Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems - Page 710: 8c

Answer

The electric potential is zero at the point $~~x = 5.5~m$

Work Step by Step

We can use Equation (24-21) to find the potential difference between two points: $\Delta V = -E~\Delta x$ The area under the $E$ versus $x$ graph is the negative of the potential difference. As we move along the x axis, if the value of $E$ is below the x axis, then the value of the electric potential continues to increase. Thus the value of the electric potential continues to increase until $x = 3.0~m$ which is the point where the electric potential is a maximum. In part (b), we found that the electric potential at $x = 3.0~m$ is $~~40.0~V$ As $x$ increases greater than $x=3.0~m$, the electric field is positive, so the electric potential decreases. We can find the point such that the area under the $E$ versus $x$ graph is $40.0~V$ from $x=3.0~m$ to this point. We can find the area under the $E$ versus $x$ graph between $x = 3.0~m$ and $x = 4.0~m$: $Area = \frac{1}{2}(20.0~N/C)(1.0~m) = 10.0~V$ We can find the area under the $E$ versus $x$ graph from $x = 4.0~m$ and $x = 5.5~m$: $Area = (20.0~N/C)(1.5~m) = 30.0~V$ The total area under the $E$ versus $x$ graph between $x = 3.0~m$ and $x = 5.5~m$ is $40.0~V$. Then the potential difference from $x = 3.0~m$ to $x = 5.5~m$ is $-40.0~V$ Therefore, the electric potential is zero at the point $~~x = 5.5~m$
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