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Gauss' Law

  1. Mar 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A long, straight wire has a linear charge density of magnitude 3.6nC/m. The wire is to be enclosed by a thin, no-conducting cylinder of ouside radius 1.5cm, coaxil witht he wire. The cylinder is to have positive charge on its outside surface with a surface charge density [tex]\sigma[/tex] such that the net external electric field is zero. Calculate the required [tex]\sigma[/tex].


    2. Relevant equations
    [tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex]


    3. The attempt at a solution

    The electric field can be found, but then how do I go about finding the required [tex]\sigma[/tex] value?
     
    Last edited: Mar 19, 2007
  2. jcsd
  3. Mar 19, 2007 #2
    Sorry, misread.
     
    Last edited by a moderator: Mar 19, 2007
  4. Mar 19, 2007 #3
    Thats what it sayss....
     
  5. Mar 19, 2007 #4
    The wire has obviously negative charge. Hasn't it?
    Apply Gauss's Law, to a gaussian surface, I recommend you to use another cylinder, and put the condition that:

    [tex]\oint E\cdot dS=0[/tex]
     
  6. Mar 19, 2007 #5
    Im still confused
     
  7. Mar 19, 2007 #6
    The total field is zero right? It is due to a superposition of the field from the wire and the field from the cylindrical shell.
     
  8. Mar 19, 2007 #7
    So Do I just find an electric field for the gaussian surface that when summed with the electric field already found wll equal zero?
     
  9. Mar 19, 2007 #8
    not sure what exactly you mean there
     
  10. Mar 19, 2007 #9
    Since the total electrical field is 0, and I can find [tex]E=\frac{\lambda}{2\pi \epsilon_0 r}[/tex], cant I just find [tex]E=\frac{\sigma}{\epsilon_0}[/tex]

    and then [tex]\frac{\lambda}{2\pi \epsilon_0 r} = -\frac{\sigma}{\epsilon_0}[/tex] and then solve for [tex]\sigma[/tex]

    Sorry if this make no sense, I dont really understand this stuff.
     
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