Gauss law application

Discussion in 'Physics' started by logearav, Nov 9, 2011.

  1. logearav

    Thread Starter Member

    Aug 19, 2011
    Revered Members,
    My attachment depicts an application of Gauss law namely Linear charge distribution. Due to + charge, the Electric field is directed outwards. But why can't the Electric field act in Upward and downward direction(i have indicated with black arrow), which is also outward direction to + charge.
  2. studiot

    AAC Fanatic!

    Nov 9, 2007
    Good question.

    The electric field does indeed also have a vertical component at the ends.

    If the book you took your example from explains things properly it will justify ignoring this end effect.

    It is not possible to obtain an analytical solution via Gauss Theorem if the end effects are included so it is usual to say that the line of charge is long in comparison to the diameter so the end effects are small in comparison so for the overall.
    Alternatively we can consider unit length of a long line.

    Thus for the greater part of the line the assumption that

    "From symmetry the electric field is everywhere directed outward and so appears as a series of disks perpendicular to the line or cylinder of charge."

    is made. This leads to the integration and solution you are referring to.
  3. logearav

    Thread Starter Member

    Aug 19, 2011
    Thanks studiot. Are you saying, since the line charge is infinitely long, the vertical component of electric field is neglected?
  4. studiot

    AAC Fanatic!

    Nov 9, 2007
    Yes that's basically what I said.

    It is neglected because it is small compared with the rest of the field.

    If it is not small then you can't neglect it, but then you don't have a long line source either.
    logearav likes this.
  5. araidland

    New Member

    Nov 12, 2011
    Your answer has units of "field" not force, which makes more sense given the information.
    To get the net field at R/2 you can find the field due to each sphere, as if the other was not there, and add them as vectors.
    The field inside the one sphere can be shown from Gauss' Law to be the field of a point charge, provided the charge used is only that amount beneath the field point.
    E1 = kq/(R/2)^2 = 4kq/R^2 (k=1/4pieo)

    You can get "q" in terms of total "Q" by ratio & proportion;
    q/(4/3)pi(R/2)^3 = Q/(4/3)piR^3
    q = Q/8

    E1 = kQ/2R^2

    The field due to the other sphere is just due like the field of a point charge "Q" a distance (3/2)R away;
    E2 = kQ/(3R/2)^2 = 4kQ/9R^2

    At "R/2" on the axis these fields are in opposite directions so you subtract to get a net field.
    Enet = E1 - E2 = kQ/18R^2 = Q/72pieoR^2