ELEC 3105 Basic EM and Power Engineering
Start Solutions to Poisson’s
and/or Laplace’s
1
Set of derivative (differential) equations


E  V
Valid for each point is space
Recall From Lecture 3
Poisson’s and Laplace’s
Equation
2

 V 
Poisson’s Equation
2
 V 0
Laplace’s Equation
o
Solutions to these equations forms part of this weeks
lectures.


E  V
38
3
Poisson’s / Laplace’s Equations
z
Consider the following system
y
x
• 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
4
Poisson’s / Laplace’s Equations
z
y
Use Laplace’s equation
since region of interest has
no charge present
2
 V 0
x
In (x, y, z)
2
2
2
2
V V V
 V  2  2  2 0
x
y
z
 2V  2V
 2 0
No change in V value in (x, y) plane then
2
x
y
5
Poisson’s / Laplace’s Equations
z
y
x
 V 



z

 0
z
V
 C1
z
2
 2V
 V  2 0
z
V  C1 z  C2
C1 and C2 are constants to be determined from Boundary conditions
6
Poisson’s / Laplace’s Equations
z
Boundary conditions given
•Bottom plate V(@ z = 0) = 0
•Top plate V(@ z = z1) = V1
y
V  C1 z  C2
x
@ z = 0, V = 0 gives C2 = 0
@ z = z1, V = V1 gives C1 = V1/z1
V1
Expression for potential between plates V 
z
z1
7
Poisson’s / Laplace’s Equations
z
Now to obtain expression
for the electric field
y


E  V
x
Recall from Lecture 3
8
Poisson’s / Laplace’s Equations
z
Now to obtain expression
for the electric field
y
x

 V
V
V
E  
xˆ 
yˆ 
y
z
 x
 V1

 z 


V
z
E   1 zˆ    1 zˆ
 z 
z1






E  V

zˆ 

V
V1
z
z1
No x or y dependence

V1
E   zˆ
z1
9
Poisson’s / Laplace’s Equations
z
Solution to problem
y
x
V1
V z
z1

V1
E   zˆ
z1
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
10
Poisson’s / Laplace’s Equations
Select V1 = 12 V
Z1 = 1 m

V
E  12 zˆ
m
V  12
V
z
m
11
Poisson’s / Laplace’s Equations
12
Poisson’s / Laplace’s Equations
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.
13
Poisson’s / Laplace’s Equations
Solution:
Poisson’s / Laplace’s Equations
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.
2
 V 0
•We will select cylindrical
coordinates for solving this
problem.
34
1   V 
r
0
r r  r 
•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
14
Poisson’s / Laplace’s Equations
1   V
r
r r  r

0

Solution:
•The first integration gives
V
r
 C1
r
•Second integration gives
V  C1 ln r   C2
15
Poisson’s / Laplace’s Equations
V  C1 ln r   C2
Solution:
•Apply boundary condition V = 0 at r = a = 1 mm
0  C1 ln 0.001  C2
•Apply boundary condition V = 150 at r = b = 20 mm
150  C1 ln 0.02  C2
Two equations with two unknowns:
C1  50.1
C2  345.9
16
Poisson’s / Laplace’s Equations
C1  50.1
C2  345.9
V  C1 ln r   C2
Solution:
• Introduce values into expression for
potential
Units are volts
V  50.1ln r   345.9
Valid only between cylinders
17
Poisson’s / Laplace’s Equations
V  50.1ln r   345.9
Solution for electric field:


E  V

V
E
rˆ
r
Units are volts / m

50.1
E
rˆ
Valid only between cylinders
r
18
Poisson’s / Laplace’s Equations
Poisson’s / Laplace’s Equations

50.1
E
rˆ
r
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.
Electric field
34
V  50.1ln r   345.9
Potential function
19
ELEC 3105 Basic EM and Power Engineering
Numerical solution to
Poisson’s and Laplace’s
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
Could be microwave waveguide
y
Outer conductor
V = 0 volts
x
2
 V 0
z
Inner conductor
V = Vin
ONE RECTANGULAR CONDUCTOR PLACED INSIDE
ANOTHER RECTANGULAR CONDUCTOR
Conductors extend to infinity along z axis
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
Find the electric field lines and
equipotentials for the square
cylindrical capacitor shown.
x
2
 V 0
V = Vin
V=0
Boundary conditions
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
By symmetry, we need only
solve for x > 0 and y > 0
quadrant.
x
2
 V 0
V = Vin
Boundary conditions
V=0
2-D problem since there are no
variations in electric field vector or
potential in the z direction. This is
obtained by symmetry .
V  V ( x, y )
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
x
2
 V 0
NOTE: In fact by symmetry
only need to solve for purple
region. Blue region is the
mirror image.
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
Technique for numerical
solution
•Establish a dense mesh or grid
x between the conducting plates.
2
 V 0
•Represent V(x, y) as a set of
discrete values Vij defined at
each grid point (i, j).
(j)
(i)
(i, j)
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
y
x

