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Electric potential on long insulating cylinder

  1. Feb 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A long, insulating cylindrical shell of radius .06 m carries a linear charge density of 8.5E-6 C spread uniformly over its outer surface. What would a voltmeter read if it were connected between the surface of the cylinder and .04 m above the surface?

    2. Relevant equations

    potential difference for a distribution of charge v = [tex]\int[/tex] dq/r


    3. The attempt at a solution

    I forget how to do a problem like this. I looked at some sources online, and for this, I first thought I could use gauss's law to find electric field, and then integrate E*dl... but I don't know what the height of my cylinder is in order to calculate it's surface area for Gauss's Law.

    If possible, I would just like a few hints before I try to attempt the problem again. Maybe I'm forgetting about something.
     
  2. jcsd
  3. Feb 15, 2009 #2

    Delphi51

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    "long" means effectively infinitely long so you don't have to worry about the ends of your Gaussian cylinder around the real cylinder.

    Express your charge with an h for height of the cylinder in it, express your Gaussian area also with an h in it and the h's should cancel out when you find the electric field.
     
  4. Feb 15, 2009 #3
    Is [tex]\int[/tex]dl the distance from the surface of the insulating cylinder to the point at which I'm measuring potential? I'm guessing not because I'm totally not getting the right answer.
     
    Last edited: Feb 15, 2009
  5. Feb 15, 2009 #4
    Instead of using V = [tex]\int[/tex] dq/r use V = [tex]\int[/tex] E-ds in which - is dot product. Using Gauss's law, find an expression for the electric field as one goes further from the cylinder. You do not need the height for the cylinder; you'll find that it will cancel out (as Delphi51 said). Use the expression for E in the integral and note that the radial displacement in your expression is along the same path as ds, thus the integral could reduce to V = [tex]\int[/tex] E ds
     
  6. Feb 15, 2009 #5

    Delphi51

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    I don't think you need to integrate because E is the same everywhere on the surface of the cylinder. Integral of E*ds is just ES, where S is the surface area of the cylinder (ignore the ends). Gauss' law says ES = Q/epsilon. Put in expressions for S and Q involving h and r and do some canceling to get a nice expression for E as a function of r. Of course you will have to integrate to get V.
     
  7. Feb 15, 2009 #6
    So...

    S = 2[tex]\pi[/tex]rh
    ES = [tex]\lambda[/tex]*h/[tex]\epsilon[/tex]0

    E(r) = [tex]\lambda[/tex]*h/([tex]\epsilon[/tex]02[tex]\pi[/tex]rh)

    h cancles....

    E(r) = [tex]\lambda[/tex]/([tex]\epsilon[/tex]02[tex]\pi[/tex]r)

    And now...

    v = [tex]\int[/tex] E(r) * dr

    Is that right?
     
  8. Feb 15, 2009 #7

    Delphi51

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    looks good to me!
     
  9. Feb 15, 2009 #8
    Nevermind. I think I just need to clear the memory out the memory in my calculator. Thanks a lot!
     
    Last edited: Feb 15, 2009
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