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 - General Problems - Page 471: 48

Answer

a. $2.64\times10^{13}\;m/s^2$ up b. $1.44\times10^{10}\;m/s^2$ down c. Electron $\frac{a}{9.8m/s^2}\approx2.7\times10^{12}$ and proton $\frac{a}{9.8m/s^2}\approx1.5\times10^{9}$

Work Step by Step

a. Assume that this is the only force acting on the electron. Use equation 16–3 to calculate the magnitude of the force, and apply Newton’s second law. $$\vec{E}=\frac{\vec{F}}{q}$$ $$\vec{F}=q\vec{E}=m\vec{a}$$ $$a=\frac{|q|}{m}E=\frac{(1.602\times10^{-19}C)}{9.11\times10^{-31}kg} (150\;N/C)$$ $$= 2.64\times10^{13}\;m/s^2$$ The electron is negatively charged, and the field is downward, so the force acting on it is opposite to the field, or upward. So is the acceleration. b. Assume that this is the only force acting on the proton. Use equation 16–3 to calculate the magnitude of the force, and apply Newton’s second law. $$\vec{E}=\frac{\vec{F}}{q}$$ $$\vec{F}=q\vec{E}=m\vec{a}$$ $$a=\frac{|q|}{m}E=\frac{(1.602\times10^{-19}C)}{1.67\times10^{-27}kg} (150\;N/C)$$ $$= 1.44\times10^{10}\;m/s^2$$ The proton is positively charged, and the field is downward, so the force acting on it is downward. So is the acceleration. c. For the electron, $\frac{a}{9.8m/s^2}\approx2.7\times10^{12}$. For the proton, $\frac{a}{9.8m/s^2}\approx1.5\times10^{9}$.
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