/*! 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 65 In this problem, you can show fr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 65

In this problem, you can show from Coulomb's law that the constant of proportionality in Gauss's law must be \(1 / \epsilon_{0} .\) Imagine a sphere with its center at a point charge q. (a) Write an expression for the electric flux in terms of the field strength \(E\) and the radius \(r\) of the sphere. [Hint: The field strength \(E\) is the same everywhere on the sphere and the field lines cross the sphere perpendicular to its surface.] (b) Use Gauss's law in the form \(\Phi_{\mathrm{E}}=c q(\text { where } c\) is the constant of proportionality) and the electric field strength given by Coulomb's law to show that $c=1 / \epsilon_{0}$

Short Answer

Expert verified
Answer: The constant of proportionality in Gauss's law derived using Coulomb's law is \(\frac{1}{\epsilon_{0}}\), where \(\epsilon_{0}\) is the vacuum permittivity.

Step by step solution

01

Find the expression for the electric flux

To find the electric flux, we must remember that it is given by the integral of the electric field over a closed surface, in this case, a sphere. Since the electric field strength, \(E\), is the same everywhere on the sphere and is perpendicular to the surface, we can write the electric flux as: \(\Phi_{E} = \int_{S} \vec{E} \cdot d\vec{A}\) As \(\vec{E}\) and \(d\vec{A}\) are both in the same direction, the dot product simplifies to \(|\vec{E}| |d\vec{A}|\). Now, we can write \(\Phi_{E}\) as: \(\Phi_{E} = E\int_{S} dA\)
02

Calculate the surface area of the sphere

The surface area of a sphere with radius r is given by the formula: \(A = 4\pi r^2\). Since E is constant, we can remove it from the integral and we get: \(\Phi_{E} = E\int_{S} dA = E \cdot (4\pi r^2)\)
03

Apply Gauss's law

According to Gauss's law, the electric flux through a closed surface is equal to the constant of proportionality (c) times the total charge (q) enclosed by the surface: \(\Phi_{E} = cq\)
04

Relate the electric field strength with Coulomb's law

Coulomb's law states that the electric field strength is given by: \(E = \frac{k q}{r^2}\), where \(k = \frac{1}{4\pi \epsilon_{0}}\)
05

Find the constant of proportionality

Now, using the expression for electric flux found in step 2 and equating it with Gauss's law, we have: \(E \cdot (4\pi r^2) = cq\) Substituting the expression for E from Coulomb's law, we get: \(\frac{k q}{r^2} \cdot (4\pi r^2) = cq\) Simplifying the equation and substituting \(k = \frac{1}{4\pi \epsilon_{0}}\): \(q\cdot \frac{1}{\epsilon_{0}} = cq\) We can now see that the constant of proportionality, \(c\), equals to \(\frac{1}{\epsilon_{0}}\). This confirms that \(c = \frac{1}{\epsilon_{0}}\) and thus, Gauss's law can be written as: \(\Phi_{E} = \frac{1}{\epsilon_{0}}q\)

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!

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

Two small metal spheres are \(25.0 \mathrm{cm}\) apart. The spheres have equal amounts of negative charge and repel each other with a force of \(0.036 \mathrm{N}\). What is the charge on each sphere?
Positive point charges \(q\) and \(2 q\) are located at \(x=0\) and \(x=3 d,\) respectively. What is the electric field at \(x=d\) (point \(P\) )?
Suppose a charge \(q\) is placed at point \(x=0, y=0 .\) A second charge \(q\) is placed at point \(x=8.0 \mathrm{m}, y=0 .\) What charge must be placed at the point \(x=4.0 \mathrm{m}, y=0\) in order that the field at the point $x=4.0 \mathrm{m}, y=3.0 \mathrm{m} \mathrm{be}$ zero?
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
A total charge of \(7.50 \times 10^{-6} \mathrm{C}\) is distributed on two different small metal spheres. When the spheres are \(6.00 \mathrm{cm}\) apart, they each feel a repulsive force of \(20.0 \mathrm{N} .\) How much charge is on each sphere?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Turning Points in Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fluids

Read Explanation

Electrostatics

Read Explanation

Physical Quantities And Units

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.