# Keyboard problem (1 Viewer)

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#### FlipStyle1308

Here is the question:

Many computer keyboards operate on the principle of capacitance. Each key forms a small parallel-plate capacitor whose separation is reduced when the key id depressed. Suppose the plates for each key have an area of 46.9 mm^2 and an initial separation of 0.550 mm. In addition, let the dielectric have a dielectric constant of 3.75. If the circuitry of the computer can detect a change in capacitance of 0.379 pF, what is the minimum distance a key must be depressed to be detected?
What I did in my attempt solve this is I substituted a lot of variables into one equation, but I still wasn't able to solve the problem. My equation was Farad = C/V = (EoA)^2k/Qd^2. I feel like I just went in a complete circle. Anyone know where I went wrong, or a different approach to solve this? Thanks in advance!

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#### FlipStyle1308

Astronuc said:
One should be able to express capacitance as a function of the known variables and the distance.

See -http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html[/QUOTE] [Broken]

Yes, I tried to solve using my known variables. By plugging in A, K, and C into C = KEoA/d, and solving for d, I got 4106.847 mm. Do I just subtract the given initial distance of 0.550 mm from this to get a final answer of 4106.2937 mm?

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#### Astronuc

Staff Emeritus
4106.847 mm = 4.1068 m, assuming mm = millimeter. This can't be right.

The answer should be fractions of mm, or on the order of microns.

One needs to check units, and it would help if one shows the forumula and substitutions.

#### FlipStyle1308

Hmm, okay, but is my number correct, except for the placement of the decimal? I keep getting the same answer :(.

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#### chroot

Staff Emeritus
Gold Member
It is "the correct answer," with the decimal point in the wrong place. Check your units. Remember that the value of epsilon-naught usually given in your textbook is in units of F/m, NOT F/mm.

- Warren

#### FlipStyle1308

Here are the values I used:

A = 0.0469 m^2
k = 3.75
C = 0.379 x 10^-12 F
Eo = 8.85 x 10^-12 C^2/Nm^2
do = 0.00055 m
d = ?

Is the decimal placement in any of these incorrect?

#### chroot

Staff Emeritus
Gold Member
You did not convert the area, 46.9 mm^2, into m^2 correctly.

- Warren

#### FlipStyle1308

Oh, should it be 46.9 mm^2 = 46.9 x 10^-9 m^2?

Staff Emeritus
Gold Member
No..

- Warren

#### FlipStyle1308

10^-6? Is that right?

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#### chroot

Staff Emeritus
Gold Member
46.9 mm^2 = 4.69 × 10-5 m^2.

(Divide by 10^3 * 10^3.)

- Warren

#### FlipStyle1308

Thanks, so I got my answer to be 4.1068 x 10^-3m. Then do I subtract 0.550mm from this to get my final answer of 3.5568mm?

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#### chroot

Staff Emeritus
Gold Member
No, it's not that simple. You're on the right track though.

First, calculate what the capacitance is at the initial spacing of 0.55 mm. Then, add 0.379 pF to it. Finally, find out the spacing that achieves that new, larger capacitance. The difference between the initial and final spacings is the distance the key must be pressed.

- Warren

#### FlipStyle1308

So the capacitance at the initial spacing is 2.83 x 10^-12 F, and plus the 0.379pF is 3.209 x 10^-12 F. This initial capacitance is 0.485 mm. 4.1068 mm - 0.485 mm = 3.6218 mm, which is my final answer?

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#### chroot

Staff Emeritus
Gold Member
Uh, what? You found the initial capacitance and the final capacitance okay, but I don't know what you mean by "This initial capacitance is 0.485 mm" or where you got 0.485 mm. Please try to show your work in more detail.

- Warren

#### FlipStyle1308

Sorry, I meant that is the distance. But is my final answer of 3.6216 mm correct?

#### chroot

Staff Emeritus
Gold Member
No, I don't believe it is. After all, the initial spacing is only 0.55 mm, and you're talking about decreasing that distance by almost 4 mm? That doesn't even make any sense.

- Warren

#### FlipStyle1308

Okay, so you helped me solve that d = 4.1068 mm (this is using the given capacitance of 0.379 pF. Then I calculated the capacitance at the initial spacing of 0.55 mm by using the equation C - kEoA/d, using k = 3.75, Eo = 8.85 x 10^-12 C^2/Nm^2, A = 0.0469 m^2, and d= 0.00055m, getting an answer of 2.83 x 10^-12 F. I then added 0.379 x 10^-12 F to this to get a total of 3.209 x 10^-12 F. Then I plugged 3.209 x 10^-12 F into the same equation I used in the first calculation, again solving for d, obtaining an answer of 0.485 mm. Lastly, I subtracted 0.485 mm from 4.1068 mm, obtaining my answer of 3.6218 mm. Please let me know if this is a simple miscalculation or if I am using a wrong equation, or whatever. Thanks.

#### chroot

Staff Emeritus
Gold Member
FlipStyle1308 said:
Okay, so you helped me solve that d = 4.1068 mm (this is using the given capacitance of 0.379 pF.
As I said, it's not that simple. What you calculated there was the spacing for a total capacitance of 0.379 pF, not the change in spacing for a change in 0.379 pF. This number is meaningless -- I was only trying to help you with your units.

Then I calculated the capacitance at the initial spacing of 0.55 mm by using the equation C - kEoA/d, using k = 3.75, Eo = 8.85 x 10^-12 C^2/Nm^2, A = 0.0469 m^2, and d= 0.00055m, getting an answer of 2.83 x 10^-12 F.
Correct.

I then added 0.379 x 10^-12 F to this to get a total of 3.209 x 10^-12 F. Then I plugged 3.209 x 10^-12 F into the same equation I used in the first calculation, again solving for d, obtaining an answer of 0.485 mm. Lastly, I subtracted 0.485 mm from 4.1068 mm, obtaining my answer of 3.6218 mm. Please let me know if this is a simple miscalculation or if I am using a wrong equation, or whatever. Thanks.
The initial spacing was 0.55 mm. The final spacing was 0.485 mm. The difference is 0.0635 mm.

- Warren

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