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Gauss's law on cylinder

  • Thread starter Acuben
  • Start date
  • #1
63
0

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

 

Answers and Replies

  • #2
1,254
3
What are you uses gauss' law ON? a wire? a uniform sheet? constant charge density? a banana?
 
  • #3
1,860
0
Mmm, gaussian banana.
 
  • #4
63
0
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.
 
  • #5
1,254
3
in that case the answer to your question is: 'it depends on the situation.'
 
  • #6
699
6
Mmm, gaussian banana.
That can only be solved using a bananal coordinate system.
 

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