Prof. David R. JacksonECE 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

20

ˆ

4

C

AB

C

C

v

V

Q d

V E dr

RE d

R

Q dV

(scalar integral,

scalar result)

(vector integral,

scalar result)

(vector integral,

vector result)

(scalar integral,

scalar re

Line

integrals :

Volume

integrals :2

0

ˆ

4v

V

RE dV

R

sult)

(vector integral,

vector result)

20

ˆ

ˆ

4

s

S

S

s

S

Q dS

I J n dS

RE dS

R

(scalar integral,

scalar result)

(vector integral,

scalar result)

(vector integral,

vector result)

Surface

integrals :

We wish to be able to perform all of these in various coordinates.

3

Rectangular Coordinates

ˆ ˆ ˆr xx yy zz

Short hand notation:

, ,r x y z

Note: Different notations are used for vectors in the books.4

Position vector:

zz

yyxx

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

dV dx dy dz

5

We increment (x, y, z) starting from an initial

point (blue dot).

dxdy

dz

dS = dxdy

dS = dxdz

dS = dydzx

y

z

Rectangular (cont.)

ˆ ˆ ˆdr x dx y dy z dz

Path Integral (we need dr)

Note on notation: The symbol dl is often used instead of dr .

ˆ ˆ ˆr xx yy zz

6

x

y

z

A

BC

dr

r r+dr

Cylindrical Coordinates

2 2

1

cos

sin

tan /

x

y

z z

x y

y x

z z

x

y

z

.z

P (, , z)

7

x

y

Cylindrical (cont.)

Unit Vectors

Note: and depend on (x, y)

x

y

x

y

z

.

z

This is why we often prefer to express them in terms of

ˆ ˆx yand

Note: A unit vector direction is defined by increasing one coordinate variable

while keeping the other two fixed.

8

1 2

1

1

ˆ ˆ ˆ ˆ ˆ ˆ

ˆ ˆ

ˆ ˆ cos

cos

x A x x A y x

A x

x

A

2 ˆ ˆ

cos2

sin

A y

Hence, we have ˆ ˆ ˆcos sinx y

x

y

1 2

ˆ ˆ ˆA x A y Assume

Similarly,Then we have:

Cylindrical (cont.)

so

Expressions for unit vectors (illustrated for )

9

ˆ ˆ ˆcos sin

ˆ ˆ ˆsin cos

ˆ ˆ

ˆˆˆ cos sin

ˆˆˆ sin cos

ˆ ˆ

x y

x y

z z

x

y

z z

Summary of Results

Cylindrical (cont.)

10

Cylindrical (cont.)

11

2 2

ˆ ˆˆ ˆ ˆcos sin cos sin cos sin

ˆ ˆcos sin

ˆ ˆ

r zz

zz

zz

ˆ ˆ ˆr xx yy zz

x

y

z

.

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

.

z

r

ˆ ˆr zz

zz

ˆˆ ˆr zz Note:

dV d d dz

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 dzdS = d dz d d

dz

x

y

d

x

y

z

dz

ˆdr d ˆdr d ˆdr z dz

Path Integrals

First, consider differential changes along any of the three coordinate directions.

y

x

d

d

Cylindrical (cont.)

14

In general:

ˆˆ ˆdr d d z dz

Cylindrical (cont.)

2 2 2d dr d d dz

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

ˆ d

d

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.)

2 2 2

1

1

sin cos

sin sin

cos

cos /

tan /

x r

y r

z r

r x y z

z r

y x

Note: = r sin

17

y

z

.z

P (r, , )

r

x

Spherical (cont.)

Note: , and depend on (x, y, z). r

Unit Vectors

18

x

y

z

r

Note: A unit vector direction is defined by increasing one coordinate variable

while keeping the other two fixed.

Spherical (cont.)

ˆ ˆ ˆ ˆsin cos sin sin cos

ˆ ˆ ˆcos cos cos sin sin

ˆ ˆ ˆsin cos

ˆˆ ˆ sin cos cos cos sin

ˆˆ ˆ sin sin cos sin cos

ˆˆ cos sin

r x y z

x y z

x y

x r

y r

z r

Transformation of Unit Vectors

19

x

y

z

r

Spherical (cont.)

