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What is the flux through one face of the cube? Is my answer correct

  1. Apr 30, 2012 #1
    1. The problem statement, all variables and given/known data

    A point charge Q is placed at the center of a cube of side L .
    What is the flux through one face of the cube?
    Express your answer in terms of some or all of the variables Q ,L and appropriate constants.



    3. The attempt at a solution

    I'm getting Q/6e0 , is this correct. Are the units Nm^2/C
     
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  3. Apr 30, 2012 #2

    HallsofIvy

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    Don't you think there should be an "L" somewhere in the formula? And what is "C"?
     
  4. Apr 30, 2012 #3

    Curious3141

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    Yes, this is correct. A simple application of Gauss' law. Your units are also correct, although it may be simpler to express it as Vm (volt*metre).
     
  5. Apr 30, 2012 #4

    cepheid

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    Since the units of electric field are N/C, your units of Nm^2/C for flux seem fine to me.

    C = coulombs

    Gauss' law says the net flux through the surface is Q/e0, and since this should be divided equally among the six faces, your answer seems plausible.

    I imagine that any L-dependence should cancel out, since surface area increases as the square of L, but electric field strength decreases as the square of L

    Edit: beaten by one minute!
     
  6. Apr 30, 2012 #5

    Curious3141

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    No, because they're asking for the electrical flux. The electric field calculation would involve L (and would not be trivial because the cube lacks spherical symmetry). C is coulomb.
     
  7. Apr 30, 2012 #6

    Curious3141

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    Haha. :biggrin:
     
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