# Homework Help: Difficulty 5 e/m problem

1. Jul 21, 2004

### imationrouter03

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

2. Jul 23, 2004

### maverick280857

Hi

The parallel plate capacitor relationship is

$$C = \frac{\epsilon_{0}A}{s}$$

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

$$C_{new} = \frac{\epsilon_{0}A}{s-x}$$

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

$$dC = \frac{\partial C}{\partial x}dx$$

I wonder if we could use this approach somehow...

3. Jul 23, 2004

### Staff: Mentor

4. Jul 23, 2004

### imationrouter03

srry for posting this twice, it was my mistake for posting it on the general physics forum.. since it involved hw help.. but w.e it doesn't matter now.. i appreciate ur help.. it won't happen again.