91Ó°ÊÓ

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 sheets 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 \(12.0 \mathrm{mm},\) instead of the paper. What area of aluminum foil will she need for her plates to get her 1.0 nF of capacitance? (c) Suppose she goes high-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 Teflon than of poster board? Explain.

Step by step solution

01

Understanding Capacitance Formula

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), \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space, \( A \) is the area of the plates, and \( d \) is the separation between the plates.

02

Calculate Number of Paper Sheets (Part a)

For part (a), we want to find the number of paper sheets needed. The given values are: capacitance \( C = 1.0 \, \text{nF} = 1.0 \times 10^{-9} \, \text{F} \), \( \varepsilon_r = 3.0 \), sheet thickness \( = 0.20 \, \text{mm} = 0.0002 \, \text{m} \), and \( A = 22 \, \text{cm} \times 28 \, \text{cm} = 0.22 \, \text{m} \times 0.28 \, \text{m} = 0.0616 \, \text{m}^2 \). Rearrange the formula to find \( d \) (thickness of multiple sheets): \[ d = \frac{\varepsilon_r \varepsilon_0 A}{C} = \frac{3.0 \times 8.85 \times 10^{-12} \times 0.0616}{1.0 \times 10^{-9}} = 1.633 \times 10^{-3} \, \text{m} \] Then, divide by the thickness of one sheet to get the number of sheets: \[ \text{Number of sheets} = \frac{1.633 \times 10^{-3}}{0.0002} \approx 8.165 \] Thus, she needs about 9 sheets of paper.

03

Calculate Aluminum Foil Area for Posterboard (Part b)

For part (b), if she uses a single sheet of posterboard with thickness \( 12.0 \, \text{mm} = 0.012 \, \text{m} \): \[ A = \frac{C \times d}{\varepsilon_r \varepsilon_0} = \frac{1.0 \times 10^{-9} \times 0.012}{3.0 \times 8.85 \times 10^{-12}} = 0.4522 \, \text{m}^2 \] Therefore, she will need an area of approximately 0.4522 m² of aluminum foil.

04

Compare Areas Required for Teflon (Part c)

For part (c), Teflon has a higher dielectric constant, approximately 2.1, resulting in \( \varepsilon_r = 2.1 \). Keeping the d (thickness of 0.012 m) the same, calculate the required area: \[ A_{\text{Teflon}} = \frac{1.0 \times 10^{-9} \times 0.012}{2.1 \times 8.85 \times 10^{-12}} = 0.646 \text{m}^2 \] Since the dielectric constant of Teflon is lower than the posterboard (3.0 < 2.1), she needs a larger area than with posterboard: approximately 0.646 m².

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Key Concepts

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

Dielectric Constant

When dealing with capacitors, the dielectric constant is a crucial concept to understand. It is a measure of a dielectric material's ability to increase the capacitance of a capacitor compared to having a vacuum between the plates.
The dielectric constant is often represented by the symbol \( \varepsilon_r \). This value is always greater than one, and the higher the value, the better the material is at storing electric charge.

• For example, paper has a dielectric constant of around 3.0, which means it can effectively increase the capacitance threefold compared to a vacuum.
• Teflon, another common dielectric material, has a dielectric constant of approximately 2.1, indicating slightly less capacitance enhancement than paper.

Understanding the dielectric constant is essential when designing or modifying capacitors to ensure that they meet the desired specifications.

Capacitance Formula

The capacitance formula for a parallel-plate capacitor is fundamental in understanding how capacitors work. It describes how the capacitance \( C \) relates to the physical properties of the capacitor:
\[ C = \frac{\varepsilon_r \varepsilon_0 A}{d} \]This equation shows that:

• \( \varepsilon_r \) is the dielectric constant and illustrates the material's ability to store electric charge.
• \( \varepsilon_0 \) is the permittivity of free space, a constant value of approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).
• \( A \) is the area of one of the plates in square meters.
• \( d \) is the distance between the two plates, typically the thickness of the dielectric material in meters.

This formula makes it clear how changes in any of these variables affect the overall capacitance. Increasing \( A \) or \( \varepsilon_r \) will boost capacitance, while increasing \( d \) will decrease it.

Parallel-Plate Capacitor

A parallel-plate capacitor is one of the most common types of capacitors you'll come across in electronics. It consists of two conducting plates separated by an insulating material known as the dielectric.
These capacitors are valued for their simplicity and efficiency in storing energy.

• The two plates are aligned parallel to each other, maximizing the electric field and, consequently, the capacitance.
• The dielectric material, placed between the plates, determines both the capacitance and the maximum voltage the capacitor can withstand.

In practical applications, parallel-plate capacitors are highly versatile and can be found in various electronic devices, from radios to televisions. Understanding how they function and how changes in their parameters affect performance is key to effectively utilizing them in electronics.