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Change in capacitence

  1. Mar 8, 2007 #1
    1. The problem statement, all variables and given/known data
    In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, air-filled capacitor. When the key is depressed, the plate separation decreases and the capacitance increases. Electronic circuitry detects the change in capacitance and thus detects that the key has been pressed. In one particular keyboard, the area of each metal plate is 50.5 mm^2, and the separation between the plates is 0.660 mm before the key is depressed.

    2. Relevant equations

    [tex] \Delta C = \frac {e_{0}A} {s_{1}} - \frac {e_{0}A} {s_{0}} [/tex]


    3. The attempt at a solution

    [tex] \Delta s = s_{0} - s_{1} => s_{0} *(1 - \frac {1} {1 + \frac {s_{0} \nabla C} {e_{0}A}}) [/tex]

    I'm not too sure if this is correct, and I'm very bad at the unit conversion and I think that also might be an issue

    is [tex] C = 0.300 * 10^{-12} [/tex]
     
    Last edited: Mar 9, 2007
  2. jcsd
  3. Mar 8, 2007 #2

    berkeman

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    Staff: Mentor

    Sorry, you haven't stated the full problem. What are you supposed to do? Are you given a final (smaller) separation distance, and asked to find the initial and final capacitances? Or given a target change in capacitance, and asked to solve for the final separation distance?

    Also, the "Delta" symbol would usually be used for "change". You've used the "del" or "nabla" symbol above in your post, and del has a different meaning than Delta. Small nitpick, but it confused me when I first saw your post.

    Delta = [tex]\Delta[/tex]

    Finally, to help you with units and unit conversions, just multiply them out like they were variables. Like, velocity is distance per time, or:

    [tex]v [m/s] = \frac{\Delta x [m]}{\Delta t }[/tex]

    And if you need to convert microseconds to seconds or something, just multiply through by one (like 1s = 10^6us), like this:

    [tex]\Delta t = \Delta t [\mu s] \frac{1 }{10^6 [\mu s]}[/tex]
     
  4. Mar 8, 2007 #3
    Sorry I left out the actual question:

    If the circuitry can detect a change in capacitance of 0.300 pF, how far must the key be depressed before the circuitry detects its depression?
     
  5. Mar 9, 2007 #4

    berkeman

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    Staff: Mentor

    So what do you get for the initial capacitance?
     
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