1 ELEC 3105 Basic EM and Power Engineering Start Solutions to Poisson’s and/or Laplace’s

1

ELEC 3105 Basic EM and Power Engineering

Start Solutions to Poisson’s and/or

Laplace’s

Set of derivative (differential) equations

VE

Valid for each point is space

3

Recall From Lecture 3

38

Poisson’s and Laplace’s Equation

Solutions to these equations forms part of this weeks lectures.

o

V

2

Laplace’s Equation02 V

Poisson’s Equation

VE

4

Poisson’s / Laplace’s Equations

x

y

z

Consider the following system

• Parallel plates of infinite extent• Bottom plate V(@ z = 0) = 0• Top plate V(@ z = z1) = V1

• Region between plates has no charge

Obtain potential and electric field for region between plates

That is: potential and electric field for a parallel plate capacitor

5


x

y

z

Use Laplace’s equation since region of interest has no charge present

02 V

In (x, y, z) 02

2

2

2

2

22

z

V

y

V

x

VV

No change in V value in (x, y) plane then 02

2

2

2

y

V

x

V

6


x

y

z

02

22

z

VV

C1 and C2 are constants to be determined from Boundary conditions

0

zzV

1Cz

V

21 CzCV

7


x

y

zBoundary conditions given• Bottom plate V(@ z = 0) = 0• Top plate V(@ z = z1) = V1

21 CzCV

@ z = 0, V = 0 gives C2 = 0

@ z = z1, V = V1 gives C1 = V1/z1

zz

VV

1

1Expression for potential between plates

8


x

y

zNow to obtain expression for the electric field

VE

E l e c t r i c p o t e n t i a l a n d t h e g r a d i e n t o p e r a t o r

VE

E

Recall from Lecture 3

9


x

y

zNow to obtain expression for the electric field

VE

zz

Vy

y

Vx

x

VE ˆˆˆ

zz

VV

1

1

No x or y dependence

zz

Vz

z

zzV

E ˆˆ1

11

1

zz

VE ˆ

1

1

10


x

y

zSolution to problem

zz

VE ˆ

1

1

zz

VV

1

1

Notice that the electric field lines are directed along the z axis and are normal to the surfaces of the plates. The electric field lines start from the upper plate and are directed towards the lower plate when V1 > 0. Lines of constant V are in the (x, y) plane and perpendicular to the electric field lines

11


14

EQUIPOTENTIAL SURFACES AND EQUIPOTENTIAL LINES

Equipotential, E field

Capacitor // platesSelect V1 = 12 VZ1 = 1 m

zm

VE ˆ12

zm

VV 12

12


13


Example: Obtain an expression for the potential and electric field in the region between the two concentric right circular cylinders. The inner cylinder has a radius a = 1 mm and is at a potential of V = 0 volts, the outer cylinder has a radius b = 20 mm and is at a potential of 150 volts. Neglect any edge effects if present.

14


34



Solution:

• We will select cylindrical coordinates for solving this problem.

• By symmetry the potential will be a function of the radial coordinate only. There is no or z dependence.

• There is not charge density between the conductors. = 0

01

r

Vr

rr

02 V

15


Solution:

• The first integration gives

• Second integration gives

01

r

Vr

rr

1Cr

Vr

21 ln CrCV

16


Solution: 21 ln CrCV

• Apply boundary condition V = 0 at r = a = 1 mm

21 001.0ln0 CC

21 02.0ln150 CC

• Apply boundary condition V = 150 at r = b = 20 mm

Two equations with two unknowns:9.345

1.50

2

1

C

C

17


Solution:

• Introduce values into expression for potential

9.345ln1.50 rV

Units are volts

1.501 C 9.3452 C

21 ln CrCV

Valid only between cylinders

18


Solution for electric field: 9.345ln1.50 rV

Units are volts / m

VE

rr

VE ˆ

rr

E ˆ1.50

Valid only between cylinders

19


9.345ln1.50 rV

rr

E ˆ1.50

34



Potential function

Electric field

ELEC 3105 Basic EM and Power Engineering

Numerical solution to Poisson’s and Laplace’s

NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION

ONE RECTANGULAR CONDUCTOR PLACED INSIDE ANOTHER RECTANGULAR CONDUCTOR

02 V

Outer conductorV = 0 volts

x

y

z

Inner conductorV = Vin

Conductors extend to infinity along z axis

Could be microwave waveguide


x

y

02 V

Find the electric field lines and equipotentials for the square cylindrical capacitor shown.

