/*! 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 87 In a cathode ray tube, electrons... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 87

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 (d鈧�): 5.0 cm - Distance traveled at a constant velocity (d鈧�): 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 d鈧� = 0.05 m and d鈧� = 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 (d鈧�), 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 (d鈧�) 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.

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Most popular questions from this chapter

In lab tests it was found that rats can detect electric fields of about $5.0 \mathrm{kN} / \mathrm{C}\( or more. If a point charge of \)1.0 \mu \mathrm{C}$ is sitting in a maze, how close must the rat come to the charge in order to detect it?
Using the results of Problem \(66,\) we can find the electric field at any radius for any spherically symmetrical charge distribution. A solid sphere of charge of radius \(R\) has a total charge of \(q\) uniformly spread throughout the sphere. (a) Find the magnitude of the electric field for \(r \geq R .\) (b) Find the magnitude of the electric field for \(r \leq R .\) (c) Sketch a graph of \(E(r)\) for \(0 \leq r \leq 3 R\)
A thin, flat sheet of charge has a uniform surface charge density \(\sigma(\sigma / 2\) on each side). (a) Sketch the field lines due to the sheet. (b) Sketch the field lines for an infinitely large sheet with the same charge density. (c) For the infinite sheet, how does the field strength depend on the distance from the sheet? [Hint: Refer to your field line sketch.J (d) For points close to the finite sheet and far from its edges, can the sheet be approximated by an infinitely large sheet? [Hint: Again, refer to the field line sketches.] (e) Use Gauss's law to show that the magnitude of the electric field near a sheet of uniform charge density \(\sigma\) is $E=\sigma /\left(2 \epsilon_{0}\right)$
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
Two tiny objects with equal charges of \(7.00 \mu \mathrm{C}\) are placed at the two lower corners of a square with sides of \(0.300 \mathrm{m},\) as shown. Find the electric field at point \(C\) the center of the square.
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