/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 (II) To get an idea how big a fa... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) To get an idea how big a farad is, suppose you want to make a 1-F air- filled parallel-plate capacitor for a circuit you are building. To make it a reasonable size, suppose you limit the plate area to 1.0 cm\(^2\). What would the gap have to be between the plates? Is this practically achievable?

Short Answer

Expert verified
The gap would be \(8.85 \times 10^{-16}\) m, which is not practically achievable.

Step by step solution

01

Understand the Formula for Capacitance

The formula for the capacitance of a parallel-plate capacitor is \( C = \frac{\varepsilon_0 A}{d} \), where \( C \) is the capacitance, \( \varepsilon_0 \) is the permittivity of free space (approximately \(8.85 \times 10^{-12} \text{ F/m}\)), \( A \) is the area of one of the plates, and \( d \) is the distance between the plates. Let us use this formula to find \( d \).
02

Rearrange the Formula to Solve for the Gap \(d\)

Rearrange the formula to solve for \( d \): \( d = \frac{\varepsilon_0 A}{C} \). This expression allows us to determine the gap between the plates given the other parameters.
03

Insert Known Values into the Formula

Insert the given values into the formula: \( C = 1 \text{ F} \) and \( A = 1.0 \text{ cm}^2 = 1.0 \times 10^{-4} \text{ m}^2 \). The formula becomes: \[ d = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (1.0 \times 10^{-4} \text{ m}^2)}{1 \text{ F}} = 8.85 \times 10^{-16} \text{ m} \].
04

Evaluate the Practicality of the Size

The calculated gap \( d = 8.85 \times 10^{-16} \text{ m} \) is extremely small, far smaller than the dimensions typically manageable with current manufacturing technology. It is many times smaller than an atomic diameter and thus practically unachievable.

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Key Concepts

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

Parallel-Plate Capacitor
A parallel-plate capacitor is a fundamental component in electrical circuits, known for its ability to store electrical energy. It consists of two metallic plates that are placed parallel to each other, separated by an insulating material known as the dielectric.
The capacity of this structure to store charge is what gives it its title of a capacitor. The capacitance of a parallel-plate capacitor is determined by the formula \[C = \frac{\varepsilon A}{d}\]where:
  • \(C\) is the capacitance measured in farads (F),
  • \(\varepsilon\) is the permittivity of the dielectric material between the plates,
  • \(A\) is the area of one of the plates,
  • \(d\) is the distance between the plates.
The parallel-plate setup is widely used because it provides a uniform electric field which is ideal for storing charge, making it an essential building block in various electrical and electronic applications.
Permittivity of Free Space
Permittivity of free space, symbolized as \(\varepsilon_0\), is a physical constant that plays a crucial role in the field of electrostatics.
It is approximately equal to \(8.85 \times 10^{-12} \text{ F/m}\) (farads per meter).
This constant arises naturally when studying the relationships between electric fields and the charges that produce them in a vacuum setting. In the context of a parallel-plate capacitor, \(\varepsilon_0\) is used to calculate the capacitance when the dielectric medium is air or vacuum.
It's a measure of how much electric field can "penetrate" a material and affects the electric force's strength per unit charge.
Understanding permittivity is key because it dictates how much charge a capacitor can store for a given electric field and area. It’s essential for students to grasp how subtle changes in permittivity can drastically affect a capacitor's ability to store energy.
Capacitor Design
Designing a capacitor involves selecting appropriate materials and dimensions to achieve the desired capacitance. The practical design revolves around balancing size, material properties, and intended application.
  • Plate Area \(A\): Larger plate area increases capacitance, hence plates are usually made as large as possible within space constraints.
  • Gap Distance \(d\): Smaller gaps increase capacitance but pose manufacturing challenges as seen in the original exercise's impractical capacitor gap.
  • Dielectric Material: It’s pivotal to choose a material with suitable permittivity and breakdown voltage to maximize efficiency and safety.
These factors illustrate that while theoretically there’s a straightforward approach to calculate desired capacitance, in practice, designing a capacitor necessitates intricate considerations of achievable physical dimensions and material characteristics. Understanding the delicate balance in capacitor design equips engineers and scientists to tailor capacitors for specific functions in technology and electronics.

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Most popular questions from this chapter

Paper has a dielectric constant \(K =\) 3.7and a dielectric strength of 15 \(\times\) 10\(^6\)V/m. Suppose that a typical sheet of paper has a thickness of 0.11 mm. You make a "homemade" capacitor by placing a sheet of 21 \(\times\) 14cm paper between two aluminum foil sheets (Fig. 17-48) of the same size. (a) What is the capacitance \(C\) of your device? (\(b\)) About how much charge could you store on your capacitor before it would break down?

(II) Two identical +9.5\(\mu\)C point charges are initially 5.3 cm from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume they have identical masses of 1.0 mg.

(I) How much charge flows from each terminal of a 12.0-V battery when it is connected to a 5.00-\(\mu\)F capacitor?

(III) How much voltage must be used to accelerate a proton (radius 1.2 \(\times\) 10\(^{-15}\)m) so that it has sufficient energy to just "touch" a silicon nucleus? A silicon nucleus has a charge of +14\(e\) and its radius is about 3.6 \(\times\) 10\(^{-15}\)m. Assume the potential is that for point charges.

In an older television tube, electrons are accelerated by thousands of volts through a vacuum. If a television set were laid on its back, would electrons be able to move upward against the force of gravity? What potential difference, acting over a distance of 2.4 cm, would be needed to balance the downward force of gravity so that an electron would remain stationary? Assume that the electric field is uniform.

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