1. Physics 2113 Lecture 12: WED 11 FEB Gauss’ Law III Physics 2113 Jonathan Dowling Carl Friedrich Gauss 1777 – 1855 Flux Capacitor (Operational)

2. What? The Flux! Constant Area: A Changing Area: dA E

3. Gauss’ Law: General Case Consider any ARBITRARY CLOSED surface S -- NOTE: this “Gaussian Surface” does NOT have to be a “real” physical object! The TOTAL ELECTRIC FLUX through S is proportional to the TOTAL CHARGE ENCLOSED! The results of a complicated integral is a very simple formula: it avoids long calculations! (One of Maxwell’s 4 equations!) S E +q

4. Examples What is Flux Through Surfaces: S1 = S2 = S3 = S4 = +q/  0 –q/ε 0 0

5. What is total charge inside?

6. Gauss’ Law: Cylindrical Symmetry Charge of q = 10 C is uniformly spread over a line of length L = 1 m. Use Gauss’ Law to compute magnitude of E at a perpendicular distance of 1 mm from the center of the line. Approximate as infinitely long line — E radiates outwards. Choose cylindrical surface of radius r, length L co-axial with line of charge. r = 1 mm E = ? L=1 m Line of Charge:  = q/L Units: [C/m]

7. Gauss’ Law: Cylindrical Symmetry Approximate as infinitely long line — E radiates outwards. Choose cylindrical surface of radius r, length L co-axial with line of charge. r = 1 mm E = ? L = 1 m dA || E so cos  = 1

8. Compare to Finite Line Example From Last Week! If the Line Is Infinitely Long (L >> r) … Blah, Blah, Blah… r With Gauss’s Law We Got Same Answer With Two Lines of Algebra! Add the Vectors! Horrible Integral! Trig Substitution! r

9. A-Rod

10. Gauss’ Law: Insulating Plate Infinite INSULATING plane with uniform charge density s E is NORMAL (perpendicular) to plane Construct Gaussian box as shown For an insulator, E=  /2  0, and for a conductor, E=  /  0. Surface Charge;  = q/A Units: [C/m 2 ] Surface Charge;  = q/A Units: [C/m 2 ]

11. Recall Disk of Charged Sheet From Last Week! Add the Vectors! Horrible Integral! Trig Substitution! So Hard We Didn’t Do It! If the Disk Has Large Radius (R>> z) … Blah, Blah, Blah… With Gauss’s Law We Got Same Answer With Two Lines of Algebra!

12. Insulating and Conducting Planes Q Insulating Plate: Charge Distributed Homogeneously. Conducting Plate: Charge Distributed on the Outer Surfaces. Electric Field Inside a Conductor is ZERO! Q/2

13. Two Insulating Sheets

14. The field from the plates cancels out so ignore them and use Coulombs inverse square law for the central charge only.

15. Two Conducting Sheets 4.86 7.68 12.54 E does not pass through a conductor Formula for E different by Factor of 2 7.68 4.86

16. Gauss’ Law: Spherical Symmetry Consider a POINT charge q & pretend that you don’t know Coulomb’s Law Use Gauss’ Law to compute the electric field at a distance r from the charge Use symmetry: –place spherical surface of radius R centered around the charge q –E has same magnitude anywhere on surface –E normal to surface r q

18. Electric Fields With Spherical Symmetry: Shell Theorem -15C +10 C A spherical shell has a charge of +10C and a point charge of –15C at the center. What is the electric field produced OUTSIDE the shell? If the shell is conducting? Field Inside a Conductor is ZERO! If the shell is conducting? Field Inside a Conductor is ZERO! E r E=k(15C)/r 2 E=k(5C)/r 2 E=0 And if the shell is insulating? Charged Shells Behave Like a Point Charge of Total Charge “Q” at the Center Once Outside the Last Shell! Charged Shells Behave Like a Point Charge of Total Charge “Q” at the Center Once Outside the Last Shell! Conducting

19. Electric Fields With Insulating Sphere

20. Summary Electric Flux: a Surface Integral (Vector Calculus!); Useful Visualization: Electric Flux Lines Like Wind Through a Window. Gauss’ Law Provides a Very Direct Way to Compute the Electric Flux.