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Homework Help: Current through parallel plate capacitor vibration sensor

  1. Sep 25, 2008 #1
    1. The problem statement, all variables and given/known data
    A paralel plate capacitor is used as a vibration sensor. The plates have an area of 100cm^2, the dieelectric is air, and the dist between the plates is given by

    d(t)=1+.01 sin(200 t) mm

    A constant voltage of 200 V is applied to the sensor. Determine the current through tthe sensor as a function of time by using the aproximation1/(1+x) = 1-x for x<<1.

    2. Relevant equations

    i(t)= C* dv/dt

    3. The attempt at a solution

    this is about as far as i have gotten because im not sure how to approach this exaclty because Im not sure what the dv/dt is since i know it has a constant voltage across it there is 200 v across the capacitor but that leaves me with dv/dt =0.
  2. jcsd
  3. Sep 25, 2008 #2
    What is the equation for electric potential (voltage) of a parallel plate capacitor? Look at your variables in the equation; something must be changing with time to keep a the potential constant.
  4. Sep 25, 2008 #3
    The only thing that seems to be changin with time is the capacitance. And the equation for voltage aross a parallel plate cap is

    v(t)=(1/c)(INTEGRAL(i(t)) + V0

    and so far that has done me no good when I use that to substitute into the equation for i(t). I have written the question down exactly as it is in the book and Im not sure where to go with this. I'm thinking about it but nothing has seemed to work yet.
  5. Sep 25, 2008 #4
    Sure, capacitance depends inversely on the distance between the plates. If the capacitance varies, the charge q must vary as well.

    So that we are on the same page, these were the equations that came to my mind.

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


    [tex] C = \frac{A\epsilon_{0}}{d} [/tex]
  6. Sep 25, 2008 #5
    yeah that makes sense i realize that something is varying and I had not seen this form of the equation yet
    V = \frac{Qd}{A\epsilon_{0}}
    and i think it might make a little more sense seeing that but Im not sure on how i would get the charge then.
  7. Sep 26, 2008 #6
    SO here is what i ended up with. I scanned my paper and hopefully this works. If anyone has any better ideas that i can implement before class tomorrow i would be more than happy for the help along with the help i have recied already.
    thanks Matt

    Attached Files:

  8. Sep 26, 2008 #7
    Yes, your dV/dt = 0, if V is constant. You said that the capacitance is changing though, so this must mean that the amount of charge Q is changing:

    [tex] C = \frac{Q}{V} [/tex]

    Since Q is changing, there must be a current:

    [tex] i\left(t\right) = \frac{dQ\left(t\right) }{dt} [/tex]

    This is a general algorithm for how I would solve the problem, which is different from your method. First, since the dielectric is air, I would assume that [tex] \kappa = 1 [/tex] (but of course you could leave it in there). Rearrange the following equation for [tex] Q\left(t\right) [/tex]

    V = \frac{Qd}{A\kappa\epsilon_{0}}

    and then substitute the time-dependent distance into the equation . Next apply your small x approximation and then take the time-derivative of [tex] Q\left(t\right) [/tex] to get your current [tex] i\left(t\right) [/tex]. The amplitude of the current is really small (on the order of femto-amps). Does this amplitude make sense to you? I am not an EE major, so I briefly looked at some commercial sites that sold sensors and found ones with this level of sensitivity.
    Last edited: Sep 26, 2008
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