Question

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 mm. (a) If the sheets of paper measure 22 \(\times\)28 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 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 posterboard? Explain.

Short answer

(a) 82 sheets. (b) Aluminum foil area: 0.452 m². (c) Larger Teflon area needed.

Step-by-step solution

Understand the Capacitance Formula

The capacitance formula for a parallel plate capacitor with a dielectric is given by\[C = \frac{\varepsilon_0 \varepsilon_r A}{d}\]where \( C \) is the capacitance, \( \varepsilon_0 \) is the vacuum permittivity \(8.85 \times 10^{-12} \text{ F/m}\), \( \varepsilon_r \) is the dielectric constant, \( A \) is the area of one plate, and \( d \) is the separation between the plates.

Calculate Required Separation (Part a)

We know the desired capacitance \( C = 1.0 \text{ nF} = 1.0 \times 10^{-9} \text{ F} \), the dielectric constant \( \varepsilon_r = 3.0 \), and the plate area \( A = 22 \text{ cm} \times 28 \text{ cm} = 0.22 \text{ m} \times 0.28 \text{ m} \). Plug these into the capacitance formula and solve for \( d \):\[1.0 \times 10^{-9} = \frac{8.85 \times 10^{-12} \times 3.0 \times (0.22 \times 0.28)}{d}\]Upon solving, \( d \approx 0.016374 \, \text{m} = 16.374 \, \text{mm}\).

Determine Number of Paper Sheets

The thickness of one sheet of paper is 0.20 mm, so the number of sheets, \( n \), is given by:\[ n = \frac{16.374 \, \text{mm}}{0.20 \, \text{mm/sheet}} = 81.87 \approx 82 \, \text{sheets}\]

Use Capacitance Formula for Single Sheet of Posterboard (Part b)

Now using posterboard with \( \varepsilon_r = 3.0 \) and thickness \( d = 12.0 \, \text{mm} = 0.012 \, \text{m} \), solve for \( A \) using the capacitance formula:\[1.0 \times 10^{-9} = \frac{8.85 \times 10^{-12} \times 3.0 \times A}{0.012}\]Solve for \( A \):\[ A \approx 0.452 \, \text{m}^2\].

Analyze the Effect of Using Teflon (Part c)

Teflon has a dielectric constant \( \varepsilon_r \approx 2.1 \), which is lower than that of posterboard. From the capacitance formula, a smaller \( \varepsilon_r \) will require a larger area \( A \) to maintain the same capacitance, assuming \( d \) remains constant at 12.0 mm. Therefore, she will need a larger area of Teflon than of posterboard.

Key concepts

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a non-conductive material called a dielectric. The function of this arrangement is to store electric charge and energy. Capacitors are fundamental components in electronic circuits, serving varied roles such as energy storage, signal filtering, and tuning circuits like the one your exercise involves.

In a parallel plate capacitor, the capacitance, which is the ability of a capacitor to store an electric charge, depends on several factors:

• The area (A) of the plates: Larger plates can store more charge.
• The separation (d) between the plates: A smaller separation allows more charge to be stored.
• The dielectric material: This affects how much electric field the capacitor can maintain.

The idea behind a parallel plate capacitor is fairly simple but crucial for understanding how many electronic circuits work. Calculating its capacitance involves applying the formula:\[C = \frac{\varepsilon_0 \varepsilon_r A}{d}\]where \(\varepsilon_0\) is the vacuum permittivity, \(\varepsilon_r\) is the dielectric constant of the material between the plates, \(A\) is the area of one plate, and \(d\) is the separation between the plates.

Dielectric Constant

The dielectric constant, denoted as \(\varepsilon_r\), is a dimensionless value that indicates how much a material can increase the capacitance of a capacitor compared to a vacuum. It's a critical factor in determining the effectiveness of a material as a dielectric in capacitors.

When a dielectric material is placed between the plates of a capacitor, it reduces the electric field, allowing the capacitor to store more charge at the same voltage. This happens because the dielectric material becomes polarized in the presence of an electric field, which in turn reduces the field between the plates.

• Each material has its unique dielectric constant. For example, common paper used in capacitors might have a dielectric constant of around 3.0.
• Higher dielectric constant means better insulating properties, allowing more capacitance with the same plate separation and plate area.

Understanding and selecting the right dielectric material is vital for designing capacitors to meet specific circuit needs and achieve the desired capacitance.

Vacuum Permittivity

Vacuum permittivity, denoted as \(\varepsilon_0\), is a fundamental physical constant that characterizes the ability of a vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12}\) F/m (farads per meter).

This constant is essential in the equation for a parallel plate capacitor's capacitance, providing a baseline measure for electric permeability in a vacuum. The dielectric constant of other materials is often defined relative to \(\varepsilon_0\), enabling comparisons of their insulating abilities. Here's why it matters:

• It's the reference point for measuring the dielectric constant of materials. \(\varepsilon_r\) for any material is essentially \(\varepsilon_x / \varepsilon_0\), where \(\varepsilon_x\) is the permittivity of the material.
• In the context of capacitors, increasing the dielectric constant above vacuum permittivity allows materials to store more charge.

Even though a vacuum itself isn't used as a practical dielectric, \(\varepsilon_0\) remains a key piece in understanding and calculating how capacitors work.

Teflon as Dielectric

Teflon is a common dielectric material with some unique properties that make it useful in certain types of capacitors, especially where stability and low loss are important. However, its dielectric constant, typically around 2.1, is lower compared to many other materials like paper or polystyrene. This affects how it would be used in designs.

To achieve a specific capacitance with Teflon, one must adjust other parameters such as the plate area or the separation between plates. Teflon is chosen for applications where particular attributes are needed:

• High chemical resistance and stability over a wide range of temperatures.
• Low dielectric loss, meaning it doesn't easily dissipate energy as heat.

In your exercise, when substituting paper or posterboard with Teflon, the lower dielectric constant means more plate area is required to achieve the same capacitance. This exemplifies the trade-offs involved in selecting materials for capacitors, balancing capacitance needs with practical and performance considerations.