Welcome to Calculus. 
I'm Professor Greist, and we're about to begin Lecture 5, Bonus Material. 
In our lesson, we saw two important series that have convergence constraints. 
These are the geometric series for 1 over 1 minus x, and the binomial series for 1 
plus x to the alpha. Both of these are guaranteed to converge, 
only when x is less than 1 in absolute value. 
Let's put these series to work. In a problem in Electrostatics. 
An electric dipole is a pair of equal and oppositely charged particles separated by 
a short distance. Now, this is a calculus class, not a 
physics class. Don't worry, you're not going to need to 
know this for the exam, but let's learn a little bit and have some fun with it. 
The electrostatic potential gives the potential energy of this dipole, this 
pair of charges, as a some of point charge potentials. 
Now what do I mean by that? Well, each charge has a potential energy 
associated to it. The pair of them together yields a 
potential energy that depends on your position. 
So let's say that you're at some location away from the dipole. 
You want to know, what is the electrostatic potential. 
It's given by the following formula, V is equal to kq over d plus, minus kq over d 
minus. What does that mean? 
Well, k is a constant, the Coulomb constant. 
Q is the amount of charge on each element of this dipole. 
And d plus and d minus are the distances, respectively, to the positive and the 
negative charge in the dipole. Alright, that's the physics that we need. 
Let's do some math. Let's compute what this electrostatic 
potential V is, in the case where you are standing at a distance d, away from the 
positive charge, and where d is much greater than r, the separation distance 
between the two charges in your dipole. Now I'm going to assume that we are 
orthogonal to the dipole, that is we're at distance d from the positive charge 
but perpendicular to the orientation of the dipole. 
That means that the distance to the negative charge is, by Pythagoras's 
theorem, the square root of d squared plus r squared. 
So that in computing V, we get kq over d, the positive charge distance, minus kq 
over square root of d squared plus r squared. 
We could stop there, but let's simplify this as much as possible. 
If we factor out kq over d, then we get a one from the left term, from the right 
term we get 1 over the square root of quantity 1 plus r over d quantity 
squared. Well, at this point we can try to expand 
that out using the binomial series. Recall the binomial series is for 
something of the form 1 plus x to the alpha. 
In this case, alpha is negative one half. We're looking at one over the square root 
of something, and x is r over d quantity squared. 
Because the first two terms in the binomial series are 1 plus alpha x, we 
get the first two terms in the series applied here to be 1 minus one half times 
r over d quantity squared. Simplifying, we see that the ones cancel 
and we're left with kq over d times one half, are over d quantity squared. 
Everything else is of higher order term in r squared over d squared. 
Let's remember that one half r over d quantity squared. 
In fact, we'll set it over here, right next to the location. 
And consider what happens when we look at distance d away from that positive 
charge, but in a different direction. Instead of being orthogonal to the 
dipole, let's look parallel to it. Well, in this case, d plus remains the 
same. It is this distance d. 
But now d minus is a bit simpler looking. It is d plus r, the separation distance 
in the dipole. Now again, factoring out a kq over d, we 
get a term on the right hand side that is of the form, one over one plus r over d. 
We can expand this out using, if we like, the binomial series with alpha equal 
negative one, or what that really is, the geometric series. 
For 1 over 1 minus x, with x being equal to negative r over d. 
Writing out the first two terms of that series, as well, we get 1 minus r over d. 
The ones cancel, just as before. And we see that up to the leading order 
term, our electrostatic potential is kq over d times r over d. 
Now, notice how different that is from before. 
By using different series, we see that the leading order terms in the 
electrostatic potential differ depending on how you are oriented with respect to 
the dipole. Inline potential is much stronger than 
orthogonal. This is just one example of how we can 
use different series in order to get a handle on various physical phenomena by 
expanding, ignoring the higher order terms and retaining the leading behavior.