/*! 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 20 (a) An electron is to be acceler... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 20

(a) An electron is to be accelerated from 3.00 \(\times 10^6\) m\(/\)s to 8.00 \(\times 10^6\) m\(/\)s. Through what potential difference must the electron pass to accomplish this? (b) Through what potential difference must the electron pass if it is to be slowed from 8.00 \(\times 10^6\) m\(/\)s to a halt?

Short Answer

Expert verified
(a) 183.75 V, (b) 181.5 V.

Step by step solution

01

Understand the Energy Change Required

The energy change associated with an electron moving between two points at different speeds is captured by its change in kinetic energy. The kinetic energy \( K \) of an electron is given by \( K = \frac{1}{2}mv^2 \) where \( m \) is the mass of the electron and \( v \) is its velocity.
02

Calculate Initial and Final Kinetic Energies (Part a)

For the electron initially moving at 3.00 \( \times 10^6 \) m/s, the initial kinetic energy \( K_i \) is \( \frac{1}{2} \times 9.11 \times 10^{-31} \times (3.00 \times 10^6)^2 \).For the final velocity of 8.00 \( \times 10^6 \) m/s, the final kinetic energy \( K_f \) is \( \frac{1}{2} \times 9.11 \times 10^{-31} \times (8.00 \times 10^6)^2 \).
03

Compute Change in Kinetic Energy (Part a)

Calculate the change in kinetic energy \( \Delta K \) using the formula:\(\Delta K = K_f - K_i.\) Substitute the expressions from Step 2 to get \( \Delta K \).
04

Relate Kinetic Energy Change to Potential Difference (Part a)

The potential difference \( V \) needed to cause this change in kinetic energy is found using:\(\Delta K = e \Delta V,\)where \( e \) is the charge of the electron \( 1.60 \times 10^{-19} \) C. Solve for \( \Delta V \) using \( \Delta V = \frac{\Delta K}{e} \).
05

Initial and Final Kinetic Energies (Part b)

In part (b), calculate the initial kinetic energy \( K_i \) using the speed 8.00 \( \times 10^6 \) m/s.Since the final speed is zero, \( K_f = 0 \).
06

Compute Change in Kinetic Energy (Part b)

The change in kinetic energy \( \Delta K \) is given by:\(\Delta K = 0 - K_i.\)Calculate \( \Delta K \) using the \( K_i \) found in Step 5.
07

Relate Kinetic Energy Change to Potential Difference (Part b)

Again use the expression \( \Delta K = e \Delta V \) to find \( \Delta V \) in this scenario:\(\Delta V = \frac{\Delta K}{e}.\)Calculate \( \Delta V \) using the result for \( \Delta K \) from Step 6.

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.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For a moving body, like an electron, the kinetic energy (\( K \)) is given by the formula \( K = \frac{1}{2}mv^2 \).
This formula highlights two key factors that affect the kinetic energy: the mass (\( m \)) and the velocity (\( v \)) of the object.
For an electron, its mass is roughly \( 9.11 \times 10^{-31} \) kg.
  • When the velocity of the electron changes, its kinetic energy changes as well.
  • The greater the speed of the electron, the more kinetic energy it possesses.
  • The change in kinetic energy can be calculated by finding the difference between the final kinetic energy and the initial kinetic energy.
This change in kinetic energy is useful when determining how much energy must be added or taken away to change the electron's speed.
Potential Difference
Potential difference, often referred to as voltage (\( V \)), is the difference in electric potential energy per unit charge between two points in an electric field.
It's a crucial concept when discussing the movement of electrons and other charged particles through different mediums.
  • The potential difference is related to the energy needed to accelerate or decelerate a charge.
  • In the context of electron acceleration, the potential difference that an electron passes through is directly related to its change in kinetic energy.
  • The relationship between kinetic energy change and potential difference is expressed by the formula \( \Delta K = e \Delta V \), where \( e \) is the electron charge.
Thus, knowing the change in kinetic energy allows us to determine the potential difference necessary for that change.
Electron Charge
The electron charge (\( e \)) is a fundamental property of electrons that quantifies their electric charge. The value of the electron charge is \( 1.60 \times 10^{-19} \) coulombs, denoting it as a negatively charged particle.
  • This charge is used in calculations relating to force, energy, and potential difference when dealing with electrons.
  • It's a constant that allows us to relate the potential difference to the change in kinetic energy of an electron.
  • Knowing the electron charge is essential for any calculation in electromagnetism that involves moving electrons.
The presence of a consistent charge value makes it possible to solve various physics problems involving electronic motion and acceleration.
Electron Mass
Electron mass is another intrinsic property of electrons, with a value of approximately \( 9.11 \times 10^{-31} \) kilograms.
This tiny mass plays a significant role in physics calculations, particularly when using the kinetic energy formula given its dependence on velocity.
  • The low mass of electrons means they can achieve very high speeds with relatively low energy input.
  • When determining the kinetic energy of an electron, the mass is necessary to apply the formula \( K = \frac{1}{2}mv^2 \).
  • In addition to kinetic energy calculations, electron mass is important for understanding the behavior of electrons in fields and potentials.
Understanding the electron mass is vital for predicting how electrons will respond to forces and fields, essential for physics and electronic 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

In a certain region of space the electric potential is given by \(V = +Ax^2y - Bxy^2,\) where \(A =\) 5.00 \(V/m^3\) and \(B =\) 8.00 \(V/m^3\). Calculate the magnitude and direction of the electric field at the point in the region that has coordinates \(x =\) 2.00 m, \(y =\) 0.400 m, and \(z = 0\).

A point charge \(q_1\) is held stationary at the origin. A second charge \(q_2\) is placed at point a, and the electric potential energy of the pair of charges is \(+5.4 \times 10^{-8} \)J. When the second charge is moved to point \(b\), the electric force on the charge does \(-1.9 \times 10^{-8}\) J of work. What is the electric potential energy of the pair of charges when the second charge is at point \(b\)?

A ring of diameter 8.00 cm is fixed in place and carries a charge of \(+\)5.00 \(\mu\)C uniformly spread over its circumference. (a) How much work does it take to move a tiny \(+\)3.00-\(\mu\)C charged ball of mass 1.50 g from very far away to the center of the ring? (b) Is it necessary to take a path along the axis of the ring? Why? (c) If the ball is slightly displaced from the center of the ring, what will it do and what is the maximum speed it will reach?

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by $$V(x) = Cx^{{4}/{3}}$$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of \(C\). (b) Obtain a formula for the electric field between the electrodes as a function of \(x\). (c) Determine the force on an electron when the electron is halfway between the electrodes.

A uniformly charged, thin ring has radius 15.0 cm and total charge \(+\)24.0 nC. An electron is placed on the ring's axis a distance 30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.
See all solutions

Recommended explanations on Physics Textbooks

Oscillations

Read Explanation

Fields in Physics

Read Explanation

Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Energy Physics

Read Explanation

Famous Physicists

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.