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Homework Help: If I don't solve these two problems, I'm gonna lose $60

  1. Mar 3, 2005 #1
    Here is the first one, dealing with parallel plate capacitors. I'm just stuck on these, if anyone could give me a push in the right direction at least that would be nice.

    The Magnitude of the electric field between the plates of a parallel plate capacitor is 480 N/C. A silver dollar is placed between the plates and oriented parallel to the plates. Ignoring the edges, find the induced charge density on each face of the coin.

    All I know is that the charge density = q/Area, and that the only other thing mentioned is the electric field magnitude. I just need some help understanding it.

    You know what, I'll do number two later.

    Thanks to anybody that can help me.
  2. jcsd
  3. Mar 3, 2005 #2
    Think of what Gauss's Law states. Net flux through a closed surface equals charge enclosed divided by epsilon nought.

    [tex] \oint \vec{E} \cdot \vec{dA} = \frac{Q_{encl}}{\epsilon_0} [/tex]

    Pick your gaussian surface to be box with faces parralel to the coin. let one of the faces lie inside the coin (between coins faces) and the other to lie between the coin and one of the plates.

    What is the electric field inside a conductor (the coin)?

    P.S. For my share, I'll only take %30 of the $$ :smile:
    Last edited: Mar 3, 2005
  4. Mar 3, 2005 #3
    The only problem with that is that this is from the section before Gauss' law. I think it's supposed to be a somewhat straightforward problem.
  5. Mar 3, 2005 #4
    Do you know [itex] E = \frac{\sigma}{\epsilon_0} [/itex]?
  6. Mar 3, 2005 #5
    This is a straight forward problem, but it is a direct result of gauss' law. Any argument you give for the answer will in some way use his result, although you may not know it yet. For example: Perhaps your book makes the claim that the electric field within a conductor is zero, but how do you know this. If it hasn't been shown yet, it will be shown as a result from gauss's law.
  7. Mar 3, 2005 #6
    I can't imagine that they would give that formula without first going through Gauss' Law. I mean how do they expect you to understand what it means :confused:
  8. Mar 4, 2005 #7
    I just realized... in the very beginning they usually show you that the electric field inside a conductor is zero because if it weren't, then the charges would fell a force and thus would not be in static equillibrium. So with this in mind, how much charge is required to be on the edges of the coin so the electric field is zero?

    think of the superposition principle of the electric field. It might help to view the coin as a small parrallel plate capacitor
  9. Mar 4, 2005 #8
    solving this problem without gauss law is a mess.... do you expect his teacher/textbook will ask him to do this problem without providing neither my formulas nor gauss law? I don't think so... I believe this formulas must be somewhere in his textbook, he just doesn't know how/when to apply it....
  10. Mar 4, 2005 #9
    BTW, you still need either my formulas or gauss' law to do this... use coulomb law doing this problem is just far too complicated...
  11. Mar 4, 2005 #10
    I totally agree (as per my earlier posts) this is a classic Gauss's Law problem. I was thinking that he could use the equation for the E of a capacitor, but I just noticed in the OP that they did not give the dimensions of the coin, so your formula would be required.
    Last edited: Mar 4, 2005
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