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

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

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

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

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

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

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

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

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

Summary of Results Cylindrical (cont.) 10

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

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

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

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

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

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

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

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

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

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

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

Spherical (cont.) Differentials dS = r 2 sin  d  d  Note: dS may be in three different forms (only one is shown). The other two are: dS = r dr d  dS = r sin  dr d  22 x y z  d  = r sin  d  dr r d   dd dd We increment (r, ,  ) starting from an initial point (blue dot).

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

Note that the formula for the dr vector never changes, no matter which direction we go along a path (we never add a minus sign!). Note on dr Vector 24 Example: integrating along a radial path in cylindrical coordinates. x y A B C x y A B C This form does not change, regardless of which limit is larger.

Example Given: Find the current I crossing a hemisphere ( z > 0 ) of radius a, in the outward direction. 25 x y z

Example (Cont.) 26

Example (Cont.) 27

Appendix 28 Here we work out some more examples.

Example P 1 (4, 60 , 1) P 2 (3, 180 , -1) d = [ m ] Find d = distance between points Given: Cylindrical coordinates ( , , z ) with distances in meters 29 This formula only works in rectangular coordinates!

Example Given:  v = -3  ( cos 2  / r 4 ) [ C/m 3 ], 2 < r < 5 [ m ] Solution: x y z b a a = 2 [ m ], b = 5 [ m ] Find Q “A sphere with a hole in it” 30 Note: The integrand is separable and the limits are fixed.

Example (cont.) Q =  [ C ] 31 Note: The average value of cos 2  is 1/2.

Example Derive Let 32 Then Dot multiply both sides with

Example (cont.) Result: x y z   x y z  x y z   33

Example Derive Let An illustration of finding the x component of 34 Dot multiply both sides with x y z   L

Example (cont.) Result: Hence Similarly, Also, 35 x y z  xx  L (  / 2) -  

Example (Part 1) Find V AB using path C shown below. x y 1 1 x y z C (0,1,0) (1,0,0). E (x,y,z) B A Top view 36 (This is not an electrostatic field.)

Example (cont.) Completing the calculus: V AB = -5/12 [ V ] 37

Example (cont.) Alternative calculation (we parameterize differently): 38 V AB = -5/12 [ V ]

Example (Part 2) Find V AB using path C shown below. (same field as in Part 1) V AB = 0 [ V ] 39 x y z C (0,1,0) (1,0,0) E (x,y,z) B A

Example V AB = -7/6 [ V ] Find V AB using an arbitrary path C in the xy plane. Note: The path does not have to be parameterized: Hence, only the endpoints are important. The integral is path independent! (This is a valid electrostatic field.) 40 x y z C (0,1,0) (1,0,0) E (x,y,z) B A

Example A x C 3 [m] B y Find V AB using path C shown below. 41

Example (cont.) V AB = 9/2 [ V ] 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. 42

Example (cont.) A x C 3 [m] B y Question: Is this integral path independent? 43 Let’s examine this same electric field once again: Note: The answer is yes because the curl of the electric field is zero, but we do not know this yet.

Example (cont.) A x C 3 [m] B y Let’s find out: Yes, it is path independent! V AB = 9/2 [ V ] 44