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 26 - The Electric Field - Exercises and Problems - Page 777: 48

Answer

$0.999455\;\rm cm$

Work Step by Step

We know that the electric field between the two plates is constant and is given by $$E=\dfrac{\eta}{\epsilon_0}=\dfrac{Q}{A\epsilon_0}\tag 1$$ This means that the electron and the proton will be affected by the same magnitude of electric field during their journey from one plate to the other. And since both particles have the same charge magnitude, the electric force exerted on them is equal. $$F=qE$$ According to Newton's law, the acceleration of each one is given by $$F=qE=ma$$ $$a=\dfrac{qE}{m}=\dfrac{q(Q/A\epsilon_0)}{m}\tag{From (1)}$$ Hence, $$a_e=\dfrac{qQ}{A \epsilon_0 m_e}\tag 2$$ and $$a_p=\dfrac{qQ}{A \epsilon_0 m_p}\tag 3$$ Now let's assume that the electron position is the origin at which $x=0$ and the proton is at $x=1$ cm. So when they pass each other $$x_e=x_p$$ Recalling the kinematic formula, $$x_{ie}+v_{ie}t+\dfrac{1}{2}a_et^2=x_{ip}+v_{ip}t-\dfrac{1}{2}a_pt^2$$ The negative sign of the $a_p$ is due to the direction. both start from rest, $$0+0+\dfrac{1}{2}a_et^2=0.01+0-\dfrac{1}{2}a_pt^2$$ $$ a_et^2=0.02 + a_pt^2$$ Plugging from (2) and (3), $$ \dfrac{qQ}{A \epsilon_0 m_e} t^2=0.02 - \dfrac{qQ}{A \epsilon_0 m_p} t^2$$ $$ \dfrac{qQ}{A \epsilon_0 m_e} t^2+ \dfrac{qQ}{A \epsilon_0 m_p} t^2=0.02$$ $$ t^2\dfrac{qQ}{A \epsilon_0 }\left[ \dfrac{1}{m_e}+ \dfrac{1}{ m_p} \right]=0.02$$ Thus, $$t=\sqrt{\dfrac{0.02}{\dfrac{qQ}{A \epsilon_0 }\left[ \dfrac{1}{m_e}+ \dfrac{1}{ m_p} \right]}}\tag 4$$ So, for the electron, $$x_e=x_{ie}+v_{ie}t+\dfrac{1}{2}a_et^2$$ Plug from (2) and (4), $$x_e=0+0+\dfrac{1}{2}a_et^2$$ $$x_e=\dfrac{1}{2}\dfrac{qQ}{A \epsilon_0 m_e}\dfrac{0.02}{\dfrac{qQ}{A \epsilon_0 }\left[ \dfrac{1}{m_e}+\dfrac{1}{ m_p} \right]}$$ $$x_e=\dfrac{1}{2m_e} \dfrac{0.02}{ \left[ \dfrac{1}{m_e}+\dfrac{1}{ m_p} \right]}$$ $$x_e=\dfrac{1}{2 } \dfrac{0.02}{ \left[ 1+ \dfrac{m_e}{ m_p} \right]}$$ Plug the known; $$x_e= \dfrac{0.02}{2 \left[ 1+\dfrac{9.11\times 10^{-31}}{ 1.67\times 10^{-27}} \right]}$$ $$x_e=0.00999455\;\rm m=\color{red}{\bf 0.999455}\;\rm cm$$
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