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 - General Problems - Page 499: 80

Answer

$1.03\times10^6m/s$

Work Step by Step

The kinetic energy of the ejected electrons becomes potential energy at the top plate. Energy is conserved. $$KE_i=PE_f$$ $$\frac{1}{2}mv^2=qV$$ $$v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-1.60\times10^{-19}C)(-3.02V)}{9.11\times10^{-31}kg}}$$ $$=1.03\times10^6m/s$$
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