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Parallel plates and force

  1. Jan 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Two parallel plates with area A and charges Q and -Q are being pulled apart by force F, so that the distance between, x, slowly increases. Express F in terms of Q, A and x.


    2. Relevant equations
    [tex]-\Delta PE=W[/tex]

    [tex]W=\int Fdx[/tex]

    [tex]\Delta PE=q_{0}\Delta V[/tex]

    [tex]Q=C\Delta V[/tex]

    [tex]C=\frac{A\epsilon_{0}}{d}[/tex]


    3. The attempt at a solution
    [tex]Q=C\Delta V[/tex]

    [tex]Q=\frac{A\epsilon_{0}}{x}\Delta V[/tex]

    [tex]\Delta V=\frac{Qx}{A\epsilon_{0}}[/tex]

    [tex]q_{0}\Delta V=\Delta PE[/tex] and in this case q0= -Q

    [tex]\Delta PE=\frac{Qx}{A\epsilon_{0}} (-Q)[/tex]

    [tex]W=-\Delta PE=-\frac{-QQx}{A\epsilon_{0}}[/tex]

    [tex]W=\frac{Q^{2}x}{A\epsilon_{0}}[/tex]

    [tex]W=\frac{Q^{2}x}{A\epsilon_{0}}=\int Fdx[/tex]

    Force is then equal to the derivative of my expression with respect to x.

    [tex]F=\frac{Q^{2}}{A\epsilon_{0}}[/tex]


    How can I get an answer independent of x? It.. just doesn't make sense to me. My friend and I both got this answer independently. Where have we gone wrong, or is it correct or.. something? I'm rather confused.

    (Ohhh the pain, did you know that Ctrl+W closes a tab [at least in Firefox]? Yeah, I was going to write W and managed to hit Ctrl instead of Shift.. towards the end of my attempt at a solution.)
     
  2. jcsd
  3. Jan 31, 2009 #2

    Delphi51

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    Homework Helper

    I don't understand the question. The force F "pulling the plates apart" must be an external applied force and there is no way to calculate it from the information given. Perhaps you are supposed to find the force of attraction between the oppositely charged plates. That would be ROUGHLY F = kQQ/x^2. The question does say "express in terms of Q, A and x" which means it is okay to have an x in the answer.

    I see you know about integration. That means you could do the force of attraction between the plates more accurately by considering a tiny charge dq in a bit of the plate area dA and the force it feels do to all the other dQ's on the other plate - integrating over dQ and dq to get the total force. A double integral. Oh, I don't think you can do all that without knowing the shape of the plate. The shape will matter.

    Sorry - I either don't know how to do it or there isn't enough information given.
     
  4. Jan 31, 2009 #3
    Ah, there was a bit more text in the question from the beginning, but there really wasn't anything important (it was on a test and I can't seem to remember the exact wording, however what I wrote was the useful part). He provided us with a little hand made drawing too, of like a parallel plate capacitor with a force F on the right plate ..

    Probably I've understood the question wrong or something.. but how on earth can both my friend and I get the same answer? And another friend of mine (who said she got Q^2x^2/Aepsilon, but then she realised she did the primitive of it instead of the derivative, so yeah..).

    And I bet you my teacher translated a question from Swedish to English. Wouldn't surprise me at all.. and he's not thet best in English. I think I'll just have to give up on this question then. Bummer, it's been nagging at me. (And a double integral doesn't really make sense to me, but then again I've only dealt with integrals from a textbook in Swedish, you never know.)
     
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