/*! 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 An electron with a speed of \(2.... [FREE SOLUTION] | 91Ó°ÊÓ

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An electron with a speed of \(2.60 \times 10^{8} \mathrm{~cm} / \mathrm{s}\) enters an electric field of magnitude \(1.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\), traveling along a field line 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 containing the electric field is \(8.00 \mathrm{~mm}\) 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?

Short Answer

Expert verified
(a) The electron will not stop within 8.00 mm, (b) Time to stop is nanoseconds, (c) Fraction of energy lost is calculated from initial and reduced speed.

Step by step solution

01

Determine the force on the electron

The electric force \( F \) on the electron is given by \[ F = eE \]where \( e \) is the charge of the electron \((1.60 \times 10^{-19} \text{ C})\), and \( E \) is the electric field strength \((1.00 \times 10^{3} \text{ N/C})\). Substitute the given values to calculate the force.
02

Calculate the acceleration of the electron

Using Newton's second law, the acceleration \( a \) of the electron is given by:\[ a = \frac{F}{m} = \frac{eE}{m} \]where \( m \) is the mass of the electron \((9.11 \times 10^{-31} \text{ kg})\). Substitute the value of force calculated in Step 1 to find the acceleration.
03

Find the stopping distance

Use the kinematic equation to find the stopping distance \(d\). The equation is:\[ v^2 = u^2 + 2ad \]where \( v \) is the final velocity \((0 \text{ m/s})\), \( u \) is the initial velocity \((2.60 \times 10^{6} \text{ m/s})\), and \( a \) is the acceleration from Step 2. Solve for \( d \).
04

Calculate the time to stop

The time \( t \) to stop can be found using the equation:\[ v = u + at \]Solve for \( t \) using the initial velocity, final velocity, and acceleration from previous steps.
05

Determine kinetic energy lost in the electric field region

First, calculate the initial kinetic energy \( KE_i \) of the electron using:\[ KE_i = \frac{1}{2}mv^2 \]Then determine the kinetic energy lost in the 8.00 mm region by finding the final kinetic energy \( KE_f \) at that distance, if the electron does not fully stop. The fraction of energy lost is\[ \frac{KE_i - KE_f}{KE_i} \].

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Key Concepts

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

Kinematics
Kinematics is the study of motion without considering the forces that cause it. In this exercise, we're interested in describing the motion of an electron as it enters an electric field.
When analyzing motion, we often use key quantities like velocity, time, acceleration, and displacement. Here, the initial velocity of the electron is \(2.60 \times 10^{6} \, \text{m/s}\). As the electron travels through the electric field, its speed is reduced until it stops momentarily.
To find how far the electron travels before stopping, we use one of the kinematic equations:
  • \(v^2 = u^2 + 2ad\)
This equation connects the initial speed \(u\), final speed \(v\), acceleration \(a\), and distance \(d\). By substituting in known values and solving for \(d\), we gain insight into how the electron behaves in the electric field.
This analysis is crucial as it helps us understand how charged particles move in electric fields, which is a fundamental concept in physics.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. The faster an object moves, the greater its kinetic energy. When an electron enters an electric field and begins to decelerate, it loses kinetic energy until it stops.
The initial kinetic energy \(KE_i\) of the electron is calculated using the formula:
  • \(KE_i = \frac{1}{2}mv^2\)
where \(m\) is the mass of the electron, and \(v\) is its initial speed.
As the electron travels through the short 8 mm electric field region, we are interested in knowing how much of this initial kinetic energy is lost.
To calculate the fraction of the energy lost, we use the difference between the initial kinetic energy and the kinetic energy when the electron has traveled 8 mm. This helps us understand how fields influence the energy of moving particles, which is essential in both theoretical and applied physics.Understanding kinetic energy changes allow scientists to design technologies manipulating electron movement, such as in particle accelerators.
Newton's Second Law
Newton’s Second Law is a central principle in physics. It describes how the acceleration of an object is related to the net force acting on it and the object's mass. This law is expressed with the equation:
  • \( F = ma \)
Here, \(a\) is acceleration, and \(m\) is the mass of the object.
In the context of this problem, we apply Newton's second law to quantify how the electric force affects the electron.
The force on the electron due to the electric field is calculated as:
  • \( F = eE \)
where \(e\) is the charge of the electron, and \(E\) is the electric field strength.
This force causes the electron to decelerate, changing its motion as predicted by the laws of kinematics.
By knowing this acceleration, we can determine how quickly the electron will stop when it enters the field. This concept shows us the unifying power of Newton's laws — they help predict the behavior of objects, from everyday items to subatomic particles, under various forces.

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Most popular questions from this chapter

Figure 22-48 shows a proton (p) on the central axis through a disk with a uniform charge density due to excess electrons. The disk is seen from an edge- on view. Three of those electrons are shown: electron \(\mathrm{e}_{c}\) at the disk center and electrons \(\mathrm{e}_{s}\) at opposite sides of the disk, at radius \(R\) from the center. The proton is initially at distance \(z=R=2.00 \mathrm{~cm}\) from the disk. At that location, what are the magnitudes of (a) the electric field \(\vec{E}_{c}\) due to electron \(\mathrm{e}_{c}\) and (b) the net electric field \(\vec{E}_{s, \text { net }}\) due to electrons \(\mathrm{e}_{s}\) ? The proton is then moved to \(z=R / 20.0\). What then are the magnitudes of (c) \(\vec{E}_{c}\) and (d) \(\vec{E}_{s, \text { net }}\) at the proton's location? (e) From (a) and (c) we see that as the proton gets nearer to the disk, the magnitude of \(\vec{E}_{c}\) increases, as expected. Why does the magnitude of \(\vec{E}_{s, \text { net }}\) from the two side electrons decrease, as we see from (b) and (d)?

An alpha particle (the nucleus of a helium atom) has a mass of \(6.64 \times 10^{-27} \mathrm{~kg}\) and a charge of \(+2 e\). What are the (a) magnitude and (b) direction of the electric field that will balance the gravitational force on the particle? (c) If the field magnitude is then doubled, what is the magnitude of the particle's acceleration?

5 Assume that a honeybee is a sphere of diameter \(1.000 \mathrm{~cm}\) with a charge of \(+60.0 \mathrm{pC}\) uniformly spread over its surface. Assume also that a spherical pollen grain of diameter \(40.0 \mu \mathrm{m}\) is electrically held on the surface of the bee because the bee's charge induces a charge of \(-1.00 \mathrm{pC}\) on the near side of the grain and a charge of \(+1.00 \mathrm{pC}\) on the far side. (a) What is the magnitude of the net electrostatic force on the grain due to the bee? Next, assume that the bee brings the grain to a distance of \(1.000 \mathrm{~mm}\) from the tip of a flower's stigma and that the tip is a particle of charge \(-60.0 \mathrm{pC}\). (b) What is the magnitude of the net electrostatic force on the grain due to the stigma? (c) Does the grain remain on the bee or move to the stigma?

An electron enters a region of uniform electric field with an initial velocity of \(30 \mathrm{~km} / \mathrm{s}\) 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 \(1.5 \mathrm{~ns}\) after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

A charged particle creates an electric field of magnitude 300 \(\mathrm{N} / \mathrm{C}\) at a point \(0.800 \mathrm{~m}\) away. What is the difference in the field magnitude between that point and one at \(0.400 \mathrm{~m}\) ?

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