 2V  0
(i, j)
(i, j+1)
h
(i+1, j)
(i-1, j)
V
x
V
x




Vi , j  Vi 1, j
h
Vi 1, j  Vi , j
h
(i, j-1)
x
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
y
V
x
V
x




(i, j+1)
x

 2V  0
(i, j)
h
Vi , j  Vi 1, j
(i+1, j)
x
(i-1, j)
h
(i, j-1)
Vi 1, j  Vi , j
h
 2V

2
x
V
x
V

x

h


Vi 1, j  Vi 1, j  2Vi , j
h2
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
Generalize in y and x
y
x

 2V  0
 2V

x 2
 2V

y 2
V
x
V
y
V

x

h


h
V
y
(i, j+1)
h




(i, j)
(i+1, j)
Vi 1, j  Vi 1, j  2Vi , j
h2
Vi , j 1  Vi , j 1  2Vi , j
h
x
(i-1, j)
(i, j-1)
2
2
 2V  2V Vi 1, j  Vi 1, j  2Vi , j Vi , j 1  Vi , j 1  2Vi , j
V 2  2 

2
x
y
h
h2
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
Generalize in y and x
y

 2V  0
x
(i, j)
(i, j+1)
h
(i+1, j)
(i-1, j)
2
 2V  2V Vi 1, j  Vi 1, j  2Vi , j Vi , j 1  Vi , j 1  2Vi , j
V 2  2 

x
y
h2
h2
(i, j-1)
2
i , j
 2V  2V Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1  4Vi , j
V 2  2 

2
x
y
h
o
Charge density present near grid point (i, j)
x
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
y
(i, j)
(i, j+1)
h
(i+1, j)
(i-1, j)
(i, j-1)
x
x

 2V  0
Finite difference representation of
Poisson’s equation
2
i , j
 2V  2V Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1  4Vi , j
V 2  2 

2
x
y
h
o
Commercial software available for solving numerical problems
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
y
y
(i, j)
(i, j+1)
h
(i+1, j)
x
(i-1, j)
x

