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Application of Gauss's Law to Charge Distribution

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A 10.0 g piece of Styrofoam carries a net charge of -0.700 [itex]\mu[/itex]C and is suspended in equilibrium 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?


    2. Relevant equations

    Fe = [itex]\frac{KqQ}{r^2}[/itex] (q is the charge of the styrofoam- given, and Q is the total charge of the plastic sheet. K is Coulumbs Constant = 8.99x10^9)
    Fg = mg
    σ = [itex]\frac{Q}{A}[/itex] (Q is the total charge of the sheet, A is its area)
    [itex]\Phi[/itex] = [itex]\oint[/itex] E[itex]\cdot[/itex]dA = qin0 (E is the electric field. dA is an infinitesimal area VECTOR hence the dot product. qin is the total charge inside a Gausian surface. ε0 is the permittivity of free space which is equal to 1/(K4[itex]\pi[/itex]) which means 1/ε0 = K4[itex]\pi[/itex]


    3. The attempt at a solution
    Obviously since the Styrofoam is floating at equilibrium Fg = Fe and I set those equal to each other, but since I don't know "r" - the height that the Styrofoam is suspended at, or "A" - the area of the plastic sheet, or "Q" (the total charge of the sheet) I don't know how to solve for σ. I couldn't come up with a Gaussian surface that would make E and A vectors parallel to simplify the surface integral, so I don't know if I can do anything with that equation. Thank you for taking the time to read and hopefully help.
     
  2. jcsd
  3. Sep 2, 2013 #2

    Doc Al

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    Staff: Mentor

    That gives the force between two point charges. Not relevant here.

    What's the field from a charged sheet? What field is required to put that Styrofoam piece in equilibrium? (Treat it as a particle.)
     
  4. Sep 2, 2013 #3
    Oh, I didn't know that only applied to particles. The formula for a force from a uniformly distributed charge on a charge q would be: F= q[itex]\oint[/itex] E[itex]\cdot[/itex]dA. Unfortunately I don't know how to solve that integral because I can't think of a Gaussian surface that would make E[itex]\cdot[/itex]da = E*dA (because the vectors are parallel), or a surface on which E is always constant (the same distance away at all points) therefore able to take out of the integral.
     
  5. Sep 2, 2013 #4

    Doc Al

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    Staff: Mentor

    The force would be F = Eq. The first job is to find the field from a uniform sheet of charge.

    Treat the charged sheet as being infinitely big. How would the field lines look? Choose a Gaussian surface that has its surfaces parallel and perpendicular to the charged surface. (Many shapes will do.)

    This is a standard exercise, so your textbook may have a discussion worth reviewing.
     
  6. Sep 2, 2013 #5
    THANK YOU. It just clicked. Much gratitude brother!
     
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