LECTURE NOTES FOR PHYSICS 308: ELECTRICITY AND MAGNETISM
TEXT: GRIFFITHS' INTRODUCTION TO ELECTRODYNAMICS
Last revised: 1/16/2007

SOME USEFUL REFERENCE STUFF: Greek alphabet, metric prefixes, conversion factors

ASSIGNMENTS: (Subject to change: check back often.)
HW #3: Due Friday, February 16 at noon: Problems 2: 23, 24, 31, 34A, plus Problem 2 from Exam I, 2005.
HW #4: Due Friday, February 23 at noon: Problems 2:39, 3:3, plus spreadsheet relaxation problem TBA.
"Dailies":
None due Tuesday, Feb. 14.
Due Thursday, Feb. 15: Verify that σ=ε0En gives the right answer for a hollow spherical shell of charge Q and radius R.
Due Tuesday, Feb. 27: Use the Method of Relaxation on an Excel spreadsheet to solve for a 5x5 grid with V=U0=100 along the top edge, V=-U0 along the right-hand side, and En=0 along the left-hand and bottom sides. Please e-mail me your Excel solution.
Condensed Story of Ms Farad by A. P. French
Miss Farad was pretty and sensual
And charged to a reckless potential;
But a rascal named Ohm
Conducted her home -
Her decline was, alas, exponential.

GO TO LECTURE 7, 8

LECTURE 7:
SECTION 2.5: CONDUCTORS
In our earlier discussion of electric fields, we assumed that we could put charge somewhere and it would stick. But there is a class of materials for which charges don't stay put: these are called conductors. We will start with the easiest conductors to deal with mathematically, namely perfect conductors, and later (Chapter 7) we will look at what happens in real conductors. The main consequence of applying electric fields near conductors it is that the conductors have free charges which will rearrange themselves until they reach some kind of steady state arrangement. By definition, since these are free charges, their steady state configuration would have to be such that the electric field acting on them can no longer move them. Inside the conductor this means that the electric field must be zero, but on the surface of the conductor it possible for there to be an electric field acting on the charge, so long as it points the charge in a direction normal to the surface. As a result of all this, we can deduce the following properties of conductors:
1. E = 0 inside.
2. E is perpendicular to the surface just outside the conductor.
3. The charge density ρ = 0 inside the conductor.
4. All excess charge in a conductor resides on the surface.
5. The conductor is an "equipotential": every part of it is at the same electric potential, V. INDUCED CHARGE, Section 2.5.2
Since it is possible for excess charge to reside on the surface, and since the electric field inside the surface is zero, the electric field just outside the surface must be related to the charge density on the surface. To see this, go back to the example we had when we applied Gauss' Law to a 2-dimensional distribution of charge. Applying this example to the general case of the surface charge on a conductor, we get an expression for the induced charge on the conductor
FARADAY CAGES AND SHIELDING
One of the most interesting applications of Gauss' law has to do with the electric field inside a charged sphere, which Gauss’ law tells us should be equal to zero. In fact we may do Problem 11 on the board and compare it to the much more difficult mathematical effort we needed to do Problem 7, which shows us the same results (this should serve as an advertisement for how useful Gauss’ law is in allowing us to actually avoid doing any heavy lifting, integrationwise).

One cool application of this is that if you wrap of aluminum foil around a portable radio, you should no longer be able to receive a signal. Try it. This should work because the signals that are received by the antenna are simply electromagnetic waves. The only thing that might gum up this prediction of mine is that these electric fields which are part of the electromagnetic waves are oscillating at about 100 MHz for FM signals, which makes them not quite electrostatic. The foil is acting as a "Faraday cage", a metal enclosure that isolates its interior from whatever electric fields are outside. Similar examples would be the chassis of a car or airplane during a thunderstorm (another reason not to fly in plastic aircraft) or the metal mesh inside the window of your microwave oven.

[Omit the rest of Section 2.5.3: The Force on a Conductor. We will not bother with it.]

