# Pre lab and Lab 1 for elec 3105 Numerical solutions to laplaces equation ```Numerical Solution to Laplace’s Equation
Carleton University, Department of Electronics
ELEC 3105 Laboratory Exercise 1
July 8 2015
PRE-LABORATORY EXERCISE
You need to complete the pre-lab and have the TA sign off your pre-lab work before starting the
computer laboratory exercise.
Pre-1: Numerical solution to Poison’s and Laplace’s equation
Please refer to the course lecture slides related to Poison’s and Laplace’s equations for additional details
on the technique. A summary is provided here. The starting equation is:
 2V   

(P-1)
which is known as Poison’s equation. It is a point function which implies that the “second derivative”
(Gradient squared here) of the potential function at a particular point in space must equal the negative
ratio of the charge density at the point divided by the dielectric constant at that same point. Should the
charge density be zero then the equation simplifies to Laplace’s form:
 2V  0
(P-2)
A considerable amount of effort goes into solving this equation. For instance, once you solve for the
potential you can determine the magnitude and direction of the electric field through:
E  V
(P-3)
Once you know the electric potential and electric field you can pretty well calculate anything else
related to electrostatic. The pre-lab will examine solving Laplace’s equation using two different
techniques. The first is a direct approach solving the second order differential equation. The second
involves a numerical solution using a finite difference approach. Both techniques are discussed in detail
in class.
Pre-1: Solving the differential equation
Laplace’s equation is a second order differential equation. In Cartesian coordinates it is:
 2V
x 2

 2V
y 2

 2V
z 2
0
(P-4)
The same function V is subjected to derivatives with respect to x, y, z  and when the second derivatives
are formed and then summed, the resultant must be zero. Only then can the original function V be a
valid solution to the equation. Under normal circumstances finding the function V that satisfies (P-4)
can be difficult and when this occurs, other approaches are used to solve the equation (such as
numerical indicated below). For this pre-lab we will consider a simple solution to (P-4).
Consider the parallel plate capacitor shown in figure Pre-1. The lower plate is at 0 volts, and resides in
the (x, y) plane. The upper plate is at 100 volts, also resides in the (x, y) plane and intersects the z axis a
distance d from the origin. We will treat d (capacitor plate separation) as small, such that we may
approximate the capacitor plates as infinite in extent in the (x, y) planes. As a result the potential
function is independent of the x and y coordinates. This statement has to do with the translational
symmetry that is present with regards to the x and y coordinates. As you move about in the (x, y) plane
KEEPING z CONSTANT the environment always looks the same. Thus in equation (P-4) the derivatives
with respect to x and y are zero as (for this geometry) the potential is independent of x and y. The
potential does vary in moving along the z direction. The potential is 0 volts at z = 0 and is 100 volts at z
= d.
Question Pre-1.1: Solve the differential equation (P-4) for the parallel plate capacitor of figure Pre-1. It
is a second order differential equation so the general solution will have two constants. Determine these
constants by making use of the know voltage values at z = 0 and z = d. Take d = 1 mm. Plot several
equipotential lines and from these draw in the electric field lines. What is the numerical value
(magnitude and direction) of the electric field? 1 mark
Figure Pre-1: Parallel plate capacitor geometry
Question Pre-1.2: Two concentric metal shells are shown in figure Pre-2: The inner shell has a radius of
1 cm and is at 100 volts, the outer shell has a radius of 2 cm and is at 200 volts. The region between the
metal surfaces is charge free and air. Express Laplace’s equation in spherical coordinates. Indicate
which derivatives of the potential function will be zero and why they are zero. Solve the remaining
differential equation and plot several equipotential lines for the region between the metal shells. Draw
the electric field lines. 1 mark
Question Pre-1.3: What approach would you use to solve the second order differential equation if the
geometry of the capacitor plates do not conform to the unit vector directions of a coordinate system?
1 mark
Figure Pre-2: Concentric metal shells geometry
Pre-2: Finite difference solution to Laplace’s equation in 1-D
At this time it is a good idea to review the course lecture slides related to the numerical solution to
Poison’s and Laplace’s equation. A review of the numerical technique is presented here for a geometry
which results in a 1-D variation in the potential function. The parallel plate capacitor geometry shown in
figure Pre-1 is such a geometry. The potential varies only the z direction and is constant in the (x, y)
plane. Now consider the parallel plate capacitor geometry redrawn in figure Pre-3. The z axis between
the capacitor plates has been segmented and each point the z axis is assigned an index (i). The spacing
between grid points is uniform and equal to h. The capacitor plate separation is d.
Figure Pre-3: Parallel plate capacitor geometry for numerical technique
Consider now any two adjacent grid points say points 4 and 5. The difference in voltage between these
two points is V54  V5  V4 . The separation along the z axis between these points is z  h . By
definition the first derivative of the potential with respect to the z axis is:
lim V z  h  V z 
V

