Many computer keyboards operate on the principle of capacitance. Each key forms

Question

Many computer keyboards operate on the principle of capacitance. Each key forms a small parallel-plate capacitor whose separation is reduced when the key is depressed.

i)Does depressing a key increase or decrease its capacitance? Explain.

ii)Suppose the plates for each key have an area of 47.5 mm^2 and an initial separation of 0.550 mm. In addition, let the dielectric have a dielectric constant 3.75. If the circuitry of the computer can detect a change in capacitance of 0.425 pF, what is the minimum distance a key must be depressed to be detected?

Explanation / Answer

Given that

C = 0.425 x 10^-12 F

d = 0.550 mm (or 0.550 x 10^-3 m)

A = 4.75 x 10^-5 m2

k = 3.75

a)The capacitance of the parallel plate capacitor with capacitor is

  C = A / d

C = capacitance
= dielectric constant
A = area of one of the plates
= 8.85 x 10^-12 farad/meter
d = distance

The capacitance of the parallel plate capacitor is inversly proportional to seperation between the plates So capacitance is increased.
C = A / d
   = (3.75)(8.85 x 10^-12 farad/meter)(4.75 x 10^-5 m2)/(0.550 x 10^-3m)

  = 2.866x10^-12 F

So total capacitance is C' = 2.866x10^-12 F+ 0.425 x 10^-12 F

                                     = 3.29x10^-12 F

d = A / C

= ( 3.75)(8.85 x 10^-12 farad/meter)( 4.75 x 10^-5 m2)/(  3.29x10^-12 F)

=0.479 x10-3m
=0.479 mm

The initial separation is 0.550 mm

the minimum distance a key must be depressed to be detected is

         d =  0.550 mm - 0.479 mm
           = 0.071mm

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