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Electric field between parallel plates

  1. Mar 28, 2014 #1
    I'm a little confused between two approaches to this problem:

    First approach (the one that makes sense to me):
    Choose a Gaussian surface over each plate. We have 2EA=σA/Eo so E=σ/2Eo for the positively charged plate and likewise for the negative plate. Within the capacitor, the fields superpose giving E=σ/Eo whilst outside they cancel.

    Second approach:
    Consider only one plate. We now seem to assume no field passes out one side of the plates, so that EA=σA/Eoand E=σ/Eo. We then don't bother superposing fields at all.

    I don't understand the second method - surely we're doing two things wrong:
    Ignoring one side of the plate
    Not superposing the fields
    and these two wrongs happen to make a right.

    Can somebody explain this to me please, thanks :)
     
  2. jcsd
  3. Mar 28, 2014 #2
    I don't understand your statement "We now seem to assume no field passes out one side of the plates" for the 2nd approach. Why are you making this assumption? And what do you hope to accomplish by only considering one plate?
     
  4. Mar 28, 2014 #3
    That's the problem though. The method I'm quoting seems to be doing that.
     
  5. Mar 28, 2014 #4

    Doc Al

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    Staff: Mentor

    The first method does not take advantage of the fact that you know there are two plates. You are taking each plate as an independent sheet of charge and then adding up the results. Perfectly OK.

    The second method looks at both plates together and takes advantage of knowing that no field passes out of one side of the plates. Also perfectly OK. That automatically incorporates the effect of both plates.
     
  6. Mar 28, 2014 #5
    In the second case, why can we assume no field passes out of one side of the plates (without using the first case). Because each plate would have some net charges, so surely there is still flux out of both sides...
     
  7. Mar 28, 2014 #6
    Okay, I see what you were saying. The 2nd approach still assumes both plates are present, but only considers one of the plates. In this case, you are making a simplification based on the symmetry of the problem. Because the plates have opposite charges but the same charge density, you know from symmetry that the fields outside of the plates will cancel each other out.
     
  8. Mar 28, 2014 #7

    Doc Al

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    Staff: Mentor

    Only if you treat each plate independently, as if the other weren't there.

    You can also put one side of your Gaussian surface within the conducting material of a plate.
     
  9. Mar 28, 2014 #8

    Doc Al

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    Staff: Mentor

    The 2nd approach tacitly considers the effect of both plates.

    Right.
     
  10. Mar 28, 2014 #9
    I just find the first method much more logical. At least I'm aware of both anyway.
     
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