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Need help in applying Binomial Aproximation when y L?

  1. Mar 21, 2015 #1
    1. The problem statement, all variables and given/known data
    So I am working on an Electric Potential problem. There is a point P that is located on top of this rod ( this rod is aligned horizontally & is length L). I've solved this problem and got an answer.

    I want to find when y>>L using Binomial Approximation except I am quite lost on how to apply it? If someone can please help.

    The answer that I got while doing this from finding the electric potential is:


    V = 2kα [(L^2 / 16) + y^2 )^(1/2) - y ]

    How would I use Binomial Approximation to get y>>L? Thanks in advance!

    2. Relevant equations

    Screen shot 2015-03-21 at 5.39.56 PM.png
     
    Last edited: Mar 21, 2015
  2. jcsd
  3. Mar 21, 2015 #2

    Matternot

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    The (L2/16)1/2 can be simplified to L/4. The equation doesn't look like that which should be binomially expanded. If y>>L, the L/4 can be neglected, resulting in -y2kα.

    Binomial expansion can be used on something of the form you have given in the "Relevant equation". I would suggest the 1/2 is maybe around the whole bracket, but then the equation isn't dimensionally correct.
     
  4. Mar 21, 2015 #3
    Oh my mistake I wrote the answer incorrectly:
    What I meant was this:

    V = 2kα [(L^2 / 16) + y^2 )^(1/2) - y ]
     
  5. Mar 21, 2015 #4

    Matternot

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    How sure are you with that equation? It doesn't seem right. You would expect it to reduce to the point particle potential for y<<L. It doesn't really seem dimensionally consistent with a potential.

    I would expect it to be divided by something like (sqrt(ay2+bL2)
     
  6. Mar 21, 2015 #5

    Simon Bridge

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    We need to see the original problem statement - how do you know your answer is correct?
     
  7. Mar 21, 2015 #6
    The problem states a rod -- nonuniformly. A point P located above center-rod (y distance away). Find e-potential at that point.
     
  8. Mar 21, 2015 #7

    Matternot

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    Aah, that can be expanded binomially:

    Fisrtly, factor out y to reduce to (L2/16y+1)1/2

    Then use (1+x)n = 1 + nx/1! + n(n-1)(x2)/2! +...

    When you see L2/y2 or higher terms, they can be ignored. This will reduce the formula down
     
  9. Mar 21, 2015 #8

    Matternot

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    Given the problem you suggested, I would really consider looking at the first part again. The solution should reduce to the typical point particle potential if y>>L... I don't see this happening with your current solution
     
  10. Mar 21, 2015 #9
    How did you factor the y out of

    V = 2kα [(L^2 / 16) + y^2 )^(1/2) - y ]

    Especially with the y^2 being in the square root still?
     
  11. Mar 21, 2015 #10

    Matternot

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    [L2+y2]1/2 = y[L2/y2+1]1/2
     
  12. Mar 21, 2015 #11

    Matternot

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    But again.... I still expect something of the form

    E=kQ/y[L2+y2]1/2

    with maybe a few more constants flying around

    Try looking at this website
     
  13. Mar 21, 2015 #12
    I do agree with you that it indeed should look like a point charge when y>>L.

    Also for the answer you provided:
    [L2+y2]1/2 = y[L2/y2+1]1/2

    Where did you L^2/4 (the 4 part in denominator go?)
     
  14. Mar 21, 2015 #13

    Matternot

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    I just ignored the constant... If I must:

    [(L2/16) + y2 )1/2 - y ] = y[[L2/16*y2+1]1/2-1]
     
  15. Mar 21, 2015 #14
    Okay I got an answer of 4kQ/y using the binomial expansion.
     
  16. Mar 21, 2015 #15

    Matternot

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    Are you sure about the 4? I expect kQ/y
     
  17. Mar 21, 2015 #16
    Yeah sorry for being so brief about the problem, but I substituted in my alpha--which I found during the process of the problem within the first step. I really appreciate your help.

    As for the problem so far, when y approaches zero, I get the same answer as when y>>L but with the denominator being L.

    For when y>>L: kQ/y.
     
  18. Mar 21, 2015 #17

    Matternot

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    Nice! Glad I could help :smile:
     
  19. Mar 21, 2015 #18
    If i were to graph both of those as E-potential vs y distance.

    Wouldn't it just be a 1/r graph on the positive side with it being a nonuniform rod? --Correct?
     
  20. Mar 21, 2015 #19
    And yeah my bad I forgot to cancel out the 4s-- got kq/y :)
     
  21. Mar 21, 2015 #20

    Matternot

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