h
0
z
h
(P-5)
If at the moment we ignore the lim as h0 we see that V ( z  h)  V ( z ) is the difference in voltage
between adjacent grid points separated by z  h . Thus an approximation to the first derivative can be
obtained by
V V
. So now we have a way to calculate the first derivative by examining voltage

z
z
values of adjacent point. But actually, Laplace’s equation is made up of second derivatives. A second
derivative is nothing more than the derivative of the derivative. So let’s first obtain the derivative
between each grid point pair as shown in figure Pre-4. Note that the derivative points are offset from
the potential points by h/2. We can now obtain the derivative of the derivative using the green grid
 V 
V ( z  h) V ( z )



2

V

z
z
z . The derivative of the derivative is also offset by h/2 in grid
points.    2 
z
z
z
point location. This brings the second derivative grid point location back on top of the original grid point
location. We are almost there, but we will start all over again. Let’s get the derivative between points 4
and 5 and also between points 5 and 6:
V54 V5  V4

z
h
V65 V6  V5

z
h
and
(P-6)
Let’s get the derivative of the derivative between points 4, 5 and 6:
 2V
z 2
V65 V5 4 V6  V5 V5  V4


V  V4  2V5
z 
h
h
 z
 6
z
h
h2
(P-7)
For the parallel plate capacitor problem there are no variations in the potential with respect to x and y
and the region between the plates is charge free. Thus
V6  V4  2V5
h
2
0
 2V
z 2
after rearranging
 0 which when using (P-7) gives:
V6  V4
 V5
2
(P-8)
This expression indicates that the voltage at grid point 5 is the average value of the voltage one grid
point up and grid point down. This expression can be turned into a numerical technique through the
following algorithm:




Divide the space into an equal number of grid points. Make certain that grid points are assigned
to surfaces that are at fixed voltages (like the plates of the capacitors, see figure Pre-3)
Assign an arbitrary voltage to each grid point that is not fixed. Try to select voltage values in the
range of the fixed values.
Update the voltage on each grid point by forming the average of its nearest neighbours.
Using the updated values for the voltages, update them again by forming the average of nearest
neighbours.


Repeat the updating process until the voltage values at each grid point no longer change.
Usually you will specify the number of decimal points for the accuracy and once the required
number of decimal points are resolved the updating process is stopped.
The final voltage values are the voltage values at the grid points.
Pre-4: Potential, first derivative and second derivative
Question Pre-2.1: For the parallel plate capacitor given in figure Pre-3 use the numerical technique to
obtain the voltages at the grid points accurate to 1 decimal place. Make a good starting guess to the
voltages. Take d = 1 mm. 1 mark
Question Pre-2.2: Develop an XL spread sheet to solve the parallel plate capacitor numerically to 3
decimal places. (If you wish you may write a MATLAB program instead). 1 mark
Question Pre-2.3: Instead of using 12 grid points use 102 grid points. Modify your program to solve
numerically Laplace’s equation for the parallel plate capacitor to 5 decimal places. 1 mark
Question Pre-2-4: Any numerical technique utilized requires an estimate of its accuracy. Examine the
course lecture slides, text books on numerical techniques, … and obtain an estimate for the error
involved in using this approach to solving Laplace’s equation. 1 mark
Pre-3: Finite difference solution to Laplace’s equation in 2-D and 3-D
The numerical approach presented above can be easily extended into 2-D and 3-D. We need to develop
the finite difference approximations to each of the second order derivatives in equation (P-4). We have
already worked out the derivative part for the z direction. We imposed a grid along the z axis and
formed the first and second derivative. Now in 3-D we need to establish grid points along the other two
axes. We thus end up with a volume of grid points with each grid point identified by the indices (i, j, k).
We then form the second derivatives for each additional direction. Figure Pre-5 shows one of the grid
points extracted (point i, j, k) and its six nearest neighbours.
Pre-6: 3-D grid points about center (i, j, k) point
The resultant combination of the three second order derivatives of equation (P-4) results in the
following expression:
 2V
x
2

