# Volume Charge Density in a long Cylinder

1. Feb 28, 2006

### MtHaleyGirl

I am never sure if I am on the right track when answering my homework problems. Anyway the problem involves a long solid cylinder of radius R and uniform charge distribution throughout its volume. We are supposed to choose a cylindrical gaussian surface of radius r and length L with r < R - so the gaussian surface is inside the cylinder. We are 1st supposed to determine q(enclosed) in terms of ρ (rho), L and r - and of course any other relevant constants. So first off - I am looking at the flux within the cylinder --- already I am nervous about this because I cannot think of how to imagine the flux within this cylinder... and then considering E ??? So to further explain where I am at - since the cylinder is "very very long" I have capped off the gaussian cylinder inside the whole. 1st off - is q(enclosed) a ratio of the whole Q ? or is it constant within the cylinder (I don't think q = Q). I want to consider q/Q - where q = π(pi) r² L then Q = π R² L - then I want to take q/Q = (π(pi) r² L )/(π R² L ) - cancel the π(pi) and L to get q =(Qr²)/R² - then I wonder if I am getting anywhere = also does q = ρL - it shows in my book that q = λL (λ is linear charge density) and I don't know if it works for volume charge density too... :uhh: - feeling like a physics flunky....

2. Feb 28, 2006

### Staff: Mentor

3. Feb 28, 2006

### MtHaleyGirl

Thanks, but I've been there - the gaussian surface is always shown outside the original cylinder, that is why I am confused...

4. Feb 28, 2006

### topsquark

Recall that all you care about is the net charge INSIDE the Gaussian surface, so we can ignore anything outside the surface. To get the charge inside, just take the volume integral over the charge density.

-Dan

5. Feb 28, 2006

### Staff: Mentor

6. Feb 28, 2006

### MtHaleyGirl

No No No!!! :rofl: I've been to hyperphysics lots - the Gaussian surface is a cylinder within another cylinder. All of the examples I have seen the cylinder is used through a plane or around another cylinder. The Charged Cylinder is of radius R and the Gaussian surface is radius r r < R... Does that make sense??? I'm no physics genius and when I am given a problem outside the regular examples I feel like my brain is being scrambled :yuck: - - - thanks though

7. Nov 19, 2007

### mjkr1988

using vectors, i make the electric field E = -lamda/2pi(sqrt.[r^2-s^2])epsilon

r^2 = s^2 +(c-a)^2, (and b is c-a), therefore rearrange, (c-a)^2 is (r^2-s^2)

does this help?