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 498: 69

Answer

$1.3\times10^{-12}V$

Work Step by Step

The electric force has the same magnitude as the electron’s weight. $$F_{electric}=F_{gravity}$$ The magnitude of the electric force is the magnitude of the electronic charge, multiplied by the electric field strength. The electric field is the potential difference divided by the distance. $$q\frac{V}{d}=mg$$ $$V=\frac{mgd}{q}=\frac{(9.11\times10^{-31}kg)(9.80m/s^2)(0.024m)}{1.60\times10^{-19}C}$$ $$=1.3\times10^{-12}V$$ This voltage is minuscule. Electrons can easily move upward against the force of gravity.
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