 2V  0
(i, j-1)
Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1  4Vi , j  0
Vi , j 
Now consider the case where i,j = 0
Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1
4
Thus the potential V at grid point (i, j) is the average of the
values of the potential at the surrounding grid points.
This suggest a simple algorithm for finding Vi,j.
NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
•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.
Now for a real example of the technique
Numerical solution parallel plate capacitor
z
V = 150 volts
z = d plane
z = 0 plane
x
y
Plates of the capacitor are
conductors extending to
infinity in the (x, y) plane.
V = 0 volts
As a result of symmetry, the potential function will vary only
in the z direction. V = V(z)
2
 V 0
Since no charge density between plates
Numerical solution parallel plate capacitor
V10 = 150 volts
z
V0 = 0 volts
2
 2V
 V  2 0
z
(i)
10
9
8
7
6
5
4
3
2
1
0
Divide region between
V10
V9
plates into a fine mesh.
V8
V7
V6
V5
V4 Select values for V1 to V9
V3
Vi
i
V2
0
0
V1
1
33
2
12
V0
Vi 1  Vi 1
Vi 
2
3
4
5
6
7
8
9
10
34
145
56
16
78
8
9
150
V10 = 150 volts
V0 = 0 volts
i
0
1
2
3
4
5
6
7
8
9
10
1
0
33
12
34
145
56
16
78
8
9
150
2
0
6
33.5
78.5
45
80.5
67
12
43.5
79
150
3
0
16.75
42.25
39.25
79.5
56
46.25
55.25
45.5
96.75
150
z
(i)
10
9
8
7
6
5
4
3
2
1
0
V10
V9
V8
V7
V6
V5
V4
V3
V2
V1
V0
4
5
6
7
8
9
10
11
0
0
0
0
0
0
0
0
21.125
14
20.5 12.95313 17.98438 12.06641 15.83984 11.47266
28
41 25.90625 35.96875 24.13281 31.67969 22.94531 28.72363
60.875 37.8125 51.4375 35.3125 45.375 33.82422 41.60742 33.22949
47.625 61.875 44.71875 54.78125 43.51563 51.53516 43.51367 50.125
62.875 51.625 58.125 51.71875 57.69531 53.20313 58.64258 55.0293
55.625 54.375 58.71875 60.60938 62.89063
65.75 66.54492 69.79688
45.875 65.8125 63.09375 74.0625 73.80469 79.88672 80.95117 84.10449
76 71.8125 89.40625
87 96.88281 96.15234 101.6641 102.0137
97.75
113 110.9063 119.7031
118.5 123.4414 123.0762 125.832
150
150
150
150
150
150
150
150
Numerical solution parallel
plate capacitor
After 18 iterations
12
0
14.36182
22.35107
39.42432
44.12939
59.96094
69.56689
85.90527
104.9683
126.0068
150
13
0
11.17554
26.89307
33.24023
49.69263
56.84814
72.93311
87.26758
105.9561
127.4841
150
14
0
13.44653
22.20789
38.29285
45.04419
61.31287
72.05786
89.44458
107.3759
127.978
150
15
0
11.10394
25.86969
33.62604
49.80286
58.55103
75.37872
89.71686
108.7113
128.6879
150
16
0
12.93484
22.36499
37.83627
46.08853
62.59079
74.13394
92.04501
109.2024
129.3557
150
17
0
11.1825
25.38556
34.22676
50.21353
60.11124
77.3179
91.66817
110.7003
129.6012
150
18
0
12.69278
22.70463
37.79955
47.169
63.76572
75.8897
94.00912
110.6347
130.3502
150
Numerical solution parallel plate capacitor
160
140
120
100
80
60
40
20
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
0
9
Grid
number
7
8
6
4
5
6
3
1
2
3
0
Iteration
Potential
Parallel plates
z
Numerical solution parallel plate
capacitor
Potential variation between plates
150
140
130
120
110
100
90
80
V
70
60
50
40
30
20
10
0
z
Almost a straight line even after only a few iterations
Numerical solution parallel plate capacitor
V10 = 150 volts
V0 = 0 volts
2
 2V
 V  2 0
z
z
(i)
29
28
…
…
…
…
…
3
2
1
0
V29
V28
Consider a finer mesh
i Vi
V3
V2
V1
V0
Select values for V1 to V28
Vi 1  Vi 1
Vi 
2
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
1
0
33
12
34
145
56
16
78
8
9
22
9
45
54
9
678
9
89
9
123
9
34
34
23
9
9
12
123
65
150
Numerical solution parallel plate capacitor
Grid after 12 iterations
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
1
0
33
12
34
145
56
16
78
8
9
22
9
45
54
9
678
9
89
9
123
9
34
34
23
9
9
12
123
65
150
2
0
6
33.5
78.5
45
80.5
67
12
43.5
15
9
33.5
31.5
27
366
9
383.5
9
106
9
78.5
21.5
28.5
21.5
16
10.5
66
38.5
136.5
150
3
0
16.75
42.25
39.25
79.5
56
46.25
55.25
13.5
26.25
24.25
20.25
30.25
198.75
18
374.75
9
244.75
9
92.25
15.25
53.5
21.5
22.25
16
41
24.5
101.25
94.25
150
4
5
6
7
8
9
10
11
12
0
0
0
0
0
0
0
0
0
21.125
14
20.5 12.95313 17.98438 12.06641 15.58984 11.19727 13.61621
28
41 25.90625 35.96875 24.13281 31.17969 22.39453 27.23242 20.75439
60.875 37.8125 51.4375 35.3125
44.375 32.72266
38.875 30.31152 34.52588
47.625
61.875 44.71875 52.78125 41.3125 46.57031 38.22852 41.81934 35.86572
62.875
51.625
54.125 47.3125 48.76563 43.73438 44.76367 41.41992 41.4668
55.625
46.375 49.90625
44.75 46.15625 42.95703 44.61133 41.11426 44.58008
29.875 48.1875
35.375
45 37.14844 45.48828 37.46484 47.74023 37.07471
40.75
