Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 16 - Electric Charge and Electric Field - Problems - Page 469: 10


$\frac{F_e}{F_g}=2.3\times 10^{39}$

Work Step by Step

We know that the electric force between electron and proton is given as $F_e=K\frac{q_1q_2}{r^2}$..........eq(1) We also know that the gravitational force between proton and electron is given as $F_g=G\frac{m_1m_2}{r^2}$.....eq(2) Dividing eq(1) by eq(2), we obtain: $\frac{F_e}{F_g}=\frac{Kq_1q_2}{Gm_1m_2}$ As charge on electron and proton is same so $q_1=q_2=q$ $\frac{F_e}{F_g}=\frac{Kq^2}{Gm_1m_2}$ We plug in the known values to obtain: $\frac{F_e}{F_g}=\frac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{6.67\times 10^{-11}(1.67\times 10^{-27})(9.11\times 10^{-31})}$ $\frac{F_e}{F_g}=2.3\times 10^{39}$
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