College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 16 - Electric Potential, Energy, and Capacitance - Learning Path Questions and Exercises - Exercises - Page 591: 4



Work Step by Step

$E=1000 V/m$ Mass = $m= 9.1\times10^{-31}kg$ The electron accelerates in vertically downward direction. $a_{e}=\frac{E}{mass}=\frac{1000}{9.1\times10^{-31}}=1.09\times10^{33}m/s^{2}$ Now from the equation, $v^{2}=u^{2}+2aS$ with $u=0$, $a$ is calculated above, $S=0.1cm$, we can calculate $v$ So, $v=1.48\times10^{15}m/s$
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