Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 36 - Relativity - Exercises and Problems - Page 1061: 57

Answer

$\Delta V = 3.1\times 10^{6}~V$

Work Step by Step

We can find $\gamma$: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{(0.99~c)^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-0.9801}}$ $\gamma = 7.09$ We can find the kinetic energy after the electron has been accelerated: $K = (\gamma -1)~mc^2$ $K = (7.09 -1) (9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $K = 5.0\times 10^{-13}~J$ We can find the required potential difference: $\Delta V~q = K$ $\Delta V = \frac{K}{q}$ $\Delta V = \frac{5.0\times 10^{-13}~J}{1.6\times 10^{-19}~C}$ $\Delta V = 3.1\times 10^{6}~V$
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