College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 18 - Problems - Page 698: 3


(a) The current flows from the Anode to the Filament. (b) $I = 9.6\times 10^{-7}~A$

Work Step by Step

(a) The electrons move from the Filament to the Anode. However, we think of the direction of current as the movement of positive charges. Since electrons have negative charges, the current flows from the Anode to the Filament. (b) We can find the current: $I = (6.0\times 10^{12}~electrons/s)(1.6\times 10^{-19}~C/electron)$ $I = 9.6\times 10^{-7}~A$
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