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Chapter 20: Problem 95

Computer Keyboards Many computer keyboards operate on the principle of capacitance. As shown in Figure \(20-16\) each key forms a small parallel-plate capacitor whose separation is reduced when the key is depressed. (a) Does depressing a key increase or decrease its capacitance? Explain. (b) Suppose the plates for each key have an area of \(47.5 \mathrm{mm}^{2}\) and an initial separation of \(0.550 \mathrm{mm}\). In addition, let the dielectric have a dielectric constant of \(3.75 .\) If the circuitry of the computer can detect a change in capacitance of \(0.425 \mathrm{pF}\), what is the minimum distance a key must be depressed to be detected?

Short Answer

Expert verified
(a) Depressing the key increases its capacitance. (b) The key must be depressed by 46.5 碌m.

Step by step solution

01

Understanding Capacitance Change

When a key is depressed, the separation between the plates in the capacitor decreases. Using the formula for capacitance \( C = \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d} \), decreasing \( d \) (the separation) while \( \varepsilon_0 \), \( \varepsilon_r \) (the dielectric constant), and \( A \) (the area) remain constant, results in an increased capacitance. Therefore, depressing a key increases the capacitance.
02

Calculating Initial Capacitance

The initial capacitance \( C_i \) when the plates are at separation \( d_i = 0.550 \) mm (or \( 0.000550 \) m), can be calculated using the formula: \[ C_i = \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d_i} \]where \( A = 47.5 \times 10^{-6} \) m虏, \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m, and \( \varepsilon_r = 3.75 \). Plug in these values to find \( C_i \).
03

Solving for Change in Distance

Given that the change in capacitance \( \Delta C = 0.425 \) pF (or \( 0.425 \times 10^{-12} \) F), we need to find the change in separation \( \Delta d \) of the capacitor plates. Using the relation \[ \Delta C = C_f - C_i = \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d_f} - \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d_i} \]where \( d_f = d_i - \Delta d \), solve for \( \Delta d \).
04

Final Calculation of \( \Delta d \)

Substitute the known values into the expression \[ \Delta C = \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d_f} - \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d_i} \]With \( \Delta C = 0.425 \times 10^{-12} \) F and initial calculations from previous steps, solve for \( d_f \), then find \( \Delta d = d_i - d_f \).

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Key Concepts

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

Capacitance
Capacitance is a measure of a capacitor's ability to store charge. In simpler terms, it tells us how much charge a capacitor can hold when a specific voltage is applied. The unit is the Farad, although capacitors often have smaller capacitance values such as microfarads (碌F) or picofarads (pF).
  • For a parallel-plate capacitor, which is commonly used in devices like keyboards, capacitance depends on the plate area, the separation between the plates, and the dielectric material between them.
  • The formula to calculate capacitance is given by: \( C = \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d} \), where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_r \) is the relative permittivity or dielectric constant, \( A \) is the plate area, and \( d \) is the distance between the plates.
Understanding this concept is crucial for analyzing how pressing keys on a keyboard changes the capacitance and thus signals a key press.
Dielectric Constant
The dielectric constant, denoted as \( \varepsilon_r \), is a measure of a material's ability to store electrical energy in an electric field. In the context of capacitors, it is vital because it helps increase capacitance without physically enlarging the capacitor.
  • Materials with higher dielectric constants can store more energy, allowing smaller capacitors to hold a significant amount of charge.
  • This property explains why keyboards use specific materials as dielectrics between the plates to efficiently detect key presses.
By increasing the dielectric constant, a keyboard can be more sensitive to the small changes in capacitance that occur when keys are pressed.
Parallel-Plate Capacitor
A parallel-plate capacitor consists of two conductive plates separated by a dielectric material. This simple structure is used in keyboards to detect the position of the keys. When you press a key, the plates move closer together, reducing the separation \( d \), and increasing the capacitance.
  • The capacitance is inversely proportional to the distance between the plates, so a small movement can create a noticeable change in capacitance.
  • By using a fixed plate area and a specific dielectric material, the keyboard can consistently measure changes in key position.
The equation \( C = \frac{\varepsilon_0 \cdot \varepsilon_r \cdot A}{d} \) clearly shows that smaller distances \( d \) result in larger capacitances.The parallel-plate design is ideal for keyboards due to its simplicity and the ease with which it can be mass-produced.

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Most popular questions from this chapter

A \(0.22-\mu \mathrm{F}\) capacitor is charged by a \(1.5-\mathrm{V}\) battery. After being charged, the capacitor is connected to a small electric motor. Assuming \(100 \%\) efficiency, (a) to what height can the motor lift a \(5.0-\mathrm{g}\) mass? (b) What initial voltage must the capacitor have if it is to lift a \(5.0-\mathrm{g}\) mass through a height of \(1.0 \mathrm{cm} ?\)

A uniform electric field of magnitude \(6.8 \times 10^{5} \mathrm{N} / \mathrm{C}\) points in the positive \(x\) direction. Find the change in electric potential between the origin and the points (a) \((0,6.0 \mathrm{m})\) (b) \((6.0 \mathrm{m}, 0)\) and (c) \((6.0 \mathrm{m}, 6.0 \mathrm{m})\)

The Sodium Pump Living cells actively "pump" positive sodium ions \(\left(\mathrm{Na}^{7}\right)\) from inside the cell to outside the cell. This process is referred to as pumping because work must be done on the ions to move them from the negatively charged inner surface of the membrane to the positively charged outer surface. Given that the electric potential is \(0.070 \mathrm{V}\) higher outside the cell than inside the cell, and that the cell membrane is \(0.10 \mu \mathrm{m}\) thick (a) calculate the work that must be done (in joules) to move one sodium ion from inside the cell to outside. (b) If the thickness of the cell membrane is increased, does your answer to part (a) increase, decrease, or stay the same? Explain. (It is estimated that as much as \(20 \%\) of the energy we consume in a resting state is used in operating this "sodium pump.")

A uniform electric field of magnitude \(4.1 \times 10^{5} \mathrm{N} / \mathrm{C}\) points in the positive \(x\) direction. Find the change in electric potential energy of a \(4.5-\mu C\) charge as it moves from the origin to the points (a) \((0,6.0 \mathrm{m}) ;\) (b) \((6.0 \mathrm{m}, 0) ;\) and (c) \((6.0 \mathrm{m}, 6.0 \mathrm{m})\).

A charge of \(3.05 \mu C\) is held fixed at the origin. A second charge of \(3.05 \mu C\) is released from rest at the position \((1.25 \mathrm{m},\) \(0.570 \mathrm{m})\). (a) If the mass of the second charge is \(2.16 \mathrm{g}\), what is its speed when it moves infinitely far from the origin? (b) At what distance from the origin does the second charge attain half the speed it will have at infinity?
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