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Electrostatics -- Laplace equation

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Hi,

I'm solving Laplace's equation for a configuration of electrodes but I'm wondering what causes the difference in the results when using a 2D vs 3D simulation. The 2D configuration is essentially the 3D config with an appropriate dimension shrunk to zero. The 3D results tend to the 2D results as I shrink the same dimension myself (while in 3D mode). Which simulation is "right" and in what cases? If this is too vague I will attach the file or I can try and explain more

Thanks
Chris

2 Replies Last Post 6 juin 2012, 02:52 UTC−4
Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 6 juin 2012, 00:49 UTC−4
Hi

For me 2D is a 3D with a depth Z equal to 1 m (it depends on the physics, some use other default depth values see the main physics node as for some you can define the depth for absolute variables, for others all variables comes out as units/meter and its up to the user to correct for the appropriate "depth Z"

in 1D its the same but with a default area of 1m^2

--
Good luck
Ivar
Hi For me 2D is a 3D with a depth Z equal to 1 m (it depends on the physics, some use other default depth values see the main physics node as for some you can define the depth for absolute variables, for others all variables comes out as units/meter and its up to the user to correct for the appropriate "depth Z" in 1D its the same but with a default area of 1m^2 -- Good luck Ivar

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Posted: 1 decade ago 6 juin 2012, 02:52 UTC−4
Hi Christopher,

Yes, you are being vague. Nevertheless, you can probably figure this out on your own. You have to ask what makes the 3d results change (and tend to the 2d case) when you shrink that dimension. Therein lies the answer to the puzzle. As to which one is correct, it is the model that reflects reality most accurately. Save for numerical complications, that should always be the 3d model. The main purpose of 2d simulations is to cut down on complexity in order to save time in geometry modeling, meshing, and solving.
Hi Christopher, Yes, you are being vague. Nevertheless, you can probably figure this out on your own. You have to ask what makes the 3d results change (and tend to the 2d case) when you shrink that dimension. Therein lies the answer to the puzzle. As to which one is correct, it is the model that reflects reality most accurately. Save for numerical complications, that should always be the 3d model. The main purpose of 2d simulations is to cut down on complexity in order to save time in geometry modeling, meshing, and solving.

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