n one type of computer keyboard, each key holds a small metal plate that serves

Question

n 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 51.0mm2 , and the separation between the plates is 0.660mm before the key is depressed.

1.Calculate the capacitance before the key is depressed.

2.If the circuitry can detect a change in capacitance of 0.290 pF, how far must the key be depressed before the circuitry detects its depression?

Explanation / Answer

Part 1)

C = eA/d

C = (8.85 X 10-12)(5.1 X 10-5)/(6.6 X 10-4)

C = 6.84 X 10-13 F (That is 684 pF)

Part 2)

The new capacitance will be 6.84 X 10-13 + .290 X 10-12 = 9.74 X 10-13 F

(9.74 X 10-13) = (8.85 X 10-12)(5.1 X 10-5)/d

d = .463 mm

The difference is .660 - .443 = .197 mm

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