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Homework Help: Capacitor charge question

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data

    On one capacitative keyboard, each key has a capacitor bottom plate 7[mm]×7[mm] with a 0.4[mm] thick sheet of mica (κ≈7) on top of it; the capacitor top plate is the same size, glued to the key bottom, usually 4[mm] from the bottom plate ; the bottom plate is 5[V] higher potential than the top .
    a) what is the effective capacitance when the key is fully depressed, so there is no air gap?
    b) . . . how much charge is on the bottom plate when the key is depressed?
    c) what's the effective capacitance when the key is not depressed? {careful-mica is still there}
    d) . . . does charge enter or leave the bottom plate while the key moves downward? How much, for the full motion?
    e) what is the Energy change during the key depression? so, what is the average Electric Force on the top plate?

    2. Relevant equations
    C= A/d κ/4pi k

    3. The attempt at a solution

    I don't think i'm understanding how this thing works, because on part a, if the key is pressed so that there is no air gap, that means there is no distance between the 2 plates? So that would mean the capacitance would be zero? But then that makes the next answer zero as well and I don't fell like that is right. So I think i'm having a problem understanding what's going on in this problem.

    Any help is greatly appreciated.
  2. jcsd
  3. Feb 16, 2010 #2


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    There's still the dielectric mica separating the plates (0.4 mm).
  4. Feb 16, 2010 #3


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    When the key isn't depressed, the plates are separated by 4 mm. In that 4 mm gap, there's a mica sheet 0.4 mm thick. The remaining 3.6 mm is filled with air. When the key is depressed, the top plate moves down until only the mica sheet is sandwiched between the two places.
  5. Feb 16, 2010 #4
    Ahhh ok, that's what I was missing. Now I feel dumb haha

    Thanks though. I'll see if I can scramble my way through it now.
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