Surface Current and Electric Field

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The discussion revolves around understanding the relationship between surface charge density and current density in the context of two infinite sheets of ideal conductive material. The user is struggling to apply Gauss's law in two dimensions to find the electric field, which is parallel to the surfaces rather than perpendicular. They express confusion about the source of the electric field and the implications for current density. The mention of a propagating electromagnetic wave suggests a consideration of dynamic fields in the analysis. The conversation highlights the complexities of applying classical physics principles to non-standard geometries.
BnayaMeir
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Hi everyone!

I'm pretty new in this forum, I found the topics here very relevant to my physics course. And here is my question:

Given the following drawing, two infinite sheets (in y and z axis) of ideal conductive material. their thickness is infinitesimal (dx->0).

Screenshot 2022-04-26 152843.png


The electric field is defined:

Screenshot 2022-04-26 160051.png

I have askes to find the surface charge and current density.

well.. I tried to apply the integral gauss law
1650978402507.png
but in 2 dimensions (didn't work).
I have also tries the derivative version of the law
1650978512678.png
which gave me zero. it looks right for me, since the electric field should be perpendicular to the surface. but the only electric field is parallel to the surfaces. Then where this field is come from?? I'm very confused..

I thought to find the current density after finding the charge density by the following equation in 2-D:
1650978895539.png


I hope you will be able to help me, thanks a lot!
 
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Have you considered that this could be the electric field of a propagating electromagnetic wave?
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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