# Homework Help: Electric Flux and Gauss's Law

1. Jan 30, 2009

### physicsfun

1. The problem statement, all variables and given/known data

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.

3. 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?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 30, 2009

### LowlyPion

3. Jan 30, 2009

### physicsfun

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. Jan 30, 2009

### LowlyPion

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. Jan 30, 2009

### physicsfun

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. Jan 31, 2009

### chrisk

Try using a cylindrical Guassian surface. First, find E as a function of r then since

$$\Phi=\oint{\vec{E}\bullet\ \mbox{d}\vec{S}$$

this should give you the flux through the side of the barrel.

7. Jan 31, 2009

### physicsfun

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. Jan 31, 2009

### LowlyPion

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. Jan 31, 2009

### Specialkick23

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 $$\mu$$C ($$10^{-6}$$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. Feb 1, 2009

### physicsfun

Thanks so much specialkick23.... much appreciated