PHY 184

1. PHY 184 Spring 2007 Lecture 6 Title: Gauss’ Law 184 Lecture 6

2. Announcements • Homework Set 2 is due Tuesday at 8:00 am. • We will have clicker quizzes for extra credit starting next week (the clicker registration closes January 19). • Homework Set 3 will open Thursday morning. • Honors option work in the SLC will start next week • Honors students sign up after class for time slots. 184 Lecture 6

3. Outline • /1/ Review • /2/ Electric Flux • /3/ Gauss’ Law 184 Lecture 6

4. Review - Point Charge ► The electric field created by a point charge, as a function of position r, ► The force exerted by an electric field on a point charge q located at position x … direction tangent to the field line through x 184 Lecture 6

5. Review – Electric Dipole • 2 equal but opposite charges, -q and +q • Dipole moment (direction: - to +) • On the axis, far from the dipole, 184 Lecture 6

6. Force and Torque on an Electric Dipole Torque vector • Assume the 2 charges (-q and +q) are connected together with a constant distance d, and put the dipole in a uniform electric field E. Net force = 0 Torque about the center: 184 Lecture 6

7. Gauss’ Law 184 Lecture 6

8. Objective • So far, we have considered point charges. But how can we treat more complicated distributions, e.g., the field of a charged wire, a charged sphere or a charged ring? • Two methods • Method #1: Divide the distribution into infinitesimal elements dE and integrate to get the full electric field. • Method #2: If there is some special symmetry of the distribution, use Gauss’ Law to derive the field. Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface. 184 Lecture 6

9. Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface. 184 Lecture 6

10. Electric Flux • Let’s imagine that we put a ring with area A perpendicular to a stream of water flowing with velocity v • The product of area times velocity, Av, gives the volume of water passing through the ring per unit time • The units are m3/s • If we tilt the ring at an angle , then the projected area isA cos, and the volume of water per unit time flowing through the ring is Av cos . 184 Lecture 6

11. Electric Flux (2) • We call the amount of water flowing through the ring the “flux of water” • We can make an analogy with electric field lines from a constant electric field and flowing water • We call the density of electric field lines through an area A the electric flux given by 184 Lecture 6

12. Math PrimerSurfaces and Normal Vectors Electric Flux • For a given surface, we define the normal unit vector n, which points normal to the surface and has length 1. 184 Lecture 6

13. Math Primer - Gaussian Surface • A closed surface, enclosing all or part of a charge distribution, is called a Gaussian Surface. • Example: Consider the flux through the surface on the right. Divide surface into small squares of area A. • Flux through surface: 184 Lecture 6

14. Electric Flux (3) • In the general case where the electric field is not constant everywhere • We define the electric flux through a closed surface in terms of an integral over the closed surface 184 Lecture 6

15. Gauss’ Law • Gauss’ Law (named for German mathematician and scientist Johann Carl Friedrich Gauss, 1777 - 1855) states • (q = net charge enclosed by S). • If we add the definition of the electric flux we get another expression for Gauss’ Law • Gauss’ Law : the electric field flux through S is proportional to the net charge enclosed by S. 184 Lecture 6

16. Theorem: Gauss’ Law and Coulomb’s Laware equivalent. r q • Let’s derive Coulomb’s Law from Gauss’ Law. • We start with a point charge q. • We construct a spherical surface with radius r surrounding this charge. • This is our “Gaussian surface” F = E A 184 Lecture 6

17. Theorem (2) • The electric field from a point charge is radial, and thus is perpendicular to the Gaussian surface everywhere. /1/ The electric field direction is parallel to the normal vector for any point. /2/ The magnitude of the electric field is the same at every point on the Gaussian surface. F = E A /1/ /2/ 184 Lecture 6

18. Theorem (3) Now apply Gauss’ Law Q. E. D. 184 Lecture 6

19. Shielding • An interesting application of Gauss’ Law: • The electric field inside a charged conductor is zero. • Think about it physically… • The conduction electrons will move in response to any electric field. • Thus the excess charge will move to the surface of the conductor. • So for any Gaussian surface inside the conductor -- encloses no charge! – the flux is 0. This implies that the electric field is zero inside the conductor. 184 Lecture 6

20. Shielding Illustration • Start with a hollow conductor. • Add charge to the conductor. • The charge will move to the outer surface • We can define a Gaussian surface that encloses zero charge • Flux is 0 • Ergo - No electric field! 184 Lecture 6

21. Cavities in Conductors • a) Isolated Copper block with net charge. The electric field inside is 0. • b) Charged copper block with cavity. Put a Gaussian surface around the cavity. The E field inside a conductor is 0. That means that there is no flux through the surface and consequently, the surface does not enclose a net charge. There is no net charge on the walls of the cavity 184 Lecture 6

22. Shielding Demonstration • We will demonstrate shielding in two ways • We will place Styrofoam peanuts in a container on a Van de Graaff generator • In a metal cup • In a plastic cup • We will place a student in a wire cage and try to fry him with large sparks from a Van de Graaff generator • Note that the shielding effect does not require a solid conductor • A wire mesh will also work, as long as you don’t get too close to the open areas 184 Lecture 6

23. Lightning Strikes a Car The crash-test dummy is safe, but the right front tire didn’t make it … High Voltage Laboratory, Technical University Berlin, Germany 184 Lecture 6