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Proof that disk of charge = point charge when very far?

  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Take the expression 21.11 (pictured below, specifically the bottom one) for the electric field above the center of a uniformly charged disk with radius R and surface charge density σ, and show that when one is very far from the disk, the field decreases with the same square of the distance as it would for a point charge, E ~ q/4πεx2, where q is the total charge on the disk.

    Screen_Shot_2016_02_25_at_9_04_14_PM.png

    2. Relevant equations


    3. The attempt at a solution
    I solved one like this pretty easily where you prove that a long wire acts as an infinite wire, but I've been looking at this one for about an hour and I'm stumped. I know I need to get x2 on the bottom of the equation without being inside a square root, but I don't know how. The only approximation I can think of is that when R is much bigger than x, √(x2 + R2 + 1) goes to 1, but then I don't have an x anymore.
     
  2. jcsd
  3. Feb 25, 2016 #2

    TSny

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  4. Feb 25, 2016 #3

    haruspex

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    I think you mean x much larger than R.
    Do you know the binomial expansion of (1+t)a for small t?

    Edit: must learn to post more quickly.
     
  5. Feb 25, 2016 #4
    Yes, I meant if x is much larger than R then √(R2/x2 + 1) goes to 0 (hopefully I said it right that time). I haven't done binomial expansions, for a while so I'll review those, thanks.
     
    Last edited: Feb 25, 2016
  6. Feb 25, 2016 #5
    Ok I'm still stuck. I used σ = q/πR2 and did the binomial expansion and got this:
    Screen_Shot_2016_02_25_at_10_19_00_PM.png

    EDIT: that should be R2/2x2
     
  7. Feb 25, 2016 #6

    TSny

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    Note that ##\frac{1}{\sqrt{1+t^2}} = (1+t^2)^{-1/2}##
     
  8. Feb 25, 2016 #7
    I see how to do it know, it's pretty simple, thanks.
     
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