A charge of Q1 = -3.0 µC is fixed in place. From a horizontal distance of S = 0.045m a particle of mass m = 7.2x103 kg and charge Q2 = -8.0 µC is fired with an initial speed of Vo = 65 m/s directly towards the fixed charge Q1. What is the minimum distance between Q1 and Q2 before Q1 flips its direction of motion? O 0.0054 m O 0.0108 m O 0.0207 m O 0.0309 m 0.0385 m O 0.0285 m

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter23: Faraday’s Law And Inductance
Section: Chapter Questions
Problem 64P
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A charge of Q1 = -3.0 µC is fixed in place. From a horizontal
distance of S = 0.045m a particle of mass m 7.2x103 kg and
charge Q2 = -8.0 µC is fired with an initial speed of Vo = 65 m/s
directly towards the fixed charge Q1.
What is the minimum distance between Q1 and Q2 before Q1 flips
%3D
its direction of motion?
O 0.0054 m
O 0.0108 m
O 0.0207 m
O 0.0309 m
0.0385 m
O 0.0285 m
Transcribed Image Text:A charge of Q1 = -3.0 µC is fixed in place. From a horizontal distance of S = 0.045m a particle of mass m 7.2x103 kg and charge Q2 = -8.0 µC is fired with an initial speed of Vo = 65 m/s directly towards the fixed charge Q1. What is the minimum distance between Q1 and Q2 before Q1 flips %3D its direction of motion? O 0.0054 m O 0.0108 m O 0.0207 m O 0.0309 m 0.0385 m O 0.0285 m
Expert Solution
Step 1

Given:

Charge, Q1 = -3.0 μC = -3.0 × 10-6 Ccharge, Q2 = -8.0 μC = -8.0 × 10-6 C horizontal distance, S = 0.045 mmass of particle, m = 7.2 ×10-3 kgInitial speed of particle, v0 = 65 m/s

Solution:

We know that, from the conservation of energy we have:

K.Ei + P.Ei = K.Ef + P.Ef12mv02 + kq1q2r = 0 +  kq1q2x12mv02 + kq1q2r=kq1q2xwhere, x is the distance of point B to charge Q1

 

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