Electric Potential, Electric Potential Energy, Electric Field

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SUMMARY

The discussion focuses on calculating the positions of a +6.2-µC charge along the x-axis where its electric potential energy (U) equals specific values, specifically 0.000124 J and 0.000217 J. The user attempted to use the equations E = -ΔV/Δx, F = Eq, and W = -ΔU but encountered difficulties in determining the correct positions (x1, x2, x3). The provided graph link was not functional, limiting the ability to visualize the electric potential changes necessary for accurate calculations.

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  • Understanding of electric potential and electric potential energy concepts
  • Familiarity with the equations E = -ΔV/Δx and W = -ΔU
  • Basic knowledge of forces in electric fields
  • Ability to interpret graphical data related to electric potential
NEXT STEPS
  • Review the relationship between electric potential and electric potential energy
  • Study how to analyze electric field graphs for potential energy calculations
  • Learn about the implications of charge movement in electric fields
  • Explore practical applications of electric potential in circuit design
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Students studying electromagnetism, physics educators, and anyone involved in electric field analysis and potential energy calculations.

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Homework Statement


As a +6.2-µC charge moves along the x-axis from x = 0 to x = 1.4 m, the electric potential it experiences is shown in the figure, below. (The horizontal axis is marked in increments of 0.1 m and the vertical axis is marked in increments of 5 V.) Find the approximate location(s) of the charge when its electric potential energy U equals the the following values. (If there are more than one answer, enter your answers from smallest to largest. If there is only one answer, enter your answer in the first answer box and enter "none" in the following boxes.)


Here is the graph: http://www.webassign.net/walker/20-22alt…

(a) U = 0.000124 J
x1 = m
x2 = m
x3 = m

(b) U = 0.000217 J
x1 = m
x2 = m
x3 = m


Homework Equations





The Attempt at a Solution



E=-[itex]\Delta[/itex]V/[itex]\Delta[/itex]x and F=Eq and W=Eqd and W= -[itex]\Delta[/itex]U. So for the a). X1 i found the slope for region one which I found to be 50, then I used the Force equation and got 3.81e8 N then plugging those values into my work equation, solving for d I got 4e-13. But this is wrong any help in the right direction would be nice.. Thanks in advance.
 
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It seems that your link is not working.
 
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20-22alt.gif
 

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