Question

In one type of computer keyboard, each key holds a small metal plate that serves as one plate of a parallel-plate, airfilled 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 mm\(^2\), and the separation between the plates is 0.700 mm 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 0.250 pF, how far must the key be depressed before the circuitry detects its depression?

Short answer

(a) Capacitance is approximately 0.531 pF. (b) Key must be depressed by about 0.128 mm.

Step-by-step solution

Understanding Capacitance Formula

Capacitance for a parallel-plate capacitor is given by the formula \( C = \frac{\varepsilon_0 A}{d} \), where \( \varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \) is the permittivity of free space, \( A \) is the area of the plate, and \( d \) is the separation between the plates.

Convert Units

The area \( A = 42.0 \ \text{mm}^2 \) and separation \( d = 0.700 \ \text{mm} \) should be converted to meters for consistency with \( \varepsilon_0 \). Thus, \( A = 42.0 \times 10^{-6} \ \text{m}^2 \) and \( d = 0.700 \times 10^{-3} \ \text{m} \).

Calculate Initial Capacitance

Insert 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 to find the initial capacitance \( C \approx 0.531 \ \text{pF} \).

Determine Depression Distance for Capacitance Change

The change in capacitance \( \Delta C \) is 0.250 pF. We have \( \Delta C = C_{new} - C_{initial} \) and \( \Delta C = \frac{\varepsilon_0 A}{d_{new}} - \frac{\varepsilon_0 A}{d_{initial}} \). Solve for the new separation distance \( d_{new} \).

Solve for New Separation Distance

Rearrange and substitute the change in capacitance into the equation: \( 0.250 \times 10^{-12} = \frac{8.85 \times 10^{-12} \times 42.0 \times 10^{-6}}{d_{new}} - 0.531 \times 10^{-12} \). Solve this for \( d_{new} \) to find that \( d_{new} \approx 0.572 \ \text{mm} \).

Calculate Depression of the Key

The distance the key must be depressed is \( d_{initial} - d_{new} = 0.700 \ \text{mm} - 0.572 \ \text{mm} = 0.128 \ \text{mm} \).

Key concepts

Parallel-plate capacitor

A parallel-plate capacitor is a fundamental component frequently used in electronic devices, like computer keyboards, to store electrical charge temporarily. This type of capacitor consists of two important elements: two conductive plates and a dielectric material in between, which in many cases is air. The arrangement is quite simple but effective. When a charge is applied to these plates, an electric field develops between them, storing energy. The amount of stored charge is what we refer to as capacitance.

Capacitance for a parallel-plate capacitor is calculated using the equation \( C = \frac{\varepsilon_0 A}{d} \), where \( C \) is capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( A \) is the plate area, and \( d \) is the separation between the plates. This equation shows that capacitance increases with a larger plate area or a smaller distance between the plates. In a keyboard, for instance, pressing a key changes the distance \( d \), altering the capacitance and allowing the keyboard circuitry to detect input.

Permittivity of free space

The concept of permittivity of free space, denoted by \( \varepsilon_0 \), is central to understanding how capacitors work. It is a physical constant that describes how electric fields interact with vacuum or free space, and it determines how much electric charge can be stored. The value of \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \ F/m \) (farads per meter).

In the context of parallel-plate capacitors, permittivity of free space influences the capacitance through the formula \( C = \frac{\varepsilon_0 A}{d} \). This means that the higher the permittivity, the greater the capacity of the capacitor to store charge for a given area and separation. Permittivity, therefore, directly affects capacitors' efficiency and effectiveness in practical applications, like detecting key presses in keyboards.

Unit conversion

Unit conversion is a crucial step in solving physics problems that involve different measurement systems. In the exercise involving keyboard capacitors, area and separation are initially given in millimeters, a common unit in device engineering due to its practicality. However, to use them in equations involving permittivity of free space, conversion to meters is required.

For example, the area of the capacitor plate, given as \( 42.0 \, \text{mm}^2 \), needs to be converted to \( \text{m}^2 \). This is done by multiplying by the factor \( 10^{-6} \), resulting in an area of \( 42.0 \times 10^{-6} \, \text{m}^2 \). Similarly, converting the separation \( 0.700 \, \text{mm} \) to meters involves multiplying by \( 10^{-3} \), yielding \( 0.700 \times 10^{-3} \, \text{m} \).

• Area: \( 42.0 \, \text{mm}^2 \rightarrow 42.0 \times 10^{-6} \text{ m}^2 \)
• Separation: \( 0.700 \, \text{mm} \rightarrow 0.700 \times 10^{-3} \text{ m} \)

These conversions ensure consistency with the units used in physical constants, facilitating accurate calculations.

Capacitance change detection

Capacitance change detection is a principle commonly used in modern electronics to sense input or environmental changes. In the context of a computer keyboard, this principle plays a vital role in its functionality. The circuit detects changes in capacitance when a key is pressed, altering the distance between the capacitor plates, \( d \).

In the original problem, circuitry is designed to detect a change in capacitance as small as \( 0.250 \, \text{pF} \). This means the keyboard must be sensitive enough to note minute changes in capacitance, triggering a response only when a key is pressed. By solving for \( d_{new} \) using the equation \( \Delta C = \frac{\varepsilon_0 A}{d_{new}} - \frac{\varepsilon_0 A}{d_{initial}} \), one can determine the extent of key depression needed for detection.

This change detection is crucial for ensuring that each key press, no matter how slight, is accurately registered by the keyboard, providing users with a reliable interface for typing and inputting data into computers.