Gauss’s Law: Quantitative Statement

1. The net electric flux through any closed surface equals the net charge enclosed by that surface divided by 0. Gauss’s Law: Quantitative Statement • How do we use this equation?? • The above equation is TRUE always but it doesn’t • look easy to use. • BUT - It is very useful in finding E when the physical situation exhibits a lot of SYMMETRY.

2.  x Again: Continuous Charge Distribution 1: Charged Line At a point P on perpendicular axis:

3. Gauss’s Law: h r Same result but much less work! Infinitely long uniformly charged line E

4. At a point P on axis: Charged Disk

5. Uniformly charged thin, infinite sheet Gauss’s Law! A h

6. tiny disk Just to left of disk Just to right of disk Thinsheet of any charge distribution

7. Physics 241 –Warm-up quiz The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A,B and C ? Note that the planes are infinitely large.  - (A). E=0 in A and C but E=to the left in B (B). E=2to the left in A and C but E=0 in B. (C). E=2to the left in A and C but E=  in B. (D). E=0 in A and C but E=to the right in B A B C

8. Charges and fields of a conductor • In electrostatic equilibrium, free charges inside a conductor do not move. Thus, E = 0 everywhere in the interior of a conductor. • Since E = 0 inside, there are no net charges anywhere in the interior.Net charges can only be on the surface(s). The electric field must be perpendicular to the surface just outside a conductor,since, otherwise, there would be currents flowing along the surface.

9. Two conducting plates In isolation, the electric field near a charged conducting plate is given by • properties of conductors • and, superposition Case (c) is not a simple superposition of cases (a) and (b), because the electric charges are re-distributed on both plates when they are brought near each other.

10. Electrostatic Shielding (Faraday Cage) • E=0 inemptycavity in a good conductor.Why? If there are E field lines, then moving a charge from one end to the other against the E line through the cavity must require some work W. On the other hand since E is perpendicular to surface, W=0 if you move the charge along the surface instead. But the work should be independent of the path used. => Contradiction! X

11. Hollow region Net charge Q ( > 0) on the conductor Net charge induced = +q Net charge –q < 0 + + + + + + Net charge (Q-q) on the outside surface Hollow region Field ? Electric Field of a Conductor: Examples

12. Total induced charge -q Cavity in a spherical conductor • With +q in the cavity of a conductor, a net charge –q is induced on the inside surface of the cavity. • If +q is at the center of the spherical cavity, E is radial by symmetry. Uniform charge distributions. • As long as the E contribution due to all the interior charges, measured at the outer surface is zero, then moving the +q charge does not affect the exterior surface charge distribution. Electrostatic shielding

13. Conducting shell electrostatically shields its exterior from changes on the inside. Conducting shell electrostatically shields its interior from changes on the outside, too. + + + + + + Electrostatic Shielding (Continued) If you move charge q in the cavity, the exterior electric fields and the exterior charge distribution are not affected. q If you now add charge Q’ to the conductor and/or Q’’ on the outside of the conductor, the interior electric fields do not change. Q’’ Add Q’

14. Physics 241 –Quiz 3a The magnitude of the electric field from a single charged plane of charge density (>0) and infinite area is E = 0). What is the magnitude of the field at point A on the right side of the two infinitely large planes of uniformly charged?  - to the right 3to the left to the right (2)to the left 3to the right A e = 1.610-19 C  = 8.8510-12 C2/(Nm2)

15. Physics 241 –Quiz 3b The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A between the two infinitely large planes of uniformly charged?   to the right 3to the left to the right (2) to the left to the left A e = 1.610-19 C  = 8.8510-12 C2/(Nm2)

16. Physics 241 –Quiz 3c The magnitude of the electric field from a single charged plane of charge density  and infinite area is E = 0). What is the magnitude of the field at point A on the left side of the two infinitely large planes of uniformly charged?   to the right 3to the left to the right (2)to the left 3to the right A e = 1.610-19 C  = 8.8510-12 C2/(Nm2)