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 29 - Potential and Field - Exercises and Problems - Page 863: 35

Answer

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Work Step by Step

$$\color{blue}{\bf [a],[c]}$$ We can use any software calculator to draw these two functions, as shown below. See the graphs below. $$\color{blue}{\bf [b]}$$ We know that $$\Delta V=V_f-V_i=-\int_i^fE_xdx$$ where $E_x=5000 x$, $$\Delta V =-5000\int_{x_i}^{x_f}x dx =\dfrac{-5000}{2}x^2\bigg|_{x_i}^{x_f}$$ $$\Delta V =-2500(x_f^2-x_i^2)$$ At $x_i=0$, $V_i=0$, $$\Delta V =-2500(x_f^2-0^2)$$ replacing $x_f$ by $x$, $$\boxed{\Delta V =-2500x^2 }$$
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