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!

Finding the linear density of charge on a cylinder

  1. Sep 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A rod with charge linear density λ is located at the long axis of a cylinder with charge linear density 2λ. With this information use Gauss's law to find (a) the charge linear density on the interior and exterior surfaces of the cylinder (b) the electric field outside of the cylinder at a distance r from the rod.

    Just to make sure: my questionnaire does refer to charge linear density, not surface density. My questionnaire does not make reference to a "long" rod nor a "long" cylinder. I've read these calculations are next to impossible if there's no reference to "longness".


    2. Relevant equations

    Θ=q/ε

    q=λL -- for the rod



    3. The attempt at a solution

    This question does not make any sense to me. So I'm not going to try to answer it, because I'm in disagreement with the question. I hope you also consider the question is wrong or weird or prove me wrong so I can understand.

    Thanks a lot.
     
  2. jcsd
  3. Sep 11, 2014 #2
    Is there a way to convert linear density to surface density?
     
  4. Sep 11, 2014 #3

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Hmmm... I have also had similar issues with poorly worded problems. In order to make sense I would have to assume that by linear charge density on the cylinder, they meant for you to take it as 2pi R dL with L being the length of (presumably) the rod and cylinder both. The charge density would then be linear with respect to L. I'm not entirely sure though. Not answering a problem, however, normally doesn't work out too well.

    If you look at the charge density this way, you should be able to treat it as linear due to symmetry about the rod for all points on the cylinder.
     
  5. Sep 11, 2014 #4
    I was thinking that if I take the wording as it is, then the linear density on the surface (interior and exterior) is just 2λ based on what their statement was.

    As far as part (b) goes, I was thinking if there is some way to convert linear density to surface density so that I can then use gauss's law to calculate the charge contained within some gaussian surface at 'r' distance from that system.

    Any ideas if this is at all possible? (converting linear density to surface density).

    Thanks.
     
    Last edited: Sep 11, 2014
  6. Sep 12, 2014 #5

    BiGyElLoWhAt

    User Avatar
    Gold Member

    I'm not really sure how I feel about that honestly, I've seen it done in reverse (unjustifiably in my opinion). It can be done in certain circumstances, and this might be one of them. After sleeping on it and reading it again, it does seem like what they mean by linear charge density is that the density doesn't vary with r or theta, only L or z or whatever you wanna call your cylindrical axis. That being the case, you shouldn't necessarily need to convert it, because your surface integral from gauss' law will merely be a function of L and constants.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted