Problem 24.52 In one type of computer keyboard, each key holds a small metal pla

Question

Problem 24.52 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 51.0 mm^2, and the separation between the plates is 0.690mm before the key is depressed. Part A Calculate the capacitance before the key is depressed. C = 9.34 . 10^-14 F Part B If the circuitry can detect a change in capacitance of 0.280pF , how far must the key be depressed before the circuitry detects its depression? h = 0.191 mm

Explanation / Answer

We'll make use the following equation to solve this problem:

C = (?*A)/d

? = Permittivity of the dielectric = 8.85 x 10-12
A = Surface area of plates = 51 mm2
d = distance between the plates = 0.690 mm

C = (8.85 x 10-12 * 51 x 10-6 ) / 0.690 x 10-3 = 6.54 x 10-13  F Ans.

b).

In this problem, the quantity ?*A will remain constant since neither the surface area of the plates nor the permittivity of air will change. Using this, you can obtain the following equation:

C1*d1 = C2*d2

The capacitance must increase by 0.280 pF to be noticeable, so C2 in this case equals (0.654 pF + 0.280 pF) which is 0.934 pF.

d = (?*A)/C = 8.854 x 10-12 * 51 x 10-6 / 0.934 x 10-12

= 0.483 mm

hence must to depress = 0.690 - 0.483 = 0.207 mm Ans.

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