Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 17 - Electric Potential - Problems - Page 496: 23

Answer

$9.15\times10^{6}m/s$.

Work Step by Step

Use the principle of energy conservation. The initial potential energy changes to kinetic energy of the electron when the electron is far away, where PE = 0. The electrical force equals the gravitational attraction. $$PE_i=KE_f$$ $$k\frac{(-e)(Q)}{r}= \frac{1}{2}mv^2$$ $$v=\sqrt{ \frac{2k(-e)Q}{mr}}$$ $$v=\sqrt{\frac{2(8.988\times10^{9}) (-1.60\times10^{-19}C)(-6.50\times10^{-9}C) }{(9.11\times10^{-31}kg)(0.245m)}}$$ $$= 9.15\times10^{6}m/s $$
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