1. Ch. 24 Electric Flux Gauss's Law
Electric flux is the amount of electric field going across a surface
It is defined in terms of a direction, or normal unit vector, perpendicular to the surface
For a constant electric field, and a flat surface, it is easy to calculate
Denoted by E
Units of Nm2/C
When the surface is flat, and the fields are constant, you can just use multiplication to get the flux
When the surface is curved, or the fields are not constant, you have to perform an integration
Here use A X normal unit vector for A vector of text.
Really when everything in integral is constant, you can just multiply.
So, if E dot n is constant along surface – you are OK.

2. Quick Quiz 24.1
JIT
Ans e

3. Warmup 03

4. JIT How does flux compare through each surface?
Number of field lines same for both surfaces. One is projection of other onto the perpendicular of field lines. Flux is same for both surfaces

5. Ans C
Is the answer independent of the shape?

6. Warmup 03

7. Solve on Board

8. Total Flux Out of Various Shapes
A point charge q is at the “center” of a (a) sphere (b) joined hemispheres (c) cylinder. What is the total electric flux out of the shape? b a q q a q

9. Electric Flux (Hard example)
A point charge q is at the center of a cylinder of radius a and height 2b. What is the electric flux out of (a) each end and (b) the lateral surface? 
top s b b  r r z a
Consider a ring of radius s and thickness ds q b a
lateral surface

10. Gauss’s Law
No matter what shape you use, the total electric flux out of a region containing a point charge q is 4keq = q/0.
Why is this true?
Electric flux is just measuring how many field lines come out of a given region
No matter how you distort the shape, the field lines come out somewhere
If you have multiple charges inside the region their effects add
However, charges outside the region do not contribute q q4
ϕ E = E ∙d A = q in ϵ 0 q3 q1 q2

11. Warmup 03

12. Using Gauss’s Law to find total charge
A cube of side a has an electric field of constant magnitude |E| = E pointing directly out on two opposite faces and directly in on the remaining four faces. What is the total charge inside the cube?
A) 6Ea20 B) – 6Ea20 C) 2Ea20 D) – 2Ea20 E) None of the above a a a
Do easy problem first – point charge,
And problem 4 points on page 696.

13. Ans C

14. Ans E (B&C)

15. Using Gauss’s Law to find flux
A very long box has the shape of a regular pentagonal prism. Inscribed in the box is a sphere of radius R with surface charge density . What is the electric flux out of one lateral side of the box?
Perspective view
The flux out of the end caps is negligible
Because it is a regular pentagon, the flux from the other five sides must be the same
End view

16. Using Gauss’s Law to find E-field
A sphere of radius a has uniform charge density  throughout. What is the direction and magnitude of the electric field everywhere?
Clearly, all directions are created equal in this problem
Certainly the electric field will point away from the sphere at all points
The electric field must depend only on the distance
Draw a sphere of radius r around this charge
Now use Gauss’s Law with this sphere
Do point charge first
Then do as point charge with q = PV r a
Is this the electric field everywhere?

17. Using Gauss’s Law to find E-field (2)
A sphere of radius a has uniform charge density  throughout. What is the direction and magnitude of the electric field everywhere? [Like example 24.3]
When computing the flux for a Gaussian surface, only include the electric charges inside the surface r a
r/a

18. Electric Field From a Line Charge
What is the electric field from an infinite line with linear charge density ? L r
Electric field must point away from the line charge, and depends only on distance
Add a cylindrical Gaussian surface with radius r and length L
Use Gauss’s Law
The ends of the cylinder don’t contribute
On the side, the electric field and the normal are parallel

19. Electric Field From a Plane Charge
What is the electric field from an infinite plane with surface charge density ?
Electric field must point away from the surface, and depends only on distance d from the surface
Add a box shaped Gaussian surface of size 2d  L  W
Use Gauss’s Law
The sides don’t contribute
On the top and bottom, the electric field and the normal are parallel
Text uses cylinder. If truly infinite, never far from it. Consider 45 degrees area pf charge increaases as r squared as does 1/E.

20. Text discusses this, page 699
Ans A

21. Serway 24-35
Solve on Board

22. Warmup 04

23. Conductors and Gauss’s Law
Conductors are materials where charges are free to flow in response to electric forces
The charges flow until the electric field is neutralized in the conductor
Inside a conductor, E = 0
Draw any Gaussian surface inside the conductor
In the interior of a conductor, there is no charge
Do all Warmup04
Conductors in equilibrium: (1) E zero inside, (2) charge on surface, (3) outside is sigma/eo, (4) greatest at point
The charge all flows to the surface

24. Warmup 04

25. Electric Field at Surface of a Conductor
Because charge accumulates on the surface of a conductor, there can be electric field just outside the conductor
Will be perpendicular to surface
We can calculate it from Gauss’s Law
Draw a small box that slightly penetrates the surface
The lateral sides are small and have no flux through them
The bottom side is inside the conductor and has no electric field
The top side has area A and has flux through it
, pg 700
The charge inside the box is due to the surface charge 
We can use Gauss’s Law to relate these

26. Where does the charge go?
A hollow conducting sphere of outer radius 2 cm and inner radius 1 cm has q = +80 nC of charge put on it. What is the surface charge density on the inner surface? On the outer surface?
A) 20 nC/cm2 B) 5 nC/cm2 C) 4 nC/cm2 D) 0 E) None of the above
80 nC
The Gaussian surface is entirely contained in the conductor; therefore E = 0 and electric flux = 0
Therefore, there can’t be any charge on the inner surface
1 cm
From the symmetry of the problem, the charge will be uniformly spread over the outer surface
2 cm
Point charge q/4Pi eps0 w/ q = 80 pi. Note at r = 2 cm, sigma/eps0 is same: 20/4 cm^2
cutaway view
The electric field:
The electric field in the cavity and in the conductor is zero
The electric field outside the conductor can be found from Gauss’s Law

27. Warmup 04

28. Ans B

29. Serway 24-47
Solve on Board

30. Solve on Board
Charges set up to neitralize field in conductor.