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Integrated Concepts An electron has an initial velocity of

Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Problem 58PE Chapter 18

Physics: Principles with Applications | 6th Edition

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Physics: Principles with Applications | 6th Edition | ISBN: 9780130606204 | Authors: Douglas C. Giancoli

Physics: Principles with Applications | 6th Edition

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Problem 58PE

Integrated Concepts An electron has an initial velocity of 5.00 × 10? m/s in a uniform 2.00 × 10? 5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point?

Step-by-Step Solution:

Solution: 6 In the above question they have given the electron has an initial velocity of 5.00 × 10 5 m/s in a uniform 2.00 × 10 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. Step 1 (a) What is the direction of the electric field The above figure diagram of the pattern of lines, sometimes referred to as electric field lines, point in the direction that a positive test charge would accelerate if placed upon the line. As the above such lines are directed away from positively charged source charges and toward negatively charged source charges. Step 2 (b) How far does the electron travel before coming to rest The coulomb’s law of electrostatic field, F = qE Where, q is charge of proton and E is the electric field Therefore Newton’s second law, (i.e) F = ma Where, m is the mass of proton and a is the acceleration of proton Step 3 In the above equation Newton’s second law and coulomb’s law of electrostatic field. We get, a = qE a = qE/m Then in the above question (b) we need to calculate the distance travelled electron before coming into rest So from Newton’s equation of motion, 2 2 v = v + 2axo Where, v is the final velocity of electron, v is the initial velocity of electron, o x is the distance travelled by electron, a is the acceleration of electron. We now need to substitute the 0 for v we get, 2 2 v = v + 2axo 2 0 = v o + 2ax In the above term the 0 is no value so we taking x in form of we get as, x = - v / 2a 2 o Then substitute q E/m for a we get as, x = - v m / oqE 5 In the above equation substitute 2.0 ×10 N/C for E , 9 10 -31 for q and 5 .0 ×10 m/s for v we get as, o x = (5.0 × 10 m/s) (9.11 ×10 = 35.58 × 10 m = 3.56 × 10 m Where the distance traveled by the electron before coming to rest is 3.56 × 10 m

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Chapter 18, Problem 58PE is Solved
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Textbook: Physics: Principles with Applications
Edition: 6th
Author: Douglas C. Giancoli
ISBN: 9780130606204

This textbook survival guide was created for the textbook: Physics: Principles with Applications, edition: 6th. The full step-by-step solution to problem: 58PE from chapter: 18 was answered by Sieva Kozinsky, our top Physics solution expert on 03/03/17, 03:53PM. This full solution covers the following key subjects: electron, Field, velocity, initial, Rest. This expansive textbook survival guide covers 35 chapters, and 3914 solutions. Since the solution to 58PE from 18 chapter was answered, more than 247 students have viewed the full step-by-step answer. The answer to “Integrated Concepts An electron has an initial velocity of 5.00 × 10? m/s in a uniform 2.00 × 10? 5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electron’s velocity when it returns to its starting point?” is broken down into a number of easy to follow steps, and 82 words. Physics: Principles with Applications was written by Sieva Kozinsky and is associated to the ISBN: 9780130606204.

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