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 42.5 mm2 , and the separation between the plates is 0.690 mm before the key is depressed.

Calculate the capacitance before the key is depressed.

If the circuitry can detect a change in capacitance of 0.270 pF , how far must the key be depressed before the circuitry detects its depression?

Explanation / Answer

Area A =42.5mm2, Initial Separation d0 = 0.69 mm

Initial capacitance C=C0

capacitance C = A/d

for air

= 8.854x10-12 C 2/N.m2

C0 = (8.854x10-12) x (42.5x10-6) / (0.69 x 10-3) = 545.36 x 10-15 C2/N.m = 0.545pF

Now after depression Capacitance increases by 0.27pF

... New Capacitance C1 = C0 + 0.27pF = 0.815 pF

And C1 = 0.815 pF = A/d1

Final separation d1 = 0.461mm

... depression = d0 - d1 = 0.229mm

= 8.854x10-12 C 2/N.m2

C0 = (8.854x10-12) x (42.5x10-6) / (0.69 x 10-3) = 545.36 x 10-15 C2/N.m = 0.545pF

Now after depression Capacitance increases by 0.27pF

... New Capacitance C1 = C0 + 0.27pF = 0.815 pF

And C1 = 0.815 pF = A/d1

Final separation d1 = 0.461mm

... depression = d0 - d1 = 0.229mm

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