Electric Flux and Gauss's Law

In summary, the electric flux through the sides of the barrel is zero because the line is much longer than the cylinder.
  • #1
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Homework Statement



A long solid nonconducting cylinder has radius r = 0.65 meters. It has a uniform charge density of 3.7 C/m3. Consider a cylinderical container concentric with the charged cylinder, with radius R = 1.5 m, and length L = 2 m.

Calculate the flux through the barrel of the cylindrical container.

The Attempt at a Solution



I thought this problem would be pretty simple. Since the charged cylinder is enclosed by the cylindercal container, I used Gauss's law to calculate the net electric flux. I used the equation

Electric Flux = Charge Enclosed / 8.85 * 10 ^ -12

The first step I took was to calculate the charge enclosed. To do this I just multiplied the charge density by the volume. 3.7 * .65^2*pi*2 = 9.8.

Then, I just divided that number by the constant Eo, 9.8/(8.*5 * 10^-12).

However this did not give me the right answer. Help! What am I doing wrong?
 
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  • #3
As far as I can tell the relationship is pretty simple... Flux is simply the charged enclosed divided by espsilon... but obviously I'm not getting something! What aren't I understanding? Or is this particular relationship not applicable to this problem?
 
  • #4
Your equation is right for the total Net flux.

But the question is asking for just the flux through the barrel end isn't it?
 
  • #5
Nope, the question is asking for the flux through the sides of the barrel. Part (a) and (b) of the question asked for the flux through the top and the bottom of the barrel, to which I correctly answered zero.
 
  • #6
Try using a cylindrical Guassian surface. First, find E as a function of r then since

[tex]\Phi=\oint{\vec{E}\bullet\ \mbox{d}\vec{S}[/tex]

this should give you the flux through the side of the barrel.
 
  • #7
I tried that and I keep getting the wrong answer, I converted sigma to lamba (charge per unit length) and used the relationship that E = lamba / (2*pi*epsilon*radius). Then I multiplied E by the surface area of the "container", 2*pi*radius*length... but this keeps yielding the wrong answer... also for the radius in calculating the electric field do I use the radius of the gaussian surface (the container) or do I use the radius of the charged cylinder? I tried both but still got the wrong answer...

What am I doing wrong here?
 
  • #8
I see so the line is much longer than the cylinder and it's not closed. This makes it much simpler. So your charge density can be expressed as what?

Figure then that the linear charge density (λ) from the volume charge number (σ) that they gave you as λ = σ*A = 3.7*103*π*(.65)²

With Qnet then being the charge inside that's just λ*ΔL isn't it?

Φnet = Qneto = λ*ΔL/εo
 
  • #9
I had the exact same problem with this question and I was convinced that I was correct with the same method you orginaly used. I have already spent an hour and a half on this single question and now I'm quite angry about what the solution was.

The problem was that they had the wrong question asked in the first place. The uniform charge density was supposed to be in [tex]\mu[/tex]C ([tex]10^{-6}[/tex]C). Unlike most of the other problems, they didn't tell you if you had a simple sign error. The professor has now since changed the question, so just go back and see if you can get the correct answer.

I was pissed too.
 
  • #10
Thanks so much specialkick23... much appreciated
 

1. What is electric flux?

Electric flux is a measure of the flow of an electric field through a given area. It is defined as the number of electric field lines passing through a surface perpendicularly.

2. What is the unit of electric flux?

The unit of electric flux is volt meters squared (V*m^2) or newton meters squared per coulomb (N*m^2/C).

3. What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric flux through a closed surface to the total charge enclosed by that surface. It states that the net electric flux through a closed surface is equal to the total charge inside that surface divided by the permittivity of free space (ε0).

4. How is Gauss's Law used to solve problems?

Gauss's Law can be used to determine the electric field of a given charge distribution by simplifying the calculations involved. It allows us to use symmetry and known values to find the electric field at a point without having to calculate the contributions from each individual charge.

5. Can Gauss's Law be applied to all situations?

No, Gauss's Law can only be applied to situations that exhibit symmetry, such as spherical, cylindrical, or planar symmetry. In cases where there is no symmetry, other methods such as Coulomb's Law or the superposition principle must be used to determine the electric field.

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