Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.3 Conservative Vector Fields - Exercises - Page 946: 28

Answer

(a) a potential function for ${\bf{F}}$ is $ - 200\ln \left( {{x^2} + {z^2}} \right)$. (b) the electron's speed at $Q$ is ${v_Q} \approx 1.59 \times {10^7}$ m/s.

Work Step by Step

(a) We have an electric field (in newtons per coulomb) given by ${\bf{F}}\left( {x,y,z} \right) = 400{\left( {{x^2} + {z^2}} \right)^{ - 1}}\left( {x,0,z} \right)$ Since the electric field is conservative, let the potential energy for the electric force ${q_e}{\bf{F}}$ be $V\left( {x,y,z} \right)$ such that ${q_e}{\bf{F}} = - \nabla V = - \left( {\dfrac{{\partial V}}{{\partial x}},\dfrac{{\partial V}}{{\partial y}},\dfrac{{\partial V}}{{\partial z}}} \right)$. So, $\nabla V = - 400{q_e}{\left( {{x^2} + {z^2}} \right)^{ - 1}}\left( {x,0,z} \right)$ 1. taking the integral of $\dfrac{{\partial V}}{{\partial x}}$ with respect to $x$ gives $V\left( {x,y,z} \right) = - 400{q_e}\smallint \dfrac{x}{{{x^2} + {z^2}}}{\rm{d}}x = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + f\left( {y,z} \right)$ 2. taking the integral of $\dfrac{{\partial V}}{{\partial y}}$ with respect to $y$ gives $V\left( {x,y,z} \right) = - 400{q_e}\smallint 0{\rm{d}}y = 0 + g\left( {x,z} \right)$ 3. taking the integral of $\dfrac{{\partial V}}{{\partial z}}$ with respect to $z$ gives $V\left( {x,y,z} \right) = - 400{q_e}\smallint \dfrac{z}{{{x^2} + {z^2}}}{\rm{d}}x = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + h\left( {x,y} \right)$ Since the three ways of expressing $V\left( {x,y,z} \right)$ must be equal, we get $ - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + f\left( {y,z} \right) = g\left( {x,z} \right) = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + h\left( {x,y} \right)$ From here we conclude that the potential energy is $V\left( {x,y,z} \right) = - 200{q_e}\ln \left( {{x^2} + {z^2}} \right) + C$, where $C$ is a constant. Thus, a potential function for ${\bf{F}}$ is $ - 200\ln \left( {{x^2} + {z^2}} \right)$. (b) By the Law of Conservation of Energy, the total energy $E$ is constant and therefore $E$ has the same value when the electron is at $P = \left( {5,3,7} \right)$ and at $Q = \left( {1,1,1} \right)$. So, $E = \dfrac{1}{2}{m_e}{v_P}^2 + {V_P} = \dfrac{1}{2}{m_e}{v_Q}^2 + {V_Q}$ where ${v_P}$ and ${V_P}$ are the velocity and potential energy of the electron at $P$, respectively. Likewise, ${v_Q}$ and ${V_Q}$ are the velocity and potential energy of the electron at $Q$, respectively. Since ${v_P} = 0$, we obtain $\dfrac{1}{2}{m_e}{v_Q}^2 = {V_P} - {V_Q}$ ${v_Q} = \sqrt {\dfrac{2}{{{m_e}}}\left( {{V_P} - {V_Q}} \right)} $ ${v_Q} = \sqrt {\dfrac{2}{{{m_e}}}\left[ { - 200{q_e}\ln \left( {{x_P}^2 + {z_P}^2} \right) + 200{q_e}\ln \left( {{x_Q}^2 + {z_Q}^2} \right)} \right]} $ $ = \sqrt {400\dfrac{{{q_e}}}{{{m_e}}}\left[ { - \ln \left( {25 + 49} \right) + \ln \left( {1 + 1} \right)} \right]} $ $ = \sqrt {400\left( { - 1.76 \times {{10}^{11}}} \right)\cdot\left[ { - \ln \left( {25 + 49} \right) + \ln \left( {1 + 1} \right)} \right]} $ $ \approx 1.59 \times {10^7}$ So, the electron's speed at $Q$ is ${v_Q} \approx 1.59 \times {10^7}$ m/s.
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