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Chapter 22: Problem 41

An electron is observed traveling at a speed of \(27.5 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) parallel to an electric field of magnitude \(11,400 \mathrm{~N} / \mathrm{C}\) How far will the electron travel before coming to a stop?

Short Answer

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Question: An electron with an initial speed of 27.5 million meters per second travels parallel to a uniform electric field of 11,400 N/C. Find the distance the electron travels before coming to a stop. Answer: The electron will travel approximately 0.000189 meters (189 micrometers) before coming to a stop.

Step by step solution

01

Find the force experienced by the electron

We are given the magnitude of the electric field, \(E=11,400 \mathrm{~N}/\mathrm{C}\), and the charge of an electron \(q=-1.6\times 10^{-19} \mathrm{~C}\). Let's calculate the force using the formula \(F=qE\): \(F=(-1.6\times 10^{-19}\mathrm{~C})(11,400\mathrm{~N}/\mathrm{C})=-1.824\times10^{-16} \mathrm{~N}\).
02

Calculate the acceleration

To find acceleration, we can use Newton's second law of motion, which states that \(F = ma\), where \(m\) is the mass of the electron, which we know is \(9.11 \times 10^{-31} \mathrm{~kg}\). So, the acceleration is: \(a = \frac{F}{m} =\frac{-1.824\times10^{-16}\mathrm{~N}}{9.11\times 10^{-31}\mathrm{~kg}} =-2\times10^{14} \mathrm{~m/s^2}\) (Note: The negative sign indicates acceleration in the opposite direction of motion)
03

Calculate the distance traveled before coming to a stop

To find the distance traveled before the electron comes to a stop, we can use the following equation of motion: \(v^2 = u^2 + 2as\). Since the final velocity is 0, the equation becomes: \(0^2 = (27.5 \cdot 10^{6} \mathrm{~m/s})^2 + 2(-2\times10^{14} \mathrm{~m/s^2})(s)\). Solving for \(s\): \(s = \frac{(27.5 \cdot 10^{6} \mathrm{~m/s})^2}{2\times 2 \times 10^{14} \mathrm{~m/s^2}} = \frac{7.56 \times 10^{13} \mathrm{~m^2/s^2}}{4 \times 10^{14} \mathrm{~m/s^2}} = 0.000189\mathrm{~m}\). So, the electron will travel approximately 0.000189 meters(189 micrometers) before coming to a stop.

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Key Concepts

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

Electron Motion
Electrons are tiny particles that carry a negative charge, and their motion in an electric field can reveal a lot about electric forces. In the context of your textbook exercise, an electron moves with a high speed of \(27.5 \times 10^6 \text{ m/s}\) in parallel to an electric field. This speed gives insight into how electrons behave under certain conditions, like fields with an intensity of \(11,400 \text{ N/C}\).
  • When placed in such an electric field, the electron experiences a force that is determined by the product of its charge and the field strength.
  • The electron's charge is known to be \(-1.6 \times 10^{-19} \text{ C}\), a fundamental constant that shows its inherent negative nature.
  • This negative charge means the force direction is opposite to the field's direction, which acts to slow the electron down when it initially moves in alignment with the field.
Understanding electron motion involves grasping both the dynamics involved and the factors affecting speed changes, such as field strength and charge polarity. This provides the foundation for later learning about electric circuits and wave dynamics.
Newton's Second Law
Newton's Second Law is a cornerstone of physics, describing how the movement of objects is influenced by forces. It is represented by the formula \( F = ma \), where \( F \) is the force applied, \( m \) is the object's mass, and \( a \) is the acceleration produced. In this exercise, this fundamental law helps calculate the electron's acceleration.
  • First, the force on the electron is determined using the electron's charge and the electric field's strength, resulting in \(-1.824 \times 10^{-16} \text{ N}\).
  • The negative sign indicates that the force direction opposes the electron's motion, due to the negative charge.
  • Given the mass of the electron is \(9.11 \times 10^{-31} \text{ kg}\), we solve for acceleration as \( a = \frac{F}{m} \).
The result, \(-2 \times 10^{14} \text{ m/s}^2\), reveals how rapidly the electron decelerates under this force. It explains the cause behind the electron eventually coming to a halt, setting the stage for understanding further electromagnetism concepts.
Acceleration Calculation
Acceleration is the rate of change of velocity, crucial for explaining how quickly an object's motion changes. In this scenario, calculating acceleration gives us insights into how the electron's velocity changes under the influence of an electric field.
  • We start with force calculation, already obtained from multiplying the electron's charge with the electric field value.
  • Newton's Second Law, \( a = \frac{F}{m} \), is used here to obtain acceleration.
  • The electron undergoes significant deceleration due to its small mass and the relatively larger force.
The calculated acceleration of \(-2 \times 10^{14} \text{ m/s}^2\) signifies how quickly the electron comes to a stop when moving in a direction parallel to the field. Understanding this calculation is vital for studying forces in electric and magnetic fields, and their effects on particles in various scientific and technological contexts.

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Most popular questions from this chapter

There is an electric field of magnitude \(150 .\) N/C, directed vertically downward, near the surface of the Earth. Find the acceleration (magnitude and direction) of an electron released near the Earth's surface.

Electric dipole moments of molecules are often measured in debyes \((\mathrm{D}),\) where \(1 \mathrm{D}=3.34 \cdot 10^{-30} \mathrm{C} \mathrm{m} .\) For instance, the dipole moment of hydrogen chloride gas molecules is \(1.05 \mathrm{D}\). Calculate the maximum torque such a molecule can experience in the presence of an electric field of magnitude \(160.0 \mathrm{~N} / \mathrm{C}\).

A thin glass rod is bent into a semicircle of radius \(R\). A charge \(+Q\) is uniformly distributed along the upper half, and a charge \(-Q\) is uniformly distributed along the lower half as shown in the figure. Find the magnitude and direction of the electric field \(\vec{E}\) (in component form) at point \(P\), the center of the semicircle.

A spherical aluminized Mylar balloon carries a charge \(Q\) on its surface. You are measuring the electric field at a distance \(R\) from the balloon's center. The balloon is slowly inflated, and its radius approaches but never reaches R. What happens to the electric field you measure as the balloon increases in radius. Explain.

Consider a uniform nonconducting sphere with a charge \(\rho=3.57 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{3}\) and a radius \(R=1.72 \mathrm{~m}\). What is the magnitude of the electric field \(0.530 \mathrm{~m}\) from the center of the sphere?
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