Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Potential above a charged shell

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A very long insulating cylindrical shell of radius 6.90 cm carries charge of linear density 8.40 microC/m spread uniformly over its outer surface. Find the potential of a point located 4.4cm above the surface of the shell


    2. Relevant equations



    3. The attempt at a solution

    I used Gaussian surface to find the electric field created by the shell outside (Q = linear charge density):
    QL/[tex]\epsilon_{0}[/tex] = E*2*pi*d*L,
    where L is length, d is distance to point from the axis of the cylinder.
    The Ls cancel out, rearranging, I got:
    E=[tex]\frac{Q}{2pi*d*\epsilon_{0}}[/tex]

    I then integrated that to get the potential:
    V=[tex]\frac{Q}{2pi*\epsilon_{0}} \int {(1/d)dd} [/tex]
    The integral's upper limit is d-r, and lower limit is r.
    I got the equation below as a result:
    V=[tex]\frac{Q}{2pi*d*\epsilon_{0}} ln(\frac{d-r}{r})[/tex]

    Substituting in the numbers I got V=-6.8*10^4, but that's wrong. Is it my method? Should I slice the cylinder into rings and do V from each one, then integrate the whole thing?

    Thanks in advance
     
    Last edited: Feb 13, 2009
  2. jcsd
  3. Feb 13, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    What's that extra 1/d doing out front of the integral?:wink:

    Why is that?
     
  4. Feb 13, 2009 #3
    First one: typo, sorry :)

    Second one: now that I think about, I'm not sure. I was thinking of using distance from the cylinder's axis and then r would be the smallest so that you're still on the outside (well, surface) and d-r would be anything beyond that. Should those limits be d and 0 instead?..

    EDIT: I got it! Instead of using d as distance to center of the cylinder, I used d as distance to surface and for gaussian surface I had:
    E=Q/(epsilon*2*pi*(d+r))
    And integrated taht with upper limit being 'd' and lower being 0. And it worked! :D Thanks so much.
     
    Last edited: Feb 13, 2009
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook