The discussion revolves around applying Gauss's Law to determine the electric field produced by an infinite plane sheet with a given surface charge density, σ0. Participants are exploring the implications of the solution provided and the reasoning behind specific components of the formula.
The discussion is active, with participants offering insights into the nature of the electric field and the flux through the Gaussian surface. There is a focus on understanding the contributions to the total flux and the reasoning behind the calculations, though no consensus has been reached on all aspects.
Participants are working from a solution that lacks detailed derivation, which has led to questions about the assumptions made regarding the Gaussian surface and the electric field's behavior around the sheet.
The solution is, where A=area:
2EA=(σ/ ϵ0)A
E=σ/2ϵ0
ap123 said:What gaussian surface did you use to get this?
ZedCar said:Possibly a cylinder?
I'm not 100% sure, as I just have the solution as shown above, and no other 'working out'.
ap123 said:Sounds a good choice.
What will the flux be through the cylinder?
ZedCar said:Not through it in one direction.
It's normal to the surface of the sheet.
As the sheet has two surfaces, the E-field is in both the two normal directions to the sheet.
ap123 said:Since the field is normal to the sheet surface, the flux through the sides of the cylinder is zero - so that only leaves the 2 cylinder ends. What's the total flux through these?
Hint : you've already got the answer :)