 2V
y
2

 2V
z
2

Vi 1, j ,k  Vi 1, j ,k  2Vi , j ,k
h
2

Vi, j 1,k  Vi , j 1,k  2Vi, j ,k
h
2

Vi, j ,k 1  Vi, j ,k 1  2Vi, j ,k
h2
(P-9)
When dealing with Laplace’s equation the above equation is equal to zero and thus can be simplified
and rearranged to yield an expression for the voltage at point (i, j, k) as the average of its nearest
neighbours (3-D Grid):
Vi 1, j ,k  Vi 1, j ,k  Vi , j 1,k  Vi , j 1,k  Vi , j ,k 1  Vi , j ,k 1
6
 Vi , j ,k
(P-10)
In the situation where the geometry can be analysed in 2-D, say x and y, the averaging would involve
only 4 nearest neighbours with the grid using indices i and j.
Vi 1, j  Vi 1, j  Vi , j 1  Vi , j 1
4
 Vi , j
(P-11)
The same numerical algorithm presented above can be applied to the 2-D and 3-D grid. The difficulty in
using this approach in 2-D and 3-D comes from the bookkeeping required to keep all the grid point
Question Pre-3.1: For the structure shown in figure Pre-7 use a 2-D numerical grid approach to obtain a
mapping of the potential inside the electrode region. To keep the problem manageable use a grid with a
10 mm spacing. Obtain the voltages on the grid points accurate to 1 decimal place and use either XL or
MATLAB to solve. 1 mark
Pre-7: Potential well electrode structure
Question Pre-3.2: From the potential values determined above draw in the electric field vectors. 1 mark
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
Lab 1: Numerical Solution of Laplace’s Equation
ELEC 3105
July 08, 2015
1. Before You Start

This lab and all relevant ﬁles can be found at the course website.

You will need to obtain an account on the network if you do not already have one from another
course.


You can work alone or with a partner.

One lab write-up per person.

Show units in all calculations, all graphs require a legend.
2. Objectives
The objective of this lab is to illustrate the use of a powerful numerical technique known as the ﬁnite
element method to solve Laplace’s equation for selected problems. The lab will run in the
Department of Electronics undergraduate laboratory, room ME4275. The software package we will
use is Maxwell SV from Ansoft Corporation. This software will enable you to visualize the electric
ﬁeld lines and equipotential lines in cross sections of structures consisting of conductors and
insulators.
3. Background
The ﬁnite element method (FEM) is a numerical technique for ﬁnding approximate solutions to
partial diﬀerential equations . Consider the example of a 2-D solution and its corresponding
mesh shown in Figure 1. The lines represent the direction and magnitude of ﬂux density simulated
using FEM in the solution image and the triangles (or sub regions) represent a single calculated
solution in the mesh image. As an analogy, compare a jpeg ﬁle with large pixels, making the image
blurry and a jpeg ﬁle with smaller pixels, allowing the image to become sharper. Therefore, the
smaller the sub region, the more accurate the entire solution. A numerical solution is always an
approximation of an analytical solution, which is based on mathematical theory.
Figure 1: The 2-D solution (left) and mesh (right) 
Consider Laplace’s equation describing the potential V in a 2-D region:
 2V  2V

0
x 2 y 2
(1)
A solution can be found using FEM by approximating the size of dV. Smaller triangles are used where
the potential V(x, y) is rapidly varying, and larger triangles are used where the potential is varying slowly.
The potential is approximated within each triangle as a polynomial expansion in x and y. A numerical
algorithm is used to solve for the coeﬃcients of the polynomial in each triangle such that the nodes of
adjacent triangles have the same potential. Conducting surfaces are constant potential surfaces - the user
initially sets the value of the potential at the conductor.
Electric energy is stored in the electric ﬁeld. The energy stored is given by the expression (units Joules).
1  
D  EdV
2
WE 


