Lecture 5 : BONUS!

This is Calculus in a Single Variable. And
I'm Robert Ghrist, professor of

Mathematics and Electrical and Systems
Engineering at the University of

Pennsylvania. We're about to begin Lecture
five, Bonus material. Let's recall what we

learned from Lecture five.
Two of the more important series we saw

were the Geometric series, one + x + x^2,
etcetera = one / one - x. And the binomial

series that generalizes this, two
functions of the form one + x to the

Alpha, where alpha is a constant. Both of
these have a limited domain of

convergence, x has to be less than one in
absolute value. Let's do an extended

example, where we apply both of these
series to answer a physical phenomenon,

something from electrostatics. An electric
dipole is a pair of equal and oppositely

charged particles that are separated by a
short distance. You have a positive and a

negative charge, each with the same but
opposite charge. One of the quantities of

interest in electrostatics is the
electrostatic potential. This tells you

about the potential energy of the dipole
as the sum of the point charge potentials.

So the positive charge has a potential and
the negative charge has a point charge

potential and the sum of these two gives
you the net potential energy of the

dipole. And here, the important thing to
remember is that this potential energy

depends on position on where you are
located relative to the charges. The

formula for the electrostatic potential V
is kq / d +, - kq / d -. Here, k is a

constant called the Coulomb constant. Q is
the charge, +q on the positive charge and

-q on the negative charge. And the
distances, d+ and d- are the distances

between where you are located on the
positive and negative charge respectively.

Now, let's look at the value of this
electrostatic potential at two different

locations. First, let's say that you are
located directly above the positive

charge. Your position vector is orthogonal
to the vector between the two charges. And

let's assume that the two charges are
separated by a distance r. If you are

located at the distan ce d above the
positive charge, then what is your

distance to the negative charge? From the
Pythagorean theorem, it's the square root

of d^2 + r^2.
Now, if we plug these in for d+ and d-

into the formula for electrostatic
potential, what do we get? We get that V =

kq / d - kq / square root d^2 + r^2.
We can factor out kq / d. And for the

positive term, we get one, for the
negative term, we get -one / square root

of one + quantity, r / d^2.2.
Now, when r / d is small, that means when

the charges are close to each other,
relative to where you are positioned,

then, we can expand this using the
binomial series for alpha =, - one-half.

Because we are looking at the square root
of one / r / d quantity squared in the

denominator. Now, applying that binomial
series gives us with a factor of kq / d on

the outside, one - quantity one.
And then what was the exponent alpha?

negative one-half.
So one minus one-half times quantity r

over d^2 plus higher order terms. So, the
leading order term for the electrostatic

potential is one-half k * q * r^2 d^3.
Now, let's see what happens when instead

of being perpendicular to the direction
between the two charges, we are parallel

to it. And again, let's say that we are d
units away from the positive charge, but

in line with the dipole, then what are the
distances? The distance to the positive

charge is d. The distance to the negative
charge is d + r. So, again, factoring out

kq / d, we have a quantity one - one / one
+ r / d. We see that we can use the

geometric series to do the expansion and
we get a factor of kq over d times

quantity one minus quantity one minus r
over d plus higher order terms, in this

quantity r / d. The leading order term is
therefore, k * q * r / d^2.

Now, compare these two. We see that the
electrostatic potential changes not only

by some numeric coefficient. There is a
one-half when you're perpendicular and a

numeric coefficient of one when you;re in
line, but as well the dependents on r and

d changes. On the left we have r^2 / d^3,
On the right we have r / d^2.

That is an interesting and illuminating
applicati on of how to use both of these

interesting series to understand and give
intuition about a complex physical

scenario. This is one of the many examples
of physical applications of calculus you

will encounter in this course.