1. a b c

2. Choose either or And E constant over surface is just the area of the Gaussian surface over which we are integrating. Gauss’ Law This equation can now be solved for E (at the surface) if we know q enclosed (or for q enclosed if we know E ). Gauss’ Law

3. Given an infinite sheet of charge as shown in the figure. You need to use Gauss' Law to calculate the electric field near the sheet of charge. Which of the Gaussian surfaces are best suited for this purpose? a cylinder with its axis along the plane? –No, E not parallel or perpendicular to the cylinder a cylinder with its axis perpendicular to the plane? –Yes a cube? –Yes a sphere? –No Gaussian Surfaces

4. Infinite sheet of charge Symmetry: direction of E = x -axis Therefore, CHOOSE Gaussian surface to be a cylinder whose axis is aligned with the x -axis.  A x EE Conclusion: An infinite plane sheet of charge creates a CONSTANT electric field. Apply Gauss' Law: The charge enclosed =  A Therefore, Gauss’ Law  On the barrel, On the ends,

5. Two Infinite Sheets (into the screen) Field outside must be zero. Two ways to see: –Superposition –Gaussian surface encloses zero charge E =0 E  + + + + + + +  + + + - - - - - - - - - - + + - - - A A 0 Field inside is NOT zero: – Superposition – Gaussian surface encloses non-zero charge

6. Question 1 The figure shows a Gaussian surface enclosing charges 2q and –q. The net flux through the surface is: a) q/  0 b) 2q/  0 c)-q/  0 d)zero 2q -q

7. Question 1 1.a 2.b 3.c 4.d

8. Question 1 The figure shows a Gaussian surface enclosing charges 2q and –q. The net flux through the surface is: a) q/  0 b) 2q/  0 c)-q/  0 d)zero 2q -q The net charge enclosed is +q. Gauss’ Law says the flux through the surface is q/  0 q of the lines from the positive charge go to the negative charge q go to infinity and thus pass through the surface

9. How to do practically all of the homework problems Gauss’ Law is ALWAYS VALID! Gauss’ Law: Help for the Problems What Can You Do With This? If you have symmetry (a) spherical, (b) cylindrical, or (c) planar AND: If you know the charge (RHS), you can calculate the electric field (LHS) If you know the field (LHS, usually because E =0 inside conductor), you can calculate the charge (RHS).

10. Gauss’ Law: Help for the Problems q = ALL charge inside radius r Spherical Symmetry: Gaussian surface = Sphere of radius r q = ALL charge inside cylinder =  A Planar Symmetry: Gaussian surface = Cylinder of area A q = ALL charge inside radius r, length L Cylindrical symmetry: Gaussian surface = cylinder of radius r

11. Example 1: spheres A solid conducting sphere is concentric with a thin conducting shell, as shown The inner sphere carries a charge Q 1, and the spherical shell carries a charge Q 2, such that Q 2 = -3 Q 1. How is the charge distributed on the sphere? How is the charge distributed on the spherical shell? What is the electric field at r < R 1 ? Between R 1 and R 2 ? At r > R 2 ? What happens when you connect the two spheres with a wire? (What are the charges?) R1R1 R2R2 Q1Q1 Q 2 =-3Q 1 A B C D

12. How is the charge distributed on the sphere? A * The electric field inside a conductor is zero. (A) By Gauss’s Law, there can be no net charge inside the conductor, and the charge must reside on the outside surface of the sphere + + + + + + + + R1R1 R2R2 Q1Q1 Q 2 =-3Q 1

13. How is the charge distributed on the spherical shell? B * The electric field inside the conducting shell is zero. (B) There can be no net charge inside the conductor, therefore the inner surface of the shell must carry a net charge of - Q 1, to cancel the inner conductor charge. The outer surface must carry the charge + Q 1 + Q 2, so that the net charge on the shell equals Q 2. The charges are distributed uniformly over the inner and outer surfaces of the shell, hence and R1R1 R2R2 Q1Q1 Q 2 =-3Q 1