V = 0

V = VinBoundary conditions


x

y

02 V

2-D problem since there are no variations in electric field vector or potential in the z direction. This is obtained by symmetry .

V = 0

V = VinBoundary conditions

By symmetry, we need only solve for x > 0 and y > 0 quadrant.

),( yxVV


x

y

02 V NOTE: In fact by symmetry

only need to solve for purple region. Blue region is the mirror image.


x

y

02 V

Technique for numerical solution

• Establish a dense mesh or grid between the conducting plates.

• Represent V(x, y) as a set of discrete values Vij defined at each grid point (i, j).

(i, j)

(j)

(i)


y

x02 V

(i, j)

(i-1, j)

(i+1, j)

(i, j-1)

(i, j+1)

hx

y

h

VV

x

V jiji ,1,

h

VV

x

V jiji ,,1


y

x02 V

(i, j)

(i-1, j)

(i+1, j)

(i, j-1)

(i, j+1)

hx

y

h

VV

x

V jiji ,1,

h

VV

x

V jiji ,,1

2

,,1,1

2

2 2

h

VVV

h

xV

xV

x

V jijiji


y

x02 V

(i, j)

(i-1, j)

(i+1, j)

(i, j-1)

(i, j+1)

hx

y

2

,,1,1

2

2 2

h

VVV

h

x

V

x

V

x

V jijiji

2

,1,1,

2

2 2

h

VVV

h

y

V

y

V

y

V jijiji

Generalize in y and x

2

,1,1,

2

,,1,1

2

2

2

22 22

h

VVV

h

VVV

y

V

x

VV jijijijijiji


y

x02 V

(i, j)

(i-1, j)

(i+1, j)

(i, j-1)

(i, j+1)

hx

y

Generalize in y and x

2

,1,1,

2

,,1,1

2

2

2

22 22

h

VVV

h

VVV

y

V

x

VV jijijijijiji

o

jijijijijiji

h

VVVVV

y

V

x

VV

,

2

,1,1,,1,1

2

2

2

22 4

Charge density present near grid point (i, j)


(i, j)

(i-1, j)

(i+1, j)

(i, j-1)

(i, j+1)

hx

y

o

jijijijijiji

h

VVVVV

y

V

x

VV

,

2

,1,1,,1,1

2

2

2

22 4

Finite difference representation of Poisson’s equation

Commercial software available for solving numerical problems

y

x02 V


(i, j)

(i-1, j)

(i+1, j)

(i, j-1)

(i, j+1)

hx

y

04 ,1,1,,1,1 jijijijiji VVVVV Now consider the case where i,j = 0

Thus the potential V at grid point (i, j) is the average of the values of the potential at the surrounding grid points.

y

x02 V

41,1,,1,1

,

jijijijiji

VVVVV

This suggest a simple algorithm for finding Vi,j.


• Guess an initial value of V at each grid point

• Traverse the mesh generating a new estimate for V at each grid point (i, j) by averaging values at surrounding points.

• Repeat until V does not change significantly.

N U M E R I C A L S O L U T I O N T O P O I S S O N ’ S / L A P L A C E ’ S E Q U A T I O NN U M E R I C A L S O L U T I O N T O P O I S S O N ’ S / L A P L A C E ’ S E Q U A T I O N

x

y

02 V

T e c h n i q u e f o r n u m e r i c a l s o l u t i o n

• E s t a b l i s h a d e n s e m e s h o r g r i d b e t w e e n t h e c o n d u c t i n g p l a t e s .

