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Plate - E - Plate (Reasoning Needed more than Answer)

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Two large metal plates of [tex]1m^2[/tex] are separated by a distance [tex].05m^2[/tex] and carry equal and opposite charges on their inner surfaces. If E between the plates is 55 N/C, what is their charge? Neglect edge effects

    2. Relevant equations

    [tex]\int \vec{E}\cdot d\vec{s}[/tex]
    [tex]\epsilon_0EA=Q[/tex]

    3. The attempt at a solution

    First of all, can I ask if the Electric Field between the plates is equal to [tex]\frac{2\sigma}{\epsilon_0}[/tex]? It might not be relevant to the problem, but I don't even know if my attempt at the calculations from this point on would be accurate if it's not. It looks like the effect of a particle from the positive plate would be [tex]\frac{\sigma}{\epsilon_0}[/tex] and the effect from the negative plate would be [tex]\frac{\sigma}{\epsilon_0}[/tex] as well. Therefore, the Electric field is the sum of the two fields, or [tex]\frac{2\sigma}{\epsilon_0}[/tex] in the direction of the negative plate.

    If I slap the values in the second equation mentioned above, the correct answer of [tex]4.9x10^-10 C[/tex] comes out. However, I don't understand where that equation comes from in terms of this problem.

    [tex]\epsilon_0EA=Q[/tex]
    [tex](8.85x10^-12)(55)(1)=4.9x10^-10 C[/tex]

    I'm completely lost on the reasoning once again or the application of Gauss's Law (the chapter this problem comes from) on the problem.
     
  2. jcsd
  3. Jul 20, 2009 #2

    Doc Al

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    Staff: Mentor

    No. There's nothing wrong with treating the field between the plates as the sum of the fields produced by the charges on each plate. But the field from a single charged plate is [itex]\frac{\sigma}{2\epsilon_0}[/itex], not [itex]\frac{\sigma}{\epsilon_0}[/itex].

    Even better is to use Gauss's law. Draw your Gaussian surface (a box, perhaps) with one end inside one of the conducting plates and the other end between the two plates. Hint: What's the field inside a conductor?
     
  4. Jul 20, 2009 #3
    I don't understand how you get the 2 in the denominator. I've attempted this like 14 times now and can't get that 2 down there. It's a charged plate, not a charged sheet, correct? If I take the Gaussian surface from the positive plate, I get [tex]EA=\frac{Q}{\epsilon_0}[/tex]. And the negative plate would return the same value. I don't understand.

    The field inside the conductor would be zero since the charge on a conductor is dispersed on its outer surface. The field emanating from the charged side would be [tex]E=\frac{Q}{A\epsilon_0}[/tex]. I don't see how this helps me. I have two plates to consider.
     
  5. Jul 20, 2009 #4

    Doc Al

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    When you use Gauss's law to find the field between the plates you automatically include the effect of both plates.

    If you use Gauss's law to find the field from a charged sheet, you'll get [itex]2EA=\frac{Q}{\epsilon_0}[/itex]. (For a charged sheet considered alone, you cannot assume the field is zero on one side.) That's where the 2 in the denominator comes from. In that case, you need to add up the field from both sheets of charge to get the total between the plates.
     
  6. Jul 20, 2009 #5
    But I'm not dealing with charged sheets in this problem. They're charged plates, with I'm guessing length, width, and height. If I take a Gaussian pillbox surface on the positive plate, with one cap between the plates, and one cap inside the positive plate, I get an E field that emanates out of the cap between the plates, but none emanating from the cap inside the conductor plate, correct?
     
    Last edited: Jul 20, 2009
  7. Jul 20, 2009 #6

    Doc Al

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    You can certainly view the problem as finding the field between two charged sheets. In which case you'd just add up the field from each sheet. (That's what I thought you were trying to do.)

    Right. But note that that gives you the total field between the plates, not just the field from one charged sheet.

    (If you had a single charged conducting plate, the charge would distribute itself so that half would be on each side. A different situation than the one you are solving, where the presence of the second charged plate makes the charges move to the inner surfaces.)
     
  8. Jul 20, 2009 #7
    Well then I'm completely lost.

    Why can I ignore the second plate completely when using a Gaussian surface? How does applying a surface to one plate give me the total field from two plates? I thought the equation uses [tex]Q_{enc}[/tex] and therefore I'd have to take a surface enclosing the left plate, and a surface using the right plate to get the total E-field.

    I'm completely in left field.
     
  9. Jul 20, 2009 #8

    Doc Al

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    You are not ignoring the second plate. Its effect is included by the fact that you are assuming all the charge is on the inner surface. (See the last paragraph of my last post.)

    Read this: http://hyperphysics.phy-astr.gsu.edu/HBASE/electric/elesht.html#c2"
     
    Last edited by a moderator: Apr 24, 2017
  10. Jul 20, 2009 #9
    I've been there a dozen times (good site) but it doesn't explain to me in a way I can grasp, obviously. I've been to multiple youtube videos, numerous homework assignments online, my textbook as well as Google Books and still don't understand it.

    I welcome the help you're trying to give, but I need another way of looking at it or going aobut it so it sinks into my stupid head.
     
  11. Jul 20, 2009 #10

    Doc Al

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    You are thinking:
    (1) Gauss's law only includes the charge enclosed within the Gaussian surface. This is true.
    (2) Therefore, when using Gauss's law to figure out the electric field in some situation, it only gives you the field produced by the enclosed charge. This is not necessarily true.

    Especially when dealing with conducting surfaces, statement (2) is often not correct.
     
  12. Jul 20, 2009 #11
    Aren't these quotes conflicting? Using Gauss's law is problematic, it seems, for a problem like this since #2 can't always be true. That effectively says that Gauss's Law can't be used when there's an effect of more than one charge because the answer derived from the Law might not be correct. I'm still confused.

    -----------------

    Anyway, if a positive sheet has a flux of 2EA (one from each side), then the plate has a flux of the same, only in one direction instead of two (the charged side). The negative plate the same. Therefore, between the plates we have:

    Positive plate:
    [tex]2EA\epsilon_0=Q[/tex]
    [tex]E=\frac{Q}{2A\epsilon_0}[/tex]

    Negative plate:
    [tex]2E(-A)\epsilon_0=(-Q)[/tex]
    [tex]E=\frac{Q}{2A\epsilon_0}[/tex]

    Therefore, the total field between the plates is [tex]E=\frac{Q}{A\epsilon_0}[/tex]

    Substituting in the values given:

    [tex]Q=EA\epsilon_0[/tex]
    [tex]Q=(55)(1)(8.85e^{-12})[/tex]
    [tex]Q=4.9e^{-10}[/tex]

    Again that's the answer, but why is the Distance between the plates given in the problem? Is it just to show that the distance between the plates is small compared to the plates themselves, mimicking the case of 'infinitely' large plates?
     
  13. Jul 20, 2009 #12

    Doc Al

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    No, why do you think they conflict?
    It doesn't say that at all. As long as the problem has sufficient symmetry, you can use Gauss's law to find the field.

    Note that by assuming this flux (same field on both sides), you are treating the sheet as if it was the only charge in the world. This will give you the field from a sheet of charge by itself. Then, since you have two such sheets, you'll have to use superposition to add the two fields.

    But you can also take advantage of the information given (that all the charge is on the inner surfaces) and apply Gauss's law with a surface that cuts through the conducting surface of the plate. That way you'll be taking everything into account in one step, without having to use superposition. You get the same answer either way, of course.
    Right.
     
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