1. 1/17/07184 Lecture 61 PHY 184 Spring 2007 Lecture 6 Title: Gauss’ Law

2. 1/17/07184 Lecture 62AnnouncementsAnnouncements  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.

3. 1/17/07184 Lecture 63OutlineOutline  /1/ Review  /2/ Electric Flux  /3/ Gauss’ Law

4. 1/17/07184 Lecture 64 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

5. 1/17/07184 Lecture 65 Review – Electric Dipole  2 equal but opposite charges, -q and +q  Dipole moment (direction: - to +)  On the axis, far from the dipole,

6. 1/17/07184 Lecture 66 Force and Torque on an Electric Dipole  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: Torque vector

7. 1/17/07184 Lecture 67 Gauss’ Law

8. 1/17/07184 Lecture 68ObjectiveObjective  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.

9. 1/17/07184 Lecture 69 Gauss’ Law The flux of electric field through a closed surface is proportional to the charge enclosed by the surface.

10. 1/17/07184 Lecture 610 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 m 3 /s  If we tilt the ring at an angle , then the projected area is A cos , and the volume of water per unit time flowing through the ring is Av cos .

11. 1/17/07184 Lecture 611 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

12. 1/17/07184 Lecture 612 Math Primer Surfaces and Normal Vectors n,  For a given surface, we define the normal unit vector n, which points normal to the surface and has length 1. Electric Flux

13. 1/17/07184 Lecture 613 Math Primer - Gaussian Surface 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:

14. 1/17/07184 Lecture 614 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

15. 1/17/07184 Lecture 615 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.

16. 1/17/07184 Lecture 616 Theorem: Gauss’ Law and Coulomb’s Law are equivalent.  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. “Gaussian surface” This is our “Gaussian surface” q r  = E A

17. 1/17/07184 Lecture 617 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. /1/ /2/  = E A

18. 1/17/07184 Lecture 618 Theorem (3) Now apply Gauss’ Law Q. E. D.

19. 1/17/07184 Lecture 619ShieldingShielding  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. Gaussian surface 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.

20. 1/17/07184 Lecture 620 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!

21. 1/17/07184 Lecture 621 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

22. 1/17/07184 Lecture 622 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

23. 1/17/07184 Lecture 623 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