A charge of Q1 = -3.0 µC is fixed in place. From a horizontaldistance of S = 0.045m a particle of mass m 7.2x103 kg andcharge Q2 = -8.0 µC is fired with an initial speed of Vo = 65 m/sdirectly towards the fixed charge Q1.What is the minimum distance between Q1 and Q2 before Q1 flips%3Dits direction of motion?O 0.0054 mO 0.0108 mO 0.0207 mO 0.0309 m0.0385 mO 0.0285 m

Given: Charge, Q1 = -3.0 μC = -3.0 × 10-6 Ccharge, Q2 = -8.0 μC = -8.0 × 10-6 C horizontal distance,…

Answered: A charge of Q1 = -3.0 µC is fixed in…

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)...

See similar textbooks

Related questions

Concept explainers

Question

100%

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:

${Charge, Q}_{1} {= -3.0 μC = -3.0 × 10}^{- 6} C charge, Q_{2} = - 8.0 μC = - 8 {.0 \times 10}^{- 6} C horizontal distance, S = 0.045 m mass of particle, m = 7.2 \times 10^{- 3} kg Initial speed of particle, v_{0} = 65 m / s$

Solution:

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

$K . E_{i} + P . E_{i} = K . E_{f} + P . E_{f} \frac{1}{2} m {v_{0}}^{2} + k \frac{q_{1} q_{2}}{r} = 0 + k \frac{q_{1} q_{2}}{x} \frac{1}{2} m {v_{0}}^{2} + k \frac{q_{1} q_{2}}{r} = k \frac{q_{1} q_{2}}{x} {where, x is the distance of point B to charge Q}_{1}$

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions