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Evaluating electric field

  1. Apr 30, 2008 #1
    [SOLVED] Evaluating electric field

    1. The problem statement, all variables and given/known data
    I have an electric field in the z-direction given by: (where sigma is charge per area and z is a distance)

    [tex]
    {\bf{E}} = \frac{{\sigma z}}{{2\pi \varepsilon _0 }}\left( {\frac{1}{z} - \frac{1}{{\sqrt {R^2 + z^2 } }}} \right){\bf{z}}
    [/tex]

    I have to evaluate this for z >> R.

    3. The attempt at a solution

    Do I just insert R=0 or what? I overheard someone talk about Taylor-expanding it, but I don't see how/why?
     
  2. jcsd
  3. Apr 30, 2008 #2

    George Jones

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    Take a factor [itex]z^2[/itex] out of the [itex]R^2 + z^2[/itex] term.
     
  4. Apr 30, 2008 #3
    Thanks, but why?

    What is the method when I am asked to evalute an expression for some variable >> some other variable?
     
  5. Apr 30, 2008 #4

    George Jones

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    After you do it, I'll give the motivation for the step.
     
  6. Apr 30, 2008 #5
    I get 1/R instead of 1/sqrt(R^2+z^2).
     
  7. Apr 30, 2008 #6

    George Jones

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    I was looking for

    [tex]R^2 + z^2 = z^2 \left( \frac{R^2}{z^2} + 1 \right).[/tex]

    What can you say about the first term inside the brackets?
     
  8. Apr 30, 2008 #7
    It goes to 0 if z >> R.
     
  9. Apr 30, 2008 #8

    George Jones

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    Right.

    Now you something of the form

    [tex]\left( 1 + a \right)^{-1/2},[/tex]

    with [itex] |a| << 1. [/itex]

    Can you write down the Taylor series expansion of the above expression?
     
  10. Apr 30, 2008 #9
    I think I get:

    1-x/2 - the first two orders. How does that sound?
     
  11. Apr 30, 2008 #10

    George Jones

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    Yes.

    Now use all this in the original expression.
     
  12. Apr 30, 2008 #11
    Then I get:

    1/z - (1-z/2).

    This is for inside the brackets. Then I multiply out and finish?
     
  13. May 1, 2008 #12
    What is the motivation for the step?
     
  14. May 2, 2008 #13

    Redbelly98

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    Not quite. Recall that the "x" in your earlier equation is actually R/z.
    Also, you have lost a factor of [tex]z^2[/tex] somewhere (the one that appears outside the brackets on the right-hand-side in George's post #6).
     
  15. May 2, 2008 #14

    Redbelly98

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    The motivation is to get something other than zero when you approximate this part of the expression:

    [tex]
    \left( {\frac{1}{z} - \frac{1}{{\sqrt {R^2 + z^2 } }}} \right)
    [/tex]
     
  16. May 3, 2008 #15
    Cool, I get it now. Thanks for being so kind.
     
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