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

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

3 Recall From Lecture 3

4 Poisson’s / Laplace’s Equations x y z Consider the following system Parallel plates of infinite extent Bottom plate z = 0) = 0 Top plate z = z 1 ) = V 1 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 Poisson’s / Laplace’s Equations x y z Use Laplace’s equation since region of interest has no charge present In (x, y, z) No change in V value in (x, y) plane then

6 Poisson’s / Laplace’s Equations x y z C 1 and C 2 are constants to be determined from Boundary conditions

7 Poisson’s / Laplace’s Equations x y z Boundary conditions given Bottom plate z = 0) = 0 Top plate z = z 1 ) = V z = 0, V = 0 gives C 2 = z = z 1, V = V 1 gives C 1 = V 1 /z 1 Expression for potential between plates

8 Poisson’s / Laplace’s Equations x y z Now to obtain expression for the electric field Recall from Lecture 3

9 Poisson’s / Laplace’s Equations x y z Now to obtain expression for the electric field No x or y dependence

10 Poisson’s / Laplace’s Equations x y z Solution to problem 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 V 1 > 0. Lines of constant V are in the (x, y) plane and perpendicular to the electric field lines

11 Poisson’s / Laplace’s Equations Select V 1 = 12 V Z 1 = 1 m

12 Poisson’s / Laplace’s Equations

13 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.

14 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 not charge density between the conductors.  = 0

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

16 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:

17 Poisson’s / Laplace’s Equations Solution: Introduce values into expression for potential Units are volts Valid only between cylinders

18 Poisson’s / Laplace’s Equations Solution for electric field: Units are volts / m Valid only between cylinders

19 Poisson’s / Laplace’s Equations 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 Outer conductor V = 0 volts x y z Inner conductor V = V in Conductors extend to infinity along z axis Could be microwave waveguide

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

NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION x y 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 = V in Boundary conditions By symmetry, we need only solve for x > 0 and y > 0 quadrant.

NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION x y 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 x y Technique for numerical solution Establish a dense mesh or grid between the conducting plates. Represent V(x, y) as a set of discrete values V ij defined at each grid point (i, j). (i, j) (j) (i)

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

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

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

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

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 Finite difference representation of Poisson’s equation Commercial software available for solving numerical problems y x

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 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 x This suggest a simple algorithm for finding V i,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.Repeat until V does not change significantly. Now for a real example of the technique

Numerical solution parallel plate capacitor x y z z = 0 plane z = d plane V = 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) Since no charge density between plates

Numerical solution parallel plate capacitor Divide region between plates into a fine mesh. Select values for V 1 to V 9 z V 10 = 150 volts V 0 = 0 volts (i) 10 V 10 9 V 9 8 V 8 7 V 7 6 V 6 5 V 5 4 V 4 3 V 3 2 V 2 1 V 1 0 V 0 ViVi i

z V 10 = 150 volts V 0 = 0 volts (i) 10 V 10 9 V 9 8 V 8 7 V 7 6 V 6 5 V 5 4 V 4 3 V 3 2 V 2 1 V 1 0 V 0 After 18 iterations i Numerical solution parallel plate capacitor

Iteration Grid number Potential Numerical solution parallel plate capacitor

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

Consider a finer mesh Select values for V 1 to V 28 z V 10 = 150 volts V 0 = 0 volts (i) 29 V V 28 … 3 V 3 2 V 2 1 V 1 0 V 0 ViVi i

Numerical solution parallel plate capacitor Grid after 12 iterations

Iteration Grid number Potential Chart after 23 iterations

Iteration Grid number Potential Chart after 50 iterations

Iteration Grid number Potential Chart after 125 iterations

Iteration Grid number Potential Chart after 250 iterations

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

Change only one number Select values for V 1 to V 28 z V 10 = 150 volts V 0 = 0 volts (i) 29 V V 28 … 3 V 3 2 V 2 1 V 1 0 V 0 ViVi i Was 678

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

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

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

Estimation of the accuracy of technique Dominant correction term 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 V in = 20 volts Outer radius b = 70 mm Outer potential V out = 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