/*! 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 62 Two capacitors have the same pla... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two capacitors have the same plate separation, but one has square plates and the other has circular plates. The square plates are a length \(L\) on each side, and the diameter of the circular plate is \(L\). The capacitors have the same capacitance because they contain different dielectric materials. The dielectric constant of the material between the square plates has a value of \(\kappa_{\text {square }}=3.00 .\) What is the dielectric constant \(\kappa_{\text {circle }}\) of the material between the circular plates?

Short Answer

Expert verified
The dielectric constant \(\kappa_{\text{circle}}\) is approximately 3.82.

Step by step solution

01

Understand the Problem

We need to find the dielectric constant \(\kappa_{\text{circle}}\) of the circular capacitor. Both capacitors have the same separation and equivalent capacitance but different shapes and materials, indicated by different dielectric constants.
02

Calculate Area of Plates

For the square capacitor, the area \(A_{\text{square}}\) is given by \(L^2\). For the circular capacitor, the area \(A_{\text{circle}}\) is given by \(\pi \left(\frac{L}{2}\right)^2 = \frac{\pi L^2}{4}\).
03

Use Capacitance Formula

The formula for capacitance is \(C = \frac{\kappa \varepsilon_0 A}{d}\), where \(\kappa\) is the dielectric constant, \(\varepsilon_0\) is the permittivity of free space, \(A\) is the area, and \(d\) is the separation. Both capacitances are the same, so \(C_{\text{square}} = C_{\text{circle}}\).
04

Set up the Equation for Capacitance

Equating the capacitance expressions for both:\[ \frac{\kappa_{\text{square}} \varepsilon_0 L^2}{d} = \frac{\kappa_{\text{circle}} \varepsilon_0 \frac{\pi L^2}{4}}{d} \]
05

Solve for \(\kappa_{\text{circle}}\)

Cancel \(\varepsilon_0\), \(d\) and \(L^2\) from both sides and solve for \(\kappa_{\text{circle}}\):\[ \kappa_{\text{square}} = \kappa_{\text{circle}} \left(\frac{\pi}{4}\right) \]\[ \kappa_{\text{circle}} = \frac{4}{\pi} \kappa_{\text{square}} \]
06

Compute \(\kappa_{\text{circle}}\)

Substitute \(\kappa_{\text{square}} = 3.00\) into the equation:\[ \kappa_{\text{circle}} = \frac{4}{\pi} \times 3.00 \approx 3.82 \]

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 91Ó°ÊÓ!

Key Concepts

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

Capacitance
Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is an essential concept in electronics and physics. The standard unit for capacitance is the farad (F). Capacitance can depend on various factors such as the size and shape of the plates, the separation between them, and the nature of the insulating material, known as the dielectric.

The formula for capacitance is given by \[C = \frac{Q}{V}\]where \(C\) is the capacitance, \(Q\) is the charge stored, and \(V\) is the voltage applied. In capacitors, using a dielectric can enhance the capacitance. The dielectric constant \(\kappa\) of a material is a factor by which the capacitance increases compared to if the material was not present.

In practical applications, capacitors can come in various shapes and configurations, leading to differences in their calculated capacitance. Despite differences in their geometry, capacitors with the same capacitance can store the same amount of electric energy under identical voltage conditions.
Dielectric Materials
Dielectric materials are insulating substances used in capacitors to separate their conductive plates while allowing them to store electric charge efficiently. The effectiveness of a dielectric material in a capacitor is determined by its dielectric constant, represented by \(\kappa\). This constant measures the ability of a material to increase the capacitance compared to the capacitance of the same capacitor with only a vacuum between the plates.

The presence of a dielectric material influences the electric field and reduces the voltage for a given amount of charge, thus increasing the overall capacitance. Different materials have different dielectric constants, influencing their suitability for specific applications.
  • A dielectric with a high \(\kappa\) can lead to increased capacitance, making it highly valuable in circuits where energy storage is crucial.
  • The material should also endure strong electric fields without breaking down, ensuring reliable performance over time.
Choosing the right dielectric material is crucial for optimizing the performance and efficiency of capacitors in electrical circuits.
Permittivity of Free Space
Permittivity of free space, often denoted as \(\varepsilon_0\), is a fundamental physical constant that characterizes the ability of a vacuum to permit electric field lines. It is an essential factor in the study of electromagnetism and affects calculations involving electric forces and fields.

The value of \(\varepsilon_0\) is approximately \[8.85 \times 10^{-12} \text{ C}^2/\text{(N}\cdot\text{m}^2)\]This constant forms part of the equations describing capacitance, especially when the dielectric between capacitor plates is a vacuum. In such a situation, the capacitance \(C\) of the capacitor is calculated using the formula:\[C = \frac{\varepsilon_0 A}{d}\]where \(A\) is the area of the plates and \(d\) is the separation between the plates.

For cases involving a dielectric, the formula is modified to:\[C = \frac{\kappa \varepsilon_0 A}{d}\]This highlights how the dielectric constant \(\kappa\) scales the permittivity of free space to increase the capacitance. Understanding \(\varepsilon_0\) is crucial for accurately modeling and predicting the behavior of capacitors and other electrostatic systems in physics and engineering.

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 positive charge \(q_{1}\) is located \(3.00 \mathrm{m}\) to the left of a negative charge \(q_{2} .\) The charges have different magnitudes. On the line through the charges, the net electric field is zero at a spot \(1.00 \mathrm{m}\) to the right of the negative charge. On this line there are also two spots where the total electric potential is zero. Locate these two spots relative to the negative charge.

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of \(1.2 \mathrm{cm},\) and the electric field within the capacitor has a magnitude of 2.1 \(\times 10^{6} \mathrm{V} / \mathrm{m}\). What is the kinetic energy of the electron just as it reaches the positive plate?

An axon is the relatively long tail-like part of a neuron. or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that the membrane acts like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{m}^{2},\) what is its capacitance?

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of \(5.29 \times 10^{-11} \mathrm{m} .\) What is \(\mathrm{EPE}_{\text {final }}-\mathrm{EPE}_{\text {initial }},\) which is the change in the electric potential energy?

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length \(0.480 \mathrm{m}\). A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right)\), with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

See all solutions

Recommended explanations on Physics Textbooks

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.