1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gauss' Law and electric fields

  1. Jul 18, 2008 #1
    A long non-conducting cylinder has a charge density ρ = α*r, where α = 4.95 C/m4 and r is in meters. Concentric around it is a hollow metallic cylindrical shell.

    [​IMG]

    1. What is the electric field at 2 cm from the central axis? Assume the length L is very long compared to the diameter of the shell, and neglect edge effects. Answer in units of N/C.
    - I used the Gauss' Law and derived the equation α*r3/(3*ε0). And got the answer 7.454e+07 N/C and it was right.

    2. What is the electric field at 5.84 cm from the central axis? Answer in units of N/C.
    - But when I use the same equation here I get the wrong answer. Can some one help me please!

    4. What is the electric field at 18.4 cm from the central axis? Answer in units of N/C.
     
  2. jcsd
  3. Jul 18, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You mean α*r^2/(3*ε0), right? Do they say anything about the diameter of the shell?
     
  4. Jul 18, 2008 #3
    Yes, sorry about the mistake, the formula I found was α*r^2/(3*ε0). No the question doesn't say anything about the diameter of the expect the values given in the drawing.
    [​IMG]
     
  5. Jul 18, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Is the value in the drawing? Your formula is only valid if r is inside of the cylinder. If it's outside, you'll need to get the charge by integrating over the whole cylinder and then applying Gauss' law.
     
  6. Jul 18, 2008 #5
    For part b of the question the value is between in the inner cylinder and the outer shell
     
  7. Jul 18, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Then when you are integrating to get the charge, only integrate out to the radius of cylinder. Not all of the way to r.
     
  8. Jul 18, 2008 #7
    Do you mean just use r = 5.48 cm in the equation α*r^2/(3*ε0) because I did that and I still got it wrong.
     
  9. Jul 18, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, no, no. You will get a different formula if you apply Gauss' law when r is greater than the radius of the charge. What is the radius of the charged cylinder??
     
  10. Jul 18, 2008 #9
    radius of the charged cylinder from the figure is 15.2 cm for outer radius and 10.5 cm for inner radius
     
  11. Jul 18, 2008 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I wish I could see your figure. I'm very confused. Are you sure those aren't diameters? You said for part b) that 5.84cm was outside of the inner cylinder.
     
  12. Jul 18, 2008 #11
    Here is a link to it.

    [​IMG]
     
    Last edited: Jul 19, 2008
  13. Jul 18, 2008 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Doh. Now it has to be approved. Hope you're not in a hurry.
     
  14. Jul 18, 2008 #13
    How about the link can you see using that?
     
  15. Jul 18, 2008 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, thanks. Ok, so when you use Gauss' law, to compute the charge only integrate out to 5.48cm. When you compute E*Area, use r=5.84cm.
     
  16. Jul 18, 2008 #15
    So you want me compute E using the equation α*r3/(3*R*ε0) where r = 5.48 cm and R = 5.84cm?
     
  17. Jul 18, 2008 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That seems right, doesn't it? It's Gauss' law. Can't go wrong with that.
     
  18. Jul 18, 2008 #17
    It still says it is wrong am I supposed to cube or square the small r. Because when I used the equation that I showed you and cubed the small r and got 5.251e+08 but it was still wrong.
     
    Last edited: Jul 18, 2008
  19. Jul 18, 2008 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I cubed the small r and used that formula and I got a different number. Be sure all of your lengths are in meters.
     
  20. Jul 18, 2008 #19
    Oh well I just ran out all my tries for that part but how about the last part when it's asking for E at 18.4 cm do I use the same equation or does it become different since it's outside the shell?
     
  21. Jul 18, 2008 #20

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You tell me. You still use Gauss' law. Does the shell carry any net charge?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gauss' Law and electric fields
Loading...