24.375 40.09375 29.54688 44.82031 31.97266 50.86914 33.03516 56.62793
18.875
32 23.71875 44.64063 26.79688
56.25 28.60547 65.51563 29.61475
23.25 23.0625 49.1875 24.04688 67.67969 25.23828 80.16211 26.19434 88.60742
27.25
66.375
24.375 90.71875 23.67969 104.0742 23.7832 111.6992 24.13281
109.5 25.6875
132.25 23.3125 140.4688 22.32813 143.2363 22.07129 143.5645
24.125 198.125
22.25 190.2188 20.97656 182.3984 20.35938 175.4297 20.18945
286.75 18.8125 248.1875 18.64063 224.3281 18.39063 207.623 18.30762 195.0078
13.5
298.25 15.03125 258.4375 15.80469 232.8477 16.25586 214.5859 16.59814
309.75
11.25 268.6875 12.96875 241.3672 14.12109 221.5488 14.88867 206.4233
9 239.125 10.90625 224.2969 12.4375
210.25 13.52148 198.2607 14.30518
168.5 10.5625 179.9063 11.90625 179.1328 12.92188 174.9727 13.72168 170.0581
12.125 120.6875 12.90625 133.9688 13.40625 139.6953 13.92188 141.8555 14.55615
72.875
15.25 88.03125 14.90625 100.2578 14.92188 108.7383 15.39063 114.5112
18.375
55.375 16.90625 66.54688 16.4375 77.78125 16.85938 87.16699 17.93506
37.875 18.5625 45.0625 17.96875 55.30469 18.79688 65.5957 20.47949 74.83154
18.75
34.75 19.03125 44.0625 21.15625 53.41016 24.09961 62.49609 27.35791
31.625
19.5 43.0625 24.34375 51.51563 29.40234 59.39648 34.23633
67.021
20.25
51.375 29.65625 58.96875 37.64844 65.38281 44.37305 71.5459 50.06934
71.125 39.8125
74.875 50.95313
79.25 59.34375 83.69531 65.90234 88.19775
59.375
98.375
72.25 99.53125 81.03906 102.0078 87.43164 104.8496 92.30908
125.625 104.6875 124.1875 111.125 124.7656 115.5195 126.0039 118.7158 127.4248
150
150
150
150
150
150
150
150
150
700
600
500
Potential
400
300
200
Chart after 23 iterations
9
6
3
23
0
21
15
17
19
Iteration
13
11
9
12
7
27
24
15
5
3
18
1
0
21
100
Grid
number
700
600
500
Potential
400
300
200
100
Chart after 50 iterations
0
49
43
5
40
37
34
31
28
10
46
Iteration
25
22
15
19
16
13
20
10
7
4
25
1
0
Grid
number
700
600
500
400
Potential
300
Chart after 125 iterations
24
28
20
16
8
12
4
0
121
113
97
0
105
Iteration
89
81
73
100
65
57
49
41
33
17
25
9
1
200
Grid
number
200
180
160
140
120
Potential
Chart after 250 iterations
22
11
0
241
0
229
217
205
193
20
181
169
157
40
145
121
Iteration
133
60
109
97
73
80
85
13
25
37
49
61
1
100
Grid
number
Numerical solution parallel plate
capacitor
z
Parallel plates
Potential variation between plates
150
140
130
120
110
100
90
V
80
70
60
50
40
30
20
10
0
z
Still quite rough, requires more iterations or better guess at
initial potential values for grid
Numerical solution parallel plate capacitor
V10 = 150 volts
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
V0 = 0 volts
z
(i)
29
28
…
…
…
…
…
3
2
1
0
V29
V28
Change only one
number
i Vi
V3
V2
V1
V0
Select values for
V1 to V28
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
1
0
33
12
34
145
56
16
78
8
9
22
9
45
54
9
99
9
89
9
123
9
34
34
23
9
9
12
123
65
150
Was
678
Estimation of the accuracy of technique
Consider a Taylor’ series expansion for i grid point direction:
Vi 1, j
V
 2V h 2  3V h3  4V h 4
 Vi , j 
h 2
 3
 4
 ...
x
x 2 x 3! x 4!
Vi 1, j
V
 2V h 2  3V h3  4V h 4
 Vi , j 
h 2
 3
 4
 ...
x
x 2 x 3! x 4!
Combine the two series:
Vi 1, j  Vi 1, j
 2V h 2
 4V h 4
 2Vi , j  2 2
2 4
 ...
x 2
x 4!
Estimation of the accuracy of technique
Consider a Taylor’ series expansion for j grid point direction:
V
 2V h 2  3V h3  4V h 4
Vi , j 1  Vi , j 
h 2
 3
 4
 ...
y
y 2 y 3! y 4!
V
 2V h 2  3V h3  4V h 4
Vi , j 1  Vi , j 
h 2
 3
 4
 ...
y
y 2 y 3! y 4!
Combine the two series:
 2V h 2
 4V h 4
Vi , j 1  Vi , j 1  2Vi , j  2 2
2 4
 ...
y 2
y 4!
Estimation of the accuracy of technique
Combining i and j grid direction results
 2V h 2
 4V h 4
Vi , j 1  Vi , j 1  2Vi , j  2 2
2 4
 ...
y 2
y 4!
+
Vi 1, j  Vi 1, j
 2V h 2
 4V h 4
 2Vi , j  2 2
2 4
 ...
x 2
x 4!
Gives :
  2V  2V  2   4V  4V  h 4
Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1  4Vi , j   2  2 h  2 4  4   ...
y 
y  4!
 x
 x
0 since V satisfies Laplace’s equation
Estimation of the accuracy of technique
Dominant correction term
Vi , j 
Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1
4


 correction term to order h 4  ...
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 mm
Inner potential Vin = 20 volts
Outer radius b = 70 mm
Outer potential Vout = 200 volts
Solve for V, as a function of the coordinates, for the region
between the cylindrical conductors.
Spherical space meshing
51
Triangular space meshing
52
Meshing
53
ELEC 3105 Basic EM and power engineering
End
Solutions
to
Poisson’s
/ Laplace’s
54