Hello, please help me with all the following parts in this question. (Electricit
ID: 3161838 • Letter: H
Question
Hello, please help me with all the following parts in this question. (Electricity, Magnetism and Waves)
In a computer keyboard design, a small metal plate is attached underneath each key to form a parallel plate capacitor. Each time a key is pressed, the gap of the parallel plate capacitor is reduced and the capacitance changes. The area of the plate is 42.0 mm^2 and the gap is 0.700 mm before a key is depressed. Calculate the capacitance before the key is depressed. Assume the dielectric is air and neglect any fringing fields. Assume the electronic circuit which detects the key press has a minimum resolution of 0.250 pF. How far does the key have to be depressed to measure a key stroke?Explanation / Answer
1) area , A = 42 mm^2
d = 0.700 mm
capacitance = A * epsilon/d
capacitance = 42 *10^-6 * 8.854 *10^-12/(0.70 *10^-3)
capacitance = 5.312 *10^-13 F
b) let the key stroke is pressed by x
the capacitance will increase by 0.250 pF
capacitance = 42 *10^-6 * 8.854 *10^-12/(0.70 *10^-3 - x)
5.312 *10^-13 + 0.250 *10^-12= 42 *10^-6 * 8.854 *10^-12/(0.70 *10^-3 - x)
solving for x
x = 0.224 mm
the key must be pressed by 0.224 mm
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