Register to reply 
Volume Charge Density in a long Cylinder 
Share this thread: 
#1
Feb2806, 03:09 AM

P: 6

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...  feeling like a physics flunky....



#2
Feb2806, 05:24 AM

Admin
P: 21,880



#3
Feb2806, 10:20 AM

P: 6

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



#4
Feb2806, 10:41 AM

P: 335

Volume Charge Density in a long Cylinder
Dan 


#5
Feb2806, 10:43 AM

Admin
P: 21,880

Is this more clear?
Gaussian surfaces  http://hyperphysics.phyastr.gsu.edu...gausur.html#c1 Or rather, what is confusing? 


#6
Feb2806, 11:24 PM

P: 6

No No No!!! 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    thanks though



#7
Nov1907, 04:30 AM

P: 7

using vectors, i make the electric field E = lamda/2pi(sqrt.[r^2s^2])epsilon
r^2 = s^2 +(ca)^2, (and b is ca), therefore rearrange, (ca)^2 is (r^2s^2) does this help? 


Register to reply 
Related Discussions  
Hieght of a volume in a cylinder on its side, with known volume.  Precalculus Mathematics Homework  2  
Volume charge density  Introductory Physics Homework  9  
Volume Charge Density  Introductory Physics Homework  1  
Volume charge density  Introductory Physics Homework  0  
Volume density of polarization charge?  General Physics  0 