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Gauss's Law and electric flux

  1. Jan 25, 2010 #1
    Four closed surfaces, S1 through S4, together with the charges -2Q, Q, and -Q are sketched in the figure below. (The colored lines are the intersections of the surfaces with the page.) Find the electric flux through each surface. (Use Q for the charge Q and epsilon_0 for 0.)
    (Picture Attached)

    2.
    Electric Flux= the integral of EdA=q/epsilon_0


    3.
    All that I could think to do was set q=the charge of the inside of the desired surface. I was not sure how I could quantitatively define how the different charges within the surfaces affected each other.
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2010 #2
    Gauss` law is your best friend. the flux through a surface will only be non zero if there is a charge inside that surface. think about putting an imaginary sphere in the middle of a river. All the water that runs into the sphere also runs out of it. Having a charge inside the surface is like putting a sprinkler inside your imaginary sphere. Now the amount of water in is 0, and the amount of water out is this analogy`s version of flux.

    So any surface with no charge inside, has 0 flux.

    This principle is displayed simply in your equations for flux. where flux is = to
    charge enclosed / epsilon naught.

    So your flux for the red surface with Q and -2Q, it will have a flux of -Q/epsilon naught
     
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