/*! 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 26 An electron is acted upon by a f... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron is acted upon by a force of \(5.00 \times 10^{-15} \mathrm{~N}\) due to an electric field. Find the acceleration this force produces in each case: (a) The electron's speed is \(1.00 \mathrm{~km} / \mathrm{s}\). (b) The electron's speed is \(2.50 \times 10^{8} \mathrm{~m} / \mathrm{s}\) and the force is parallel to the velocity.

Short Answer

Expert verified
(a) \(5.49 \times 10^{15} \mathrm{~m/s^2}\); (b) \(5.49 \times 10^{15} \mathrm{~m/s^2}\) with classical mechanics.

Step by step solution

01

Identify the given values

For both parts (a) and (b), the force acting on the electron is given as \( F = 5.00 \times 10^{-15} \mathrm{~N} \). The mass of an electron is a constant \( m = 9.11 \times 10^{-31} \mathrm{~kg} \).
02

Recall the equation for acceleration

Use Newton's second law, which states \( F = ma \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. We can rearrange this to find acceleration: \( a = \frac{F}{m} \).
03

Calculate acceleration for case (a)

In part (a), the speed of the electron does not affect the calculation because the force and mass determine the acceleration. Using the equation from Step 2, substitute the values: \( a = \frac{5.00 \times 10^{-15} \mathrm{~N}}{9.11 \times 10^{-31} \mathrm{~kg}} \). Calculate the result to find the acceleration.
04

Solve for acceleration in part (a)

Perform the calculation: \[ a = \frac{5.00 \times 10^{-15} \mathrm{~N}}{9.11 \times 10^{-31} \mathrm{~kg}} = 5.49 \times 10^{15} \mathrm{~m/s^2} \].
05

Analyze part (b) requirements

In part (b), the speed \( 2.50 \times 10^8 \mathrm{~m/s} \) is a significant fraction of the speed of light (\( c = 3.00 \times 10^8 \mathrm{~m/s} \)), so relativistic effects are important. However, since we are asked to use classical mechanics initially, calculate like in part (a).
06

Calculate acceleration for case (b) using classical mechanics

Repeat the calculation using \( a = \frac{F}{m} \) without considering relativistic effects for the initial solution: \[ a = \frac{5.00 \times 10^{-15} \mathrm{~N}}{9.11 \times 10^{-31} \mathrm{~kg}} = 5.49 \times 10^{15} \mathrm{~m/s^2} \].
07

Note on relativistic effects

For completeness, note that in reality, at speeds close to the speed of light, relativistic mechanics would need to be used. Acceleration calculated in Step 6 is an approximation valid for non-relativistic situations.

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Key Concepts

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

Newton's second law
Newton's second law of motion is a fundamental concept in physics, which helps us understand how objects move. It states that the net force acting upon an object is equal to the product of its mass and acceleration. Mathematically, this is expressed as:
  • \( F = ma \)
Here, \( F \) is the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration produced. This equation tells us that the acceleration of an object is directly proportional to the net force acting and inversely proportional to its mass. When you apply a greater force on an object, its acceleration increases. Likewise, for a given force, an object with less mass will accelerate more than one with greater mass.
In our exercise, an electron experiences a force of \( 5.00 \times 10^{-15} \mathrm{~N} \). Using the electron's known mass of \( 9.11 \times 10^{-31} \mathrm{~kg} \), we can calculate its acceleration by rearranging the equation to \( a = \frac{F}{m} \). This helps us find how quickly the electron speeds up due to the electromagnetic force applied to it.
Relativistic effects
As objects move closer to the speed of light, classical mechanics fall short, and relativistic effects become significant. According to Einstein's theory of relativity, as an object's speed approaches the speed of light, its mass effectively increases, making further acceleration harder to achieve.
In the context of the exercise, when the electron moves at \( 2.50 \times 10^8 \mathrm{~m/s} \), which is a substantial fraction of the speed of light, traditional Newtonian mechanics no longer suffice. Under these conditions, relativistic mechanics must be applied to get accurate results. This involves using the concept of relativistic mass or modifying the force equation to consider the speed of the electron.
The relativistic mass \( m_r \) can be expressed as:
  • \( m_r = \frac{m_0}{\sqrt{1-\left(\frac{v^2}{c^2}\right)}} \)
Where \( m_0 \) is the rest mass, \( v \) is the velocity of the object, and \( c \) is the speed of light. This increase in mass leads to a decrease in calculated acceleration when using \( F = m_r \, a \). That's why relativistic dynamics must be considered for precise calculations at such high speeds.
Electron dynamics
Understanding electron dynamics is crucial for grasping how electrons behave under various electromagnetic forces. Electrons, being subatomic particles with a negative charge, interact strongly with electric and magnetic fields. This interaction is what accelerates them in our exercise.
  • Electrons have a very small mass of \( 9.11 \times 10^{-31} \mathrm{~kg} \).
  • They respond quickly to applied forces due to their small mass and high charge-to-mass ratio.
In scenarios where electrons are subjected to external forces such as those from electric fields, they accelerate according to Newton's second law, as explored in the exercise. Their acceleration depends on the force and inversely on their small mass, making them ideal candidates for studying dynamics at high speeds.
When calculating electron dynamics in electric fields, it's important to consider the linearity of motion unless speeds reach relativistic values where more complex equations are necessary. The key takeaway is that electron dynamics are significantly influenced by electromagnetism, dictating how and why electrons move in specific ways, making them a vital component in the study of electromagnetism and its applications, like in electric circuits and particle accelerators.

