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 688: 88

Answer

Please see the work below.

Work Step by Step

(a) We can find the number of electrons as follows: $N_p=\frac{M}{m_p}$ $\implies N_p=\frac{4.56\times 10^{-23}}{1.673\times 10^{-27}}=2.73\times 10^4 protons$ $q_e=Q-q_p$ $\implies q_e=1.84\times 10^{-15}-(2.73\times 10^4)(1.6\times 10^{-19})=-2.53\times 10^{-15}C$ Now $N_e=\frac{q_e}{-e}$ We plug in the known values to obtain: $N_e=\frac{-2.53\times 10^{-15}}{-1.6\times 10^{-19}}$ $N_e=1.58\times 10^4electrons$ (b) We can find the number of protons as $N_p=\frac{M}{m_p}$ $\implies N_p=\frac{4.56\times 10^{-23}}{1.673\times 10^{-27}}=2.73\times 10^4 protons$
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