1. Fall 2014 Notes 23 ECE 2317 Applied Electricity and Magnetism Prof. David R. Jackson ECE Dept. 1

2. Boundary Value Problem Uniqueness theorem: On boundary: Goal: Solve for the potential function inside of a region, given the value of the potential function on the boundary. (Please see the textbooks for a proof of the uniqueness theorem.) (no charges) 2 As long as our solution satisfies the Laplace equation and the B.C.s, it must be correct!

3. Example: Faraday Cage Effect Guess:  B = V 0 = constant Check: Hollow PEC shell Prove that E = 0 inside a hollow PEC shell (Faraday cage effect). Therefore: The correct solution is V S 3 Note: We can make any guess that we wish, as long as our final solution satisfies Laplace’s equation and the boundary conditions.

4. Example (cont.) Hollow PEC shell V S Hence E = 0 everywhere inside the hollow cavity.  B = V 0 = constant 4 (Faraday cage effect)

5. Example Solve for  ( x, y, z ) Assume: + -+ - rr h x V0V0 5 Ideal parallel-plate capacitor Note: We can make any assumptions that we wish, as long as our final solution satisfies the boundary conditions.

6. Example (cont.) Hence Solution: 6

7. Example (cont.) Hence we have The solution is then 7

8. Example (cont.) Calculate the electric field: From previous notes: so 8 Hence

9. Example Assume 9 + - V0V0 Wedge Insulating gap

10. Example (cont.) Hence 10

11. Example (cont.) Hence Find the electric field: We then have 11

12. Example (cont.) 12 Flux plot + - V0V0 Insulating gap

13. Example (Four unknowns) + - r1r1 h1h1 h2h2 x r2r2 V0V0 h 13 Two-layer capacitor

14. Example (cont.) Hence: so We need two more equations: use interface boundary conditions. BC #1 The potential is continuous across the boundary. 2 1 Now there are two unknowns ( c 1 and d 1 ). (The path length is zero!) 14

15. Example (cont.) BC #2: To calculate E x, use Hence we have Therefore 15 The normal component of flux density is continuous.

16. Example (cont.) Therefore or Hence we have Also, 16

17. Example (cont.) We now find the electric fields and flux density: 17

18. Example (cont.) We now find the surface charge densities on the plates. Use 18 + - r1r1 h1h1 h2h2 x r2r2 V0V0 h ++++++++++++++++ - - - - - - - -