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Homework Help: Crazy e/m problem

  1. Jul 20, 2004 #1
    Thank you for viewing this problem, hope u can help me.

    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 A , and the separation between the plates is "s" before the key is depressed.

    If the circuitry can detect a change in capacitance of DeltaC, how far must the key be depressed before the circuitry detects its depression? Use epsilon_0 for the permittivity of free space.

    The correct answer involves the variable "s" adn it does not depend on the variable DeltaS

    Thanks again for your time and concern

  2. jcsd
  3. Jul 22, 2004 #2
    Well it really depends on how sensitive the circuitry is, but u'd need to know the minium change (delta C) for the keyboard to detect a key pressed then you can just...

    DeltaC = (Epislion_0 * A)(1/(deltaS))

    (Epislion_0 *A)/ (DeltaC) = deltaS, idk if i helped hopefully..
  4. Jul 22, 2004 #3

    Doc Al

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    parallel plate capacitor

    Model the key as a parallel plate capacitor, the formula for which is: [itex]C = \frac{\epsilon_0 A}{d}[/itex]. Now find the change in capacitance when d changes from s to s - [itex]\Delta x[/itex]. Solve for [itex]\Delta x[/itex].
  5. Jul 22, 2004 #4
    I've tried the following of deltax=s-(epsilon_0*A)/(C) but the correct answer didn't involve the variable C =/
    but y would u be solving for deltax?

    I've also tried (epsilon_0*A)/DeltaC but the correct answer involves the variable "s"
    I've also tried (epsilon_0*A)/(s-DeltaS) but the correct answer doesn't depend on DeltaS

    And i've still haven't been able to solve this problem... any feedback will be appreciated..thanks
  6. Jul 23, 2004 #5
    Doc Al:

    I posted a half solution to this problem a while back right here...it is missing strangely :-(


    EDIT: Not quite...I posted it here: https://www.physicsforums.com/showthread.php?t=36180 (hope that helps)
  7. Jul 23, 2004 #6

    Doc Al

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    I'm not sure what that equation is supposed to be. We need to solve for [itex]\Delta x[/itex] because that's what the problems asks us to find: "how far must the key be depressed". (I just happen to call the distance the key is depressed [itex]\Delta x[/itex].)

    In any case, maverick280857 gave you some good advice on solving this problem (in the other thread! please don't post the same question twice), but here's a bit more. Starting with the equation for capacitance:
    [itex]C = \frac{\epsilon_0 A}{d}[/itex]
    now find [itex]\Delta C[/itex]:
    [itex]\Delta C = \frac{\epsilon_0 A}{s - \Delta x} - \frac{\epsilon_0 A}{s}[/itex]
    Now rearrange and solve this equation for [itex]\Delta x[/itex] in terms of s and [itex]\Delta C[/itex]. (It's easy.)

    (Note: in maverick280857's other post, he calls the distance x instead of [itex]\Delta x[/itex]; it doesn't matter--take your pick!)
  8. Jul 23, 2004 #7
    Retrospectively this is quite an interesting situation since the circuit--which you're not concerned with if you're solving this problem per se--has been made to detect a minimum threshold of DeltaC (or exactly DeltaC, as the case may be). Another twist of the problem would be to consider the capacitor to be filled with a dielectric and to have the student/problem-solver find the minimum distance that the key would have to be pressed so that the dilectric would break down. (Of course, you are less likely to have dielectrics for keyboards in computers still...) But this "twist" as I have simply put in words here, cannot be a reasonable question for starters though it does look like an interesting situation. Ah...one of the beauties of physics...somewhat easy to visualize, difficult to model mathematically :-D

    Coming back to this problem and DocAl's situation, it is clear now that DeltaC is a definite quantity and so after rearranging the terms so as to get an expression for x (or [tex]\Delta x[/tex]) you will discover that it is a function of DeltaC and s only (and more specifically, a function of s since DeltaC is constant preconfigured for your detection circuitry).

    Cheers (and sorry for this rather verbose post...)
  9. Jul 23, 2004 #8
    thank u all for ur help.. the problem is finally resolved it can out to be the following:

    thanks =)
  10. Jul 24, 2004 #9

    Doc Al

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    Exactly right.
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