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Charged plate

  1. Jun 21, 2006 #1
    obvious question IMO, but just need to make sure before test.

    if you have two parallel conducting plates, and apply charge to one of them, then the field between them is half, compared to if the plates were acting as an air capacitor ie. both plates oppositely charged?

    thing is, when I use a simplified version of gauss, the field on an infinite sheet of charge is

    charge density/permittivity

    which is the exact same as if I use

    Q/V=[permittivity*plate area]/plate seperation.

    i tried to use latex, but i've never really used it, and it came out all wrong due bad syntax. sorry.
    Last edited: Jun 21, 2006
  2. jcsd
  3. Jun 21, 2006 #2


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    So what is your question?
  4. Jun 21, 2006 #3
    whether the 2nd paragraph is correct....thought the question mark might've given it away. :P
  5. Jun 21, 2006 #4


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    Sorry. You said it like it was a true statement, which it is, and I didn't see the question mark. The second plate has no effect in this case, and the field is just that of a single charged plane.
  6. Jun 21, 2006 #5
    ok, from common sense POV thats fine.

    the further question i suppose, is why the two (bold) word equations simplify to the same thing, when one is for a single charged plane, and the other is for 2 opposing ones. (presuming I've made no mistake)
  7. Jun 21, 2006 #6


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    You're missing a factor of 1/2 in the first equation.
  8. Jun 21, 2006 #7
    theeen my gauss is wrong....may have been the way i've learned it - via the "number of lines" instead of a formal mathematical integral.

    when I learn my latex (after tomorrows exam then) I'll clarify what i meant.

    ed: actually i think i figured out what I did wrong now, post tomorrow :D
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