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

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Homework Help Overview

The discussion revolves around a problem related to the detection of key presses in a computer keyboard, specifically focusing on the change in capacitance of a parallel-plate capacitor formed by the key's metal plate. The original poster seeks to determine how far a key must be depressed to achieve a detectable change in capacitance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the change in capacitance to the distance the key is depressed, exploring various equations but encountering confusion regarding the variables involved. Some participants clarify the relationship between capacitance and plate separation, suggesting the use of derivatives to express changes in capacitance.

Discussion Status

The discussion is ongoing, with participants providing clarifications on the capacitance formula and raising questions about the assumptions made in the problem. There is a focus on understanding the relationship between the variables involved, but no consensus or resolution has been reached yet.

Contextual Notes

Participants note the importance of accurately defining the variables and the conditions under which the capacitance changes, as well as the implications of neglecting certain factors like fringing effects.

imationrouter03
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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

[tex] C = \frac{\epsilon_{0}A}{s}[/tex]

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 [tex]C_{new}[/tex] is given by

[tex] C_{new} = \frac{\epsilon_{0}A}{s-x}[/tex]

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

Also could you consider C as a general function so that

[tex] dC = \frac{\partial C}{\partial x}dx[/tex]

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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