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 46.0mm2 , and the separation between the plates is 0.680mm before the key is depressed.

(B) If the circuitry can detect a change in capacitance of 0.230

pF , how far must the key be depressed before the circuitry detects its depression?

*I have already solved for the capacitance which equals:

Explanation / Answer

dC=E0a(1d2 -1/d1)

0.230*10^(-12)=8.854*10^(-12)*46*10^(-6)(1/d2 -1000/680)

5.64+100/68=1/d2

d2=0.140 mm

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