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 47.0mm2 , and the separation between the plates is 0.670mm before the key is depressed.

Calculate the capacitance before the key is depressed.

If the circuitry can detect a change in capacitance of 0.230pF , how far must the key be depressed before the circuitry detects its depression?

Explanation / Answer

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

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 48.0 mm^2 , and the separation between the plates is 0.680mm before the key is depressed.

a. Calculate the capacitance before the key is depressed.
b. If the circuitry can detect a change in capacitance of 0.300 pF, how far must the key be depressed before the circuitry detects its depression?

Answer

if u look at the formula for calculating capacitance ,
u can observe than capacitance is inversely proportional to the distance between the plates
in other words,

C1 / C2 = d2 / d1 ................. (1)

u have found that original capacitance is 6.25X 10-13 F , lets denote it by C0

now , the circuit can only detect key if the capacitance diffrence is 0.3000 pF

which is equal to 0.3 X10-12 F or 3 X 10-13 F

so the new C is 6.25 + 3 = 9.25 X 10 -13 F

putting it in equation (1)

6.25 / 9.25 = new seperation / 0.680 mm

this gives ,

new seperation = 0.459 mm

so the amount till which it needs to be pressed = old seperation - new seperation

= 0.680 mm - 0.459 mm

= 0.221 mm

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