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Surface charge density

  1. Feb 16, 2008 #1
    A 10.0g piece of styrofoam carries a net charge of -.700 x 10^-6 C and floats above the center of a large horizontal sheet of plastic that has a uniform charge density on its surface. What is the charge per unit area on the plastic sheet?

    I've been trying to think a way to model this with a Gaussian surface (without using integrals)...but doesn't seem to work.
     
  2. jcsd
  3. Feb 16, 2008 #2

    Doc Al

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    Hint: What electric field must the sheet of charge produce to support the styrofoam?

    You can certainly use Gauss's law to find the field from a uniform sheet of charge, if you don't happen to know it.
     
  4. Feb 16, 2008 #3
    The electric force should balance out the force of gravity. Dividing that by the charge of the styrofoam ball should give the electric field. But I keep bumping into the distance squared quantity. How would I eliminate it?
     
  5. Feb 16, 2008 #4

    Doc Al

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    Good.
    Where does the distance squared come from? (That appears in the force between two point charges, but that's not relevant here.)

    What's the electric field from an infinite sheet of charge with some given surface charge density?
     
  6. Feb 16, 2008 #5
    The electric field should be E = o/2e, in which o is the surface charge density and e is the permittivity of free space. Then that would also imply that o = E*2e. The Electric field is equal to the electric force divided by the charge of the styrofoam ball....oh. I see. I got it now. Thanks for the help.
     
  7. Oct 25, 2008 #6
    Doc Al - PLEASE help me.

    So I understand that the electric F equals the gravitational force. Then I divide that by the charge given in the problem to get the electric field. Now, to get the charge per unit area we use the equation E=sigma/2e0. We know the electric field so we rearrange the equation to solve for sigma because that is the charge/unit area factor we are looking for.

    So now I'm stuck with this equation:

    sigma = E * 2e0

    I know the electric field and 2e0 but when I multiply these 2 together I get a wrong answer and a wrong SIGN as well. What am I doing wrong?
     
  8. Oct 25, 2008 #7

    Doc Al

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    Everything you've said makes sense. To see what you've done wrong, show step by step exactly what you did.

    FYI: Don't use equations to figure out the sign--use the rule that opposite charges repel.
     
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