CAPACITORS, Section 2.5.4
Consider two conductors, with a charge of +Q on one, and a charge of -Q on the other. These charges will result in there being an electric field between the two conductors. There will also be a potential difference between them. It is much easier to describe them in terms of this voltage than the E-field, because it is a single scalar value, while the field is a vector which varies throughout space. Let's consider this potential difference. If we double the charge from +Q and -Q to +2Q and -2Q, the electric field will double as well, and since the potential difference is related to the electric field by , the potential difference will double as well. In other words, the voltage between the two conductors is proportional to the charge. In other, other words, the ratio of charge to voltage, Q/V, is a constant for any particular set of two conductors. We call this ratio the capacitance, C:
with units of Farads (F=C/V). Careful: V stands for voltage and Volts, respectively, in the definition of capacitance above and the definition of Farads to the left, while C stands for capacitance and Coulombs (which are not interchangeable), respectively, in the same two equations.

There are a few systems of conductors ("geometries") that come up again and again, so we should document how capacitance is related to the dimensions of the capacitor in each case. Also, these systems will give us some practice (a) in calculating the electric field using Gauss' law, and (b) in calculating the potential difference by line integration. A little practice is probably a good thing, no?

PARALLEL PLATE CAPACITORS
The granddaddy of all capacitors is the parallel-plate capacitor. This is simply two similarly-shaped flat sheets of metal, with areas A, placed a distance d apart. If A>>d2, then we can approximate the electric field between the two plates by the field between two infinite sheets having uniform surface charge distributions. Remember from our section on Gauss' law examples that the field is uniform on either side of such a sheet, with E=2πkσ, where σ is the charge per area on the sheet. If two such sheets are placed near each other, their fields cancel out outside their "sandwich", but add inside. To get the voltage between the two sheets, note that the integral we just mentioned reduces to V=Ed. And so,
ISOLATED SPHERE CAPACITORS
A more difficult system to visualize as a capacitor is a single spherical piece of metal. Think of this as a two-conductor system with one conductor an infinite distance away. Even if it were not at infinity, if it were at a distance much larger than the radius of the sphere, it is a useful approximation to imagine it infinitely far away. Using Gauss' law to get E, and integrating to get V, we find
You can consider your finger, reaching for a doorknob on a dry winter's day, as such a capacitor. Given that air breaks down (sparks) at about 3MV/m, you can now calculate the amount of charge built up in your finger if you can cause a spark to fly 1 cm. (V=Ed=30kV, assume r=5mm, which gives you C=0.55pF, so Q=CV=17nC.)

CONCENTRIC SPHERE CAPACITORS
Another good exercise is the case of two nested capcitors, one inside the other, both centered about the same point, with radii a and b, b>a.
Convince yourself that if b>>a, then this expression reduces to the case of the single spherical conductor above, with d=b-a and A=4πa2.

LECTURE 8:
ENERGY STORED IN A CAPACITOR, Still Section 2.5.4
It is easy to calculate the energy density between two parallel plates of charge and multiply by the volume between the plates, to get
It turns out that this result is true for all capacitors, so knowing the capacitance of a pair of conductors can save us a lot of time when it comes to calculating how much energy is stored inside the electric field between them. (Otherwise, we would have to calculate the energy density, u, throughout space, and then integrate it.)

By the way, now we can calculate the amount of energy dissipated when that spark flies between your finger and the doorknob. Since V=30kV and C=0.55pF, the total energy is (1/2)CV2= 0.25mJ, not very large, but enough to get your attention. Obviously, the further the spark flies, the larger the amount of energy released.

SECTION 3.1: LAPLACE'S EQUATION
1D:
solution V(x)=mx+b
V(x)=1/2[V(x-a)+V(x+a)]
No local maxima or minima
2D:
 
V(x,y)=average on a circle
No local maxima or minima
3D:
 
V(x,y,z)=average on a sphere
No local maxima or minima


BOUNDARY CONDITIONS, Section 3.1.5
There are two classes of boundary conditions that commonly exist, as well as hybrid combinations of the two. Physicists commonly refer to these two classes by name. If you need to learn more about them after this course, it helps to recognize these names. Besides, I have to use these names myself in a few minutes. Also, there is a wonderful theorem or two that says that if you have either one, or the other of these types of boundary conditions covering all of your boundary, then there is exactly one solution to Laplace's equation inside those boundaries. In other words, once you've found a solution, you can quit: it is the only solution.

The two conditions are called "Dirichlet" conditions, wherein the electric potential, V, is specified on the boundary, and "Neumann" conditions, in which the derivative of the potential normal to the surface is given. An example of Dirichlet conditions would be if the four walls of the lecture room were held at four separate "voltages". A Neumann example would be if a certain constant current were made to flow between two opposing walls, with the other walls, floor, and ceiling electrically insulated and passing no current.