•• R e p r e s e n t R e p r e s e n t V ( x , y )V ( x , y ) a s a s e t o f a s a s e t o f d i s c r e t e v a l u e s d i s c r e t e v a l u e s VV i ji j d e f i n e d a t d e f i n e d a t e a c h g r i d p o i n t ( i , j ) .e a c h g r i d p o i n t ( i , j ) .

( i , j )

( j )

( i )

Now for a real example of the technique

Numerical solution parallel plate capacitor

x

y

z

z = 0 plane

z = d planeV = 150 volts

V = 0 volts

Plates of the capacitor are conductors extending to infinity in the (x, y) plane.

As a result of symmetry, the potential function will vary only in the z direction. V = V(z)

02 V

Since no charge density between plates


Divide region between plates into a fine mesh.

Select values for V1 to V9

zV10 = 150 volts

V0 = 0 volts

(i)10 V10

9 V9

8 V8

7 V7

6 V6

5 V5

4 V4

3 V3

2 V2

1 V1

0 V0

0 01 332 123 344 1455 566 167 788 89 9

10 150

Vii

02

22

z

VV

211

iii

VVV

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 180 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 01 33 6 16.75 21.125 14 20.5 12.95313 17.98438 12.06641 15.83984 11.47266 14.36182 11.17554 13.44653 11.10394 12.93484 11.1825 12.692782 12 33.5 42.25 28 41 25.90625 35.96875 24.13281 31.67969 22.94531 28.72363 22.35107 26.89307 22.20789 25.86969 22.36499 25.38556 22.704633 34 78.5 39.25 60.875 37.8125 51.4375 35.3125 45.375 33.82422 41.60742 33.22949 39.42432 33.24023 38.29285 33.62604 37.83627 34.22676 37.799554 145 45 79.5 47.625 61.875 44.71875 54.78125 43.51563 51.53516 43.51367 50.125 44.12939 49.69263 45.04419 49.80286 46.08853 50.21353 47.1695 56 80.5 56 62.875 51.625 58.125 51.71875 57.69531 53.20313 58.64258 55.0293 59.96094 56.84814 61.31287 58.55103 62.59079 60.11124 63.765726 16 67 46.25 55.625 54.375 58.71875 60.60938 62.89063 65.75 66.54492 69.79688 69.56689 72.93311 72.05786 75.37872 74.13394 77.3179 75.88977 78 12 55.25 45.875 65.8125 63.09375 74.0625 73.80469 79.88672 80.95117 84.10449 85.90527 87.26758 89.44458 89.71686 92.04501 91.66817 94.009128 8 43.5 45.5 76 71.8125 89.40625 87 96.88281 96.15234 101.6641 102.0137 104.9683 105.9561 107.3759 108.7113 109.2024 110.7003 110.63479 9 79 96.75 97.75 113 110.9063 119.7031 118.5 123.4414 123.0762 125.832 126.0068 127.4841 127.978 128.6879 129.3557 129.6012 130.3502

10 150 150 150 150 150 150 150 150 150 150 150 150 150 150 150 150 150 150

zV10 = 150 volts

V0 = 0 volts

(i)10 V10

9 V9

8 V8

7 V7

6 V6

5 V5

4 V4

3 V3

2 V2

1 V1

0 V0

After 18 iterations

i


0

3

6

9

1

2

3

4

56

78

9 10 1

1 12 1

3 14 1

5 16 1

7 18 1

9 20 2

1 22 2

3

0

20

40

60

80

100

120

140

160

IterationGridnumber

Potential


0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

Parallel plates

Potential variation between plates

z

Almost a straight line even after only a few iterations


V

z

Numerical solution parallel plate capacitorConsider a finer mesh


zV10 = 150 volts

V0 = 0 volts

(i)29 V29

28 V28

……………3 V3

2 V2

1 V1

0 V0

Vii

02

22

z

VV

211

iii

VVV

10 01 332 123 344 1455 566 167 788 89 9

10 2211 912 4513 5414 915 67816 917 8918 919 12320 921 3422 3423 2324 925 926 1227 12328 6529 150