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Most popular questions from this chapter

A spaceship makes the long trip from earth to the nearest star system, Alpha Centauri, at a speed of \(0.955 c .\) The star is about 4.37 light-years from earth, as measured in earth's frame of reference. (a) How many years does the trip take according to an observer on earth? (b) How many years does the trip take according to a passenger on the spaceship? (c) How many light-years distant is Alpha Centauri from earth, as measured by a passenger on the speeding spacecraft? (Note that, in the ship's frame of reference, the passengers are at rest while the space between earth and Alpha Centauri goes rushing past at \(0.955 c .\) ) (d) Use your answer from part (c) along with the speed of the spacecraft to calculate another answer for part (b). Do your two answers for that part agree? Should they?

A particle has a rest mass of \(6.64 \times 10^{-27} \mathrm{~kg}\) and a momentum of \(2.10 \times 10^{-18} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). (a) What is the total energy (kinetic plus rest energy) of the particle? (b) What is the kinetic energy of the particle? (c) What is the ratio of the kinetic energy to the rest energy of the particle?

A rocket ship flies past the earth at \(85.0 \%\) of the speed of light. Inside, an astronaut who is undergoing a physical examination is having his height measured while he is lying down parallel to the direction the rocket ship is moving. (a) If his height is measured to be \(2.00 \mathrm{~m}\) by his doctor inside the ship, what would a person watching this from earth measure for his height? (b) If the earth-based person had measured \(2.00 \mathrm{~m},\) what would the doctor in the spaceship have measured for the astronaut's height? Is this a reasonable height? (c) Suppose the astronaut in part (a) gets up after the examination and stands with his body perpendicular to the direction of motion. What would the doctor in the rocket and the observer on earth measure for his height now?

A nuclear physicist measures the momentum and corresponding relativistic energy of an atomic particle. A table of her data is shown here. $$\begin{array}{ll}\hline \text { Momentum }(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}) & \text { Energy }(\mathrm{J}) \\\\\hline 1.42 \times 10^{-18} & 7.36 \times 10^{-10} \\\2.00 \times 10^{-18} & 8.50 \times 10^{-10} \\\2.45 \times 10^{-18} & 9.51 \times 10^{-10} \\\2.83 \times 10^{-18} & 1.04 \times 10^{-9} \\\3.17 \times 10^{-18} & 1.13 \times 10^{-9} \\\\\hline\end{array}$$ Make a plot of \(E^{2}\) as a function of \(p^{2} c^{2}\). Using a linear "best fit" to the data and Equation \(27.24,\) determine the mass of the particle. Express your answer in amu. What is the particle?

In an experiment, two protons are shot directly toward each other, each moving at half the speed of light relative to the laboratory. (a) What speed does one proton measure for the other proton? (b) What would be the answer to part (a) if we used only nonrelativistic Newtonian mechanics? (c) What is the kinetic energy of each proton as measured by (i) an observer at rest in the laboratory and (ii) an observer riding along with one of the protons? (d) What would be the answers to part (c) if we used only nonrelativistic Newtonian mechanics?

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