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Application of Taylor Series

  1. Nov 3, 2006 #1
    The electric potential V at a distance R along the axis perpendicular to the center of a charged disc with radius a and constant charge density d is give by

    V = 2pi*d*(SQRT(R^2 +a^2) - R)

    Show that for large R

    V = pi*a^2*d / R



    This is what I have done so far...

    V = 2pi*d * (SQRT(R^2 * (1+ a^2/R^2)) - R)
    V = 2pi*d*R (SQRT(1 + a^2/R^2) - R)
    V = 2pi*d*R * ( 1 + (1/2)(a^2/R^2) + (1/2)(1/2 -1)/2! * (a^2/R^2)^2 + ... - R)

    V = 2pi*d*R * (1/2) (1/2 + a^2/r^2 - (1/2)/2! * (a^2/R^2)^2 + ... - R)
    V = pi*d* R ( 1/2 + a^2/R^2 - (1/4)*a^4/R^4 + ...-R)
    V = pi * d * R/R^2 (1/2 + a^2 - (1/4)a^4/R^2 + ...-R)
    V = pi * d *R (1/2 + a^2 - (1/4)a^4/R^2 + ...-R)
    V = pi *d * a^2/R (1/2a^2 + 1 - (1/4) + ...-R)

    Am I doing this correctly. How do I simplify what is in the parenthesis to get 1, which multiply to give me pi*d*a^2/R?
     
  2. jcsd
  3. Nov 3, 2006 #2
    How did you get from your first line to your second? :smile:

    Other than that, when you are using Taylor series to create approximations, you usually just handwave away all but the constant and linear terms (and perhaps the quadratic term if there is no linear term).
     
  4. Nov 3, 2006 #3

    quasar987

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    Hi, I haven't read all your steps, but this is the way to do it,

    You know that for small [itex]\epsilon[/itex], [itex]\sqrt{1+\epsilon}\approx 1+\frac{1}{2}\epsilon[/itex], right? Then you want to get V under this form. so take R² out of the root and set [itex]\epsilon =(a/R)^2[/itex].

    In physics, when doing Taylor expansions, you almost always only want to keep just two terms. And 90% of taylor approximations you make are binomial. So just remember that (1+x)^n ~ 1+(x/n).
     
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