Areas and Stuff January 29, 2010

What’s Up? We are finishing up the chapter on Electric Field and Gauss. Quiz Today Monday we will begin Potential .. Read first two sections in Chapter 19. WebAssign due Sunday Night Short WA is also due Tuesday Night (2 problems) Best Guess – EXAM #! – Chapters 18,19 Wednesday February 10th. 1 hour After this, we begin to move a bit faster through the material.

I am free 8:30-9:30 9:30-10:30 Fridays 12:30-1:30 None of these

VALUES ON “N” REPORTED BY CLASS RESEARCHERS Results to date VALUES ON “N” REPORTED BY CLASS RESEARCHERS We found the value of n to be 2.92 Received a n value of -2.08 The number we calculated for N was 1.13 we got n = 1.73 I found that n=1.8. What the … ?? Mr. Coulomb

The Area Vector This area vector is defined as being POSITIVE because it points OUTWARD from a CLOSED surface

18.8 The Electric Field Inside a Conductor: Shielding At equilibrium under electrostatic conditions, any excess charge resides on the surface of a conductor. At equilibrium under electrostatic conditions, the electric field is zero at any point within a conducting material. The conductor shields any charge within it from electric fields created outside the condictor.

18.8 The Electric Field Inside a Conductor: Shielding The electric field just outside the surface of a conductor is perpendicular to the surface at equilibrium under electrostatic conditions.

18.8 The Electric Field Inside a Conductor: Shielding Conceptual Example 14 A Conductor in an Electric Field A charge is suspended at the center of a hollow, electrically neutral, spherical conductor. Show that this charge induces a charge of –q on the interior surface and (b) a charge of +q on the exterior surface of the conductor.

 Today’s Stuff

But first … QUIZ

New Topic Gauss’s Law

What would you guess is inside the cube?

What about now?

How about this?? Positive point charge Negative point charge Large Sheet of charge No charge You can’t tell from this

Let’s Do Gauss Stuff

Putting it Together

18.9 Gauss’ Law

18.9 Gauss’ Law

The electric flux through a Gaussian 18.9 Gauss’ Law GAUSS’ LAW The electric flux through a Gaussian surface is equal to the net charge enclosed in that surface divided by the permittivity of free space: SI Units of Electric Flux: N·m2/C

Example 15 The Electric Field of a Charged Thin Spherical Shell 18.9 Gauss’ Law Example 15 The Electric Field of a Charged Thin Spherical Shell A positive charge is spread uniformly over the shell. Find the magnitude of the electric field at any point (a) outside the shell and (b) inside the shell.

18.9 Gauss’ Law

Outside the shell, the Gaussian surface encloses all of the charge. 18.9 Gauss’ Law Outside the shell, the Gaussian surface encloses all of the charge. (b) Inside the shell, the Gaussian surface encloses no charge.

Continuous Charge Distributions Volume r = charge per unit volume C/m3 Area s = charge per unit area C/m2 Line m or l = charge per unit length C/m

Which Way?? D D

Line of Charge Gaussian Surface It is not a REAL surface, but it is imagined. It has multiple surfaces. In this case it has a cylindrical surface as well as two circular end-caps. For each of the surfaces, the E field must be either normal to the surface (flux) or parallel to the surface (no flux).

Line of Charge

Suprising Answer

Two infinite planes = capacitor!! s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 s/2e0 E=0 E=s/e0 E=0 The Field between two charges capacitor plates is s/e0

Charged Conductors s E=0 E Charge Must reside on the SURFACE - - - - - Very SMALL Gaussian Surface

May the FLUX be with you!