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 777: 23.7


$F = 1.94 \times 10^{-5} \mathrm{N} $

Work Step by Step

The kinetic energy for the two protons equals the potential energy at the maximum value \begin{gather*} 2 K=U \\ 2\left(\frac{1}{2} m V^{2}\right)= \frac{1}{4 \pi \epsilon_{0}} \frac{\left|qq\right|}{r}\\ m V^2 = \frac{1}{4 \pi \epsilon_{0}} \frac{\left|e^{2}\right|}{r}\\ r =\frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{m V^{2}}\\ r =\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(1.602 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(1.67 \times 10^{-27} \mathrm{kg}\right)\left(2.00 \times 10^{5} \mathrm{m} / \mathrm{s}\right)^{2}}\\ r= 3.45 \times 10^{-12} \mathrm{m} \end{gather*} The maximum electric force between the two protons is obtained at $r$ by \begin{aligned} F_{max} &=\frac{1}{4 \pi \epsilon_{\mathrm{o}}} \frac{\left|e^{2}\right|}{r^{2}} \\ &=\left(9.0 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left|\left(1.602 \times 10^{-19} \mathrm{C}\right)^{2}\right|}{\left(3.45 \times 10^{-12} \mathrm{m}\right)^{2}} \\ &=\boxed{1.94 \times 10^{-5} \mathrm{N} } \end{aligned}
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