Electrodynamics/Tutorials/4/1/2

Video Intro

Hi, this is Jonathan Gardner.

We're covering [section reference] of Griffiths Introduction to Electrodynamics.

I'm going to move fast, but you can always rewind.

Thumbs up and share if you appreciate my effort.

As always, questions in a video response or comments.

Let's get started.

4.1.2 Induced Dipoles

What is an Atom (Electrically speaking?)

Take a single atom, any atom from Hydrogen to Livermorium.

As you probably know by now, there is a tiny, dense, positive charge in the center, called the nucleus. It is made up of tiny dense, massive protons and neutrons. Around this is a cloud of electrons, which are much less massive but negatively charged.

What happens in no electric field?

If you put this where there is no electric field, you get the two lining up exactly with each other. You can think of the electrons as spheres around the center, like so. (Draw it.) They come in different sizes and have different charges, but in the absence of a field, all by itself, the total charge is 0. That means it creates no potential or electric field on its own. There is no monopole term, no dipole term, no quadrupole term, etc...

What happens in an electric field?

Now, what happens when you put this neutral particle into an electric field? You'd think that nothing would happen, but that is wrong. The positive core is pulled along with the electric field, the negative of the gradient of the potential. The negative shells are pulled in the opposite direction. They're still attracted to each other, so they don't go very far.

Ionization

That is, unless you increase the strength of the electric field such that it overpowers the attraction. Then the negative charges are ripped away from the core, and you are left with two particles, one with a strong positive charge, and another with a strong negative charge. Now you have free-roaming charges, which turn the dielectric into a conductor.

This process is called ionization. It produces two ions. We'll talk more about that later.

Polarization

If the electric field isn't so strong as to rip the electrons off, then you have created a tiny dipole. The dipole moment p will point in the same direction as the electric field. This is nearly a perfect "pure" dipole. As long as we don't think about the electric field very near the atom, this is a really good assumption.

Many Atoms

Now think of what happens when we have many atoms bouncing around and off each other. Let's ignore molecules for now.

Even though they are constantly in motion, the average of a group of these are going to produce a net dipole moment. Remember, dipole moments add up as long as the net charge is zero.

= Atomic Polarizability

The constant alpha describes how big the dipole moment is given an electric field. Some atoms, H, He, Ne, etc... really resist being polarized. Others don't mind being pushed around at all, and have comparatively large polarizabilities.

Example 1

Given a primitive model of an atom, a positive point charge surrounded by a solid spherical negative charge, calculate the atomic polarizibility of the atom.

We simply want to find the point of equilibrium here. That's where the force caused by the electric field exactly cancels out the force of the attraction between the core and the shell.

When the two forces cancel, the atom is in equilibrium. That's also where the two electric fields will cancel, the external field and the field caused by the nucleus moving outside of the center of the shell.

What is the electric field inside this solid spherical shell at the distance d? Using Gauss's Law:

<math>\int_\text{surface} \vec{E} \cdot \vec{da} = {1 \over \epsilon_0} Q_\text{enclosed}\;</math>

If we take our Gaussian Surface to be a spherical shell centered on the center of the negative charge, then we can assume that the electric field is constant everywhere, pointing inside because of the charge contained within it. We're ignoring the electric field we're in -- we want to balance that with the electric field due to the negative charge.

This changes the equation to be:

<math>E \int da = 4 \pi d^2 E = {1 \over \epsilon_0} Q_\text{enclosed}\;</math>

How much charge is enclosed? We find the charge density by dividing the total charge by the volume of the sphere.

<math>\rho = {q \over {4 \over 3} \pi a^3}\;</math>

And multiply that by the volume of the sphere we just too the surface of:

<math>Q_\text{enclosed} = \rho {4 \over 3} \pi d^3 = { q {4 \over 3} \pi d^3 \over {4 \over 3} \pi d^3} = {q d^3 \over a^3}</math>

And that's going to be equal to the left hand side of Gauss's Law in integral form:

<math>

\begin{align} 4 \pi d^2 E &= {q d^3 \over \epsilon_0 a^3} \\

         E &= {1 \over 4 \pi \epsilon_0} {q d \over a^3}\\

\end{align} </math>

I just solved problem 2.12 for you, by the way.

That gives us the electric field due to the negative charge. Now we just have to balance that with the electric field the atom is subjected to.

Remember, we wanted to find the atomic polarizibility alpha. Let's find our p and we're almost there.

<math>

\begin{align} p &= q d \\

 &= {4 \pi \epsilon_0 a^3} E \\

\end{align} </math>

There's our alpha.

Griffiths notes that this is about 4 times the value for atoms that roughly have this shape. I guess we got pretty close with a primitive model. (TODO)

Molecules

Molecules are simply multiple atoms joined together in a particular shape. Many of the simplest molecules have a rod shape. Others have a 2-D or 3-D shape.

When you have these kinds of shapes, the polarizibility will depend on the direction of the electric field. In one direction, things will slide along easily. In another, the charges are held tightly in place.

T describe the polarizibility of these configurations, a single constant is not enough. For simple shapes like CO2, we can use two constants describing the polarizibility along the perpendicular or parallel axes. The induced dipole moment is not likely to be parallel to E anymore. (We'll talk about how that affects the orientation of the molecule in the next video.)

(Equations)

For more complex shapes, a 3x3 matrix of 9 total values is needed. This is the polarizibility tensor. In practice, you can orient the axis to eliminate all but the 3 diagonal terms.

(Equations)

Next

Next, we'll talk about how the electric field can cause the particles to rotate.

Next