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Homework Help: Gauss's Law to Symmetric Charge Distribution

  1. Apr 5, 2008 #1
    A 10.0 gram piece of styrofoam carries a net charge of -0.700[tex]\mu[/tex]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?

    2. Relevant equations
    [tex]\Phi[/tex] = E[tex]\int[/tex]dA = [tex]\frac{q}{\epsilon}[/tex]

    3. The attempt at a solution

    I wasn't even sure how to start this problem. The fact that mass is given threw me off and I can't see how it would be used in the problem. I thought that possibly it could be used to find the electric force between the two objects since it must overcome the downward gravitational pull on the styrofoam (right?). From there, I used the equation that states F=qE to get that the magnitude of the electric field of the styrofoam is -1.4E5. Still, I have no idea how to find the charge per unit area of the plastic sheet and don't think my approach is anywhere near correct. Can someone please help me?
  2. jcsd
  3. Apr 5, 2008 #2
    I think you're on the right track. Think about what kind of electric field a flat, charged surface creates.
  4. Apr 5, 2008 #3
    Since the styrofoam floats above the sheet of charge, its technically at rest there (with 0 velocity). In that case you could set the electric force its feeling equal to the gravitational force. Solve for the electric force and use that to solve for the electric field at that point. Use the electric field to find the surface charge density by applying Gauss's Law.
  5. Apr 5, 2008 #4
    Thank you for all your help. I believe I have solved the problem. It wasn't as difficult as I had made it seem. The most tedious step was probably noticing that the electric force on the styrofoam is equal to the gravitational force on it. In addition to this, the electric force on the plastic sheet is also equal to the electric force on the styrofoam due to Newton's Third Law. After that, the surface charge density is simple to calculate. Thanks again for the help.
  6. Oct 24, 2008 #5
    This is how far I got...


    sigma = E * 2e0

    because sigma is charge/area. What am I doing wrong?
  7. Oct 25, 2008 #6
    I did this problem so long ago, I'm not sure if I remember everything correctly.

    However, I did notice you say E=mg. Actually, since F=mg and F=qE, your equation would actually be qE=mg. You have the charge of the object, therefore, you can solve for the electric field. I'm not sure how you got 0.735.

    0.01 kg * 9.8 m/s2 = E
    -0.0000007 C

    You are obviously right that E= sigma/2(epsilon0).

    So rearranging the equation, sigma is simply = 2E(epsilon0).

    I'm not sure if that helps, but you have the right idea.
  8. Oct 25, 2008 #7
    Thank you very much, I really appreciate it.
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