Problem 52 - University Physics with Modern Physics

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

Step-by-Step Solution

Answer

Capacitance before depression: \(5.31\,\mathrm{pF}\). Depression to detect: \(21.0\,\mu\mathrm{m}\).

Step 1: Calculate Capacitance Before Key is Depressed

The capacitance of a parallel-plate capacitor can be calculated using the formula: \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \mathrm{F/m} \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the plate separation. First, convert area from \( \mathrm{mm}^2 \) to \( \mathrm{m}^2 \): \( A = 42.0 \times 10^{-6} \mathrm{m}^2 \). The separation \( d = 0.700 \times 10^{-3} \mathrm{m} \). Plug these values into the formula to get the capacitance.

Step 2: Perform the Calculation

Substitute the values into the capacitance formula: \[ C = \frac{8.85 \times 10^{-12} \times 42.0 \times 10^{-6}}{0.700 \times 10^{-3}} \]Calculate this to find the initial capacitance.

Step 3: Capacitance Detected Change Analysis

The circuit detects a change in capacitance \( \Delta C = 0.250 \mathrm{pF} = 0.250 \times 10^{-12} \mathrm{F} \). We need to calculate at what plate separation \( d' \) the capacitance changes by \( \Delta C \) from its initial value.

Step 4: Calculate New Plate Separation

Set the new capacitance \( C + \Delta C = \frac{\varepsilon_0 A}{d'} \) equal to the initial capacitance \( + \Delta C \). Solve for the new separation \( d' \).\[ \frac{\varepsilon_0 A}{d'} = C + \Delta C \]Solve for \( d' \).

Step 5: Determine Depressed Distance

The distance the key must be depressed is the change in separation \( d - d' \). Calculate this value.

Key Concepts

Capacitance

Capacitance is a key property of capacitors that measures their ability to store an electric charge. It is expressed in farads (F) and given by the formula: \[ C = \frac{\varepsilon_0 A}{d} \] where:

• \(C\) is the capacitance,
• \(\varepsilon_0\) is the permittivity of free space,
• \(A\) is the area of one of the plates,
• \(d\) is the separation between the plates.

In the context of a parallel-plate capacitor, when the area of the plates or the permittivity of the medium between them increases, the capacitance also increases.
Conversely, increasing the distance between the plates decreases the capacitance. Understanding how to calculate capacitance is crucial as it is a direct measure of how effective a capacitor will be in its application.

Permittivity of Free Space

Permittivity of free space, often denoted as \(\varepsilon_0\), is a fundamental physical constant that describes how electric fields interact with the vacuum. In calculations involving capacitors, it influences how much electric charge a particular setup can store. The permittivity of free space is expressed as:\[ \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \]This value is crucial in determining the capacitance of capacitors. It reflects how the electric field lines are dispersed in a vacuum and acts as a calibration factor for equations describing electromagnetism. Using \(\varepsilon_0\) helps to distinguish how capacitors might behave in different media, although for air or vacuum, the permittivity remains constant, allowing consistent calculations.

Capacitance Change Detection

Detecting changes in capacitance is an essential function of electronic circuits where accurate input is necessary, like in keyboard designs. As presented in this exercise, each key acts as a plate in a parallel-plate capacitor, with another plate fixed below it. When a key is pressed, the separation between these plates decreases, thereby increasing the overall capacitance. The system is set to recognize a specific change in capacitance, as seen in the given problem where a change of 0.250 pF can be detected. For accurate detection:

• One must calculate the baseline capacitance before the key is pressed.
• Then, any subtle shift in the separation that results in a measurable difference above a certain threshold is detected as a key press.

Understanding how to detect these changes is critical for designing efficient touch-responsive systems and ensures the precision and reliability necessary in devices demanding sensitive inputs.