Keyboard problem

1. Sep 6, 2006

FlipStyle1308

Here is the question:

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!

2. Sep 7, 2006

Staff: Mentor

Last edited: Sep 7, 2006
3. Sep 7, 2006

FlipStyle1308

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?

Last edited: Sep 7, 2006
4. Sep 7, 2006

Staff: Mentor

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.

5. Sep 7, 2006

FlipStyle1308

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

Last edited: Sep 7, 2006
6. Sep 7, 2006

chroot

Staff Emeritus
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

7. Sep 7, 2006

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?

8. Sep 7, 2006

chroot

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

- Warren

9. Sep 7, 2006

FlipStyle1308

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

10. Sep 7, 2006

chroot

Staff Emeritus
No..

- Warren

11. Sep 7, 2006

FlipStyle1308

10^-6? Is that right?

Last edited: Sep 7, 2006
12. Sep 7, 2006

chroot

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

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

- Warren

13. Sep 7, 2006

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?

Last edited: Sep 7, 2006
14. Sep 7, 2006

chroot

Staff Emeritus
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

15. Sep 7, 2006

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?

Last edited: Sep 7, 2006
16. Sep 7, 2006

chroot

Staff Emeritus
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

17. Sep 7, 2006

FlipStyle1308

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

18. Sep 7, 2006

chroot

Staff Emeritus
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

19. Sep 7, 2006

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.

20. Sep 7, 2006

chroot

Staff Emeritus
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.

Correct.

The initial spacing was 0.55 mm. The final spacing was 0.485 mm. The difference is 0.0635 mm.

- Warren