1. Prof. D. Wilton ECE Dept. Notes 24 ECE 2317 Applied Electricity and Magnetism Notes prepared by the EM group, University of Houston.

2. Stored Energy Note: Please see the text book or supplementary notes for a derivation. charge formula: electric-field formula: (volume charge density) (surface charge density)

3. Example + -+ - rr h x V0V0 A [m 2 ] Method #1 Find the stored energy

4. Example (cont.) Or, Recall that Hence

5. Example (cont.) Method #2 (since we have surface charge in this problem) rr h x A [m 2 ] A B + + + + + + + + + + + + + - - - - - - - - -

6. Example (cont.)

7. Capacitance from Stored Energy

8. Example  v0 [C/m 3 ] 00 a Method #1: Gauss’s Law: r < a r > a Find the stored energy Solid sphere of uniform volume charge density

9. Example (cont.)  v0 [C/m 3 ] 00 a

10. Example (cont.) Result: Let: Note:

11. KCL Law i2i2 i3i3 iNiN i4i4 i1i1 Wires meet at a “node”.

12. KCL Law (cont.) A C +Q ground C is the stray capacitance between the “node” and ground + - v

13. KCL Law (cont.) 1)In “steady state” (no time change) 2) As A (area of node)  0, C  0 Two cases for which the KCL law is valid:

14. KCL Law (Differential Form) Differential form: J VV (valid for D.C. currents) For D.C. : (circuit form) Integral form:

15. DC Current Formulas Ohm’s Law Charge-Current Formula (an experimental law) (this was derived earlier in the semester)

16. Resistor Formula + - V J L x A Hence I

17. Joule’s Law  W = work done on a small volume of charge as it moves a small distance inside the conductor from point A to point B. This goes to heat! VV A B vv E + v conducting body

18. Joule’s Law (cont.) Since, we also have

19. Power Dissipation by Resistor Hence, Note: passive sign convention appears in the final result. Resistor - V I A + L x

20. RC Analogy Goal: Assuming we know how to solve the C problem (find C ), can we solve the R problem (find R )? “C Problem” A + - V AB B ECEC “R Problem” A + - V AB B ERER I

21. RC Analogy (cont.) Theorem: E C = E R (same field) “C Problem” A + - V AB B ECEC “R Problem” A + - V AB B ERER IAIA

22. Hence: E C = E R “C Problem” “R Problem” Same D. E. since  (r) =  (r) Same B. C. since the same voltage is applied RC Analogy (cont.)

23. “C Problem” A + - V AB B ECEC RC Analogy (cont.)

24. “R Problem” A + - V AB B ERER IAIA RC Analogy (cont.)

25. Hence RC Analogy (cont.)

26. RC Analogy Recipe for calculating resistance: 1.Calculate the capacitance of the C problem. 2.Replace  everywhere with  to obtain G. 3.Take the reciprocal to obtain R. In equation form:

27. Special case: A single homogeneous medium of conductivity  surrounds the two conductors. RC Formula Hence, or

28. Example A h Find R C problem: A h Method #1 (RC analogy)

29. Example Find R C problem: Method #2 (RC formula) A h A h

30. Example Find resistance h1h1 h2h2  2 2  1 1 Note: We cannot use the RC formula, since there is more than one region.

31. Example h1h1 h2h2 22  1 1 C problem:

32. Example (cont.) h1h1 h2h2  2 2  1 1

33. h1h1 h2h2  2 2  1 1 Hence Example (cont.)

34. Example Find equivalent circuit C v (t) R i (t) + - AQ h + + + + + + + + + - - - - - - - - - - - - - E v(t)v(t) + - i (t)