/*! 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 30 Charge is uniformly distributed ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 30

Charge is uniformly distributed around a ring of radius \(R=4.60 \mathrm{~cm}\), and the resulting electric field magnitude \(E\) is measured along the ring's central axis (perpendicular to the plane of the ring). At what distance from the ring's center is \(E\) maximum?

Short Answer

Expert verified
The maximum electric field along the axis occurs at \(z = \frac{R}{\sqrt{2}}\). Here, \(z = \frac{4.60 \, \text{cm}}{\sqrt{2}}\).

Step by step solution

01

Understanding the problem

We need to find the distance from the center of a charged ring at which the electric field magnitude is maximum. The ring has a given radius, and the electric field is measured along its central axis.
02

Setting up the Electric Field Equation

The electric field at a point on the central axis of a ring with charge can be described by the formula: \[ E(z) = \frac{kQz}{(z^2 + R^2)^{3/2}} \]where:- \(E(z)\) is the electric field at distance \(z\) along the axis,- \(k\) is Coulomb's constant,- \(Q\) is the total charge,- \(R\) is the radius of the ring, and- \(z\) is the distance from the center along the axis.
03

Differentiating the Electric Field Equation

To find the maximum electric field, we take the derivative of \(E(z)\) with respect to \(z\) and set it to zero:\[ \frac{dE}{dz} = 0 \]First, let's find the derivative:\[ \frac{dE}{dz} = \frac{d}{dz}\left(\frac{kQz}{(z^2 + R^2)^{3/2}}\right) \]
04

Applying the Quotient Rule

Using the quotient rule, \( \frac{d}{dz}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \), where:- \(f = kQz, \quad f' = kQ\)- \(g = (z^2 + R^2)^{3/2}, \quad g' = \frac{3z(z^2 + R^2)^{1/2}}{(z^2 + R^2)}\)Therefore, the derivative becomes:\[ \frac{dE}{dz} = \frac{kQ \times (z^2 + R^2)^{3/2} - kQz \times \frac{3z(z^2 + R^2)^{1/2}}{(z^2 + R^2)}}{(z^2 + R^2)^3} \]

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.

Charged Ring
A **charged ring** is essentially a circular loop that has an even distribution of electric charge along its circumference. This is a fundamental concept in electrostatics, where symmetrical distribution allows for simpler calculations.

The main characteristic of a charged ring is:
  • It has a constant radius, in this case, given as 4.60 cm.
  • The charge is uniformly spread out, meaning each segment of the ring contributes equally to the electric field.
This uniformity simplifies the calculation of electric fields along the ring's axis using symmetry arguments. The charged particles create an electric field because they exert a force on other charges, which becomes an essential aspect in determining the electric field at any point along the axis of the ring.
Coulomb's Law
**Coulomb's Law** is a cornerstone in understanding electric forces. It describes how the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It is encapsulated in the formula:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

where:
  • \( F \) is the magnitude of the force between the charges,
  • \( q_1 \) and \( q_2 \) are the amounts of the two charges,
  • \( r \) is the distance between the centers of the two charges, and
  • \( k \) is Coulomb's constant \((8.99 \times 10^9 \, ext{Nm}^2/ ext{C}^2)\).
Coulomb's Law is used here to derive the electric field created by a charged ring, simplifying the computations with symmetry by considering its effect at various points on its central axis.
Derivatives in Physics
In physics, **derivatives** are used to understand how a quantity changes with respect to another variable. For this exercise, we need to determine at what distance along the axis of a charged ring the electric field is at its maximum.

The electric field is described by a function of distance, \( E(z) \). To find maximum values of such a function:
  • The first derivative of the function \( \frac{dE}{dz} \) is calculated.
  • This derivative is set equal to zero \( \frac{dE}{dz} = 0 \) to find critical points.
Once these points are found, we can determine if they correspond to maxima, minima, or points of inflection using additional tests. This derivative process helps us determine the specific position along the axis where the electric field reaches its maximum value.
Electric Field Maximum
The concept of an **electric field maximum** involves finding the point along a line (such as the central axis of a charged ring) where the field strength is at its highest.

To find this maximum for a charged ring, you:
  • Set up the formula for the electric field along the axis, \( E(z) = \frac{kQz}{(z^2 + R^2)^{3/2}} \).
  • Find the derivative \( \frac{dE}{dz} \), using calculus rules such as the quotient rule, to handle the complex fraction appropriately.
  • Solve \( \frac{dE}{dz} = 0 \) for \( z \) to find critical points where the electric field changes.
  • Verify whether this point is a maximum using further analysis.
Through these steps, students learn how to determine where the electric field is strongest, providing not only the mathematical understanding but also a visualization of how charge distributions interact with space.

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 charged particles are attached to an \(x\) axis: Particle 1 of charge \(-4.00 \times 10^{-7} \mathrm{C}\) is at position \(x=-5.00 \mathrm{~cm}\) and particle 2 of charge \(+4.00 \times 10^{-7} \mathrm{C}\) is at position \(x=10.0 \mathrm{~cm}\). Midway between the particles, what is their net electric field in unit-vector notation?

5 Assume that a honeybee is a sphere of diameter \(1.000 \mathrm{~cm}\) with a charge of \(+60.0 \mathrm{pC}\) uniformly spread over its surface. Assume also that a spherical pollen grain of diameter \(40.0 \mu \mathrm{m}\) is electrically held on the surface of the bee because the bee's charge induces a charge of \(-1.00 \mathrm{pC}\) on the near side of the grain and a charge of \(+1.00 \mathrm{pC}\) on the far side. (a) What is the magnitude of the net electrostatic force on the grain due to the bee? Next, assume that the bee brings the grain to a distance of \(1.000 \mathrm{~mm}\) from the tip of a flower's stigma and that the tip is a particle of charge \(-60.0 \mathrm{pC}\). (b) What is the magnitude of the net electrostatic force on the grain due to the stigma? (c) Does the grain remain on the bee or move to the stigma?

Figure 22-36 shows two concentric rings, of radii \(R\) and \(R^{\prime}=4.00 R\), that lie on the same plane. Point \(P\) lies on the central z axis, at distance \(D=2.00 R\) from the center of the rings. The smaller ring has uniformly distributed charge \(+Q .\) In terms of \(Q\), what is the uniformly distributed charge on the larger ring if the net electric field at \(P\) is zero?

An electron enters a region of uniform electric field with an initial velocity of \(30 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5 \mathrm{~ns}\) after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

An electron is released from rest in a uniform electric field of magnitude \(2.00 \times 10^{4} \mathrm{~N} / \mathrm{C}\). (a) Calculate the acceleration of the electron. (Ignore gravitation.) (b) How much time does the electron take to reach \(1.00 \%\) of the speed of light?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Quantum Physics

Read Explanation

Electricity

Read Explanation

Measurements

Read Explanation

Fields in Physics

Read Explanation

Astrophysics

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.