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

  • Thread starter Thread starter imationrouter03
  • Start date Start date
  • Tags Tags
    Difficulty
AI Thread Summary
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.
imationrouter03
Messages
11
Reaction score
0
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
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top