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Much greater than

  1. Sep 9, 2006 #1
    After getting through the set up and integration of this problem I am stuck on using the >> operator. My objective is to show what happens when r, a constant, is >> than a, another constant.

    The original equation: (1/r)-(1/(a + r)
    The intended restul: (1/r^2)

    My initial thought was that because r is much greater than a, it would essentially make a close to zero in importance; however, this leads to (1/r) - (1/r). Any help on where I am going wrong with this?
     
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  3. Sep 9, 2006 #2

    StatusX

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    You're right, it's approximately zero. But we almost never say this, instead always trying to find the largest non-zero correction to the approximation. In this case, try subtracting the fractions before dropping a. I think the intended result you have is wrong.
     
    Last edited: Sep 9, 2006
  4. Sep 9, 2006 #3
    Well the intended result is necessary because without it Coulumb's Law would not hold. Unless this means I need to work with the rest of the equation.

    The full final equation: F = ((q*Q)/a*4*pi*epsilon-naught)*(original equation)

    where q is a small fixed charge, Q being the large fixed charge, and a as previously define. I don't see how that part would affect it but it does have to come out so that there is a r^2 on the bottom.
     
  5. Sep 10, 2006 #4

    StatusX

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    The dimensions are wrong. Did you try subtracting the fractions like I suggested?
     
  6. Sep 10, 2006 #5
    1/r - 1/(a+r) = 1/[ (r^2)/a + r] < 1/(r^2 + r) < 1/(r^2) lol
     
  7. Sep 10, 2006 #6

    StatusX

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    You can't compare things like r2/a and r2 or r and r2. They have different dimensions. It's like asking, what's larger: 3 m or 1 m2? One is an area and one is a length. And even if you say, "fine, but 3>1, so can't we just say 3 m > 1 m2?", again we can't, because 3 m = 300 cm and 1 m2= 10000 cm2.

    What you can say is r/a>>1, so r (r/a) >> r.
     
  8. Sep 10, 2006 #7
    I'm not so sure, you have:

    [tex]F=\frac{qQ}{4\pi\epsilon_{0}a}(\frac{1}{r}-\frac{1}{(a+r)})[/tex]

    Which is ok as [tex]\frac{1}{a}*(\frac{1}{r}-\frac{1}{(a + r)})[/tex] has units 1/m^2. You need to use the a when you do your approximation I think. It cancels when you subtract the two fractions, letting you take out r^2 underneath. Is a the radius of Q? I think that would make sense.

    I hope the tex works!
     
  9. Sep 10, 2006 #8

    arildno

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    It is quite simple, really. Rewrite your expression in terms of the tiny parameter a/r<<1:
    [tex]\frac{1}{r}-\frac{1}{a+r}=\frac{1}{r}(1-\frac{1}{1+\frac{a}{r}})[/tex]
    Take the MacLaurin series wrt. to a/r:
    [tex]\frac{1}{r}(1-\frac{1}{1+\frac{a}{r}})=\frac{1}{r}(1-(1-\frac{a}{r}+(\frac{a}{r})^{2}-(\frac{a}{r})^{3}+-+-)=\frac{a}{r^{2}}(1-\frac{a}{r}+(\frac{a}{r})^{2}-+-)[/tex]
    The expression contained in the bracket goes to 1 as [itex]\frac{a}{r}\to{0}[/itex], hence, we have:
    [tex]\frac{1}{r}(1-\frac{1}{1+\frac{a}{r}})\approx\frac{a}{r^{2}}, a<<r[/tex]
     
  10. Sep 10, 2006 #9

    Gokul43201

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    If r>>a, then a/r<<1. This is the only thing that can be stated as directly resulting from the given condition. This result then allows us to neglect term like a/r (and/or its higher powers) in comparison to terms ~ 1. So, for instance, a term like 1-(a/r) can be approximated to 1, if you neglect the first order correction.

    The standard trick is to express all terms as either dimensionless constants or terms in powers of a/r. In this case, you first write the given difference as a single fraction, by computing the LCM (and so on) and then you divide numerator and denominator by r^2. That leaves you with the appropriate form, from which you can throw away the correction terms and get the desired result.

    Edit: was writing this before I saw arildno's post...but since it proposes a route that avoids a series expansion, I'm leaving it standing.
     
    Last edited: Sep 10, 2006
  11. Sep 10, 2006 #10
    funny if I try getting it under common denominator I end up with

    1/r - 1/(a+r) = a/r(r + a) = which is 0/r^2 also known as 0 o_O
     
  12. Sep 10, 2006 #11

    arildno

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    Well, it wasn't STRICTLY necessary of me to indicate the whole series..
    but it is sort of illuminating to show how straightforward it is to get higher-order approximations.
     
  13. Sep 10, 2006 #12

    arildno

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    True enough.
    What do you get if you set a=0 in the original expression?
     
  14. Sep 10, 2006 #13
    Alright, after reading the posts I was able to do a few things and hope that I got it right. I distributed (1/a) and then found the LCD of the two fractions. Took out the (1/r) that was in the fraction, distributed a in the denominator, factored out a squared in the denominator, canceled the a squared and found myself left with (1/(a+r) to which I factored out (1/r) again and got 1 inside the parenthesis. Thank you all for your help.Some reminded me and some gave me ways of getting through fractions in variables in the future.

    One thing I do wish is that I was able to understand the post using the mcclaurin series. Too bad my teacher last semester ran out of time and never got to the point where you start using series in functions. Will have to come back to this method once I actually learn the power, mcclaurin, and taylor series. By then I should have learned the latex system as well.
     
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