University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 23 - Electric Potential - Problems - Exercises - Page 779: 23.29

Answer

$v = 150 \mathrm{~m/s}$

Work Step by Step

From the conservation law, we could get the speed \begin{gather} K_{a}+U_{a} =K_{b}+U_{b} \\ K_{a} =U_{b} \\ \frac{1}{2} m v^{2} =\frac{1}{4 \pi \epsilon_{0}} \frac{Q q}{r}\tag{Solve for $v$}\\ v =\sqrt{\frac{1}{4 \pi \epsilon_{0}}\left[\frac{2 Q q}{m r}\right]} \\ v = \sqrt{\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left[\frac{2\left(5.00 \times 10^{-6} \mathrm{C} \times 3.00 \times 10^{-6} \mathrm{C}\right)}{\left(6.0 \times 10^{-5} \mathrm{kg}\right)(0.20 \mathrm{m})}\right]} \\ v = \boxed{150 \mathrm{~m/s}} \end{gather}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.