Physics 212 Lecture 6

1. Physics 212 Lecture 6 Electric Potential

2. BIG IDEA • Last time we defined the electric potential energy of charge q in an electric field: • The only mention of the particle was through its charge q. • We can obtain a new quantity, the electric potential, which is a PROPERTY OF THE SPACE, as the potential energy per unit charge. V in Volts = Joules/Coulomb • Note the similarity to the definition of another quantity which is also a PROPERTY OF THE SPACE, the electric field. 40

3. Electric Potential from E field Dx D • What is the sign of DVAC = VC - VA ? (A) DVAC < 0 (B) DVAC = 0 (C) DVAC > 0 • Remember the definition: • Consider the three points A, B, and C located in a region of constant electric field as shown. • Choose a path (any will do!) 40

4. Checkpoint 2 V is constant !! A B C D • Remember the definition 08

5. E from V • In Cartesian coordinates: • We obtain the potential by integrating the electric field: • So, we can obtain the electric field by differentiating the potential. 40

6. Checkpoint 1a • How do we get E from V?? Look at slopes !!! 08

7. Checkpoint 1b Look at slopes !!! A B C D • How do we get E from V?? 08

8. Equipotentials Equipotentials produced by a point charge • Equipotentials are the locus of points having the same potential. Equipotentials are ALWAYS perpendicular to the electric field lines The SPACING of the equipotentials indicates The STRENGTH of the electric field 40

9. Checkpoint 3a A B C D 08

10. Checkpoint 3b A B C D A B C D 08

11. HINT • What is work done by E field to move negative charge from A to C ? (A) WAC < 0 (B) WAC = 0 (C) WAC > 0 WAC = 0 E - FIELD LINES A B C D EQUIPOTENTIALS A B C D A and C are on the same equipotential Equipotentials are perpendicular to the E field: No work is done along an equipotential. 08

12. Checkpoint 3b Again? A B C D A B C D • A and C are on the same equipotential • B and D are on the same equipotential • Therefore the potential difference between A and B is the SAME as the potential between C and D 08

13. Checkpoint 3c A B C D A B C D 08

14. cross-section a4 a3 Point charge q at center of concentric conducting spherical shells of radii a1, a2, a3, and a4.The inner shell is uncharged, but the outer shell carries charge Q. What is V(r) as a function of r? +Q a2 a1 +q metal metal - Charges q and Q will create an E field throughout space • Spherical symmetry: Use Gauss’ Law to calculate E everywhere • Integrate E to get V 40

15. r Why? Gauss’ law: a4 a3 r > a4: What is E(r)? (A) 0 (B) (C) +Q a2 a1 +q (D) (E) metal metal

16. r Applying Gauss’ law, what is Qenclosedfor red sphere shown? (A) q (B)–q(C)0 charge at r=a4 surface = Q+q a4 a3 a3 < r < a4: What is E(r)? (A) 0 (B) (C) +Q a2 a1 +q (D) (E) metal metal How is this possible??? -q must be induced at r=a3 surface

17. a4 a3 +Q a2 a1 +q metal metal

18. a4 a3 +Q a2 a1 +q metal metal For r  a4 Just like a point charge ! So… 0 First find V(r) for r  a4

19. a4 How about V(r) for ? a3 +Q a2 a1 +q metal metal 0 V( r ) constant inside a conductor !

20. ? V(r) for a4 a3 +Q a2 a1 +q metal metal 0

21. ? V(r) for a4 a3 +Q a2 a1 +q metal metal 0 0

22. a4 V(r) for ? a3 +Q a2 a1 +q metal metal 0 0

23. Summary V a1 a2 a3 a4 0 r