(2)

where D  E is the electric ﬂux density (C/m2), E is the electric ﬁeld intensity (V/m), and the dot
product is used in the integrand. The energy stored in a capacitor C is given by (units J, Joules):
WE 
1
C (V ) 2
2
(3)
where ∆V is the potential diﬀerence between the conductors of the capacitor. The capacitance of a
structure can be evaluated as (units F, Farads):
C
2WE
V 2
(4)
Maxwell 2D Field Simulator can calculate the energy WE over the 2-D cross-section and then calculate
the approximate value of the capacitance C per unit length (F/m) of the structure. You will be analyzing
five diﬀerent structures:
Problem 1 - Field at a sharp or raised point
Problem 2 - Field in a hollow
Problem 3 - Parallel wire transmission line
Problem 4 - Parallel wire transmission line with plastic coating
Problem 5- Rectangular potential well
You will be asked to plot the voltage and electric ﬁeld lines for these structures. The relation between
electric ﬁeld and voltage is found by using the relation below (units J/C or V).  (pg.60)
B
V AB
B
W

   E  dl    E dl cos
Qunit
A
A
(5)
which describes the potential, V, of point A with respect to point B, deﬁned as the work done, W, in
moving a unit charge Qunit, from A to B. The electric ﬁeld and the potential are perpendicular. In the case
of the structures in this lab, equation 5 can be simpliﬁed by choosing a path integral such that cos(θ) = 1.
If the electric ﬁeld is constant in the region of integration, then all that is left to calculate is the integral
with respect to the displacement l . Based on these special circumstances, the resulting equation is
E
V
l
(6)
where ∆𝑉 is the diﬀerence in potential between two points and l is the distance between the points. The
structures in this lab have pre-deﬁned voltages. Keep track of their values as you go through the lab.
4. Running Maxwell 2D Field Simulator
1. Go to the course website, and click on the link for the lab material. Follow the instructions
provided.
3. Start Maxwell Control Panel on your desktop (go to the ANSOFT directory if the icon isn’t
available). If a message pops up asking if you would like to create a working directory, say Yes.
4. Choose Projects in the Maxwell Control Panel
5. In the lower left hand corner, look for a Project Directories heading. Click Change Dir...
6. Go to Sub Directories and double click on “../” to go up one level to the Maxwell directory that
the zipped ﬁles were extracted to.
7. You should see the directory ﬁles named prob1c, prob2a, prob3, and prob4 show up under the
Projects ﬁle list.
8. Click OK to go back to Maxwell Projects.
9. Again, you should see the directory ﬁles named prob1c, prob2a, prob3, and prob4 show up
under the Projects ﬁle list, only this time as you single click on each directory a drawing should be
displayed in the graphics window to the right. The drawings, boundary conditions, materials, and post
processes have been completed but make sure you check each setting manually.
5. Problem 1: Field at a Raised Point
This problem models a parallel plate capacitor in which one plate is dented toward the other as shown
in Figure 2. The top plate is at 1 V and the bottom plate is at 0 V. You would set these values by
clicking Setup Boundaries/Sources, but for this lab the values have been set for you. The material of
both plates is copper. The material around the plates is air. You would set these values by clicking
Setup Materials, but again, this has been done already.
Figure 2: Conductor structure for Problem 1
1. Click on prob1c in the Maxwell Projects window and click open under the graphics window. If there
is a version mismatch between the ﬁle and the software you will see the message in Figure 3. Click OK.
2. Choose Post Process
Figure 3: Version mismatch message
3. To plot the electric potential go to Plot in the tool bar and run through Plot → Field → Phi - surface all - all.
4. Ensure the values are similar to those in the form shown Figure 4. To modify this form anytime, go to
Plot → modify in the ﬁle menu.
Figure 4: Phi scalar surface plot form for Problem 1
5. To plot the electric ﬁeld: Plot → Field → E vector - surface - all - all.
6. Ensure the values are similar to those in the form shown in Figure 5.
Figure 5: E vector surface plot form for Problem 1
7. Answer the following questions for Problem 1.
(a) Plot the equipotential lines (contours of constant voltage) and the electric ﬁeld lines of your
structure together in one printout, or individually. Display at least 10 voltage contours and don’t
forget to clearly include the legends. 2 marks
(b) Where is the location of the maximum electric ﬁeld strength? What is the value of the maximum
ﬁeld strength? Use the coloured electric ﬁeld intensity plot and the accompanied legend. Don’t
forget units. 2 marks
(c) Insulating materials will break down or become conducting if the electric ﬁeld strength exceeds
the breakdown strength of the material. For air, the breakdown strength is about 3 x 106 V/m. If
the gap is reduced to 1 mm, estimate the maximum voltage that could be applied to the top plate.
Answer this question using theory and include units. You may use Maxwell to check your
calculation (note: Maxwell does not actually simulate the dielectric breakdown). 1 mark
6. Problem 2: Field in a Hollow
This problem models a parallel plate capacitor with one plate dented away from the other as shown in
Figure 6. The top plate is at 1 V and the bottom plate is at 0 V source. The material of both plates is
copper and the dielectric is air.
Figure 6: Conductor structure for Problem 2
1. Click on prob2a in the Maxwell Projects window and click open under the graphics window.
2. Manually check Setup Boundaries/Sources and Setup Materials.
3. Choose Post Process
4. To plot the electric potential go to Plot in the tool bar and run through Plot → Field → Phi - surface all - all.
5. Ensure the values are similar to those in the form shown in Figure 7.
Figure 7: Phi scalar surface plot form for Problem 2
6. To plot the electric ﬁeld: Plot → Field → E vector - surface - all - all.
7. Ensure the values are similar to those in the form shown in Figure 8.
Figure 8: E vector surface plot form for Problem 2
8. Answer the following questions for Problem 2.
a) Plot the equipotential and electric ﬁeld lines of your structure as in Problem 1. 2 marks
b) Consider the region between the two plates. Why is the electric ﬁeld diﬀerent in the hollow?
2 marks
7. Problem 3: Parallel Wire Transmission Line
VHF and UHF antennas are usually connected to TV sets by transmission lines consisting of two parallel
wires of ﬁxed separation, as shown in Figure 9. To design the transmission line, we need to ﬁnd the
capacitance per unit length between the wires. The capacitance per unit length is given analytically by
(units F/m)
C

