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Equipotential Lines

  1. Jul 4, 2008 #1
    The figure shows the equipotential lines for another electric field. What is the approximate strength of the electric field at points A and B?

    https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1202/summer/homework/Ch-20-Potential/equipotential_lines/equ-lines-4.jpg

    I used the equations:
    Ex = -V/X
    = -(25V-40V)/(0.005m-0m)
    = 3000 V/m
    Ey = -V/X
    = ( 25V-40V)/(0.004m-0m)
    = 3750 V/m
    EA= sqrt (Ex squared + Ey squared)
    = 4802 V/m

    I'm not sure what else to do. Any pointers?
     
  2. jcsd
  3. Jul 4, 2008 #2

    Kurdt

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    Its mainly graph reading I believe with no calculation involved. What do you think A and B will be accoriding to the graph you were given?
     
  4. Jul 4, 2008 #3
    I thought point A would be 4802 V/m. The hint the problem gave me was: Remember that although the electrical potential, V, is a scalar, the electric field, E is a vector. You can break the vector E down into it's components. That is Ex = ΔV/Δx and Ey = ΔV/Δy. Now just add up the components of the electric field to get the total strength. Remember you have to add them as you normally would any vector components.

    I broke up the problem into x and y components like it told me to do, but I still can't get the right answer.
     
  5. Jul 4, 2008 #4

    Kurdt

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    Sorry I thought it said potential.

    I'm not sure why you've chosen those potentials as they're seemingly not related to the distances you have. Have a think about that again.
     
  6. Jul 4, 2008 #5
    I'm not really sure what you're saying but... I chose 25V-40V because point A is in the middle of 30V and 20V and the graph starts at 40V. Point A also seems to be at (0.005m, 0.004m)
     
  7. Jul 5, 2008 #6

    Kurdt

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    The graph doesn't seem to start at 40V from what I'm looking at.
     
  8. Jul 5, 2008 #7
    The x-axis starts at 40V.
     
  9. Jul 6, 2008 #8

    G01

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    That doesn't make any sense. There is no equipotential line for that 40 V marker. Also, the x-axis can't be an equipotential line since multiple equipotentials of different values pass through it.

    I think that 40V marker is a mistake since there is no equipotential line corresponding to it. Try to work out your values using numbers for which there are obvious equipotential lines. Avoid using that number 40V and see what answer you get.
     
  10. Jul 6, 2008 #9
    I tried it again for point A and this is what I got....

    Ex = -V/X
    = -(25V-0V)/(0.005m-0m)
    = 5000 V/m
    Ey = -V/X
    = (25V-0V)/(0.004m-0m)
    = 6250 V/m
    EA= sqrt (Ex squared + Ey squared)
    = 8003.905 V/m

    Am I picking the right values this time? I still can't get the right answer.
     
  11. Jul 6, 2008 #10

    Kurdt

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    At x=0 and the same y value as the point, what is the potential. It isn't zero.
     
  12. Jul 6, 2008 #11
    I'm not really sure what you're asking me....
     
  13. Jul 7, 2008 #12

    Kurdt

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    I'm saying if you move a little x-distance from the point then what is the difference in potential.
     
  14. Jul 7, 2008 #13
    Nevermind...I solved it. Thanks. =)
     
  15. Jul 2, 2011 #14
    I have the same problem how do you know what numbers to use?
     
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