation results is wrong.
Thus, simulation set up has to be the issue, but I can't find what is the problem in set up.

Thank you.




5 Replies Last Post 16 mar 2016, 15:02 GMT-4
Edgar J. Kaiser Certified Consultant
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Posted: 9 years ago 16 mar 2016, 07:10 GMT-4
Hi,

the formula gives a potential difference between ri and r. ri has to be >0, so what ri where you using?

In COMSOL the results for r -> 0 will be quite mesh dependent. The line charge creates a singularity at r = 0.

Cheers
Edgar

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Hi, the formula gives a potential difference between ri and r. ri has to be >0, so what ri where you using? In COMSOL the results for r -> 0 will be quite mesh dependent. The line charge creates a singularity at r = 0. Cheers Edgar -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com
Woojin Ahn
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Posted: 9 years ago 16 mar 2016, 10:09 GMT-4
Thanks for the reply!

For understanding, I attached a picture.
Please find the attached JPG file

'Vs' is the potential at the equipotential line (which is the cylinder surface)
I set 'ri' as the distance between line charge (rho l)and the cylinder surface,
and 'r' as the distance between the other line charge(-rho l) and the cylinder surface.
both 'r' and 'ri' has no relationship with line geometry itself.

I think if the set up is reasonable, mesh or other comsol simulation set up may has not been properly set.
But the problem is that I can't figure it out at all..

Thanks!
Thanks for the reply! For understanding, I attached a picture. Please find the attached JPG file 'Vs' is the potential at the equipotential line (which is the cylinder surface) I set 'ri' as the distance between line charge (rho l)and the cylinder surface, and 'r' as the distance between the other line charge(-rho l) and the cylinder surface. both 'r' and 'ri' has no relationship with line geometry itself. I think if the set up is reasonable, mesh or other comsol simulation set up may has not been properly set. But the problem is that I can't figure it out at all.. Thanks!

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Edgar J. Kaiser Certified Consultant
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Posted: 9 years ago 16 mar 2016, 12:39 GMT-4
Ok, I see. I think the issue is that the zero charge BCs in your model generate a mirror ground plane (in z-direction) and mirror line charges (in +/-y-directions). Check the zero Charge BC and its symmetry properties in the documentation. Your model is different from a line charge over a ground plane.

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Ok, I see. I think the issue is that the zero charge BCs in your model generate a mirror ground plane (in z-direction) and mirror line charges (in +/-y-directions). Check the zero Charge BC and its symmetry properties in the documentation. Your model is different from a line charge over a ground plane. -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com
Woojin Ahn
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Posted: 9 years ago 16 mar 2016, 13:34 GMT-4
I see. What your saying is that in the case of ground plane at z=a,
if I set zero charge BC at the bottom of the xy plane (z=0), that will assume another ground plane at (z=-a)
so that I am solving the problem literally with two ground plane not one.

Instead of setting bottom plane as zero charge BC, I extended bottom part with long length not to effected by two ground plane, result seems matching well!

at this point, my final question is that instead of extending bottom part, (instead of setting zero charge BCs, at the bottom) what else BCs I should apply then?

Thank you.
I see. What your saying is that in the case of ground plane at z=a, if I set zero charge BC at the bottom of the xy plane (z=0), that will assume another ground plane at (z=-a) so that I am solving the problem literally with two ground plane not one. Instead of setting bottom plane as zero charge BC, I extended bottom part with long length not to effected by two ground plane, result seems matching well! at this point, my final question is that instead of extending bottom part, (instead of setting zero charge BCs, at the bottom) what else BCs I should apply then? Thank you.
Edgar J. Kaiser Certified Consultant
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Posted: 9 years ago 16 mar 2016, 15:02 GMT-4
You could try to place the line charge in the interior of the air box and add infinite element domains where the zerocharge BCs are now.

--
Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
You could try to place the line charge in the interior of the air box and add infinite element domains where the zerocharge BCs are now. -- Edgar J. Kaiser emPhys Physical Technology http://www.emphys.com

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