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Find Force From A Graph Of Potential

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data
    The electric potential along the x-axis (in kV) is plotted versus the value of x, (in m). Evaluate the x-component of the electrical force on a charge of 4.05 C located on the x-axis at -13.0 m.


    2. Relevant equations
    [tex] E_{x}= -\frac{\partial V}{\partial X} [/tex]

    3. The attempt at a solution

    Ok So i drew a tangent line at x=-13 and kv[tex]\approx[/tex]62.5

    I know I must perform Eq=F for the x-axis.
    So I did:
    [tex] E_{x}=-\frac{\partial V}{\partial X}=\frac{f(x+h)-f(x)}{h}[/tex]
    [tex]\rightarrow\frac{f(-13+h)-f(-13)}{h}[/tex] I broke each cube into 5's vertically and thought 1 would be good to equall h so :
    [tex]\frac{f(-12)-f(-13)}{1}[/tex] [tex]\rightarrow \frac{-125-62.5}{1}[/tex]
    Sound good so far? My online HW system says I have to be really accurate with these measurements and such
     

    Attached Files:

  2. jcsd
  3. Sep 9, 2007 #2

    learningphysics

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    I'm seeing voltage as 0 at x=-12.5, and a positive value at x=-13.
     
  4. Sep 9, 2007 #3
    so :
    [tex] \frac{f(-12.5)-f(-13)}{.5}[/tex]
     
    Last edited: Sep 9, 2007
  5. Sep 9, 2007 #4

    Astronuc

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    Staff: Mentor

    Correct. V(-13) > 0 and dV/dx (-12.5) < 0.

    But for accuracy, one wants to minimize dx, or [itex]\Delta x[/itex] in this case.
     
    Last edited: Sep 9, 2007
  6. Sep 9, 2007 #5

    learningphysics

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    Yeah, what you had before was ok, but h=0.5 is better than h=1... smaller the h the better...
     
  7. Sep 9, 2007 #6
    So how about :
    [tex] \frac{f(-12.9)-f(-13)}{.9}= 34.7222kv/m \longrightarrow 34722.22 V/m[/tex]
     
  8. Sep 9, 2007 #7

    learningphysics

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    Although you want h to be small, you want it to be big enough to accurately discriminate between two points... don't make h smaller than 1 division... 1 division on your graph should be good... I'm seeing 14 divisions between -15 and -10... so that's 0.357 per division.

    so maybe use h = 0.357 which is 1 division horizontally.

    -13 + 0.357 = -12.643
     
  9. Sep 9, 2007 #8
    does f(-12.643)=125 sound accurate?
     
  10. Sep 9, 2007 #9

    learningphysics

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    f(-13) should be greater than f(-12.643)....

    from your first post you measured f(-13) at 62.5...

    so one of them isn't right.

    Looking at your first post... you drew the tangent at x=-13 right? Did you get the slope of the line?
     
  11. Sep 9, 2007 #10
    Oops. I meant -125
     
  12. Sep 9, 2007 #11

    learningphysics

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    Hmmm... that doesn't seem right to me either... don't you think f(-12.5) = 0 ? So f(-12.643) would be positive but less than f(-13)...

    The best way might just be to draw the tangent at x = -13, and then get the slope of that line as you had initially mentioned...
     
  13. Sep 9, 2007 #12
    But I don't think it is accurate enough
     
  14. Sep 9, 2007 #13

    learningphysics

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    what do you get as the slope?
     
  15. Sep 9, 2007 #14
    from the top -1.875E2 kv/m
     
  16. Sep 9, 2007 #15

    learningphysics

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    Hmmm... that doesn't look right... where does the line intersect the y- axis. where does it intersect the x - axis?
     
  17. Sep 9, 2007 #16
    ok -250Kv/m is my slope
     
  18. Sep 9, 2007 #17

    learningphysics

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    That sounds much better.
     
  19. Sep 9, 2007 #18
    so I convert -250kv/m to -250000v/m times 4.05E-6 C and get -1.0125 N.
    but it is wrong
     
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