Prof. David R. Jackson ECE Dept.

1. ECE 2317 Applied Electricity and Magnetism Spring 2014 Prof. David R. Jackson ECE Dept. Notes 6 Notes prepared by the EM Group University of Houston

2. Review of Coordinate Systems z P (x, y, z) y x An understanding of coordinate systems is important for doing EM calculations.

3. Kinds of Integrals That Often Occur We wish to be able to perform all of these in various coordinates.

4. Rectangular Coordinates Position vector: z Short hand notation: P (x,y,z) r z P (x,y,z) y r x y Note: A unit vector direction is defined by increasing one coordinate variable while keeping the other two fixed. x Note: Different notations are used for vectors in the books.

5. Rectangular Coordinates z dS = dxdy dy dx dz dS = dxdz y dS = dydz x We increment (x, y, z) starting from an initial point (blue dot).

6. Rectangular (cont.) Path Integral (we need dr) z B C dr r A r+dr y x Note on notation: The symbol dlis often used instead of dr .

7. Cylindrical Coordinates z  P (, , z) z . y  x y   x

8. Cylindrical (cont.) z . y y x Note: and depend on (x, y)  x Unit Vectors Note: A unit vector direction is defined by increasing one coordinate variable while keeping the other two fixed. This is why we often prefer to express them in terms of

9. Cylindrical (cont.) y  x Expressions for unit vectors (illustrated for ) Assume Similarly, Then we have: so Hence, we have

10. Cylindrical (cont.) Summary of Results

11. Cylindrical (cont.) z Example: Express the r vector in cylindrical coordinates. . r y x Substituting from the previous tables of unit vector transformations and coordinate transformations, we have

12. Cylindrical (cont.) z . r y x Note:

13. Cylindrical (cont.) dS =  d d z  dz dS =  d dz  d d dS = d dz y x Differentials Note: dS may be in three different forms. We increment (,  , z) starting from an initial point (blue dot).

14. Cylindrical (cont.) d y y z  d d dz   x x y x Path Integrals First, consider differential changes along any of the three coordinate directions.

15. Cylindrical (cont.) Note: A change is z is not shown, but is possible. In general: C y dr x If we ever need to find the length along a contour:

16. Spherical Coordinates z z  P (r, , ) P (r, , )   z z . . r r y y   x x Note: 0 <  <  Note:  = r sin 

17. Spherical (cont.) z  P (r, , )  . z r y  x Note: =r sin

18. Spherical (cont.) Unit Vectors z Note: A unit vector direction is defined by increasing one coordinate variable while keeping the other two fixed. y x Note: ,and depend on (x, y, z).

19. Spherical (cont.) Transformation of Unit Vectors z y x

20. Spherical (cont.) z Example: Express the r vector in spherical coordinates. r y Substituting from the previous tables of unit vector transformations and coordinate transformations, we have x

21. Spherical (cont.) z After simplifying: r y x Note:

22. Spherical (cont.) Differentials We increment (r, , ) starting from an initial point (blue dot).  d =r sin d z  d dS = r2 sin d d d y Note: dS may be in three different forms (only one is shown). The other two are: dr r d x dS = rdrd dS = rsin dr d

23. Spherical (cont.) z z dr dr r d y y x x Path Integrals z  d dr r y x

24. Note on dr Vector Note that the formula for the dr vector never changes, no matter which direction we go along a path. Example: integrating along a radial path in cylindrical coordinates. y y A C C B A B x x

25. Example P1 (4, 60, 1) P2 (3, 180, -1) Cylindrical coordinates (, , z) Given: with distances in meters Find d= distance between points This formula only works in rectangular coordinates! d=6.403 [m]

26. Example Derive Let Dot multiply both sides with z  y  L An illustration of finding the x component of x

27. Example (cont.) z  ( / 2) -   Hence y L x  Similarly, x Also, Result:

28. Example Derive Let Dot multiply both sides with Then

29. Example (cont.) z z z      y y y x x x Result:

30. Example z a y b x a = 2 [m], b = 5 [m] Given:v= -310-8(cos2 / r4) [C/m3] , 2 < r < 5 [m] Find Q Note: The integrand is separable and the limits are fixed. Solution: “A sphere with a hole in it”

31. Example (cont.) Note: The average value of cos2 is 1/2. Q = -5.65510-8[C]

32. Example Given: Find the current I crossing a hemisphere (z > 0) of radius a, in the outward direction. z y x

33. Example (Cont.)

34. Example (Cont.)

35. Example (Part 1) (This is not an electrostatic field.) Find VAB using path C shown below. z E (x,y,z) . (0,1,0) y B C (1,0,0) A y x Top view 1 x 1

36. Example (cont.) Completing the calculus: VAB= -5/12 [V]

37. Example (cont.) Alternative calculation (we parameterize differently): VAB= -5/12 [V]

38. Example (Part 2) (same field as in Part 1) Find VAB using path C shown below. z E (x,y,z) (0,1,0) y B (1,0,0) C A x VAB= 0[V]

39. Example (This is a valid electrostatic field.) Find VAB using an arbitrary path C in the xy plane. z E (x,y,z) (0,1,0) y B (1,0,0) C A x Note: The path does not have to be parameterized: Hence, only the endpoints are important. VAB= -7/6[V] The integral is path independent!

40. Example y B x A 3 [m] C Find VAB using path C shown below.

41. Example (cont.) Note: The angle  must change continuously along the path. If we take the angle  to be  / 2 at point B, then the angle  must be - at point A. VAB= 9/2[V]

42. Example (cont.) y B x A 3 [m] C Let’s examine this same electric field once again: Question: Is this integral path independent? Note: The answer is yes because the curl of the electric field is zero, but we do not know this yet.

43. Example (cont.) y B x A 3 [m] C Let’s find out: Yes, it is path independent! VAB= 9/2[V]