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Chapter 24: Problem 40

24.40. A budding electronics hobbyist wants to make a simple 1.0 -nF capacitor for tuning her crystal radio, using two sheets of aluminum foil as plates, with a few shects of paper between them as a dielectric. The paper has a dielectric constant of 3.0, and the thickness of one sheet of it is 0.20 \(\mathrm{mm}\) . (a) If the sheets of paper measure \(22 \times 28 \mathrm{cm}\) and she cuts the aluminum foil to the same dimensions, how many sheets of paper should she use between her plates to get the proper capacitance? (b) Suppose for convenience she wants to use a single sheet of posterboard, with the same dielectric constant but a thickness of \(120 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 \(\mathrm{nF}\) of capacitance? (c) Suppose she goes hight-tech and finds a sheet of Teflon of the same thickness as the posterboard to use as a dielectric. Will she need a larger or smaller area of Tellon than of posterboard? Explain.

Short Answer

Expert verified
(a) 9 sheets of paper, (b) 4.52 m虏 of aluminum foil, (c) Larger area for Teflon.

Step by step solution

01

Determine Formula for Capacitance

The capacitance of a parallel plate capacitor is given by the formula: \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \), where \( C \) is the capacitance, \( \varepsilon_r \) is the relative permittivity (dielectric constant), \( \varepsilon_0 \) is the permittivity of free space \( (8.854 \times 10^{-12} \, \text{F/m}) \), \( A \) is the area of the plates, and \( d \) is the separation distance between the plates (thickness of the dielectric).
02

Calculate Number of Paper Sheets (Part a)

For part (a), we need to find the thickness \( d \) such that the capacitance \( C = 1.0 \, \text{nF} = 1.0 \times 10^{-9} \, \text{F} \). The area \( A \) is \( 22 \, \text{cm} \times 28 \, \text{cm} = 0.22 \, \text{m} \times 0.28 \, \text{m} = 0.0616 \, \text{m}^2 \). Solving for \( d \), we have:\[ d = \frac{\varepsilon_r \varepsilon_0 A}{C} = \frac{(3.0)(8.854 \times 10^{-12})(0.0616)}{1.0 \times 10^{-9}} \approx 1.6357 \times 10^{-3} \, \text{m} = 1.6357 \, \text{mm} \]Each paper sheet is 0.20 mm thick, so the number needed is \( \frac{1.6357 \, \text{mm}}{0.20 \, \text{mm/sheet}} \approx 8.18 \), or 9 sheets (rounding up).
03

Calculate Area for Posterboard (Part b)

For part (b), use a single sheet of posterboard of thickness \( d = 120 \, \text{mm} = 0.12 \, \text{m} \). Rearranging the capacitance formula to solve for the required area \( A \):\[ A = \frac{C \cdot d}{\varepsilon_r \varepsilon_0} = \frac{(1.0 \times 10^{-9})(0.12)}{(3.0)(8.854 \times 10^{-12})} \approx 4.52 \, \text{m}^2 \] You will need an area of about 4.52 m虏 for the aluminum foil.
04

Compare Dielectric Constants for Teflon and Posterboard (Part c)

For Teflon, use the same thickness \( d = 120 \, \text{mm} = 0.12 \, \text{m} \). Assuming \( \varepsilon_r = 2.1 \) for Teflon:\[ A = \frac{C \cdot d}{\varepsilon_r \varepsilon_0} = \frac{(1.0 \times 10^{-9})(0.12)}{(2.1)(8.854 \times 10^{-12})} \approx 6.47 \, \text{m}^2 \]Since Teflon has a lower dielectric constant than the posterboard, a larger area of Teflon is needed for the same capacitance.

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Key Concepts

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

Dielectric Constant
The dielectric constant, also known as the relative permittivity, is a material property that measures its ability to store electrical energy in the presence of an electric field. It is denoted as \( \varepsilon_r \), and it compares the permittivity of a material to the permittivity of free space \( \varepsilon_0 \). The formula that relates these is \( \varepsilon_r = \frac{\varepsilon}{\varepsilon_0} \), where \( \varepsilon \) is the permittivity of the material.

