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Explaining Gauss Law

  1. Feb 10, 2008 #1
    i hav read gauss law chapter from my textbook but still i m facing problems based on it.
    plz explain me GAuss law in an easy language.
  2. jcsd
  3. Feb 10, 2008 #2
    The whole idea behind gauss's law is to quantify the amount of the electric field flowing across a boundary. Picture the a charge as a water fountain the sprays water up, down, left, right, upright, upleft, etc. Now you put bubble around that water fountain, and you want to know how much water is leaving that bubble, will it matter whether that bubble is a sphere, a cube, a pyramid, some incredibly odd surface? No, because the amount of water leaving will always be the same. I could put the bubble a couple inches away, or put it a few feet away (barring gravity), and the same amount of water will always be leaving it. Physics calls this concept flux, how much stuff flows out of a surface area. Does that help?
  4. Feb 10, 2008 #3
    thank you,
    yes i am understanding flux but can u explain gauss law with cavity example ,i wanted some more important things on gauss law which is needed when solving questions.
  5. Feb 11, 2008 #4
    What kind of cavity do you mean? Basically any time you have symmetry in a problem, and you can ignore fringing effects then you should use Gauss' law.
  6. Feb 11, 2008 #5
    Lets say that you had a point charge q within an enclosed surface, such as a sphere and that the point charge was in the center of the sphere. Because the radius is uniform, the electric flux could be calculated as q/e, in which e = permittivity of free space. Also remember that electric flux is the distribution of an electric field over an area. If the same point charge was taken and put in another enclosed object of irregular shape and not necessarily in the center of this enclosed object, then regardless of the shape, the same number of electric field lines are going to pass through th object as for the sphere, therefore the net electric flux through any enclosed surface surrounding point charge q is given by q/e.
    If the enclosed object was placed outside of the charge, then the number of lines entering the object would equal the number leaving the object, therefore the net electric flux for such a condition would be 0.
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