**Educational Transcription:**The figure illustrates an electron traveling between two charged metal plates that produce a \(108 \, \text{N/C}\) vertical electric field, which is perpendicular to the electron's initial horizontal velocity. This setup is commonly used to alter the electron's trajectory, such as in an oscilloscope. The electron's initial speed is \(3.1 \times 10^6 \, \text{m/s}\), and it moves a horizontal distance of \(4.8 \, \text{cm}\) in the uniform field.**Diagram Explanation:**- The diagram shows two parallel plates with a positive charge on the top plate and a negative charge on the bottom plate.- An electron (\(e^-\)) is moving from left to right, entering between the plates with an initial horizontal velocity \(v_0\).- As the electron passes through the plates, its path curves upward due to the influence of the vertical electric field.- The vector \(v_1\) represents the electron's final velocity, making an angle \(\theta\) with the horizontal.**Question:**What is the vertical component of its final velocity in \(\text{m/s}\)?**Equation:**\(V_{y,f} =\) [to be calculated]This material is designed for educational use to help students understand the motion of charged particles in electric fields.

Answered: Image /qna-images/answer/0911eece-b7da-410b-90e4-91cae51c4239.jpg

The figure shows an electron passing between two charged metal plates that create a 108 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 3.1 x 10 m/s, and the horizontal distance it travels in the uniform field is 4.8 cm. What is the vertical component of its final velocity in m/s?

The figure shows an electron passing between two charged metal plates that create a 108 N/C vertical electric field perpendicular to the electron's original horizontal velocity. (These can be used to change the electron's direction, such as in an oscilloscope.) The initial speed of the electron is 3.1 x 10 m/s, and the horizontal distance it travels in the uniform field is 4.8 cm. What is the vertical component of its final velocity in m/s?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

Inside a radiation therapy device, an electron is in the presence of a uniform electric field with a magnitude of 295 N/C. (a) What is the magnitude of the acceleration of the electron (in m/s2)? m/s² (b) The electron is initially at rest. What is its speed (in m/s) after 9.00 x 10-9 s? m/s
An electron is traveling through a uniform electric field. The field is constant and given by E=(2.00×10−11N/C) i^− (1.20×10−11N/C) j^. At t=0 , the electron is at the origin and traveling in the x direction with a speed of 2.20 m/s. Part A) What is its position 2.50 s later? x= ? (in m) Part B) y=?
Inside a cathode ray tube, an electron is in the presence of a uniform electric field with a magnitude of 275 N/C. (a) What is the magnitude of the acceleration of the electron (in m/s2)? m/s2 (b) The electron is initially at rest. What is its speed (in m/s) after 1.20 ✕ 10−8 s? m/s
Charge q1 = 1.50 nC is at x1 = 0 and charge q2 = 5.00 nC is at x2 = 4.00 m. At what point between the two charges is the electric field equal to zero? (Enter the x coordinate in m.)
Inside a vacuum tube, an electron is in the presence of a uniform electric field with a magnitude of 305 N/C. (a) What is the magnitude of the acceleration of the electron (in m/s2)? (b) The electron is initially at rest. What is its speed (in m/s) after 1.05 ✕ 10−8 s?
A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude of the electric field is 1.81 105 N/C. (a) What is the speed of the proton after it has traveled 7.00 cm? (b) What is the speed of the proton after it has traveled 70.0 cm?
An electron is moving east in a uniform electric field of 1.50 N/C directed to the west. At point A, the velocity of the electron is 4.55×105 m/s pointed toward the east. What is the speed of the electron when it reaches point B, which is a distance of 0.400 m east of point A?
An electron with a speed of 5.63 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 1.81 × 103 N/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 4.23 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?