Gauss's law, cylinder of length L

In summary, the conversation is about a physics problem involving a cylinder with a length of L and various variables such as r, R, \Phi, and \Phi_0. The goal is to calculate the charge of the cylinder by using the equation Q=integral of /rho(1-r/R) times rdr d/phi dl. However, there is confusion and disagreement about the problem statement and approach to solving it.
  • #1
lidl
10
0
http://img404.imageshack.us/img404/4148/gauss3j.png (infinitly long)

[itex]\Phi=\Phi_{0}(1-{r \over R}) {C \over m^2}[/itex]

What's the charge of the cylinder of length L?

Again, almost everything was wrong...
 
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  • #2
anyone? :(
 
  • #3
Here's link to my paper: http://img202.imageshack.us/img202/5889/physics.jpg
 
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  • #4
You need to be much clearer in the problem statement and your work.

Is the cylinder hollow? What do r, R, [tex]\Phi[/tex], and [tex]\Phi_0[/tex] represent?

Also, it's probably OK to write notes in your homework like "the electric field points radially outward so the flux through the top and bottom of the gaussian cylinder is zero" rather than relying on vague drawings
 
  • #5
This is from the test, as you can see, and everything I knew was written down...
 
  • #6
please delete this post
 
  • #7
ok, now I got the following:

Q=integral of /rho(1-r/R) times rdr d/phi dl

Which is the answer for the question of what's the charge of the cylinder L.

How to calculate it?
 

What is Gauss's law?

Gauss's law is a fundamental law of electromagnetism that relates the electric flux through a closed surface to the charge enclosed by that surface.

How is Gauss's law applied to a cylinder of length L?

In the case of a cylinder of length L, Gauss's law states that the electric flux through the curved surface of the cylinder is directly proportional to the charge enclosed by the cylinder. This can be represented mathematically as Φ = q/ε0, where Φ is the electric flux, q is the enclosed charge, and ε0 is the permittivity of free space.

What is the significance of a cylinder of length L in Gauss's law?

A cylinder of length L is often used as a simplified model for other more complex geometries, as it is a relatively simple shape with a well-defined surface area. This allows for easier application of Gauss's law and can help in understanding the concept behind it.

Can Gauss's law be used to calculate the electric field within a cylinder of length L?

Yes, Gauss's law can be used to calculate the electric field within a cylinder of length L if the charge distribution is known. This is done by taking the derivative of the electric flux with respect to the distance from the center of the cylinder.

Are there any limitations to Gauss's law when applied to a cylinder of length L?

One limitation of Gauss's law is that it assumes a vacuum or uniform medium between the charges. In reality, there may be other factors that affect the electric field within the cylinder, such as nearby conductors or dielectrics. Additionally, the cylinder must have a uniform charge distribution for Gauss's law to accurately calculate the electric field.

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