Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 25 - Electric Charges and Forces - Exercises and Problems - Page 747: 57

Answer

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Work Step by Step

$$\color{blue}{\bf [a]}$$ Let's assume that the electron accelerates from rest, so $v_i=0$ and $K_i=0$. And then its final speed is given by the kinetic energy it gains $$K_f=\frac{1}{2}m v_f^2$$ Hence, $$v_f^2=\dfrac{2K_f}{m }\tag 1$$ This is the final velocity of the electron and if we assume that it is speeding up at a constant rate, $a=$constant. Hence, we can use the kinematic formula of $$v_f^2=v_i^2+2ad=0+2ad$$ Thus, its acceleration $$a=\dfrac{v_f^2}{2d}$$ Plugging from (1); $$a=\dfrac{2K_f}{2md}=\dfrac{K_f}{md}$$ Plugging the known; $$a =\dfrac{(2\times 10^{-18})}{(9.11\times 10^{-31})(2\times 10^{-6})}$$ $$a=\color{red}{\bf 1.1 \times 10^{18}}\;\rm m/s^2$$ $$\color{blue}{\bf [b]}$$ According to Newton's second law, the force exerted on the electron to give it this acceleration is given by $$ F=ma=(9.11\times 10^{-31})(1.1 \times 10^{18})$$ $$F=\color{red}{\bf 1.0 \times 10^{-12}}\;\rm N$$ $$\color{blue}{\bf [c]}$$ We know that the electric field is given by $$E=\dfrac{F}{q}=\dfrac{1.0 \times 10^{-12}}{1.6\times 10^{-19}}$$ $$E=\color{red}{\bf 6.25\times 10^6}\;\rm N/C$$ $$\color{blue}{\bf [d]}$$ We need the point charge to give the electron a force as we calculated in part (b). So according to Coulomb's law, $$F=\dfrac{kq_1q_2}{r^2}$$ where $q_1=q_e$, and $q_2 $ is the charge of the point charge. $$F=\dfrac{kq_eq_2}{r^2}$$ Hence, $$q_2=\dfrac{Fr^2}{kq_e}$$ Noting that we are neglecting the signs of the charges and focusing on the charge magnitude. Plug the known; $$q_2=\dfrac{ (1\times 10^{-12})(1\times 10^{-2})^2}{(8.99\times 10^{9})(1.6\times 10^{-19})}$$ $$q_2=6.95\times 10^{-8}\;\rm C\approx\color{red}{ \bf 70}\;nC$$
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