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Gauss's Law - field of infinite plane sheet

  1. Mar 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Use Gauss's Law to calculate the field of an infinite plane sheet of surface charge density σ0.


    2. Relevant equations



    3. The attempt at a solution


    The solution is, where A=area:

    2EA=(σ/ ϵ0)A

    E=σ/2ϵ0


    Why is there a '2' in the solution? I know it's something to do with twice the electric field. Is it because there are two sides to the sheet surface?

    Thank you.
     
  2. jcsd
  3. Mar 28, 2013 #2
    What gaussian surface did you use to get this?
     
  4. Mar 28, 2013 #3
    Possibly a cylinder?

    I'm not 100% sure, as I just have the solution as shown above, and no other 'working out'.
     
  5. Mar 28, 2013 #4
    Sounds a good choice.
    What will the flux be through the cylinder?
     
  6. Mar 28, 2013 #5
    Not through it in one direction.

    It's normal to the surface of the sheet.

    As the sheet has two surfaces, the E-field is in both the two normal directions to the sheet.
     
  7. Mar 28, 2013 #6
    Since the field is normal to the sheet surface, the flux through the sides of the cylinder is zero - so that only leaves the 2 cylinder ends. What's the total flux through these?
    Hint : you've already got the answer :)
     
  8. Mar 28, 2013 #7
    2EA=(σ/ ϵ0)A

    Where A, area, is the area of one cylinder end. i.e. ∏r^2
     
  9. Mar 28, 2013 #8
    Yes, so the 2 comes from the fact that the flux through each cylinder end is EA
     
  10. Mar 28, 2013 #9
    Thanks very much ap123..!
     
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