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

Recall From Lecture 3

Poisson’s / Laplace’s Equations z Consider the following system y Parallel plates of infinite extent Bottom plate V(@ z = 0) = 0 Top plate V(@ 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

Poisson’s / Laplace’s Equations z 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

Poisson’s / Laplace’s Equations z y x C1 and C2 are constants to be determined from Boundary conditions

Poisson’s / Laplace’s Equations z Boundary conditions given Bottom plate V(@ z = 0) = 0 Top plate V(@ 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

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

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

Poisson’s / Laplace’s Equations z 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

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

Poisson’s / Laplace’s Equations

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.

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

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

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:

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

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

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

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 Inner conductor V = Vin z 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 V = Vin Boundary conditions V = 0

NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION y 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 .

NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION y x 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 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)

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

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

NUMERICAL SOLUTION TO POISSON’S / LAPLACE’S EQUATION y x Generalize in y and 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 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)

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

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.

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 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

Numerical solution parallel plate capacitor z 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

Numerical solution parallel plate capacitor z 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

Numerical solution parallel plate capacitor Potential Grid number Iteration

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

Numerical solution parallel plate capacitor z 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

Numerical solution parallel plate capacitor Grid after 12 iterations

Potential Grid number Iteration Chart after 23 iterations

Potential Grid number Iteration Chart after 50 iterations

Potential Iteration Grid number Chart after 125 iterations

Potential Iteration Grid number Chart after 250 iterations

Numerical solution parallel plate capacitor z 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

Numerical solution parallel plate capacitor z 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

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. More grip points = higher accuracy = greater computation time = more computer memory

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

Triangular space meshing

Meshing

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