METHOD OF RELAXATION (not in book: See professor's handout instead, based on a student's project in this class years ago.)
As we saw last class, the electric potential in 2D or 3D, assuming Laplace's equation, is equal to the average of the potential at points equidistant from the point of interest, that is, the points on a circle in 2D, or on a sphere in 3D. This allows us to use a very powerful approximation technique. Let's consider 2D, since it is the dimensionality of your notepad, the blackboard, and a computer screen.

Consider a boundary-value problem defined in two dimensions -- a square, for example. We can replace the continuous values of x and y coordinates inside the square with gridpoints, as long as we make the points close enough. Laplace's equation then tells us that the electric potential at a point on our grid will equal the average of the values of potential at neighboring points. For the grid I've displayed below, that means that V1,1=(V0,1+V2,1+V1,0+V1,2)/4.

To apply this technique, create a spreadsheet in Excel or create a matrix in some programming language. Place your initial guess for the potential (the initial "guess" can be all zeros: not a very creative guess.) in the matrix/spreadsheet, then apply the formula above to re-evaluate the potential in every nook and cranny of the matrix/spreadsheet. Eventually, your values of Vi,j will converge or "relax" to values approximating the function you were trying to solve for. This is called the relaxation technique.

Simple, no? I have to add a couple more details before you can actually get it to fly right. First, you need to apply boundary conditions, or else your Vi,j=0 initial guess will be your final answer as well. What if V1,0 is a boundary point, and what if the boundary condition at that point is that the normal component of the E-field is zero (a Neumann boundary condidition)? Imagine a phantom gridpoint, V1,-1, to the left of our nine gridpoints. The boundary condition I've just imposed is equivalent to saying that V1,-1=V1,1. (Think about why that is.) Consequently, averaging the four neighbors, as we did above becomes:
V1,0=(V0,0+V2,0+V1,-1+V1,1)/4 =(V0,0+V2,0+2V1,1)/4. In other words, on the boundary cell you still average the neighbors, but you count the neighbor interior to the boundary cell twice.

There are ways of speeding up the convergence of the relaxation technique. One is to recruit "neighbors" from further afield. In other words, instead of just averaging the nearest neighbors, you include second-nearest neighbors, third-nearest neighbors, etc. There are various schemes for figuring out how to weight each set of neighbors in the averaging, but this is clearly beyond the scope of this course, and besides, this is not nearly as elegant as our much simpler approach.

A second, much-easier-to-program approach is called "over-relaxation". It makes use of the fact that relaxation converges monotonically in a single direction. In other words, successive values for V1,1 might equal 1.000V, 1.100V, 1.110V, 1.111V. Why not anticipate the direction of change in V? One way to do this is to define a quantity α, usually between 1 and 2, and changing our original equation from V1,1=(V0,1+V2,1+V1,0+V1,2)/4          to         V1,1=(1-α)V1,1+α(V0,1+V2,1+V1,0+V1,2)/4 Notice that the (1-α) term in the last expression is necessary, in case the average of the neighbors already does equal the potential V1,1.
YSBATS FOR CHAPTER 2:
Memorize:
Coulomb's law
Units for Q, F, E, V, u, U, etc.
Conversions between ρ, E, V
Divergence and curl for E
Definition of electric flux
Gauss' law

Understand:
Difference between test and source charge
Superposition of E, V
When you really need to integrate, and when to avoid it
How to calculate flux for standard simple systems

You should be able to:
Integrate standard charge distributions
Conversions between ρ, E, V other than the formal definitions
Calculate div, grad, curl for any function
Determine whether a vector function is a valid E-field
Find the appropriate Gaussian surface for simple systems
Calculate the enclosed charge for simple systems

Keep for future reference:
Equations for E for simple systems
Standard charge distributions:
linear, λ
surface, σ
volume, ρ
Standard volume integrals:
dτ   for
spherical shells
hollow cylinders
infintesimal cubes
Simple systems
0. Point charge
0a. Collection of point charges
1. Line charge
1a. Concentric cylinders
2. Plane of charge
2a. Collection of [nonparallel] planes
2b. Hollow charged sphere
3. Sphere of uniform charge density
3a. Sphere of nonuniform charge density