Why is the flux on top and bottom of a cylinder ignored in Gauss's law?

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Homework Help Overview

The discussion revolves around the application of Gauss's law to a cylindrical surface and the reasoning behind ignoring the flux through the top and bottom surfaces of the cylinder. Participants are exploring the implications of this approach in the context of electric fields and charge distributions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the conditions under which the flux through the top and bottom of the cylinder can be considered zero, with some suggesting that it may depend on the specific charge distribution being analyzed, such as a long wire.

Discussion Status

The conversation is ongoing, with participants sharing thoughts on the applicability of Gauss's law in different scenarios. Some guidance has been offered regarding the dependence on the situation, but no consensus has been reached.

Contextual Notes

There is a lack of specificity regarding the charge distribution being considered, which may affect the interpretation of the electric field and the resulting flux calculations.

Acuben
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Homework Statement



(first of all, let me know if I am wrong in any parts)
gauss's law can be used on cylinder.
Why is it that flux on top and flux on bottom of the cylinder, with area pi*r^2 ignored? (aka, equal to 0)
such that
Flux= Integral(E*da)=E2(pi)rl=q/(Eo)
where (Eo) is permitivity of/in vacuum.

Homework Equations



Cylinder
Flux= Integral(E*da)=E2(pi)rl=q/(Eo)
Sphere
Flux= Integral(E*da)=E2(pi)r^2=q/(Eo)

The Attempt at a Solution

 
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What are you uses gauss' law ON? a wire? a uniform sheet? constant charge density? a banana?
 
Mmm, gaussian banana.
 
zhermes said:
What are you uses gauss' law ON? a wire? a uniform sheet? constant charge density? a banana?

nothing in particular. I meant making an gaussian cylinder in general did not define what I am using it on.
Although if it's a long wire, I suppose there is no Electric field towards top and bottom and it cancels out.
 
in that case the answer to your question is: 'it depends on the situation.'
 
Mindscrape said:
Mmm, gaussian banana.

That can only be solved using a bananal coordinate system.
 

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