/*! 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 57 For a Teflon \(^{\mathrm{TM}}\)-... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 57

For a Teflon \(^{\mathrm{TM}}\)-filled, parallel-plate capacitor, the area of the plate is \(50.0 \mathrm{cm}^{2}\) and the spacing between the plates is \(0.50 \mathrm{mm}\). If the capacitor is connected to a 200 -V battery, find (a) the free charge on the capacitor plates, (b) the electrical field in the dielectric, and (c) the induced charge on the dielectric surfaces.

Short Answer

Expert verified
The free charge on the capacitor plates is \(Q = 1.87\times10^{-8}\,\mathrm{C}\). The electric field in the dielectric is \(E = 4.0\times10^5\,\mathrm{V/m}\). The induced charge on the dielectric surfaces is \(Q_{induced} = 8.9\times10^{-9}\,\mathrm{C}\).

Step by step solution

01

Calculate the capacitance of the capacitor

Using the formula for the capacitance of a parallel-plate capacitor, C: \(C = \frac{\epsilon A}{d}\), where \(\epsilon\) is the permittivity of the dielectric material, A is the area of the plates and d is the distance between the plates. For Teflon, the relative permittivity (\(\epsilon_r\)) is 2.1. The permittivity of a dielectric material is given by \(\epsilon = \epsilon_0\epsilon_r\), where \(\epsilon_0\) is the vacuum permittivity, with a value of \(8.85\times10^{-12} \mathrm{F/m}\). First, convert the given area and distance to meters: Area \(A = 50.0 \mathrm{cm^2} = 5.0\times10^{-3}\mathrm{m^2}\) Distance \(d = 0.50 \mathrm{mm} = 5.0\times10^{-4}\mathrm{m}\) Now, let's calculate the capacitance: \(C = \frac{\epsilon A}{d} = \frac{(8.85\times10^{-12} \mathrm{F/m})(2.1)(5.0\times10^{-3}\mathrm{m^2})}{5.0\times10^{-4}\mathrm{m}}\)
02

Find the free charge on the capacitor plates

The charge on the capacitor is given by the formula, Q: \(Q = CV\) where Q is the free charge on the plates, C is the capacitance and V is the voltage. With the capacitance from step 1 and the given voltage of 200V, we can now calculate the charge on the plates: \(Q = (C)(200\mathrm{V})\)
03

Calculate the electric field in the dielectric

The electric field in the dielectric can be calculated using the formula, E: \(E = \frac{V}{d}\) where E is the electric field, V is the voltage and d is the distance between the plates. With the given voltage of 200V and distance between the plates from step 1, we can now calculate the electric field: \(E = \frac{200\mathrm{V}}{5.0\times10^{-4}\mathrm{m}}\)
04