1 2 3 4 5 6 7 8 9 10 11 120 0 0 0 0 0 0 0 0 0 0 0 01 33 6 16.75 21.125 14 20.5 12.95313 17.98438 12.06641 15.58984 11.19727 13.616212 12 33.5 42.25 28 41 25.90625 35.96875 24.13281 31.17969 22.39453 27.23242 20.754393 34 78.5 39.25 60.875 37.8125 51.4375 35.3125 44.375 32.72266 38.875 30.31152 34.525884 145 45 79.5 47.625 61.875 44.71875 52.78125 41.3125 46.57031 38.22852 41.81934 35.865725 56 80.5 56 62.875 51.625 54.125 47.3125 48.76563 43.73438 44.76367 41.41992 41.46686 16 67 46.25 55.625 46.375 49.90625 44.75 46.15625 42.95703 44.61133 41.11426 44.580087 78 12 55.25 29.875 48.1875 35.375 45 37.14844 45.48828 37.46484 47.74023 37.074718 8 43.5 13.5 40.75 24.375 40.09375 29.54688 44.82031 31.97266 50.86914 33.03516 56.627939 9 15 26.25 18.875 32 23.71875 44.64063 26.79688 56.25 28.60547 65.51563 29.61475

10 22 9 24.25 23.25 23.0625 49.1875 24.04688 67.67969 25.23828 80.16211 26.19434 88.6074211 9 33.5 20.25 27.25 66.375 24.375 90.71875 23.67969 104.0742 23.7832 111.6992 24.1328112 45 31.5 30.25 109.5 25.6875 132.25 23.3125 140.4688 22.32813 143.2363 22.07129 143.564513 54 27 198.75 24.125 198.125 22.25 190.2188 20.97656 182.3984 20.35938 175.4297 20.1894514 9 366 18 286.75 18.8125 248.1875 18.64063 224.3281 18.39063 207.623 18.30762 195.007815 678 9 374.75 13.5 298.25 15.03125 258.4375 15.80469 232.8477 16.25586 214.5859 16.5981416 9 383.5 9 309.75 11.25 268.6875 12.96875 241.3672 14.12109 221.5488 14.88867 206.423317 89 9 244.75 9 239.125 10.90625 224.2969 12.4375 210.25 13.52148 198.2607 14.3051818 9 106 9 168.5 10.5625 179.9063 11.90625 179.1328 12.92188 174.9727 13.72168 170.058119 123 9 92.25 12.125 120.6875 12.90625 133.9688 13.40625 139.6953 13.92188 141.8555 14.5561520 9 78.5 15.25 72.875 15.25 88.03125 14.90625 100.2578 14.92188 108.7383 15.39063 114.511221 34 21.5 53.5 18.375 55.375 16.90625 66.54688 16.4375 77.78125 16.85938 87.16699 17.9350622 34 28.5 21.5 37.875 18.5625 45.0625 17.96875 55.30469 18.79688 65.5957 20.47949 74.8315423 23 21.5 22.25 18.75 34.75 19.03125 44.0625 21.15625 53.41016 24.09961 62.49609 27.3579124 9 16 16 31.625 19.5 43.0625 24.34375 51.51563 29.40234 59.39648 34.23633 67.02125 9 10.5 41 20.25 51.375 29.65625 58.96875 37.64844 65.38281 44.37305 71.5459 50.0693426 12 66 24.5 71.125 39.8125 74.875 50.95313 79.25 59.34375 83.69531 65.90234 88.1977527 123 38.5 101.25 59.375 98.375 72.25 99.53125 81.03906 102.0078 87.43164 104.8496 92.3090828 65 136.5 94.25 125.625 104.6875 124.1875 111.125 124.7656 115.5195 126.0039 118.7158 127.424829 150 150 150 150 150 150 150 150 150 150 150 150

