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

3. Set of derivative (differential) equations
Valid for each point is space

4. Recall From Lecture 3

5. Poisson’s / Laplace’s Equationsz
Consider the following system y
Parallel plates of infinite extent
Bottom plate z = 0) = 0
Top plate z = z1) = V1
Region between plates has no charge x
Obtain potential and electric field for region between plates
That is: potential and electric field for a parallel plate capacitor

6. Poisson’s / Laplace’s Equationsz
Use Laplace’s equation since region of interest has no charge present y x
In (x, y, z)
No change in V value in (x, y) plane then

7. Poisson’s / Laplace’s Equationsz y x
C1 and C2 are constants to be determined from Boundary conditions

8. Poisson’s / Laplace’s Equationsz
Boundary conditions given
Bottom plate z = 0) = 0
Top plate z = z1) = V1 y x
@ z = 0, V = 0 gives C2 = 0
@ z = z1, V = V1 gives C1 = V1/z1
Expression for potential between plates

9. Poisson’s / Laplace’s Equationsz
Now to obtain expression for the electric field y x
Recall from Lecture 3

10. Poisson’s / Laplace’s Equationsz
Now to obtain expression for the electric field y x
No x or y dependence

11. Poisson’s / Laplace’s Equationsz
Solution to problem y
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 x

12. Poisson’s / Laplace’s Equations
Select V1 = 12 V
Z1 = 1 m

13. Poisson’s / Laplace’s Equations

14. 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.

15. Poisson’s / Laplace’s Equations
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 no charge density between the conductors.  = 0

16. Poisson’s / Laplace’s Equations
Solution:
The first integration gives
Second integration gives

17. Poisson’s / Laplace’s Equations
Solution:
Apply boundary condition V = 0 at r = a = 1 mm
Apply boundary condition V = 150 at r = b = 20 mm
Two equations with two unknowns:

18. Poisson’s / Laplace’s Equations
Solution:
Introduce values into expression for potential
Units are volts

19. Poisson’s / Laplace’s Equations
Solution for electric field:
Units are volts / m

20. Poisson’s / Laplace’s Equations
Electric field
Potential function

21. ELEC 3105 Basic EM and Power Engineering
Numerical solution to Poisson’s and Laplace’s

22. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
Could be microwave waveguide y
Outer conductor
V = 0 volts x
Inner conductor
V = Vin z
ONE RECTANGULAR CONDUCTOR PLACED INSIDE ANOTHER RECTANGULAR CONDUCTOR
Conductors extend to infinity along z axis

23. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy
Find the electric field lines and equipotentials for the square cylindrical capacitor shown. x
V = Vin
Boundary conditions
V = 0

24. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy
By symmetry, we need only solve for x > 0 and y > 0 quadrant. x
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 .

25. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy x
NOTE: In fact by symmetry only need to solve for purple region. Blue region is the mirror image.

26. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy
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). x
(j)
(i, j)
(i)

27. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy x y
(i, j)
(i, j+1) h
(i+1, j) x
(i-1, j)
(i, j-1)

28. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy x y
(i, j)
(i, j+1) h
(i+1, j) x
(i-1, j)
(i, j-1)

29. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy x
Generalize in y and x
(i, j)
(i-1, j)
(i+1, j)
(i, j-1)
(i, j+1) h x y

30. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy x
Generalize in y and x
(i, j)
(i-1, j)
(i+1, j)
(i, j-1)
(i, j+1) h x y
Charge density present near grid point (i, j)

31. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATIONy x
(i, j)
(i-1, j)
(i+1, j)
(i, j-1)
(i, j+1) h x y
Finite difference representation of Poisson’s equation
Commercial software available for solving numerical problems

32. NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION
(i, j)
(i-1, j)
(i+1, j)
(i, j-1)
(i, j+1) h x y y x
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.
This suggest a simple algorithm for finding Vi,j.

33. 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

34. Numerical solution parallel plate capacitorz
V = 150 volts
z = d plane
Plates of the capacitor are conductors extending to infinity in the (x, y) plane.
z = 0 plane y
V = 0 volts x
As a result of symmetry, the potential function will vary only in the z direction. V = V(z)
Since no charge density between plates

35. Numerical solution parallel plate capacitorz
V10 = 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
Divide region between plates into a fine mesh.
Select values for V1 to V9 i Vi

36. Numerical solution parallel plate capacitorz
V10 = 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

37. Numerical solution parallel plate capacitor
Potential
Grid
number
Iteration

38. Numerical solution parallel plate capacitorz
Numerical solution parallel plate capacitor
Parallel plates
Potential variation between plates V z
Almost a straight line even after only a few iterations

39. Numerical solution parallel plate capacitorz
V10 = 150 volts
(i)
29 V29
28 V28 …
3 V3
2 V2
1 V1
0 V0
Consider a finer mesh i Vi
V0 = 0 volts
Select values for V1 to V28

40. Numerical solution parallel plate capacitor
Grid after 12 iterations

41. Potential
Grid
number
Iteration
Chart after 23 iterations

42. Potential
Grid
number
Iteration
Chart after 50 iterations

43. Potential
Iteration
Grid
number
Chart after 125 iterations

44. Potential
Iteration
Grid
number
Chart after 250 iterations

45. Numerical solution parallel plate capacitorz
Numerical solution parallel plate capacitor
Parallel plates
Potential variation between plates V z
Still quite rough, requires more iterations or better guess at initial potential values for grid

46. Numerical solution parallel plate capacitorz
V10 = 150 volts
(i)
29 V29
28 V28 …
3 V3
2 V2
1 V1
0 V0
Change only one number i Vi
Was
678
V0 = 0 volts
Select values for V1 to V28

48. Estimation of the accuracy of technique
Consider a Taylor’ series expansion for i grid point direction:
Combine the two series:

49. Estimation of the accuracy of technique
Consider a Taylor’ series expansion for j grid point direction:
Combine the two series:

50. Estimation of the accuracy of technique
Combining i and j grid direction results +
Gives :
0 since V satisfies Laplace’s equation

51. Estimation of the accuracy of technique
Dominant correction term
This correction term becomes very small as the grid point spacing h becomes small.
More grip points
= higher accuracy
= greater computation time
= more computer memory

52. 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.

53. Spherical space meshing

54. Triangular space meshing

55. Meshing

57. 97.315 Basic EM and power engineering
End Solutions to Poisson’s / Laplace’s