Find the induced charge on the dielectric surfaces

The induced charge on the dielectric surfaces can be calculated using the formula, Q_induced: \(Q_{induced} = Q(1 - \frac{1}{\epsilon_r})\) where \(Q_{induced}\) is the induced charge on the dielectric surfaces, Q is the charge on the plates from step 2, and \(\epsilon_r\) is the relative permittivity of Teflon, which is 2.1. Now, let's calculate the induced charge: \(Q_{induced} = Q(1 - \frac{1}{2.1})\) By solving each step and performing the calculations, we will find the free charge on the capacitor plates, the electric field in the dielectric, and the induced charge on the dielectric surfaces.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Capacitance
Capacitance is a fundamental concept in electrostatics that describes a capacitor's ability to store electric charge. A capacitor typically consists of two conductive plates separated by an insulating material, known as the dielectric.
A key formula to remember here is the capacitance formula for a parallel-plate capacitor:
  • \( C = \frac{\epsilon A}{d} \)
Here, \( \epsilon \) represents the permittivity of the dielectric material, \( A \) is the area of the plates, and \( d \) is the distance between them.
For a Teflon-filled capacitor, this equation helps us calculate its capacitance by considering both the physical dimensions and the material properties involved. The dielectric's permittivity, including both the vacuum permittivity (\( \epsilon_0 \)) and the relative permittivity (\( \epsilon_r \)) of Teflon, plays a crucial role.
In this exercise, adjusting the given dimensions into meters first ensured accuracy in the capacitance calculation, providing us with the capacity to store charge, further used to determine other electrical properties.
Electric Field
The electric field is a concept that describes how a voltage difference between two points can create a force that acts on a charged particle. It鈥檚 measured in volts per meter (V/m) and signifies the strength of the force available in the space around charged objects.
For a capacitor, this field exists between the plates. It can be calculated using:
  • \( E = \frac{V}{d} \)
Where \( E \) is the electric field, \( V \) is the voltage applied across the plates, and \( d \) is the separation distance.
The narrower the gap between the plates for a set voltage, the stronger the electric field. Teflon鈥檚 use as a dielectric doesn鈥檛 change the field's basic calculation but does influence the overall behavior of the field within the capacitor.
Dielectric Materials
Dielectric materials are insulating substances inserted between the plates of a capacitor. They are vital because they affect the capacitor's ability to store charge.
Dielectrics increase capacitance because they allow the capacitor to hold more charge at the same voltage compared to a vacuum or air. They achieve this by reducing the electric field's strength due to their polarizing effect.
  • Their impact is quantified through the relative permittivity (\( \epsilon_r \)).
For instance, Teflon, deployed in the problem here, has a \( \epsilon_r \) of 2.1, doubling the capacitance more than with air alone. This added capacity is owing to the alignment of the material鈥檚 internal charge structure in response to the applied electric field, storing energy by balancing out charge disparity across the dielectric surface.
Permittivity
Permittivity is a measure of how easily a material can polarize in response to an electric field, contributing to the material's ability to store electrical energy. It's denoted by \( \epsilon \) in equations and has units of farads per meter (F/m).
Two components describe permittivity:
  • Vacuum permittivity (\( \epsilon_0 \)), a constant foundational value.
  • Relative permittivity (\( \epsilon_r \)), which varies for different materials.
For calculations, the absolute permittivity \( \epsilon \) is the product of these two. So, the permittivity of Teflon is \( \epsilon = \epsilon_0\epsilon_r \).
This defines how effective the material will be in supporting the electric field and storing energy. It affects how much charge can accumulate on a capacitor's plates under a given voltage, driving the outcomes derived, such as capacitance and induced charge discussed in the initial solution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel-plate capacitor has charge of magnitude \(9.00 \mu \mathrm{F}\) on each plate and capacitance \(3.00 \mu \mathrm{C}\) when there is air between the plates. The plates are separated by \(2.00 \mathrm{mm} .\) With the charge on the plates kept constant, a dielectric with \(\kappa=5\) is inserted between the plates, completely filling the volume between the plates. (a) What is the potential difference between the plates of the capacitor, before and after the dielectric has been inserted? (b) What is the electrical field at the point midway between the plates before and after the dielectric is inserted?

What is the capacitance of a large Van de Graaff generator's terminal, given that it stores \(8.00 \mathrm{mC}\) of charge at a voltage of \(12.0 \mathrm{MV}\) ?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit the flash lasts for \(1 / 675\) fraction of a second with an average light power output of \(270 \mathrm{kW}\). (a) If the conversion of electrical energy to light is \(95 \%\) efficient (because the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 V when the stored energy equals the value stored in part (a). What is the capacitance?

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of \(1.60 \times 10^{7} \mathrm{V} / \mathrm{m} .\) The capacitor has to have a capacitance of \(1.25 \mathrm{nF}\) and must be able to withstand a maximum potential difference \(5.5 \mathrm{kV}\). What is the minimum area the plates of the capacitor may have?

An anxious physicist worries that the two metal shelves of a wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. (a) What is the capacitance of the empty shelves if they have area \(1.00 \times 10^{2} \mathrm{m}^{2}\) and are \(0.200 \mathrm{m}\) apart? (b) What is the voltage between them if opposite charges of magnitude \(2.00 \mathrm{nC}\) are placed on them? (c) To show that this voltage poses a small hazard, calculate the energy stored. (d) The actual shelves have an area 100 times smaller than these hypothetical shelves. Are his fears justified?
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Kinematics Physics

Read Explanation

Magnetism

Read Explanation

Electricity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.