Physics 51

1. Physics 51 Tues. 2/11/14 Finish Gauss’ Law Start Ch. 23

2. HW Question Approximately how many hours did you spend working on HW #3 • Less than 1 hour 2) between 1 and 3 3) Between 3 and 5 4) more that 5 hours

3. HW grading problem Often I see a beginning equation and then an answer with few or no steps in between. This is because: • Answer is given in book and it has been copied • Additional steps have been done on scratch paper • Student is quicker than me (duh!) and the answer is obvious

4. HW grading problem 2 I have not had time to grade all homework in detail. I should: • Assign less homework • Keep same amount but just check off if it was done • Keep same amount but grade one problem in a detailed way, just check off the rest • Suck it up, just grade it all

5. HW problem My perception: HW gives you practice with both easy and difficult problems Your Perception: HW is important to get points toward a better grade (my guess only)

6. Misc. Items • HW #4 • Chapter 23 Exercises: #4, #28, #39, #62 • Due next Tuesday, Feb 18

7. Quiz #4 • Maybe a bit too hard (damn that MIT) • But a few of you did a great job! • I am worried that for many of you by skipping to a hard problem we did not reinforce the basic idea

8. Electric Flux Definition FE=òE . dA Idea The electric flux is the amount of electric field passing Through a surface of area A

9. Gauss’ Law What is says • Electric flux through a closed surface equals the net charge inside that surface (extra e0 to make units work) To be useful • Need to be able to calculate flux • Need to calculate charge

10. Gauss’s Law can be used to calculate the magnitude of the E field vector

11. In the diagram below there is a spherical ball of radius R with total charge Q, uniformly distributed through its volume. The red dotted line represents a spherical Gaussian surface of radius r, outside the ball r R What is the electric flux through the Gaussian surface: 1) Q/e0 2) 4pr2E 3) 4pR2E 4) 4/3pr3E

12. In the diagram below there is a spherical ball of radius R with total charge Q, uniformly distributed through its volume. The red dotted line represents a spherical Gaussian surface outside the ball r R What is the Electric charge inside the the Gaussian surface: 1) Q/e0 2) Q 3) Q (r3/R3) 4) Q 4/3pr3

13. In the diagram below there is a spherical ball of radius R with total charge Q, uniformly distributed through its volume. The red dotted line represents a spherical Gaussian surface inside the ball R r What is the Electric flux through the Gaussian surface: 1) Q/e0 2) 4pr2E 3) 4pR2E 4) Q(r3/R3)E

14. In the diagram below there is a spherical ball of radius R with total charge Q, uniformly distributed through its volume. The red dotted line represents a spherical Gaussian surface inside the ball R r What is the Electric charge inside the Gaussian surface: 1) Q/e0 2) 4pr2E 3) Q(r2/R2) 4) Q(r3/R3)

15. In the diagram below there is a long wire with a uniform charge/length of l. The red dotted figure is a Gaussian surface shaped like a flat can with two ends and an outer surface L r l What is the Electric flux through the two ends of the Gaussian surface: 1) 2pr2E 2) 4pr2E 3) 0 4) lL

16. In the diagram below there is a long wire with a uniform charge/length of l. The red dotted figure is a Gaussian surface shaped like a flat can with two ends and an outer surface L r l What is the Electric flux through the outer surface of the Gaussian surface: 1) 2pr2E 2) 2prLE 3) 0 4) lL

17. In the diagram below there is a long wire with a uniform charge/length of l. The red dotted figure is a Gaussian surface shaped like a flat can with two ends and an outer surface L r l What is the charge enclosed by this Gaussian surface: 1) 2pr2E 2) lE/e0 3) 0 4) lL

18. Simple Concept Question on Flux A cylindrical piece of insulating material with no charge is placed in an external electric field, as shown. The net electric flux passing through the surface of the cylinder is 1. positive. 2. negative. 3. zero.

19. Conductors and Gaussian Surfaces Gaussian surface Inside Conductor E = 0 Therefore Qin-side = 0 Under electrostatic conditions, any excess charge resides entirely on the surface of a solid conductor.

20. Conductors and Gaussian Surfaces A charge inside a hole in a conductor gives rise to an opposite charge on the inner surface

21. A Gaussian surface drawn inside the conducting material of which the box is made must have zero electric field on it (field inside a conductor is zero).  If the Gaussian surface has zero field on it, the charge enclosed must be zero per Gauss's Law. The E field inside a conducting box (a “Faraday cage”) in an electric field.

22. 2 ideas I skipped • Charge conservation • Charging by induction

23. Charge Conservation • When two objects come in contact charge is neither created or destroyed • The total net charge is the same • When conductors come in contact charge flows until they have no field between them

24. Two identical metal balls have charges on them as shown on the left. They are brought in contact then separated. The charges on the balls after this process are - 3C Q2 + 5C Q1 • Q1= 2C, Q2 = 0 2) Q1=1C , Q2=1C • 3) Q1 = 5C, Q2 = -3C 4) Q1 = 5C, Q2 = 0

25. CHARGING METAL SPHERE BY INDUCTION Charges are free to move in a conductor but are tightly bound in an insulator. The earth (ground) is a large conductor having many free charges.

26. Clicker Question A positively charged object is placed close to a conducting object attached to an insulating glass pedestal (a). After the opposite side of the conductor is grounded for a short time interval (b), the conductor becomes negatively charged (c). Based on this information, we can conclude that within the conductor 1. both positive and negative charges move freely. 2. only negative charges move freely. 3. only positive charges move freely. 4. We can’t really conclude anything.