/*! 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 (a) An electron is moving east i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 30

(a) An electron is moving east in a uniform electric field of \(1.50 \mathrm{~N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{~m}\) east of point \(A ?\) (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{~m} / \mathrm{s},\) east. What is the speed of the proton at point \(B ?\)

Short Answer

Expert verified
The final speed of the electron is approximately \(4.47 \times 10^5 m/s\), and the final speed of the proton is approximately \(1.92 \times 10^4 m/s\).

Step by step solution

01

Determining the force on the electron

First, remember that the force on a charged particle in an electric field is given by \( F = qE \), where \( q \) is the charge of the particle and \( E \) is the force of the electric field strength. For an electron, \( q = -1.6 \times 10^{-19} C \) and \( E = 1.5 N/C \), so the force is \( F = -1.6 \times 10^{-19} C \times 1.5 N/C = -2.4 \times 10^{-19} N. \) The negative sign indicates that the force is acting in a direction opposite to the motion of the electron.
02

Determining the work done on the electron

Work is given by the equation \( W = Fd \), where \( F \) is the force and \( d \) is the distance. So the work done on the electron by the electric field is \( W = -2.4 \times 10^{-19} N \times 0.375 m = -9 \times 10^{-20} J \). The negative sign indicates that the work done is reducing the electron's kinetic energy.
03

Calculating the final speed of the electron

The work done by the electric field corresponds to the change in the electron's kinetic energy, which is given by \( \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \), where \( m \) is the electron's mass and \( v_f \) and \( v_i \) are the final and initial speeds, respectively. Rearranging this equation for \( v_f \), we get \( v_f = \sqrt{(2 \Delta KE/m) + v_i^2} \). Substituting the given values and the computed work \( \Delta KE = -9 \times 10^{-20} J \), \( m = 9.11 \times 10^{-31} kg \), and \( v_i = 4.5 \times 10^5 m/s \), we find \( v_f \approx 4.47 \times 10^5 m/s \).
04

Determining the force on the proton

The force on a proton moving in the same electric field as the electron is also given by \( F = qE \). However, the charge of a proton is positive, so \( q = 1.6 \times 10^{-19} C \), and this gives \( F = 1.6 \times 10^{-19} C \times 1.5 N/C = 2.4 \times 10^{-19} N \). The force is now acting in the same direction as the proton's motion.
05

Determining the work done on the proton

The work done on the proton is calculated similarly as for the electron, yielding \( W = 2.4 \times 10^{-19} N \times 0.375 m = 9 \times 10^{-20} J \). In this case, the work done contributes to an increase in the proton's kinetic energy.
06

Calculating the final speed of the proton

The final speed of the proton can be found using the same formula as for the electron. This time, however, the initial speed of the proton \( v_i = 1.9 \times 10^4 m/s \), and the kinetic energy change \( \Delta KE = 9 \times 10^{-20} J \). Substituting these values, the result is \( v_f \approx 1.92 \times 10^4 m/s \).

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.

Force on Charged Particles
When charged particles like electrons and protons are placed in an electric field, they experience a force. This force is calculated using the equation \( F = qE \), where \( q \) is the charge of the particle and \( E \) is the electric field strength. For an electron, the charge \( q \) is a negative value, \(-1.6 \times 10^{-19} C\). When in an electric field directed opposite to its movement, the force on the electron will also oppose its motion. This force can cause the particle to either accelerate or decelerate depending on whether it is in the direction of the particle's motion or against it.
  • For an electron moving east in an electric field directed west, the electron experiences a decelerating force.
  • For a proton under the same conditions, the positive charge means it experiences an accelerating force.
Understanding how forces act on charged particles in electric fields is crucial, as this phenomenon underlies many physical processes in electromagnetism and electronic devices.
Work-Energy Principle
The work-energy principle states that the work done by forces on an object results in a change in its kinetic energy. Work done by a force is calculated as \( W = Fd \), with \( F \) being the force applied and \( d \) the distance over which it acts. In this context, when an electric field exerts a force on a charged particle:
  • For the electron, the work done is negative, reducing its kinetic energy.
  • For the proton, the work done is positive, increasing its kinetic energy.
The change in kinetic energy \( \Delta KE \) relates to the change in speed using the equation \( \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \). Here \( m \) represents the particle's mass, \( v_f \) the final speed, and \( v_i \) the initial speed.
This principle helps us understand how energy transfer occurs in charged systems and is a fundamental concept in physics, employed in analyzing motion and energy transformations.
Kinematics of Particles
Kinematics deals with the motion of particles without considering the forces that cause the motion. To analyze the electron and proton traveling through an electric field, we use kinematic equations to relate speed, distance, and acceleration. Given that the force from the electric field causes a change in speed, these equations are valuable in computing the final speed after moving a certain distance:
  • Initial velocity \( v_i \) and the final velocity \( v_f \) change due to work done.
  • We use \( v_f = \sqrt{(2 \Delta KE/m) + v_i^2} \) to find the final velocity.
This helps to calculate how far and how fast the particles will travel under the influence of an electric field. By understanding the kinematic behavior of charged particles, we delve deeper into the principles that govern particle dynamics and prepare the ground for more advanced studies in electromagnetism and mechanics.

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 uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, \(1.60 \mathrm{~cm}\) distant from the first, in a time interval of \(3.20 \times 10^{-6} \mathrm{~s} .\) (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x=0, y=-1.35 \mathrm{~m}\) (b) at \(x=12.0 \mathrm{~cm}, y=12.0 \mathrm{~cm} ;\) (c) at \(x=-1.10 \mathrm{~m}, y=2.60 \mathrm{~m} ?\) Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\)

A proton is traveling horizontally to the right at \(4.50 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of \(3.20 \mathrm{~cm}\). (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

Point charge \(q_{1}=-5.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=-0.400 \mathrm{~m}\). Point \(P\) is on the \(x\) -axis at \(x=+0.200 \mathrm{~m}\). Point charge \(q_{2}\) is at the origin. What are the sign and magnitude of \(q_{2}\) if the resultant electric field at point \(P\) is zero?

An average human weighs about \(650 \mathrm{~N}\). If each of two average humans could carry \(1.0 \mathrm{C}\) of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their \(650 \mathrm{~N}\) weight?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Kinematics Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Modern Physics

Read Explanation

Electrostatics

Read Explanation

Particle Model of Matter

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.