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.

BONUS!

Calculus: Single Variable Part 1 - Functions

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