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Chapter 25: Problem 74

You have two plates of copper, a sheet of mica (thickness = \(0.10 \mathrm{~mm}, \kappa=5.4\) ), a sheet of glass (thickness \(=2.0 \mathrm{~mm}, \kappa=7.0\) ), and a slab of paraffin (thickness \(=1.0 \mathrm{~cm}, \kappa=2.0\) ). To make a parallel-plate capacitor with the largest \(C\), which sheet should you place between the copper plates?

Short Answer

Expert verified
Mica should be placed between the copper plates for the largest capacitance.

Step by step solution

01

Understanding the Problem

We have a task to determine which material, when placed between two parallel copper plates, results in the maximum capacitance for a parallel-plate capacitor. We have three different dielectric materials available: mica, glass, and paraffin. The dielectric constant or permittivity (\(\kappa\)) and thickness of each material are provided.
02

Formula for Capacitance

The formula for the capacitance \(C\) of a parallel-plate capacitor with a dielectric is given by: \[ C = \frac{\varepsilon_0 \kappa A}{d} \]where \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area of the plates, \(\kappa\) is the dielectric constant, and \(d\) is the thickness of the dielectric material.
03

Calculate Capacitance for Each Material

We use the capacitance formula to calculate \(C\) for each material. We note that \(\kappa/d\) ratio determines the effective capacitance.- For mica: thickness \(d = 0.10\,\mathrm{mm} = 0.0001\,\mathrm{m}\), \(\kappa = 5.4\).- For glass: thickness \(d = 2.0\,\mathrm{mm} = 0.002\,\mathrm{m}\), \(\kappa = 7.0\).- For paraffin: thickness \(d = 1.0\,\mathrm{cm} = 0.01\,\mathrm{m}\), \(\kappa = 2.0\).Computing \(\kappa/d\) for each:- Mica: \(\frac{5.4}{0.0001} = 54000\)- Glass: \(\frac{7.0}{0.002} = 3500\)- Paraffin: \(\frac{2.0}{0.01} = 200\)
04

Determine Maximum Capacitance

Based on the \(\kappa/d\) calculation, the material yielding the highest capacitance value is mica, since it has the largest \(\kappa/d\) ratio of 54000. Therefore, placing mica between the plates results in the largest possible capacitance for the given choices.

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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, often denoted as \(\kappa\), is a measure of a material's ability to increase the capacitance of a capacitor compared to a vacuum. It is a dimensionless quantity that indicates how much electric charge a material can store compared to the charge that can be stored in a vacuum.
  • A higher dielectric constant means the material is better at storing charge.
  • It's a crucial factor in selecting materials for capacitors as it directly affects the capacitance鈥攈igher \(\kappa\) values lead to higher capacitance.
In exercises involving parallel-plate capacitors, manufacturers look for materials with high dielectric constants to maximize the capacitor's efficiency. The exercise provided shows how this constant plays a pivotal role in determining the most effective material for maximizing capacitance.
Parallel-plate capacitor
A parallel-plate capacitor is one of the simplest capacitor types. It consists of two conductive plates separated by a dielectric material. Capacitance is determined by several factors including the area of the plates, the distance between them, and the dielectric material used. The formula for capacitance \(C\) in a parallel-plate setup is given by: \[C = \frac{\varepsilon_0 \kappa A}{d}\]where \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area, \(\kappa\) is the dielectric constant, and \(d\) is the distance between plates.
  • Using different dielectric materials between the plates allows us to increase capacitance without changing plate size or distance.
  • The materials' dielectric properties directly influence how much charge the capacitor can hold.
Parallel-plate capacitors are widely used in electronics for their simplicity and effectiveness in circuit applications.
Dielectric Material
Dielectric materials are insulators placed between the plates of a capacitor to store electric energy. They become polarized in the presence of an electric field, which enhances the capacitor's ability to hold charge.
  • Common dielectric materials include mica, glass, and paraffin, each with different properties impacting their suitability for various applications.
  • In the given exercise, the different thicknesses and dielectric constants of materials such as mica, glass, and paraffin show their diverse capabilities in storing charge.
Dielectrics affect the capacitor's performance significantly, thus choosing the right material is crucial for optimizing capacitance as shown by their varying \(\kappa/d\) ratios.
Permittivity of Free Space
The permittivity of free space, denoted as \(\varepsilon_0\), is a fundamental physical constant. It quantifies how an electric field affects and is affected by a vacuum. It is the baseline measure against which all other dielectric materials are evaluated, influencing the calculated capacitance in the capacitor's formula.
  • The value of \(\varepsilon_0\) is approximately \(8.854187817 \times 10^{-12} \ F/m\).
  • It is crucial for understanding how electrical interactions occur in a vacuum and for calculating capacitance values in physics and engineering problems.
Understanding \(\varepsilon_0\) helps comprehend how different materials can alter electric field strengths within capacitors, as illustrated by its role in the formula used in the problem to assess capacitance changes with different materials.

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Most popular questions from this chapter

You have many \(2.0 \mu \mathrm{F}\) capacitors, each capable of withstanding \(200 \mathrm{~V}\) without undergoing electrical breakdown (in which they conduct charge instead of storing it). How would you assemble a combination having an equivalent capacitance of (a) \(0.40 \mu \mathrm{F}\) and (b) \(1.2 \mu \mathrm{F}\), each combination capable of withstanding \(1000 \mathrm{~V}\) ?

Assume that a stationary electron is a point of charge. What is the energy density \(u\) of its electric field at radial distances (a) \(r=\) \(1.00 \mathrm{~mm}\), (b) \(r=1.00 \mu \mathrm{m},(\mathrm{c}) r=1.00 \mathrm{~nm}\), and \((\mathrm{d}) r=1.00 \mathrm{pm} ?\) (e) What is \(u\) in the limit as \(r \rightarrow 0\) ?

Two parallel-plate capacitors, \(6.0 \mu \mathrm{F}\) each, are connected in series to a \(10 \mathrm{~V}\) battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the total charge stored on the capacitors (the charge on the positive plate of one capacitor plus the charge on the positive plate of the other capacitor)?

A \(2.0 \mu\) F capacitor and a \(4.0 \mu \mathrm{F}\) capacitor are connected in parallel across a \(300 \mathrm{~V}\) potential difference. Calculate the total energy stored in the capacitors.

A certain substance has a dielectric constant of \(2.8\) and a dielectric strength of \(18 \mathrm{MV} / \mathrm{m}\). If it is used as the dielectric material in a parallel-plate capacitor, what minimum area should the plates of the capacitor have to obtain a capacitance of \(7.0 \times 10^{-2} \mu \mathrm{F}\) and to ensure that the capacitor will be able to withstand a potential difference of \(4.0 \mathrm{kV} ?\)
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