Gauss's Law - field of infinite plane sheet

  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. What gaussian surface did you use to get this?
     
  4. Possibly a cylinder?

    I'm not 100% sure, as I just have the solution as shown above, and no other 'working out'.
     
  5. Sounds a good choice.
    What will the flux be through the cylinder?
     
  6. 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. 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. 2EA=(σ/ ϵ0)A

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