Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 683: 10

Answer

$21cm$

Work Step by Step

We know that $L=\frac{N_e(e)}{q/l}$ We plug in the known values to obtain: $L=\frac{(1.8\times 10^{13})(1.6\times 10^{-9})}{0.14\times 10^{-6}}=21cm$
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