1. 1/16/2008 J.Velkovska 1 PHYS117B: Lecture 4 Last lecture: We used  Coulomb’s law  Principle of superposition To find the electric field of continuous charge distributions Today: I’m going to teach you the easy way!

2. 1/16/2008J.Velkovska2 The key word is SYMMETRY

3. 1/16/2008 J.Velkovska 3 We will: Recognize and use symmetry to determine the shape of electric fields Calculate electric flux through a surface Use Gauss’s law to calculate the electric field of symmetric charge distributions Use Gauss’s law to understand the properties of conductors in electrostatic equilibrium

4. 1/16/2008 J.Velkovska 4 Suppose we know only 2 things about electric fields: 1. An electric field points away from + charges and towards negative charges 2. An electric field exerts a force on a charged particle What can we deduce for the electric field of an infinitely long charged cylinder ?

5. 1/16/2008 J.Velkovska 5 The charge distribution has cylindrical symmetry What does this mean ?  There is a group of geometrical transformations that do not cause any physical change  Let’s try it (I have a cylinder here)

6. 1/16/2008 J.Velkovska 6 Translate, rotate, reflect

7. 1/16/2008 J.Velkovska 7 If you can’t tell that the charge distribution (the cylinder) was transformed geometrically, then the electric field should not change either ! The symmetry of the electric field MUST match the symmetry of the charge distribution!

8. 1/16/2008 J.Velkovska 8 Can the electric field of the cylinder look like this ?

9. 1/16/2008 J.Velkovska 9 Can the electric field of the cylinder look like this ?

10. 1/16/2008 J.Velkovska 10 What about this situation?

11. 1/16/2008 J.Velkovska 11 What about this situation ?

12. 1/16/2008 J.Velkovska 12 Three basic symmetries: planar, cylindrical, spherical We will make heavy use of these three basic symmetries.

13. 1/16/2008 J.Velkovska 13 We learned how to use symmetry to determine the direction of the electric field, if we know the geometry of the charge distribution. Can we reverse the problem ? Can we use the symmetry of the electric field configuration to determine the geometry of the charge distribution that causes this field ?

14. 1/16/2008 J.Velkovska 14 The mystery box Take a test charge, move it around, measure the force on it, figure out the field configuration => deduce the symmetry of the charge distribution inside

15. 1/16/2008 J.Velkovska 15 Imagine a situation like this: No field around the box, or same “amount of field” going in and out

16. 1/16/2008 J.Velkovska 16 Or you can have: Same box, same geometry of the field, but “more field” going out of the box

17. 1/16/2008 J.Velkovska 17 We need some way to measure “how much” field goes in or out of the box To measure the volume of water that passes through a loop per unit time, we use FLUX: the dot product of the velocity vector and the area vector gives volume/time

18. 1/16/2008 J.Velkovska 18 We can define electric field flux:

19. 1/16/2008 J.Velkovska 19 If the field is non-uniform: This is trouble, I need to do an integral

20. 1/16/2008 J.Velkovska 20 The surface is NOT flat : ooh bother !

21. 1/16/2008 J.Velkovska 21 Relax ! We are going to deal with two EASY situations: the field is UNIFORM in both. The field is EVERYWHERE tangent the surface: FLUX = 0 ! The field is EVERYWHERE perpendicular to the surface: FLUX = EA

22. 1/16/2008 J.Velkovska 22 What good is the electric field flux ? Gauss’s law gives a relation between the electric field flux through a closed surface and the charge that is enclosed in that surface. From symmetry: determine the shape (direction in every point in space) of the electric field From Gauss’s law: determine the magnitude of E

23. 1/16/2008 J.Velkovska 23 What surface are we talking about ? This is NOT a physical surface It is an IMAGINARY surface If we want to make life simple, we have to figure out what type of surface to choose, so that the integral in Gauss’s law is EASY to do We need to choose a surface that has the same symmetry as the charge distribution Then the field will be either tangent to the surface or perpendicular to the surface => no angles to deal with in the dot product!

24. 1/16/2008 J.Velkovska 24 Let’s do it for an infinitely long wire with uniform linear charge density Last time we used Coulomb’s law: this is the HARD way Today: or sweet simplicity – we’ll use Gauss’s law to get the same result !

25. 1/16/2008 J.Velkovska 25 Field of a line of charge: use Coulomb’s law, superposition and symmetry ! For an infinite line of charge i.e. x<< a line of charge: length 2a and linear charge density

26. 1/16/2008 J.Velkovska 26 See now how to do it using symmetry and Gauss’s law