1. Think of flowing water as an analogy where the flow lines would be mass flow (mass/time) per area. Here E is field lines per area. We define a quantity called electric flux,  E which is EA in this simple case. Electric flux is total number of lines through the area. In the water analogy we would have mass flow. Ch 24: Gauss’s Law 24.1 Electric Flux

2. The total electric flux  E which goes through the vertical plane A also goes through the diagonal plane. Noting that EA = EAcos , we see that  E = EA, where A is a vector normal to the area of value equal to the area, is a consistent definition for  E. For small area patches d  E = EdA 24.1 Electric Flux

3.  E = E  A = E  Acos  i

4.  E is positive. Outflow is positive.

5.  E is negative. Inflow is negative.

6.  E is zero. Inflow equals outflow.

7. Ch 24: Gauss’s Law 24.2 Gauss’s Law

8. -3Q CT1:  S1 equals CT2:  S2 equals CT3:  S3 equals CT4:  S4 equals A.-3Q/  0 B.-2Q /  0 C.-Q /  0 D.0 E.Q /  0 F.2Q /  0 G.3Q /  0

9. CT5: A icosahedron has 20 equal triangular faces as pictured above. Assume a charge q is placed at the center of the icosahedron (an equal distance from each face). Using the symmetry of the situation, determine how much electric flux goes through each face. A. q/ 0 B. q/4 0 C. q/6 0 D. q/20 0 E. none of the above

10.  E = = q in /  0 A.Symmetries: 1) spherical 2) cylindrical (linear) 3) planar B. Method: 1) note symmetry 2) draw appropriate Gaussian surface 3) calculate electric flux  E 4) set  E = q in /  0 5) solve for E Gauss's law: using superposition q in is the sum of the charges enclosed by the Gaussian surface.

11. 1.E = 0 inside perfect conductors 2.The charge must reside on the surface of a perfect conductor 3.E =  /  0 n where n is a unit vector normal to the surface and pointing outward. 4.  is greatest at points of least radii of curvature (i.e. pointy) 24.4 Perfect Conductors in Electrostatic Equilibrium

12. Before Class Assignment 2/6/08 13 correct 1 I don’t know 1 no explanation

13. P24.44 (p. 689)

14. 2Q -3Q CT6: At a distance r >c from the common center, the electric field is A. 2Q/4 0 r 2 B. -3Q/4 0 r 2 C. -Q/4 0 r 2 D. -Q/4 0 r E. constant 24.3 Application of Gauss’s Law to Various Charge Distributions

15. P24.39 (p. 689)

16. CT7: The electric field in the conducting cylinder is A. 0 B. /2 0 C./4 0 D. /2 0 r E. /4 0 r F. r/4 0 G. r/4 0

17. CT8: The electric field between the wire and the cylinder is A. 0 B. /2 0 C./4 0 D. /2 0 r E. /4 0 r F. r/4 0 G. r/4 0

18. P24.35 (p. 688) Gaussian Surface 1: A squat cylinder Gaussian Surface 2: A squat cylinder E  Left  Right 1 2 3 1 2 3

19. A. B. E. D. C. conducting plates so  on each side

20. Table 24-1, p.754

21. Summary Gauss’s Law:  E = = q in / 0 Spherical Symmetry: use concentric, spherical Gaussian surface Cylindrical Symmetry: use concentric, cylindrical Gaussian surface, assume far from ends of long thin cylinder Planar Symmetry: use a disk shaped Gaussian surface, assume far from edges of planar surface E will be constant over the surface and either parallel or normal to the area so will either be EA or 0. You may have to integrate to get q in.