cosh 1 (
D
)
2a
(7)
where ∆V is the diﬀerence in potential between the two wires,  is the dielectric constant of the
homogeneous material surrounding the wires, D is the center to center wire spacing, and a is the radius of
the wires, as shown in Figure 9. The dielectric constant of air is  0 = 8.854×10−12 F/m. For other
materials, we multiply this value by the relative dielectric constant  r of the material (that is  =  0 r ).
1
The function cosh is found using the hyp button on any scientiﬁc calculator. The object of problem 3 is
to ﬁnd the capacitance numerically and compare with the theoretical value. We will assume that the
radius of the wire is always 1 mm, but will allow for diﬀerent spacing between the wires. The basic
drawing of the two wires has already been completed, and is in directory prob3. Edit the drawing as
explained below and use a center to center wire spacing of D = 6 mm. The material of both wires is
copper and one wire is at 1 V while the other is at -1 V. If we assume that the parallel wires can be
estimate by two parallel plates, then the capacitance, neglecting fringing, can also be written as (units F),
 (pg. 96)
C
 0 r A Q Q
 
l
V El
(8)
where A is the area of the plates, Q is the charge on the plates, l is the distance between the plates,  0 is
the dielectric constant of air, and  r is the relative dielectric constant of the material between the plates.
This relation indicates that the electric ﬁeld is related to the dielectric properties of the material in
between the plates.
Figure 9: Transmission wire structure for Problem 3
1. Click on prob3 in the Maxwell Projects window and click open under the graphics window.
2. In the Maxwell SV window, choose Solve.
3. To plot the electric potential go to Plot in the tool bar and run through Plot → Field → Phi surface - all - all.
4. Ensure the values are similar to those in the form shown Figure 10.
Figure 10: Phi scalar surface plot from Problem 3
5. To plot the electric ﬁeld: Plot → Field → E vector - surface - all - all.
6. Ensure the values are similar to those in the form shown in Figure 11.
Figure 11: E vector surface plot form for Problem 3
7. Answer the following questions for Problem 3.
a) Plot the equipotential and electric ﬁeld lines of your structure. 2 marks
b) What do you notice about the direction of the electric ﬁeld at any point in relation to the
equipotential lines? 1 mark
c) Specify the region at which the electric ﬁeld is maximum and state the maximum value. Use
the legend to guide you. Theoretically you will ﬁnd that the maximum should not be one
point, but several points. 3 marks
d) Estimate the capacitance per unit length of the transmission line using the Post Processor
Field Calculator and Equation 2 and 4. You can ﬁnd help in the Maxwell SV manual. The
link to the manual is provided in the “Objectives” section of this lab. Hint: as an example of
how to use the calculator, the manual calculates the capacitance of a test electrode set to 1 V.
In our case, we are using two wires with ∆V = (1 V − (−1 V)) = 2 V. Therefore, C 
where V = 1 V becomes C 
you must take the
2U
V2
2U
where ∆V = 2 V. If you follow the instructions exactly,
V 2
1
fraction into account in your ﬁnal result. 3 marks
4
e) Calculate the theoretical value of the capacitance per unit length as explained in the
introduction to Problem 3. Compare to the estimated value of d) and explain any
discrepancy. Remember that you are comparing 2 diﬀerent methods of solving for
capacitance: numerical and analytical. 3 marks
8. Problem 4: Transmission Line with Plastic Coating
Now modify the structure in Problem 3 so that the wires are coated with a plastic (dielectric) layer of
