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Field behind metal shield
Posted Oct 12, 2012, 5:42 a.m. EDT Low-Frequency Electromagnetics Version 4.3a 1 Reply
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Dear Users
I want to model the following simple case.
Within vacuum, a 2D body carries a surface charge. Next to it is a metal body. I want to obtain the quality of the field direction behind the metal. From theory we know that the metal is not polarizable, i.e. when subject to a field, the charge within separates in order to build up an opposing field within the metal that completely cancels the net field within the metal. The separated charge then produces a field of its own on the opposite side and so the field exits on the opposite side. I tried to model this and show it in this figure:
picturepush.com/public/11036949
My question is:
How do I define the right square to be a metal, i.e. such that the field lines terminate ON the metal on the close side and not "go around it" and then exit on the far side again?
Currently, due to a lack of knowing how to define the metal, I subtract the right rectangle from the vacuum box, but that's not giving the right solution.
Thanks for help.
I want to model the following simple case.
Within vacuum, a 2D body carries a surface charge. Next to it is a metal body. I want to obtain the quality of the field direction behind the metal. From theory we know that the metal is not polarizable, i.e. when subject to a field, the charge within separates in order to build up an opposing field within the metal that completely cancels the net field within the metal. The separated charge then produces a field of its own on the opposite side and so the field exits on the opposite side. I tried to model this and show it in this figure:
picturepush.com/public/11036949
My question is:
How do I define the right square to be a metal, i.e. such that the field lines terminate ON the metal on the close side and not "go around it" and then exit on the far side again?
Currently, due to a lack of knowing how to define the metal, I subtract the right rectangle from the vacuum box, but that's not giving the right solution.
Thanks for help.
1 Reply Last Post Oct 12, 2012, 7:53 a.m. EDT