1. 1 Chapter 22

2. 2 Flux Number of objects passing through a surface

3. 3 Electric Flux,  is proportional to the number of electric field lines passing through a surface Assumes that the surface is perpendicular to the lines If not, then we use a cosine of the angle between them to get the components that are parallel Mathematically:

4. 4 Simple Cases E  =EA E A A  =0  =EAcos  E A E cos  

5. 5 From  to A  represents a sum over a large a collection of objects Integration is also a sum over a collection of infinitesimally small objects, in our case, small areas, dA So

6. 6 Gauss’s Law The field lines emitted by a charge are proportional to the size of the charge. Therefore, the electric field must be proportional to the size of the charge In order to count the field lines, we must enclose the charges in some geometrical surface (one that we choose)

7. 7 Mathematically Charge enclosed within bounding limits of this closed surface integral

8. 8 Fluxes, Fluxes, Fluxes

9. 9 3 Shapes Sphere Cylinder Pillbox

10. 10 Sphere When to use: around spherical objects (duh!) and point charges Hey! What if an object is not one of these objects? Closed surface integral yields: r is the radius of the geometrical object that you are creating

11. 11 Sphere Example What if you had a sphere of radius, b, which contained a material whose charge density depend on the radius, for example,  =Ar 2 where A is a constant with appropriate units? At r=b, both of these expressions should be equal

12. 12 Cylinder When to use: around cylindrical objects and line charges Closed surface integral yields: r is the radius of the geometrical object that you are creating and L is the length of the cylinder L

13. 13 Cylinder Example What if you had an infinitely long line of charge with a linear charge density, ?

14. 14 Pillbox When to use: around flat surfaces and sheets of charge Closed surface integral yields: A is the area of the pillbox

15. 15 Charge Isolated Conductor in Electrostatic Equilibrium If excess charge is placed on an isolated conductor, the charge resides on the surface. Why? If there is an E-field inside the conductor then it would exert forces on the free electrons which would then be in motion. This is NOT electrostatic. Therefore, if there is no E-field inside, then, by Gauss’s Law, the charge enclosed inside must be zero If the charges are not on the outside, you are only left with the surface A caveat to this is that E-field lines must be perpendicular to the surface else free charges would move.

16. 16 Electric field on an infinitely large sheet of charge +++++++++++++++++++++++++++++++++++++++++++++ A E E

17. 17 Electric field on a conducting sheet +++++++++++++++++++++++++++++++++++++++++++++ A E So a conductor has 2x the electric field strength as the infinite sheet of charge

18. 18 A differential view of Gauss’s Law Recall the Divergence of a field of vectors Div=0 Div=+large How much the vector diverges around a given point

19. 19 Divergence Theorem (aka Gauss’s Thm or Green’s Thm) Suspiciously like LHS of Gauss’s Law Sum of the faucets in a volume = Sum of the water going thru the surface A place of high divergence is like a faucet Bounded surface of some region

20. 20 Div(E) So how the E-field spreads out from a point depends on the amount of charge density at that point