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 28 - The Electric Potential - Exercises and Problems - Page 835: 42

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Work Step by Step

$$\color{blue}{\bf [a]}$$ We know that the potential inside a parallel-plate capacitor is given by $$V=Es$$ where $s$ is measured from the negative plate. Hence, $$V_{max}=Ed=500\;\rm V$$ And since the proton is fired from the middle point between the two plates, $$V_i=250\;\rm V\tag 1$$ And since the energy is conserved, we need to find the initial speed that makes the proton reach the positive plate. The energy is conserved, so $$E_1=E_2$$ $$U_1+K_1=U_2+K_2$$ where $_1\rightarrow$ is at the middle point between the two plates and $_2\rightarrow$ is at the positive plate at which the full energy of the proton now is potential energy. $$qV_i+K_1=qV_{max}+0$$ where $q=e$ which is the charge of the proton. $$eV_i+\frac{1}{2}mv_1^2=eV_{max} $$ $$ v_1^2=2\left[\dfrac{eV_{max} -eV_i}{m}\right]$$ $$ v_1 =\sqrt{2e\left[\dfrac{ V_{max} - V_i}{m}\right]}$$ Plug from (1) and the rest of the known; $$ v_1 =\sqrt{2(1.6\times 10^{-19})\left[\dfrac{500 -250}{1.67\times 10^{-27}}\right]}$$ $$v_1=\bf 218,870\;\rm m/s$$ which means that the speed of 200,000 m/s is not enough for the proton to reach the positive plate. $$\color{blue}{\bf [b]}$$ The energy is conserved, so $$E_1=E_3$$ $$U_1+K_1=U_3+K_3$$ where $_1\rightarrow$ is at the middle point between the two plates and $_3\rightarrow$ is at the negative. $$eV_i+\frac{1}{2}mv_1^2=eV_{\rm min}+\frac{1}{2}mv_3^2$$ $$eV_i+\frac{1}{2}mv_1^2=0+\frac{1}{2}mv_3^2$$ Solving for $v_3$; $$v_3=\sqrt{\dfrac{2eV_i}{m}+ v_1^2}$$ Plug the known; $$v_3=\sqrt{\dfrac{2(1.6\times 10^{-19})(250)}{1.67\times 10^{-27}}+ (200,000)^2}$$ $$v_3=\color{red}{\bf 2.96\times 10^5}\;\rm m/s$$
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