How Far Must a Key Be Depressed to Detect a Change in Capacitance in a Keyboard?

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In a specific computer keyboard design, each key functions as part of a parallel-plate capacitor, where pressing the key reduces the distance between metal plates, thereby increasing capacitance. The challenge is to determine how far a key must be depressed to achieve a detectable change in capacitance, denoted as DeltaC. The relationship for capacitance is given by C = (epsilon_0*A)/s, and when the key is pressed, the new capacitance can be expressed as C_new = (epsilon_0*A)/(s-x). The discussion emphasizes the need to clarify the relationship between DeltaC and the changes in capacitance due to key depression. Overall, the conversation revolves around understanding the mathematical implications of capacitance changes in relation to key movement.
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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.

I've tried (epsilon_0*A)/DeltaC but the correct answer involves variable "s"
I've also tried (epsilon_0*A)/(s-Deltas) but the correct answer does not depend on the variable: Deltas
I've ran out of idea.. please ill take any feedback =/

Thanks again for your time and concern
 
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Hi

The parallel plate capacitor relationship is

<br /> C = \frac{\epsilon_{0}A}{s}<br />

where A is the crosssectional area and s is the distance between the plates (here of course, we neglect fringing).

When you press a key you decrease s and therefore increase the capacitance, as you correctly mentioned. Supposing the key is depressed a certain distance 'x' so that the new capacitance C_{new} is given by

<br /> C_{new} = \frac{\epsilon_{0}A}{s-x}<br />

The DeltaC is the difference between C_{new} and C, as I understand from your problem. Please clarify all this first...

Also could you consider C as a general function so that

<br /> dC = \frac{\partial C}{\partial x}dx<br />

I wonder if we could use this approach somehow...
 
srry for posting this twice, it was my mistake for posting it on the general physics forum.. since it involved homework help.. but w.e it doesn't matter now.. i appreciate ur help.. it won't happen again.
 
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