watc x x Bb Wee WAH CBS x M H Sign x EL Netf X Your X Acac X C https:// dep 1361
ID: 1409560 • Letter: W
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watc x x Bb Wee WAH CBS x M H Sign x EL Netf X Your X Acac X C https:// dep 13617636#Q1 For this assignment, you submit answers by question parts. The number of Write out your solutions on paper to these questions. They will be turned in during recitation to Fatimeh during week 2 and graded for completion submissions remaining for each question part only changes if you submit or change the answer. Ed Instructions Assignment Scoring Your last submission is used for your score. Write out your solutions on paper to these questions. They will be turned in during recitation to Fatimeh during week 2 and graded for completion. 1. C+D -6 points SerPSE9 23. P.012 My Notes Three point charges lie along a straight line as shown in the figure below, where q 6.12 HC, q2 1.4 HC, and q3 2.12 HC. The separation distances are d 3.000 cm and d 2.00 cm Calculate the magnitude and direction of the net electric force on each of the charges. 93 (a) q1 magnitude direction Select (b) q2 magnitude direction Select C) 93 magnitude direction Select Need Help? Read it Submit Answer Save Progress Practice Another Version 2. C+ 2/2 points I Previous Answers SerPSE9 23.P.029.MI.FB My Notes In the figure below, determine the point (other than infinity) at which the electric field is zero (Let q1 1.60 HC and q2 7.00 HC.) 0.92 m to the left of qExplanation / Answer
(a)
Net Force on q1,
F1 = - k*q1q2/d1^2 + k*q1q3/(d1+d2)^2
F1 = - (8.9 * 10^9 * 6.12 * 10^-6 * 1.41 * 10^-6)/ (0.03)^2 + (8.9 * 10^9 * 6.12 * 10^-6 * 2.12 * 10^-6)/ (0.05)^2
F1 = - 39.14 N
Magnitude, F1 = 39.14 N
Direction = West !!
(b)
Net Force on q2,
F2 = k*q1q2/d1^2 + k*q2q3/(d1+d2)^2
F2 = (8.9 * 10^9 * 6.12 * 10^-6 * 1.41 * 10^-6)/ (0.03)^2 + (8.9 * 10^9 * 1.41 * 10^-6 * 2.12 * 10^-6)/ (0.02)^2
F2 = 151.84 N
Magnitude, F2 = 151.84 N
Direction = East !!
(c)
Net Force on q3,
F3 = - k*q1q3/d1^2 + k*q2q3/(d1+d2)^2
F3 = (8.9 * 10^9 * 6.12 * 10^-6 * 2.12 * 10^-6)/ (0.05)^2 + (8.9 * 10^9 * 1.41 * 10^-6 * 2.12 * 10^-6)/ (0.02)^2
F2 = 112.7 N
Magnitude, F2 = 112.7 N
Direction = West !!
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