20

ˆˆ sin cos cos cos sin sin cos

ˆˆ sin sin cos sin cos sin sin

ˆ cos sin cos

r r r

r r

r r

ˆ ˆ ˆr xx yy zz

x

y

z

r

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

ˆr rr

After simplifying:

ˆ ˆˆr rr Note:

x

y

z

r

r

Spherical (cont.)Differentials

dS = r2 sin d d

2 sindV r dr d d

Note: dS may be in three different forms (only one is shown). The other two are:

dS = r dr ddS = r sin dr d

22

x

y

z d = r sin d

drr d

d

d

We increment (r, , ) starting from an initial point (blue dot).

Spherical (cont.)

ˆdr r dr

x

y

z drr

x

y

z

d

dr

ˆdr r d ˆ sindr r d

ˆ ˆˆ sindr r dr rd r d

Path Integrals

23

x

y

z

drd

r

sind r d

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

ˆˆ ˆdr d d z dz

Example: integrating along a radial path in cylindrical coordinates.

ˆdr d

B

AB

A

V E dr

x

y

A

B

dr

0d

C

x

y A

B

dr

0d

C

B

A

ABV E d

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

2ˆ [A/m ]J x x

x

y

z

ˆ ˆn r

J

Example (Cont.)

26

2 2

ˆ

ˆ

ˆ ˆ

sin cos

sin cos

sin cos sin cos

sin cos

S

S

x

S

x

S

S

S

S

I J n dS

J r dS

x J r dS

J dS

x dS

a dS

a dS

ˆ ˆ ˆ ˆsin cos sin sin cos

ˆ ˆ ˆcos cos cos sin sin

ˆ ˆ ˆsin cos

ˆˆ ˆ sin cos cos cos sin

ˆˆ ˆ sin sin cos sin cos

ˆˆ cos sin

r x y z

x y z

x y

x r

y r

z r

sin cos

sin sin

cos

x r

y r

z r

Example (Cont.)

27

2 2

2 /22 2 2

0 0

2 /23 2 2

0 0

/23 2

0

/23 3

0

3

sin cos

sin cos sin

sin cos sin

sin sin

sin

2

3

S

I a dS

a a d d

a d d

a d

a d

a

2

2

0

1cos 2

2d

Note :

/23

0

2sin

3d

Note :

32[A]

3I a

Appendix

28

Here we work out some more examples.

ExampleP1 (4, 60 , 1)

P2 (3, 180 , -1)

d = 6.403 [m]

2 2 2

1 2 1 2 1 2d x x y y z z

cos

sin

x

y

z z

1

1

1

4cos 60 2

4sin 60 3.4641

1

x

y

z

2

2

2

3cos 180 3

3sin 180 0

1

x

y

z

Find d = distance between points

Given: Cylindrical coordinates (, , z)with distances in meters

29

This formula only works in rectangular

coordinates!

Example

Given: v = -310-8 (cos2 / r4) [C/m3] , 2 < r < 5 [m]

Solution:

2

2

0 0

28 2

20 0

28 2

20 0

, sin

13 10 cos sin

13 10 cos sin

v

V

b

v

a

b

a

b

a

Q dV

r r drd d

drd dr

d dr dr

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.)

2 22

0 0

2

0

2

00

1 cos 2cos

2

1 sin 2

2 4

1 1

1 13/10

sin cos 2

bb

aa

d d

drr r

a b

d

Q = -5.65510-8 [C]

31

Note: The average value of cos2 is 1/2.

22

0

1cos 2

2d

Example

Derive ˆˆ cos sinz r

1 2 3

ˆˆz r A A A

1 ˆˆA z r

2 ˆA z

3ˆˆA z

Let

32

Then

Dot multiply both sides with ˆˆ, ,r

Example (cont.)

