Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In a cathode ray tube, electrons initially at rest are accelerated by a uniform electric field of magnitude $4.0 \times 10^{5} \mathrm{N} / \mathrm{C}\( during the first \)5.0 \mathrm{cm}$ of the tube's length; then they move at essentially constant velocity another \(45 \mathrm{cm}\) before hitting the screen. (a) Find the speed of the electrons when they hit the screen. (b) How long does it take them to travel the length of the tube?

Short Answer

Expert verified
(a) The speed of the electrons when they hit the screen is the final speed after acceleration, given by the equation \(v^2 = 2ad\). To find the actual speed (v), take the square root of the result. (b) The total time it takes for the electrons to travel the length of the tube can be found by adding the time it takes to travel the accelerated distance (\(t_1\)) and the time it takes to travel the constant-velocity distance (\(t_2\)). This is given by the equation \(T = t_1 + t_2\).

Step by step solution

01

Identify given values

We are given the following values: - Electric field (E): \(4.0 \times 10^5 \mathrm{N} / \mathrm{C}\) - Distance traveled under acceleration (d₁): 5.0 cm - Distance traveled at a constant velocity (d₂): 45 cm - Electron charge (e): \(1.6 \times 10^{-19} \mathrm{C}\) - Electron mass (m): \(9.1 \times 10^{-31} \mathrm{kg}\) Note that we need to convert the distances into meters, so d₁ = 0.05 m and d₂ = 0.45 m.
02

Find the acceleration of the electrons

Using the electric field (E) and the charge of the electron (e), we can find the force experienced by the electrons: \(F = e \times E\) Now, we can find the acceleration (a) using the mass of the electron (m) and the force (F): \(a = F/m\)
03

Find the final speed of the electron after acceleration

Since the electrons are initially at rest, we can use the equation of motion for constant acceleration: \(v^2 = u^2 + 2ad\) As the initial velocity (u) is 0, this simplifies to: \(v^2 = 2ad\) This equation allows us to find the final speed of the electron (v) after being accelerated by the electric field.
04

Calculate the time it takes to travel the accelerated distance

Using the final speed (v) and the distance traveled during acceleration (d₁), we can find the time it takes for the electrons to travel this distance using the equation of motion for constant acceleration: \(v = u + at\) As the initial velocity (u) is 0, this simplifies to: \(t_1 = d_1/a\)
05

Calculate the time it takes to travel the constant-velocity distance

Next, we need to find the time it takes for the electrons to travel the remaining distance (d₂) at the constant velocity (v). Using the distance formula: \(d = vt\) We can find the time \(t_2\) it takes to travel this distance: \(t_2 = d_2/v\)
06

Find the total time and answer the question

To find the total time (T) it takes for the electron to travel the length of the tube, we add the time it takes to travel the accelerated distance (\(t_1\)) and the time it takes to travel the constant-velocity distance (\(t_2\)): \(T = t_1 + t_2\) Now we can answer the problem: (a) The speed of the electrons when they hit the screen is v. (b) The total time it takes for the electrons to travel the length of the tube is T.

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 Vaia!

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 raindrop inside a thundercloud has charge - \(8 e\). What is the electric force on the raindrop if the electric field at its location (due to other charges in the cloud) has magnitude $2.0 \times 10^{6} \mathrm{N} / \mathrm{C}$ and is directed upward?

A metal sphere A has charge \(Q\). Two other spheres, B and \(\mathrm{C},\) are identical to \(\mathrm{A}\) except they have zero net charge. A touches \(\mathrm{B}\), then the two spheres are separated. B touches \(C\), then those spheres are separated. Finally, C touches A and those two spheres are separated. How much charge is on each sphere?

An electron is placed in a uniform electric field of strength $232 \mathrm{N} / \mathrm{C}$. If the electron is at rest at the origin of a coordinate system at \(t=0\) and the electric field is in the positive \(x\) -direction, what are the \(x\) - and \(y\) coordinates of the electron at \(t=2.30 \mathrm{ns} ?\)
Some forms of cancer can be treated using proton therapy in which proton beams are accelerated to high energies, then directed to collide into a tumor, killing the malignant cells. Suppose a proton accelerator is \(4.0 \mathrm{m}\) long and must accelerate protons from rest to a speed of $1.0 \times 10^{7} \mathrm{m} / \mathrm{s} .$ Ignore any relativistic effects (Chapter 26) and determine the magnitude of the average electric field that could accelerate these protons.
What are the magnitude and direction of the acceleration of a proton at a point where the electric field has magnitude \(33 \mathrm{kN} / \mathrm{C}\) and is directed straight up?
See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free