Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 26 - Potential and Field - Exercises and Problems - Page 737: 4


The potential at $x = 3.0~m$ is $-550~V$

Work Step by Step

We can write an expression for the potential difference: $\Delta V = -\Delta x~E_x$ The potential difference is equal to the the negative of the area under the E versus x graph. We can find the area under the E versus x graph between x = 0 m and x = 3.0 m: $area = (2.0~m)(200~V/m)+\frac{1}{2}(1.0~m)(200~V/m)$ $area = 500~V$ The potential difference between $x_i = 0~m$ and $x_f = 3.0~m$ is $-500~V$. Since the potential at the origin is $-50~V$, then the potential at $x = 3.0~m$ is $-550~V$
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