1. For a non-uniform electric field, we integrate along the path: which we can write as the sum of three separate integrals:

2. Your turn…

3. Uniformly charged rod. To calculate V C – V A we should use: 1) 1 path segment, uniform field 2) 2 path segments, uniform fields 3) 3 path segments, uniform fields 4) 1 path segment, integral 5) 2 path segments, integrals 6) 2 path segments: 1 integral, 1 unif. field

4. Example: A negatively charged metal sphere

8. Potential difference around a closed loop is zero.

9. Potential at one location + B Let r A go to infinity…

10. Potential at one location The potential near a point charge, relative to infinity: ++

11. Potential energy of two charges The potential energy of two point charges, relative to infinity: ++

12. Summary The potential difference between two points: The potential energy difference for a charge q, moved between two points: The potential near a point charge, with respect to infinity:

13. Potential in a conductor + At equilibrium, the field inside the conductor must be zero. A B

14. Capacitor

15. Does ΔV change when we insert a slab of metal?

16. Your turn… Now what is the magnitude of the electric field in the air gaps? 1) Same as originally 2) Less than originally 3) Greater than originally 4) Not enough information to tell In a capacitor E = (Q/A)/  0

17. Your turn… Without the metal slab, V B – V A was –1000 volts. Then a metal slab was inserted into the capacitor. Now  V = V B – V A = 1) + 1000 volts 2) +500 volts 3) 0 volts 4) –500 volts 5) –1000 volts 6) Not enough information to tell In a capacitor E = (Q/A)/  0

18. Originally ΔV was –1000 volts. A plastic slab is inserted into the capacitor. Now ΔV = V B – V A = 1) –1000 volts 2) +1000 volts 3) between 0 and –1000 volts 4) between 0 and +1000 volts 5) 0 volts 6) not enough information to tell Your turn… In a capacitor E = (Q/A)/  0

21. MaterialK Vacuum1 Air1.0006 Plastic5 Water80 dielectric constant 介电常数 （相对电容率）

22. Your turn… In a capacitor E = (Q/A)/  0 Originally  V was – 1000 volts. A plastic slab is inserted into the capacitor. The dielectric constant of the plastic is 2. Now  V = V B – V A = 1) – 1000 volts 2) – 750 volts 3) – 250 volts 4) 0 volts 5) +250 volts

23. - - - - - - - + + + + + + + Try to pull the plates apart…

24. - - - - - - - + + + + + + + Work done by you: Electrical force on plate: This equals the change in potential energy of the system, ΔU.

25. I will now re-write this in a different way…

26. This is the electric field E between the plates. This is the change in volume ( 体积 ) of space between the plates.

29. Energy density in an electric field This is just another way of thinking about the potential energy U of the system of charges. It is not a ‘new’ energy.

30. Energy density in an electric field It is true for any electric field, not just in a capacitor! Electric fields contain energy. (Also momentum!)

31. Can you think of a distribution of static charges to make this field? Electrostatic forces are conservative. The change in potential around a loop must be zero. for fields made by charges at rest.

32. means: No curly electric fields. BUT: This is only true for “Coulomb” fields (fields caused by stationary charges).