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Chrome File Edit View History Bookmarks People Window Help A3 22% D Fri 11:09 AM Bryan O'Connor a E x Bb Reading Handouts 2016 x Chegg Study I Guided Sol x Bryan Chegg Study I Guided Sol Wiley PLUS edugen.wileyplus.com /edugen/student/mainfr.un Google If Facebook D this song is sick g Goodreads l Recent G5 reddit: the front page Blackboard Learn App Other Bookmarks Times M Assignment Open Assignment FULL scREEN PRINTER vERSION BACK NEXT ASSIGNMENT RESOURCES Chapter 21, Problem 063 Assignment 5. Your answer is partially correct. Try again. Chapter 21 S16 M Chapter 21, Concept Question 01 Two point charges of 30 nC and -72 nC are held fixed on an x axis, at the origin and at x 84 cm, respectively. A particle with a charge of 42 HC is M Chapter 21, Concept 02 released from rest at x 42 cm. If the initial acceleration of the particle has a magnitude of 120 km/s what is the particle's mass? M Chapter 21, Concept 03 Number 2.186e-6 Units M Chapter 21, Concept kg Question 06 M Chapter 21, Concept the tolerance is +/-2% 09 M Chapter 21 Problem 004 Chapter 21 Problem SHOW HINT 040 SN Ch Prob LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE MATH HELP LINK TO TEXT 010 Chapter 21 Problem 061 M Chapter 21 Problem 062 Question Attempts: 4 of 6 used SAVE FOR LATER SUBMIT ANSWER M Chapter 21 Problem 069 M Chapter 21 Problem 052 Copyright C 2000-2016 by John Wiley & Sons, Inc. or related companies. All rights reserved 21, Prob 063 M Chapter 21 Problem 050 License Agreement l Privacy Policy I G 2000-2016 John Wiley & Sons Inc. All Rights Reserved. A Division of John Wiley & Sons Inc. 4.17.3.3 Show All ModAmerica. Chapt ...doc

Explanation / Answer

We have a charge 30 nC at the origin , - 72 nC is at 84 cm
and third charge is 42 uC is at the 42 cm .
Now the force between the 30 nC and the 42 uC will be repulsive as they are both positive charge.
So force will be
F1 = 9*109*(30*10-9)*(42*10-6) / (42*10-2)2 = 6.428*10-2 N
Now the force between the charge 42 uC and the cahrge -72 nC , force between these two will be attractive
therefore
F2 = 9*109*(72*10-9)*(42*10-6) / (42*10-2)2 = 15.428*10-2 N
Both forces will have same direction therefore net force will be
F = F1 + F2
F = 21.856*10-2 N
We know that Force = mass*Acceleration
Mass = Force/ acceleration = 21.856*10-2 / (120*103) = 1.8213*10-6 kg

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