The image displays an electron navigating between two charged metal plates that produce a 108 N/C vertical electric field, perpendicular to the electron's initial horizontal velocity. This setup can alter the trajectory of an electron, similar to an oscilloscope.Key parameters:- Initial speed of the electron: \(3.1 \times 10^6 \, \text{m/s}\)- Distance traveled in the uniform field: \(4.8 \, \text{cm}\)The accompanying diagram shows:- Horizontal path of the electron entering between the plates.- Vertical electric field direction denoted by blue arrows pointing from the positive to the negative plate.- Trajectory of the electron (green path), illustrating deflection due to the field.- Initial horizontal velocity \(v_0\).- Final velocity \(v_f\), making an angle \(\theta\) with the horizontal.Questions included:1. What is its vertical deflection in meters? - \(y =\)2. What is the vertical component of its final velocity in m/s? - \(v_{yf} =\)3. At what angle does it exit in degrees? Neglect any edge effects. - \(\theta =\)

The Given That: Electric Field, E=108 N/C The Horizontal Velocity, Vx=3.1×106 m/s Horizontal…

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 its vertical deflection in m? yo What is the vertical component of its final velocity in m/s? At what angle does it exit in degrees? Neglect any edge effects.

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 its vertical deflection in m? yo What is the vertical component of its final velocity in m/s? At what angle does it exit in degrees? Neglect any edge effects.

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)...

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

The image displays an electron navigating between two charged metal plates that produce a 108 N/C vertical electric field, perpendicular to the electron's initial horizontal velocity. This setup can alter the trajectory of an electron, similar to an oscilloscope.

Key parameters:
- Initial speed of the electron: \(3.1 \times 10^6 \, \text{m/s}\)
- Distance traveled in the uniform field: \(4.8 \, \text{cm}\)

The accompanying diagram shows:
- Horizontal path of the electron entering between the plates.
- Vertical electric field direction denoted by blue arrows pointing from the positive to the negative plate.
- Trajectory of the electron (green path), illustrating deflection due to the field.
- Initial horizontal velocity \(v_0\).
- Final velocity \(v_f\), making an angle \(\theta\) with the horizontal.

Questions included:
1. What is its vertical deflection in meters?
- \(y =\)
2. What is the vertical component of its final velocity in m/s?
- \(v_{yf} =\)
3. At what angle does it exit in degrees? Neglect any edge effects.
- \(\theta =\)

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