Numerical solution parallel plate capacitorGrid after 12 iterations

0

3

6

9

12

15

18

21

24 2

7

1

3

5

7

9

11

13

15

17

19

21

23

0

100

200

300

400

500

600

700

Iteration

Gridnumber

Potential

Chart after 23 iterations

1 4 7

10

13

16

19

22

25

28

31

34

37

40

43

46

49

0

5

10

15

20

250

100

200

300

400

500

600

700

Iteration

Gridnumber

Potential


1

9

17

25

33

41

49

57

65

73

81

89

97

10

5

11

3

12

1 0 4 8 12 16 20 24 28

0

100

200

300

400

500

600

700

IterationGridnumber

Potential


1

13

25

37

49

61

73

85

97

10

9

12

1

13

3

14

5

15

7

16

9

18

1

19

3

20

5

21

7

22

9

24

1

0 11 2

2

0

20

40

60

80

100

120

140

160

180

200

IterationGridnumber

Potential


Parallel platesPotential variation between plates

z

Still quite rough, requires more iterations or better guess at initial potential values for grid


V

z0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

Numerical solution parallel plate capacitorChange only one number


zV10 = 150 volts

V0 = 0 volts

(i)29 V29

28 V28

……………3 V3

2 V2

1 V1

0 V0

Vii1

0 01 332 123 344 1455 566 167 788 89 9

10 2211 912 4513 5414 915 9916 917 8918 919 12320 921 3422 3423 2324 925 926 1227 12328 6529 150

Was678

0

10

20

30

40

50

60

70

80

90

100

110

120

130

140

150

Estimation of the accuracy of technique

Consider a Taylor’ series expansion for i grid point direction:

...!4!32

4

4

43

3

32

2

2

,,1

h

x

Vh

x

Vh

x

Vh

x

VVV jiji

...!4!32

4

4

43

3

32

2

2

,,1

h

x

Vh

x

Vh

x

Vh

x

VVV jiji

Combine the two series:

...!4

22

224

4

42

2

2

,,1,1

h

x

Vh

x

VVVV jijiji


Consider a Taylor’ series expansion for j grid point direction:

...!4!32

4

4

43

3

32

2

2

,1,

h

y

Vh

y

Vh

y

Vh

y

VVV jiji

...!4!32

4

4

43

3

32

2

2

,1,

h

y

Vh

y

Vh

y

Vh

y

VVV jiji

Combine the two series:

...!4

22

224

4

42

2

2

,1,1,

h

y

Vh

y

VVVV jijiji


Combining i and j grid direction results

Gives :

...!4

22

224

4

42

2

2

,1,1,

h

y

Vh

y

VVVV jijiji

...!4

22

224

4

42

2

2

,,1,1

h

x

Vh

x

VVVV jijiji

+

...!4

244

4

4

4

42

2

2

2

2

,1,1,,1,1

h

y

V

x

Vh

y

V

x

VVVVVV jijijijiji

0 since V satisfies Laplace’s equation


Dominant correction term

...order to termcorrection4

41,1,,1,1,

h

VVVVV jijijiji

ji

This correction term becomes very small as the grid point spacing h becomes small.

Problem not to try yet

Cylindrical capacitor

Inner radius a = 10 mmInner potential Vin = 20 volts

Outer radius b = 70 mmOuter potential Vout = 200 volts

Solve for V, as a function of the coordinates, for the region between the cylindrical conductors.

51

Spherical space meshing

52

Triangular space meshing

53

Meshing

54

ELEC 3105 Basic EM and power engineering

End Solutions

to Poisson’s

/ Laplace’s

Documents

1 ELEC 3105 Basic EM and Power Engineering Start Solutions to Poisson’s and/or Laplace’s