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Gauss's Law and E fields

  1. May 10, 2008 #1
    So I'm trying to understand a couple things regarding this.

    lets say we have a plate thats charged to Q. and we want to find the E field of a point .1 meters above the plate. when i looked at the solution, it never took the .1 meters into account when calculating the strength of the E field.

    my roommate says that E field should be the same and thus the distance was irrelevant. so whether it was .1 meters or 1 meter above the plate, the E field would be the same.

    but that doesnt make sense to me when you think about the equation for an electric field E= kq/r.

    can anyone resolve that contradiction for me? :(
  2. jcsd
  3. May 10, 2008 #2

    Doc Al

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    Staff: Mentor

    He's right, at least if you're talking about the field from an infinite sheet of charge. Or a finite sheet of charge at distances much smaller than the dimensions of the sheet. Of course, if the sheet of charge is only 10 cm square, the field 10 m away certainly will depend on distance.
    That's the field from a point charge--a very different configuration from an infinite sheet of charge. (Of course, as you get far enough away from a finite sheet of charge the field begins to look more and more like the field from a point charge.)

    You might want to browse through this site: Applications of Gauss' Law
  4. May 11, 2008 #3
    Yep, because the e-field lines will be 100% perpendicular to the charged (supposedly infinitely wide) plate.
    i.e. going straight up or straight down, so if you bound a gaussian cylinder above and through the plate, the field will go through only the circular discs at the end of the cylinder (and not the 'side walls' of the cylinder) - so it doesn't matter how high the cylinder is (drawing a picture would help heaps)
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