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Chapter 24: Problem 54

24.54. 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 \(42.0 \mathrm{mm}^{2},\) and the separation between the plates is 0.700 \(\mathrm{mm}\) before the key is depressed. (a) Calculate the capaci- tance before the key is depressed. (b) If the circuitry can detect a change in capacitance of 0.250 \(\mathrm{pF}\) , how far must the key be depressed before the circuity detects its depression?

Short Answer

Expert verified
(a) 0.531 pF; (b) 0.224 mm needed to detect key press.

Step by step solution

01

Understand the Problem

We are given a parallel-plate capacitor with the area of each metal plate as \( A = 42.0 \, \text{mm}^2 \) and the plate separation \( d = 0.700 \, \text{mm} \). We need to find the initial capacitance and the required change in separation for the change in capacitance to be \( 0.250 \, \text{pF} \).
02

Convert Units

First, convert the area and separation to meters. For area, \( 42.0 \, \text{mm}^2 = 42.0 \times 10^{-6} \, \text{m}^2 \). For separation, \( 0.700 \, \text{mm} = 0.700 \times 10^{-3} \, \text{m} \).
03

Calculate Initial Capacitance

Use the formula for a parallel-plate capacitor: \[ C = \frac{\varepsilon_0 A}{d} \]where \( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \). Substitute the values: \( C = \frac{(8.854 \times 10^{-12})(42.0 \times 10^{-6})}{0.700 \times 10^{-3}} \)Evaluate to find \( C \approx 0.531 \times 10^{-12} \text{F} \) or \( 0.531 \, \text{pF} \).
04

Equation for Change in Capacitance

The change in capacitance \( \Delta C = C' - C = 0.250 \, \text{pF} \) means the new capacitance \( C' = C + 0.250 \, \text{pF} \). Use the equation:\[ C' = \frac{\varepsilon_0 A}{d'} = C + 0.250 \times 10^{-12} \text{F} \]where \( d' \) is the new separation.
05

Solve for New Separation

Set \( C' \) from Step 4 equal to:\[ C + 0.250 \times 10^{-12} = \frac{\varepsilon_0 A}{d'} \]Solve for \( d' \):\( d' = \frac{\varepsilon_0 A}{C + 0.250 \times 10^{-12} } \).Substitute values and solve \( d' = \frac{(8.854 \times 10^{-12})(42.0 \times 10^{-6})}{0.781 \times 10^{-12}} \)Calculate to find \( d' \approx 0.476 \times 10^{-3} \text{m} \).
06

Calculate Required Depression

Now calculate how much the key is depressed: \( \Delta d = d - d' = 0.700 \times 10^{-3} - 0.476 \times 10^{-3} = 0.224 \times 10^{-3} \text{m} \, \text{or} \, 0.224 \, \text{mm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parallel-plate Capacitor
A parallel-plate capacitor is a simple yet fascinating device found in many electronic applications. It consists of two conductive plates separated by a small gap.
These plates are often made of metal and can be filled with a dielectric, like air. When connected to a power source, one plate becomes positively charged, and the other becomes negatively charged. This creates an electric field between them.
The primary function of a parallel-plate capacitor is to store electrical energy. The ability to store this energy is called capacitance and is a crucial feature in electronic circuits. Capacitors can filter signals, store power temporarily, and even help in timing applications. The simple design makes parallel-plate capacitors easy and cost-effective to produce, yet incredibly useful.
Capacitor Formula
To understand how capacitors work and calculate their capacitance, we rely on a fundamental formula derived from their geometric properties. The capacitance (\(C\)) of a parallel-plate capacitor is easily calculated using the formula:
\[ C = \frac{\varepsilon_0 A}{d} \]
Here:
  • \(C\) is the capacitance in farads (F).

  • \(\varepsilon_0\) (epsilon naught) is the permittivity of free space, approximately \(8.854 \times 10^{-12} \, \text{F/m}\).

  • \(A\) is the area of one of the plates in square meters (\(\text{m}^2\)).

  • \(d\) is the separation between the plates in meters (m).

The formula shows that the greater the plate area and the smaller the distance between them, the higher the capacitance. This tells us how efficient the capacitor is at storing charge.
Capacitance Change Detection
Detecting changes in capacitance is a powerful mechanism for sensing in various applications, such as touch-sensitive devices. When a key on a keyboard presses down, the separation between capacitor plates decreases, altering the capacitance.
Electronic circuitry is often used to detect these subtle changes in capacitance, signaling some action. The example exercise illustrates how a small change in capacitance, such as \(0.250 \, \text{pF}\), can be detected as significant in computerized systems.
This is particularly relevant in modern devices where precision is critical. By carefully designing the circuitry, it can detect very small movements or changes, making touch sensors accurate and reliable. This concept forms the basis for various user interfaces and control systems today.

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Most popular questions from this chapter

24.24. A parallel-plate air capacitor has a capacitance of 920 \(\mathrm{pF}\) . The charge on each plate is 2.55\(\mu \mathrm{C}\) . (a) What is the potential difference between the plates? (b) If the charge is kept constant,what will be the potential difference between the plates if the sep ration is doubled? (c) How much work is required to double the separation?

24.25. A \(5.80-\mu F\) , parallel-plate, air capacitor has a plate separation of 5.00 \(\mathrm{mm}\) and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3}\) .

24.38. A parallel-plate capacitor has capacitance \(C_{0}=5.00 \mathrm{pF}\) when there is air between the plates. The separation between the plates is 1.50 \(\mathrm{mm}\) (a) What is the maximum magnitude of charge \(Q\) that can be placed on each plate if the eloctric field in the region between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{V} / \mathrm{m} ?\) ( b) A dielec- tric with \(K=270\) is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed \(3.00 \times 10^{4} \mathrm{V} / \mathrm{m} ?\)

24.10. A cylindrical capacitor consists of a solid inner conducting core with radius \(0.250 \mathrm{cm},\) surrounded by an outer hollow con- ducting tube. The two conductors are separated by air, and the length of the cylinder is 12.0 \(\mathrm{cm}\) . The capacitance is 36.7 \(\mathrm{pF}\) . (a) Calculate the inner radius of the hollow tabe. (b) When the capacitor is charged to 125 \(\mathrm{V}\) , what is the charge per unit length \(\lambda\) on the capacitor?

24.35. A cylindrical air capacitor of length 15.0 m stores \(3.20 \times 10^{-9} \mathrm{J}\) of energy when the potential difference between the two conductors is 4.00 \(\mathrm{V}\) . (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.
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