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

Question

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 50.0 mm2 , and the separation between the plates is 0.750 mm before the key is depressed.

part 1) Calculate the capacitance before the key is depressed. I got C=5.9*10^-13 F which is correct.

part 2) If the circuitry can detect a change in capacitance of 0.300 pF , how far must the key be depressed before the circuitry detects its depression?

Explanation / Answer

Given area of each plate A = 50.0 mm2 = 50.0*10^-6 m2

separatio of the plates d = 0.750 mm = 0.75*10^-3 m

we know that the capacitance C = epsilon not A /d epsilon not is permitivity of free space = 8.85*10^-12 C2/Nm2

before pressign the key C1 = 8.85*10^-12*50.0*10^-6/0.750*10^-3

C1 = 5.90*10^-13 F = 0.590 pF

change in capacitance after pressing Dc = C2-C1= epsilon not A /d2 d2= epsilon not A /c2-c1

= (8.85*10^-12*50*10^-6)/0.300 *10^-12 = 0.001475 m = 1.475 mm

for the given change in cpacitance key must be pressed 1.475 mm

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