/*! 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 49 Entering a Field An electron ent... [FREE SOLUTION] | 91影视

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Chapter 23: Problem 49

Entering a Field An electron enters a region of uniform electric field with an initial velocity of \(40 \mathrm{~km} / \mathrm{s}\) 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 \(1.5\) ns after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

Short Answer

Expert verified
The speed of the electron after 1.5 ns is 53170 m/s, and it travels a distance of 6.99 x 10^{-5} m.

Step by step solution

01

- Given Information

Identify the given values and constants.Initial velocity, \( u = 40 \text{ km/s} = 40 \times 10^3 \text{ m/s} \)Electric field, \( E = 50 \text{ N/C} \)Charge of an electron, \( q = 1.6 \times 10^{-19} \text{ C} \)Mass of an electron, \( m = 9.11 \times 10^{-31} \text{ kg} \)Time interval, \( t = 1.5 \text{ ns} = 1.5 \times 10^{-9} \text{ s} \)
02

- Calculate the acceleration

The acceleration \(a\) can be found using the formula from Newton's second law, \( F = ma \), where \( F = qE \):\[ a = \frac{F}{m} = \frac{qE}{m} \]Substitute the given values:\[ a = \frac{(1.6 \times 10^{-19} \text{ C})(50 \text{ N/C})}{9.11 \times 10^{-31} \text{ kg}} \]\[ a = 8.78 \times 10^{12} \text{ m/s}^2 \]
03

- Compute the final velocity

Use the formula \( v = u + at \) to find the final velocity after time \( t \):\[ v = 40 \times 10^3 \text{ m/s} + (8.78 \times 10^{12} \text{ m/s}^2)(1.5 \times 10^{-9} \text{ s}) \]\[ v = 40,000 \text{ m/s} + 13.17 \times 10^3 \text{ m/s} \]\[ v = 53170 \text{ m/s} \]
04

- Calculate the distance traveled

Use the formula \( s = ut + \frac{1}{2}at^2 \) to calculate the distance traveled:\[ s = (40 \times 10^3 \text{ m/s})(1.5 \times 10^{-9} \text{ s}) + \frac{1}{2}(8.78 \times 10^{12} \text{ m/s}^2)(1.5 \times 10^{-9} \text{ s})^2 \]Simplify the terms:\[ s = 6.0 \times 10^{-5} \text{ m} + \frac{1}{2}(8.78 \times 10^{12} \text{ m/s}^2)(2.25 \times 10^{-18}) \]\[ s = 6.0 \times 10^{-5} \text{ m} + 9.8775 \times 10^{-6} \text{ m} \]\[ s = 6.98775 \times 10^{-5} \text{ m} \]\[ s = 6.99 \times 10^{-5} \text{ m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Dynamics
Electron dynamics studies the behavior of electrons under various forces and fields. In this exercise, an electron moves through a uniform electric field.
Uniform Electric Field
A uniform electric field has the same strength and direction at every point. In this problem, the electric field has a magnitude of 50 N/C.
Physics Problem-Solving Steps
When solving physics problems, follow a systematic approach. This problem followed specific steps starting with identifying known values and calculating acceleration.

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Most popular questions from this chapter

High-Speed Protons Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. (a) What acceleration would a proton experience if the gun's electric field were \(2.00 \times 10^{4} \mathrm{~N} / \mathrm{C} ?(\mathrm{~b})\) What speed would the proton attain if the field accelerated the proton through a distance of \(1.00 \mathrm{~cm}\) ?

Velocity Components At some instant the velocity components of an electron moving between two charged parallel plates are \(v_{x}=\) \(1.5 \times 10^{5} \mathrm{~m} / \mathrm{s}\) and \(v_{y}=3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Suppose that the electric field between the plates is given by \(\vec{E}=(120 \mathrm{~N} / \mathrm{C}) \hat{\mathrm{j}} .\) (a) What is the acceleration of the electron? (b) What will be the velocity of the electron after its \(x\) coordinate has changed by \(2.0 \mathrm{~cm}\) ?

Charged Cloud A charged cloud system produces an electric field in the air near Earth's surface. A particle of charge \(-2.0 \times\) \(10^{-9} \mathrm{C}\) is acted on by a downward electrostatic force of \(3.0 \times\) \(10^{-6} \mathrm{~N}\) when placed in this field. (a) What is the magnitude of the electric field? (b) What are the magnitude and direction of the electrostatic force exerted on a proton placed in this field? (c) What is the gravitational force on the proton? (d) What is the ratio of the magnitude of the electrostatic force to the magnitude of the gravitational force in this case?

Field Retards Motion An electron with a speed of \(5.00 \times\) \(10^{8} \mathrm{~cm} / \mathrm{s}\) enters an electric field of magnitude \(1.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\), traveling along the field lines in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily and (b) how much time will have elapsed? (c) If the region with the electric field is only \(8.00 \mathrm{~mm}\) long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region?

Frequency of Oscillation Find the frequency of oscillation of an electric dipole, of dipole moment \(\vec{p}\) and rotational inertia \(I\), for small amplitudes of oscillation about its equilibrium position in a uniform electric field of magnitude \(|\vec{E}|\)
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