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Gaussian surface line charge

  1. Sep 24, 2011 #1
    1. The problem statement, all variables and given/known data
    A uniformly charged, straight filament 7.00 m in length has a total positive charge of 2.00 [itex]\mu[/itex]C. An uncharged cardboard cylinder 2.00 cm in length and 10.0 cm in radius surrounds the filament at its center, with the filament as the axis of the cylinder. Using reasonable approximations, find (a) the electric field at the surface of the cylinder and (b) the total electric flux through the cylinder.


    2. Relevant equations

    Gauss's Law
    Area of a cylinder
    [itex]\Phi[/itex]=EA

    3. The attempt at a solution

    a) Flux is zero at the ends. The area of the gaussian surface is 2[itex]\pi[/itex]rL
    EA = [itex]\frac{q_en}{\epsilon_0}[/itex]

    The charge enclosed is the total charge of the 7.00 m rod which is given.

    E = [itex]\frac{Q}{\epsilon_0 2\pi rL}[/itex]

    The length of the cylinder does not cover the whole rod. Since the electric field is uniform, the electric field at the surface of a larger cylinder is the same and the charge enclosed is proportional to the length of the cylinder.

    E = [itex]\frac{Q}{2 \pi rL \epsilon_0}[/itex]

    plugging in E = 51.382k N/C

    B)
    Flux is
    [itex]\Phi[/itex]= EA
    =E(2[itex]\pi[/itex]rl
    = 645 [itex]\frac{Nm^2}{C}[/itex]
     
  2. jcsd
  3. Sep 24, 2011 #2
    Did you take [itex]Q[/itex] to be the whole charge of [itex]2.00 \mu C[/itex]? You have to use the amount of charge inside your Gaussian surface.
     
  4. Sep 24, 2011 #3
    Yes, but also I used L as the length of the rod. Is it OK? Thank you.
     
    Last edited: Sep 24, 2011
  5. Sep 24, 2011 #4
    Yes, it is OK if you used L for the length of your Gaussian surface.

    I think I should mention this here even thought it doesn't change the numerical answer. It seems that they want you to have your Gaussian surface 2.00 cm long and not 7.00 m long. That is because 7.00 m is much larger than 2.00 cm, so the filament can be thought of as one of infinite length. It says in the beginning "Using reasonable approximations...". The approximation you make here is that the electric field vector is radially outward from the filament. That is only true for a line charge of infinite length. Near the ends of the filament the electric field vector wouldn't point radially outward.
     
  6. Sep 25, 2011 #5
    Thank you for letting me know.
     
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