Much greater than

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?

StatusX
Homework Helper
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.

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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.

StatusX
Homework Helper
The dimensions are wrong. Did you try subtracting the fractions like I suggested?

1/r - 1/(a+r) = 1/[ (r^2)/a + r] < 1/(r^2 + r) < 1/(r^2) lol

StatusX
Homework Helper
dragonlorder said:
1/r - 1/(a+r) = 1/[ (r^2)/a + r] < 1/(r^2 + r) < 1/(r^2) lol

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.

StatusX said:
The dimensions are wrong. Did you try subtracting the fractions like I suggested?

I'm not so sure, you have:

$$F=\frac{qQ}{4\pi\epsilon_{0}a}(\frac{1}{r}-\frac{1}{(a+r)})$$

Which is ok as $$\frac{1}{a}*(\frac{1}{r}-\frac{1}{(a + r)})$$ 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!

arildno
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Gold Member
Dearly Missed
It is quite simple, really. Rewrite your expression in terms of the tiny parameter a/r<<1:
$$\frac{1}{r}-\frac{1}{a+r}=\frac{1}{r}(1-\frac{1}{1+\frac{a}{r}})$$
Take the MacLaurin series wrt. to a/r:
$$\frac{1}{r}(1-\frac{1}{1+\frac{a}{r}})=\frac{1}{r}(1-(1-\frac{a}{r}+(\frac{a}{r})^{2}-(\frac{a}{r})^{3}+-+-)=\frac{a}{r^{2}}(1-\frac{a}{r}+(\frac{a}{r})^{2}-+-)$$
The expression contained in the bracket goes to 1 as $\frac{a}{r}\to{0}$, hence, we have:
$$\frac{1}{r}(1-\frac{1}{1+\frac{a}{r}})\approx\frac{a}{r^{2}}, a<<r$$

Gokul43201
Staff Emeritus
Gold Member
Sculptured said:
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.
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.

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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

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.

arildno