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Why doesn't gauss law work ?

  1. Sep 13, 2015 #1
    1. The problem statement, all variables and given/known data

    A line charge exists in air along the z-axis between z=0 and z=5 cm. It has a uniform charge density given by:
    ρl = 4(uC/m)
    Determine E at (0,10 cm, 0)

    2. Relevant equations


    3. The attempt at a solution

    I am using gauss law so I draw a cylinder around the wire

    Qenc = ρl * L = ∫ D.ds , Gauss law

    Qenc = ρl * L , where L is the length of the wire

    ds = r*dθ*dz , where r is the distance from the wire to the point, 10 cm

    ρl * L = D ∫ 0 to L ∫ 0 to 2pi r*dθ*dz

    D = ρl/2pi*r = E*ε

    E = ρl / 2pi*ε*r

    = (4*10^-6)/(2pi*0.1*8.854*10^-12)
    = 7.19*10^5
     
  2. jcsd
  3. Sep 13, 2015 #2

    blue_leaf77

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    Gauss method works only if the electric field at any point on the enclosing surface is uniform, which is not in this problem because the wire is of finite length. Use the integration method to integrate the electric field from each line segment along the wire.
     
  4. Sep 13, 2015 #3
    I'm a bit confused. Why is the electric field uniform for an infinite wire and non uniform for a finite length wire ?
     
  5. Sep 13, 2015 #4

    blue_leaf77

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    For the case of finite wire, imagine two points, one located near the left end of the wire, the other located at the center, both point lie on your cylindrical surface The electric field in the first point will tend to be directed to the left or right depending on the charge of the wire. For the point in the middle, it will feel the same amount of electric field from the left and right parts of the wire and hence will be directed perpendicularly to the wire.
    For the case of infinite wire, no matter where you translate your observation point you will always end up in the same configuration as the wire has no end on both sides.
     
  6. Sep 13, 2015 #5

    rude man

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    Another view: your Gaussian cylinder has flux coming out the sides in addition to the length. Gauss's theorem still holds but you can't ignore the side flux.
     
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