/*! 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 55 The space between two concentric... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The space between two concentric conducting spherical shells of radii \(b=1.70 \mathrm{~cm}\) and \(a=1.20 \mathrm{~cm}\) is filled with a substance of dielectric constant \(\kappa=23.5 . \mathrm{A}\) potential difference \(V=73.0 \mathrm{~V}\) is applied across the inner and outer shells. Determine (a) the capacitance of the device, (b) the free charge \(q\) on the inner shell, and (c) the charge \(q^{\prime}\) induced along the surface of the inner shell.

Short Answer

Expert verified
The capacitance is \(1.02 \times 10^{-10} \text{ F}\), the free charge is \(7.446 \times 10^{-9} \text{ C}\), and the induced charge is \(7.129 \times 10^{-9} \text{ C}\).

Step by step solution

01

Understand the Problem

We are given two concentric spherical shells with radii \(b = 1.70 \text{ cm}\) and \(a = 1.20 \text{ cm}\), separated by a dielectric material with dielectric constant \(\kappa = 23.5\). A potential difference \(V = 73.0 \text{ V}\) is applied across these shells. We need to find the capacitance, the free charge on the inner shell, and the induced charge on the inner shell's surface.
02

Calculate Capacitance

The capacitance of a spherical shell capacitor with a dielectric is given by:\[C = 4\pi\varepsilon_0\kappa\left(\frac{ab}{b-a}\right)\]where \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}\). Substituting the values:\[a = 0.012 \text{ m}, \quad b = 0.017 \text{ m}, \quad \kappa = 23.5\]\[C = 4 \pi (8.85 \times 10^{-12})(23.5)\left(\frac{0.012 \times 0.017}{0.017 - 0.012}\right)\]Calculate the expression inside the brackets first:\[\frac{0.012 \times 0.017}{0.017 - 0.012} = \frac{0.000204}{0.005} = 0.0408\]Now calculate the capacitance:\[C = 4 \times 3.1416 \times 8.85 \times 10^{-12} \times 23.5 \times 0.0408\]\[C \approx 1.02 \times 10^{-10} \text{ F}\]
03

Calculate Free Charge

Free charge \(q\) can be calculated using the formula:\[q = CV\]Substitute the values:\[q = (1.02 \times 10^{-10} \text{ F})(73 \text{ V})\]\[q = 7.446 \times 10^{-9} \text{ C}\]
04

Calculate Induced Charge

The induced charge \(q'\) on the inner shell's surface considering the dielectric constant is:\[q' = q(1 - \frac{1}{\kappa})\]Substitute the values:\[q' = 7.446 \times 10^{-9} \text{ C} \left(1 - \frac{1}{23.5}\right) \]\[q' = 7.446 \times 10^{-9} \text{ C} \left(1 - 0.0426\right)\]\[q' \approx 7.129 \times 10^{-9} \text{ C}\]

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 refers to the ability of a system to store electrical charge. In this exercise, we have two concentric spherical shells acting as a capacitor with a dielectric medium between them. The capacitance (\( C \)) of such a configuration depends on several factors:
  • The radii of the inner (\( a \)) and outer (\( b \)) spheres.
  • The dielectric constant (\( \kappa \)) of the material between the spheres.
  • Permittivity of free space (\( \varepsilon_0 \)), a constant value approximately equal to \( 8.85 \times 10^{-12} \, \text{F/m} \).
The formula to calculate capacitance is \[C = 4 \pi \varepsilon_0 \kappa \left(\frac{ab}{b-a}\right)\]where \( a \) and \( b \) are the radii converted into meters. By understanding this relationship, you can see how different factors like size and material properties affect the capacitance of a system.
Dielectric Constant
The dielectric constant, symbolized as \( \kappa \), is a measure of a substance's ability to insulate charges from each other. It is a crucial factor in determining the capacitance of a capacitor when a dielectric material separates the conductors. A higher \( \kappa \) implies that the material can store more electrical energy for the same electric field.

In the context of the concentric spheres, the dielectric constant increases the effective capacitance compared to having just air in-between. The greater the dielectric constant, the less electric field required to store a particular amount of charge, showcasing its insulating property. In this exercise, the dielectric constant is given as \( 23.5 \), indicating a strong insulating capability which contributes to a higher capacitance than if the space were filled with air.
Concentric Spheres
Concentric spheres refer to a configuration where one sphere is placed inside another with both sharing the same center. This setup is a common way to design spherical capacitors in electrostatic studies.
  • The inner sphere serves as one conductor, and the outer sphere serves as another, holding potential differences between them.
  • The space between the spheres can be vacant or filled with a dielectric material.
For this exercise, the radii of the two spheres are given as \( b = 1.70 \, \text{cm} \) and \( a = 1.20 \, \text{cm} \). Understanding this formation helps visualize how electric fields interact within the spherical geometry, and how capacitance is calculated based on the radii and the intervening dielectric.
Induced Charge
Induced charge refers to the redistribution of free charge on the surface of a conductor in response to a nearby electric field. This is a fundamental concept in electrostatics, playing a pivotal role when a dielectric is involved.

In this problem, the inner shell of the spherical capacitor experiences an induced charge due to the presence of a dielectric. The induced charge \( q' \) is calculated by the formula:\[q' = q \left( 1 - \frac{1}{\kappa} \right)\]where \( q \) is the free charge on the shell, and \( \kappa \) is the dielectric constant.
  • The induced charge impacting the surface influences the overall electrostatic behavior of the system.
  • This calculation elucidates how not only free charges contribute to the electric field, but also how the induced charges due to polarization of the dielectric material align themselves.
Comprehending induced charges helps in understanding the efficiency and charge storage capability in capacitors enhanced by dielectric materials.

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

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?

As a safety engineer, you must evaluate the practice of storing flammable conducting liquids in nonconducting containers. The company supplying a certain liquid has been using a squat, cylindrical plastic container of radius \(r=0.20 \mathrm{~m}\) and filling it to height \(h=10 \mathrm{~cm}\), which is not the container's full interior height (Fig. \(25-44\) ). Your investigation reveals that during handling at the company, the exterior surface of the container commonly acquires a negative charge density of magnitude \(2.0 \mu \mathrm{C} / \mathrm{m}^{2}\) (approximately uniform). Because the liquid is a conducting material, the charge on the container induces charge separation within the liquid. (a) How much negative charge is induced in the center of the liquid's bulk? (b) Assume the capacitance of the central portion of the liquid relative to ground is \(35 \mathrm{pF}\). What is the potential energy associated with the negative charge in that effective capacitor? (c) If a spark occurs between the ground and the central portion of the liquid (through the venting port), the potential energy can be fed into the spark. The minimum spark energy needed to ignite the liquid is \(10 \mathrm{~mJ} .\) In this situation, can a spark ignite the liquid?

A charged isolated metal sphere of diameter \(10 \mathrm{~cm}\) has a potential of \(8000 \mathrm{~V}\) relative to \(V=0\) at infinity. Calculate the energy density in the electric field near the surface of the sphere.

What capacitance is required to store an energy of \(10 \mathrm{~kW} \cdot \mathrm{h}\) at a potential difference of \(1000 \mathrm{~V} ?\)

A parallel-plate capacitor has a capacitance of \(100 \mathrm{pF}\), a plate area of \(100 \mathrm{~cm}^{2}\), and a mica dielectric \((\kappa=5.4)\) completely filling the space between the plates. At \(50 \mathrm{~V}\) potential difference, calculate (a) the electric field magnitude \(E\) in the mica, (b) the magnitude of the free charge on the plates, and (c) the magnitude of the induced surface charge on the mica.

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.