**Educational Text for Website:****Understanding Electron Motion in an Electric Field**In this example, we study how an electron, initially at rest, moves through a potential difference and how it interacts with an electric field between two parallel plates.**Scenario Description:**An electron is accelerated from rest through a potential difference \( V_1 = 1.13 \, \text{kV} \) before it enters the gap between two parallel plates. These plates have a separation of \( d = 24.3 \, \text{mm} \) and hold a potential difference of \( V_2 = 74.2 \, \text{V} \). The lower plate is at the lower potential. For simplicity, fringing effects are neglected. Here, we need to determine the magnetic field that will ensure the electron travels in a straight line between the plates.**Diagram Explanation:**The diagram shows the electron entering from the left through \( V_1 \), represented as a dashed horizontal line. The plates are parallel and separated by distance \( d \), with the electric field acting vertically between them. The electron enters horizontally, indicating that its velocity vector is perpendicular to the electric field vector.**Objective:**Using unit-vector notation, identify the correct magnetic field (expressed in milliteslas, mT) that allows the electron to travel in a straight line in the presence of the described electric field.Below the scenario, fields allow for the input of the magnetic field components along the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) axes. The focus is particularly on determining the correct values needed to counteract the electric forces being exerted on the electron, ensuring a linear trajectory.

V1 = 1.13 kV = 1.13 x 103 V d = 24.3 mm = 0.0243 m V2 = 74.2 V

In the figure, an electron accelerated from rest through potential difference V1=1.13 kV enters the gap between two parallel plates having separation d = 24.3 mm and potential difference V2= 74.2 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? Number k) Units mT

In the figure, an electron accelerated from rest through potential difference V1=1.13 kV enters the gap between two parallel plates having separation d = 24.3 mm and potential difference V2= 74.2 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? Number k) Units mT

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

In the figure, an electron accelerated from rest through potential difference V1-1.26 kV enters the gap between two parallel plates having separation d = 19.6 mm and potential difference V2- 52.4 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? Number ( o ĵ+ i R) Units mT 124.7
As shown in the figure, an electron is fired with a speed of 3.43 x 106 m/s through a hole in one of the two parallel plates and into the region between the plates separated by a distance of 0.20 m. There is a magnetic field in the region between the plates and, as shown, it is directed into the plane of the page (perpendicular to the velocity of the electron). Determine the magnitude of the magnetic field so that the electron just misses colliding with the opposite plate. electron
Alpha particles (m= 6.7 × 10-27 kg, q= + 2e) are accelerated from rest through a potential difference of 6.7 kV. Then they enter a magnetic field B = 0.2 T perpendicular to their direction of motion. The radius of the path described by them is
A uniform magnetic field of 2.00×10^(−3) T is applied to put an electron into a circular orbit with angular momentum 5.00×10^(−25) kg m^2/s. Calculate the orbital radius and the speed of the electron.
In the figure, an electron accelerated from rest through potential difference V₁-1.03 kV enters the gap between two parallel plates having separation d-17.2 mm and potential difference V₂-108 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit- vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? Number (0 7+0 3.30e-4 A) Units mT
In the figure, an electron accelerated from rest through potential difference V₁=1.07 kV enters the gap between two parallel plates having separation d = 19.5 mm and potential difference V₂= 177 V. The lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity vector is perpendicular to the electric field vector between the plates. In unit-vector notation, what uniform magnetic field allows the electron to travel in a straight line in the gap? Number (i î+ i V₂ k) Units
Problem 5: An electron (q= -e = -1.60 × 10-19 C and me = 9.11 × 10-31 kg) in a cathode-ray tube is accelerated through a potential difference of 10 kV. Thus, the electron leaves the cathode-ray tube with a speed v = 5.93 × 107 m/sec. The electron then passes through the 2.0 cm wide region of uniform magnetic field (shown in the Figure above on the right). What field strength will deflect the electron by 10°? Answer: B = 0.0029 T. OV 10 kV 2.0 cm 10°
An electron of kinetic energy 1.74 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 31.2 cm. Find (a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion. (a) Number i Units (b) Number i Units (c) Number i Units (d) Number i Units
An electron initially at rest is accelerated through a potential difference of 3 kV in the positive x- direction. It enters a region where a uniform magnetic field of 5.0-mT is in a positive y-direction (perpendicular to the velocity of the electron). Calculate the radius of the path this electron will follow in the magnetic field. (e = 1.60 × 10-19 c, m electron = 9.11 × 10¯31 kg)
A proton accelerated to 500 keV enters a uniform transverse magnetic field of 0.51 T. The field extends over a region of space d = 10 cm thickness. Find the angle a through which the proton deviates from the initial direction of its motion.
An electron that has velocity т i = (2.0 x 106)i+ (3.0 × 106 10", %3D moves through the uniform magnetic field B = (0.030 T)î – (0.15T)f. (a) Find the force on the electron due to the magnetic field. (b) Repeat your calculation for a proton having the same velocity.
An electron is accelerated through 2500 V from rest and then enters a region where there is a uniform 1.60 T magnetic field. What are the maximum and minimum magnitudes of the magnetic force acting on this electron?