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.