A higher dielectric constant means a material can store more electrical energy, which is valuable in designing capacitors. In our exercise, paper and posterboard are used as dielectrics, with a constant of 3.0, while Teflon has a constant of 2.1. This is why Teflon requires a greater area to achieve the same capacitance as materials with higher constants like posterboard.

A dielectric material in a capacitor allows it to hold more charge at a given voltage, by essentially reducing the electric field between the plates. By using materials with higher dielectric constants, electronics developers can design compact capacitors for devices.
Parallel Plate Capacitor
A parallel plate capacitor is a simple structure used to store electrical energy. It consists of two conductive plates separated by an insulating material called a dielectric. The capacitance, \( C \), which is the ability to store charge, is given by the formula \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \).

- \( A \) is the area of one of the plates.- \( d \) is the separation (or thickness of the dielectric) between the plates.- \( \varepsilon_r \) is the dielectric constant of the material between the plates.- \( \varepsilon_0 \) is the permittivity of free space (approximately \( 8.854 \times 10^{-12} \text{ F/m} \)).

In the given situation, sheets of aluminum and materials like paper or Teflon act as plates and dielectrics. By adjusting the area of the plates or the thickness of the dielectric, the capacitance can be precisely tuned. For instance, larger plate areas or higher dielectric constants result in greater capacitance.

The parallel plate capacitor is fundamental in electronics as it is used in tuning circuits, energy storage, and in filtering applications.
Relative Permittivity
Relative permittivity, another term for the dielectric constant, indicates how much electric field is reduced in a material compared to a vacuum. It is crucial in the formula for calculating the capacitance of capacitors, \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \). Higher relative permittivity means better storage of electrical energy.

In our example, using different materials such as paper, posterboard, or Teflon as dielectrics shows how relative permittivity affects capacitance. Paper and posterboard with a relative permittivity of 3.0 enable a certain capacitance level with a smaller area compared to Teflon with a relative permittivity of 2.1.

This emphasizes the importance of selecting materials with the appropriate relative permittivity based on the desired size and capacitance of the capacitor in the design of electronic components.

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Most popular questions from this chapter

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?

24.34. A 0.350 -m-long cylindrical capacitor consists of a solid conducting core with a radius of 1.20 \(\mathrm{mm}\) and an outer hollow conducting tube with an inner radius of 2.00 \(\mathrm{mm}\) . The two conductors are separated by air and charged to a potential difference of 6.00 \(\mathrm{V}\) . Calculate (a) the charge per length for the capacitor; (b) the total charge on the capacitor; (c) the capacitance; (d) the energy stored in the capacitor when fully charged.

24.50. A parallel-plate air capacitor is made by using two plates 16 \(\mathrm{cm}\) square, spaced 4.7 \(\mathrm{mm}\) apart. It is connected to a \(12-\mathrm{V}\) bat- tery. (a) What is the capacitance? (b) What is the charge on each plate? (c) What is the electric field between the plates? (d) What is the energy stored in the capacitor? (e) If the battery is disconnected and then the plates are pulled apart to a separation of \(9.4 \mathrm{mm},\) what are the answers to parts \((\mathrm{a})-(\mathrm{d}) ?\)

24.3. A parallel-plate air capacitor of capacitance 245 \(\mathrm{pF}\) has a charge of magnitude 0.148\(\mu \mathrm{C}\) on each plate. The plates are 0.328 \(\mathrm{mm}\) apart. (a) What is the potential difference between the plates? (b) What is the area of each plate? (c) What is the electric-field magnitude between the plates? (d) What is the surface charge density on each plate?

24.27. A \(450-\mu\) F capacitor is charged to 295 V. Then a wire is con- nected between the plates. How many joules of thermal energy are produced as the capacitor discharges if all of the energy that was stored goes into heating the wire?
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