Problem 24.47 13 of 15> Part A In one type of computer keyboard, each key holds

Question

Problem 24.47 13 of 15> Part A 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 51.0 mim, and the separation between the plates is 0.750 mn before the key is depressed. Calculate the capacitance before the key is depressed. C6.018 10-14 Submit Previous Answers Request Answer X Incorrect. Try Again; 5 attempts remaining Part B 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? Submit

Explanation / Answer

According to the concept of the electric potential and capacitance

Given that

Area A=51*10^-6 m^2

Distance d=0.75*10^-3 m

Now we find the capacitance of the circuit

Capacitance C=€A/d

=8.85*10^-12*51*10^-6/(0.75*10^-3)

=601.8*10^-15 F

=0.6018 pF

Now we find the distance b/w plates where the capacitance change

C/C'=d'/d

0.6018/0.270=d'/0.75*10^-3

d'=1.672 mm

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