I wasted so, so much of this day…but I got a little better at dancing 😉

1. Kennedy HW
2. Kidonakis
3. Andy reading
5. Laguna stuff
40m on 1.
1.5h on 2.
30m on 3.
20m in 1. (progress!)
30m on 3. 

as of ~1:30

As of ~2:00

30m on 5.

I guess I’ll see how that work pans out quite a bit more tomorrow. Lots and lots to be done there. 

Notes on Andy’s stuff:

rho known everywhere? Just integrate or sum.
rho not-known? Then matter must be present and hence polarization…the problem becomes a boundary value problem.
Boundary value problems are not going to be relevant on the test, probably.

Charge cannot exert a force on itself.

Learn matching conditions?

Kind of cool: DNA can be approximated by a 1D line of charge

Poisson’s eq can be used if rho is specified for every point…but it HAS to be used for matter…I guess I’ll see when I hit Ch. 7?

In electrostatics work is path independent.
W = Int[q1,q2, qE(r)dl]

Coulomb conservative due to central nature of the force…cool.

Potentials are continuous at charged surfaces.

Earnshaw’s theorem’s proof is trivial at the level of rigor we consider.

The line integral between any two points on an equipotential surface is zero
…if E is nonzero, then E must be perpendicular to dl at all points on the surface.

Two electric field lines can only cross at a null point.

I don’t like example 3.3 …form a more concrete opinion when more awake

To use Gauss’ law effectively we need a surface where dS * E is constant.
Invariance of a charge distribution under rotations implies the same for E…how much does this generalize?
The invariance of rho tells us an invariance of E, but E is unique under the Hemholtz theorem
…Note that the infinite charge distributions uniqueness can be established in a limiting process.

sigma are invariant to translations along a fixed direction…not so sure.
super impose line charges waited by charge per length for other shapes?
Ugh…this is not clear at all.

stopped at 3.4.2




Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s