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Gauss's Law for Line of Charge

  1. Sep 29, 2012 #1
    Just refreshing on Gauss's Law.

    For a line of charge, we choose a cylindrical Gaussian surface. But if the E field is radially outwards, why is there no flux through the ends of the cylinder and only through the sides?

    I know that the field is only assumed to be perpendicular to the sides of the cylinder, but isn't there field through the ends as well?

    Or is the flux through the ends zero because the line of charge is infinitely long and there are field lines in equal and opposite direction at the ends of the cylinder so they cancel out, hence there is no flux?
     
  2. jcsd
  3. Sep 29, 2012 #2

    Dale

    Staff: Mentor

    If the line is straight and infinite then, by symmetry, the E field must be purely radial. Therefore the flux on the ends is zero since the field has no component normal to the surface.
     
  4. Sep 29, 2012 #3
    Yes, I understand that, but does radially outward not mean outward in all directions?

    That includes going through the ends of the cylinder
     
  5. Sep 29, 2012 #4

    Matterwave

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    Science Advisor
    Gold Member

    Just think about it from a symmetry point of view. Could there be a net field going one way or the other?
     
  6. Sep 30, 2012 #5

    Dale

    Staff: Mentor

    An infinite line is axisymmetric, or in other words it can best be represented using cylindrical coordinates. Radially outward here means in the direction of the r coordinate, which is always perpendicular to the line itself.

    Consider this. Either the field can be purely in the r direction, or it could tilt a little in the z direction. If it were to tilt in the z direction, how could it choose whether to tilt in the +z or the -z direction? Everything is symmetric, the charge goes out to infinity in both directions. Since there is no way for the field to choose +z or -z then it must be purely r.
     
  7. Sep 30, 2012 #6
    Yes and we can also consider the two infinitesimal point charges where a given point in the field lies on their perpendicular bisector. The field strengthes in the z direction clearly cancel each other out. And this holds true for any points in the field if the line charge is infinitely long so we can say E is radiating on the r direction only.
     
  8. Sep 30, 2012 #7

    jtbell

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    Staff: Mentor

    It depends on the context. In a situation with spherical symmetry, e.g. a point charge, it means "outwards in all directions from the center." In a situation with cylindrical symmetry, e.g. a line charge, it means "perpendicularly outwards from the axis of symmetry."
     
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