1

1

ˆˆ

ˆˆ cos

cos

A z r

z r

A

2

2

ˆ

ˆ cos2

sin

A z

z

A

3

3

ˆˆ

ˆ 0

0

A z

z

A

ˆˆ cos sinz r Result:

x

y

z

rz

z

x

y

z

x

y

z

z

33

1 2 3

ˆˆz r A A A

Example

Derive ˆ ˆ ˆ ˆsin cos sin sin cosr x y z

1

2

3

ˆ ˆ ˆ

ˆ ˆ ˆ

ˆ ˆˆ

A r x x r

A r y y r

A r z z r

component of

component of

component of

Let 1 2 3ˆ ˆ ˆ ˆr x A y A z A

An illustration of finding the x component of r

34

Dot multiply both sides with ˆ ˆ ˆ, ,x y z

x

y

z

r

L

Example (cont.)

ˆ ˆ cos sin cosr x L

ˆ ˆ ˆ ˆsin cos sin sin cosr x y z Result:

ˆ ˆ sin sin sinr y L

Hence

cos sin2

L

Similarly,

ˆ ˆ cosr z

Also,

35

x

y

z

x

L

r

( / 2) -

Example (Part 1)

Find VAB using path C shown below.

2ˆ ˆ ˆ3 2 1E x xy y xy z z

2

02

1

3

1

3 1 1

B B

AB x y z

A A

B

A

AB

V E dr E dx E dy E dz

xy dx xy dy

y x dy dx

V x x x x dx

x

y

1

11y x

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.)

02

1

02 2 3

1

12 2 3

0

13 2

0

3 1 1

3 3 2

3 3 2

2

1 1 1 3 4 12 52

4 3 2 12 12

ABV x x x x dx

x x x x x dx

x x x x x dx

x x x dx

Completing the calculus:

VAB = -5/12 [V]

37

Example (cont.)Alternative calculation (we parameterize differently):

38

0 1

2

1 0

3

B B

AB x y z

A A

V E dr E dx E dy E dz

xy dx xy dy

0 12

1 0

0 12 2 3

1 0

3 1 1

3 3

3 1 11

2 3 4

ABV x x dx y y dy

x x dx y y dy

1y x

VAB = -5/12 [V]

Example (Part 2)

Find VAB using path C shown below.

2ˆ ˆ ˆ3 2 1E x xy y xy z z

2

02 2

0

0 1

1 0

3

3 3

0 0

0

B B

x y z

A A

B

A

B

A

E dr E dx E dy E dz

xy dx xy dy

xy dx xy dy xy dx xy dy

dx dy

(same field as in Part 1)

VAB = 0 [V]39

x

y

z

C

(0,1,0)

(1,0,0)

E (x,y,z)

B

A

Example

VAB = -7/6 [V]

Find VAB using an arbitrary path C in the xy plane.

2ˆ ˆ ˆ3 2E x x y y z z

2

0 12

1 0

1 12

0 0

3

3

3

1 13

2 3

7 / 6

B B

x y z

A A

B

A

AB

E dr E dx E dy E dz

x dx y dy

x dx y dy

V x dx y dy

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 ˆ ˆ 2E x x y y

ˆˆ ˆ

ˆ 3

cos 3cos

sin 3sin

ˆˆˆ cos sin

ˆˆˆ sin cos

B

AB

A

V E dr

dr d d zdz

d

x

y

x

y

Ax

C

3 [m]

B

y

Find VAB using path C shown below.

41

Example (cont.)

VAB = 9/2 [V]

ˆ ˆˆ ˆcos sin 3cos sin cos 2 3sinE ˆ 3dr d

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

/2

/2

/2

9sin cos

9sin 2

2

cos 29

2 2

9 1 1 9

2 2 2 2

ABV d

d

9 18 sin cos 9sin cosE dr d d

ˆ ˆ 2E x x y y

Example (cont.)

ˆ ˆ 2E x x y y

Ax

C

3 [m]

B

y

Question: Is this integral path independent?

B

AB

A

V E dr

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.) ˆ ˆ 2E x x y y

Ax

C

3 [m]

B

y

Let’s find out:B

AB

A

V E dr

0 3

3 0

2

2

99

2

9 / 2

B B

x y z

A A

B

A

E dr E dx E dy E dz

x dx y dy

x dx y dy

Yes, it is path independent!

VAB = 9/2 [V]

44