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Keyboard problem |
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| Sep6-06, 11:42 PM | #1 |
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Keyboard problem
Here is the question:
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| Sep7-06, 06:26 AM | #2 |
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One should be able to express capacitance as a function of the known variables and the distance.
See - http://hyperphysics.phy-astr.gsu.edu...ic/pplate.html |
| Sep7-06, 04:54 PM | #3 |
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| Sep7-06, 06:33 PM | #4 |
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Keyboard problem
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. |
| Sep7-06, 06:42 PM | #5 |
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Hmm, okay, but is my number correct, except for the placement of the decimal? I keep getting the same answer :(.
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| Sep7-06, 06:49 PM | #6 |
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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 |
| Sep7-06, 06:52 PM | #7 |
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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? |
| Sep7-06, 06:58 PM | #8 |
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You did not convert the area, 46.9 mm^2, into m^2 correctly.
- Warren |
| Sep7-06, 07:00 PM | #9 |
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Oh, should it be 46.9 mm^2 = 46.9 x 10^-9 m^2?
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| Sep7-06, 07:02 PM | #10 |
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No..
- Warren |
| Sep7-06, 07:03 PM | #11 |
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10^-6? Is that right?
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| Sep7-06, 07:11 PM | #12 |
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46.9 mm^2 = 4.69 × 10-5 m^2.
(Divide by 10^3 * 10^3.) - Warren |
| Sep7-06, 07:15 PM | #13 |
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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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| Sep7-06, 08:02 PM | #14 |
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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 |
| Sep7-06, 08:15 PM | #15 |
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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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| Sep7-06, 09:46 PM | #16 |
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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 |
| Sep7-06, 09:54 PM | #17 |
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Sorry, I meant that is the distance. But is my final answer of 3.6216 mm correct?
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