14. (C) r < R 1 : Inside the conducting sphere * The electric field inside a conductor is zero. (C) Between R 1 and R 2 : R 1 < r < R 2 Charge enclosed = Q 1 (C) r > R 2 Charge enclosed = Q 1 + Q 2 What is the Electric Field at r < R 1 ? Between R 1 and R 2 ? At r > R 2 ? C R1R1 R2R2 Q1Q1 Q 2 =-3Q 1

15. D What happens when you connect the two spheres with a wire? (What are the charges?) After electrostatic equilibrium is reached, there is no charge on the inner sphere, and none on the inner surface of the shell. The charge Q 1 + Q 2 on the outer surface remains. - - - - - - - - - - - - - - - - - - - - - - Also, for r < R 2 and for r > R 2 R1R1 R2R2 Q1Q1 Q 2 =-3Q 1

16. Are Gauss’ and Coulomb’s Laws Correct? The table shows the results of such experiments looking for a deviation from an inverse-square law:

17. Are Gauss’ and Coulomb’s Laws Correct? One problem with the above experiments is that they have all been done at short range, 1 meter or so. Other experiments, more sensitive to cosmic-scale distances, have been done, testing whether Coulomb’s law has the form: No evidence for a nonzero μ has been found.

18. Example 2: Cylinders An infinite line of charge passes directly through the middle of a hollow, charged, CONDUCTING infinite cylindrical shell of radius R. We will focus on a segment of the cylindrical shell of length h. The line charge has a linear charge density, and the cylindrical shell has a net surface charge density of  total.  outer R  inner h  total

19.  outer h R  total  inner How is the charge distributed on the cylindrical shell? What is  inner ? What is  outer ? What is the electric field at r < R ? What is the electric field for r > R ? A B C

20.  outer h R  total  inner The electric field inside the cylindrical shell is zero. Choose as our Gaussian surface a cylinder, which lies inside the cylindrical shell. Then the net charge enclosed is zero. Therefore, there will be a surface charge density on the inside wall of the cylinder to balance out the charge along the line. A1A1 What is  inner ?

21.  outer h  total  inner R The total charge on the enclosed portion (of length h) of the line charge is  h Therefore, the charge on the inner surface of the conducting cylindrical shell is The total charge is evenly distributed along the inside surface of the cylinder. Therefore, the inner surface charge density  inner is just Q inner divided by total area of the cylinder: Notice that the result is independent of h. A1A1 What is  inner ?

22.  outer h R  total  inner We know that the net charge density on the cylinder is  total. The charge densities on the inner and outer surfaces of the cylindrical shell have to add up to  total. Therefore,  outer =  total –  inner =  total + /(2  R ). What is  outer ? A2A2

23. h Gaussian surface What is the Electric Field at r < R ? Whenever we are dealing with electric fields created by symmetric charged surfaces, we must always first chose an appropriate Gaussian surface. In this case, for r < R, the appropriate surface surrounding the line charge is a cylinder of radius r. r R h B

24. h Gaussian surface What is the Electric Field at r < R ? Using Gauss’ Law, the E field is given by q enclosed is the charge on the enclosed line charge h, (2  rh ) is the area of the barrel of the Gaussian surface. The Field at radius r is: r R h B

25. Chose a Gaussian surface as indicated above. Charge enclosed in our Gaussian surface Net charge enclosed on the line: h Net charge on the cylindrical shell: = 2  Rh  total Therefore, net charge enclosed is The surface area of the Gaussian surface is 2  rh Gauss’ Law: h R r Gaussian surface  total What is the Electric field for r > R ? C Solve for E r to find

26. Summary Practice Gauss’ Law problems Remember SYMMETRY, CONSTANT FIELD Choose appropriate Gaussian surface Next week Electric Potential. Read Chapter 24 Try Chapter 23 problems 25, 33, 47, 51, 56 Next week: Quiz on Thursday and Friday will be on chapters 21,22,23