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Electric Field along bisector of charged line (w/error)

  1. Apr 17, 2016 #1
    1. The problem statement, all variables and given/known data
    You wish to determine the electric field magnitude along the perpendicular bisector of a 230-mm line along which35 nC of charge is distributed uniformly. You want to get by with a minimal amount of work, so you need to know when it is sufficient to approximate the line of charge as a charged particle.

    At what distance along the perpendicular bisector does your error in E reach 5 % when you use this approximation?

    2. Relevant equations
    dE=kλdx/r2
    E=kQ/r2
    λ=Q/L


    3. The attempt at a solution
    If I do the approximation part first, I get:

    1. Eapprox=kQ/y2
    Plugging in the values (minus units at this time):
    2. Eapprox=315/y2

    So now I work on the Eline-charge.

    Because point P at distance d is along the bisector of the line, I know that Ex cancels itself out. All I need to worry about is Ey.

    3. dE=((kλdx)/(x2+y2)2)*(y/(x2+y2)1/2)
    4. dE=(kλydx)/(x2+y2)3/2)
    =>
    5. E=kλ∫(ydx/(x2+y2)3/2
    I punch that through an integral calculator, and I get
    6. E=kλ[x/y(x2+y2)1/2] ---- evaluated from -L/2 to L/2

    Now I start to plug in all the values (minus the units for the time being) and I get

    7. E=1370[0.23/(y(.013+y2)1/2)]
    8. E=315.1/(y(.013+y2)1/2)

    Okay.

    Now I'm working on finding the approximate error.

    error=(|Eapprox-Eexact|)/Eexact

    I know the error (5%) and now I want to solve for y at that error.

    9. 0.05=|315/y2-315.1/(y(.013+y2)1/2)|/315.1/(y(.013+y2)1/2)

    This is pretty nasty, so I plugged it in the Mathway calculator and it promptly informed me there are no solutions.

    That sums up what I've done. Where am I going wrong?

    Thanks,

    SOG
     
  2. jcsd
  3. Apr 17, 2016 #2

    haruspex

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    I suspect you are going a much more elaborate path than intended. The objective is to avoid such complications.
    Where does the error come from in pretending the charge is a point particle? Can you think of a simple geometric argument for deciding when that error will be less than 5%? I don't think you need to be too precise about it, just able to say "when the distance is more than .... the error will be less than 5%".
     
  4. Apr 17, 2016 #3
    Well... I could probably compare the radii for both. In which case, I would have:

    0.05 = |d2-(d2+0.013)|/d2
    0.05=0.013/d2
    d=0.51 m

    Am I on the right track there?
     
  5. Apr 17, 2016 #4

    haruspex

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    That's what I had in mind, except I'm not sure how you arrived at 0.013m. That's a lot less than half 230mm.
     
  6. Apr 17, 2016 #5
    0.013m2 is ~ (230mm/2)2.
     
  7. Apr 17, 2016 #6
    I switched some numbers around in my second post, so it screw some more things up. I think it's supposed to be:

    0.05=|d2-(d2+0.013)|/(d2+0.013)

    in which case it gives d=0.501m,

    which is wrong.
     
  8. Apr 17, 2016 #7

    haruspex

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    Ah, sorry - didn't notice it was not squared in the equation you wrote. So, yes, I think that's fine.
    If you wanted to refine it some, you could show that 1/r2 at the point half way from the middle of the charge to one end is more than the average of the values at the middle and at the end respectively. But that's probably overkill.

    [Edit: I've not checked this carefully, but I think this argument might allow you to use 0.013/4.]
     
  9. Apr 17, 2016 #8

    haruspex

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    Oh - so you know what the answer is supposed to be?
     
  10. Apr 17, 2016 #9
    I don't, unfortunately. My online homework only tells me that it's wrong.
     
  11. Apr 17, 2016 #10

    haruspex

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    The refinement I suggested in post #7 gets it down to 0.26m. The exact value is about 0.21m.
     
  12. Apr 17, 2016 #11
    Ok, the correct answer is d=0.36m. I don't know how they come up with that answer.
     
  13. Apr 17, 2016 #12

    haruspex

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    I just realised what we both forgot, to take only the component in the x direction. So we should use 1/r3, not 1/r2.
     
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