1. Areas and Stuff
January 29, 2010

2. 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.

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

4. 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

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

6. 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.

7. 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.

8. 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.

9. 
Today’s
Stuff

10. But first …
QUIZ

11. New Topic
Gauss’s Law

12. What would you guess is inside the cube?

13. What about now?

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

15. Let’s Do Gauss Stuff

16. Putting it Together

17. 18.9 Gauss’ Law

18. 18.9 Gauss’ Law

19. 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

20. 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.

21. 18.9 Gauss’ Law

22. 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.

23. 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

24. Which Way?? D D

25. 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).

26. Line of Charge

27. Suprising Answer

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

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

30. May the FLUX be with you!