relative permittivity  r = 2.1 and radius 2.0 mm. The plastic material is Teﬂon and when drawing, the
plastic layer with the Object/Circle/2 Point tool the center of the plastic should be the same as the center
of the copper wire.
1. Click on prob4 in the Maxwell Projects window and click open under the graphics window.
2. Choose Post Process
3. Edit the drawing as previously explained.
4. Choose Solve
5. To plot the electric potential go to Plot in the tool bar and run through Plot→ Field → Phi surface - all - all.
6. Ensure the values are similar to those in the form shown Figure 12.
Figure 12: Phi scalar surface plot form for Problem 4
7. To plot the electric ﬁeld: Plot → Field → E vector - surface - all - all.
8.
Ensure the values are similar to those in the form shown in Figure 13.
Figure 13: E vector surface plot form for Problem 4
9. Answer the following questions for Problem 4.
a) Plot the equipotential and electric ﬁeld lines of your structure. 2 marks
b) State the maximum value of the electric ﬁeld and state why it is greater or less than the
maximum values found in Question 3. 2 marks
c) Estimate the capacitance per unit length of the transmission line using the Post Processor
Field Calculator. 2 marks
d) Is the capacitance greater or less than the one estimated in Problem 3? Explain. 3 marks
9.
Problem 5: Rectangular potential well
Figure 14: Conductor Structure for Problem 5
1. Choose Projects in the Maxwell Control Panel. In the Project menu choose New. In the Alias
box enter Prob5 then click on the Make New Directory circle and hit OK
2. Specify Solver Type: Select Electrostatic
3. Specify Drawing Plane: Select XY Plane
4. Choose Define Model → Draw Model
5. Choose Model → Drawing Units → mm
6. Choose Model → Drawing Size and set the Minima to (0, 0) and the Maxima to (125, 125)
7. Choose Window → Grid and set dU and dV smaller for there to be less space between each
point (1 mm is required).
8. Select Object→ Polyline and match the example provided. Hint: Objects can be moved after
they have been drawn by selecting them and then choosing Arrange→Move (i.e. objects can be
centered after drawn).
9. Click save and exit the window (File → Exit)
10. Choose Setup Materials. Select an object and material and press Assign. The background
material will be air and the conducting plates will be copper.
11. Click save and exit the window (File → Exit)
12. Choose Set Up Boundaries → Sources
13. Choose Edit→ Select → Object → By Clicking and left click on an object so it is highlighted.
Then right-click anywhere in the display area to stop selecting objects
14. Choose Assign → Source → Solid then select Voltage and Assign a corresponding voltage to
the conductor. The top plate has a voltage of 100 V and the other plate will be 0 V.
15. For the background once it is highlighted choose Assign → Boundary → Balloon then
select Charge and Assign it. Selecting the Charge option for the balloon boundary models an
electrically insulated system. That is, the charge at infinity balances the charge in the problem
region, forcing the net charge to be zero. [The Voltage option models an electrically grounded
system, i.e. the voltage at infinity is zero, but the charge at infinity may not exactly balance the
charge in the problem region]
16. When all objects are assigned their source or boundary, save and exit
17. Choose Solve and wait for the “Solution Process is Complete” message
18. Choose Post Process... and wait for a “2D Post Processor” window
19. To plot the E Vector and the Voltage together for best display do the following: choose Plot →
Field...
20. Select Phi → Surface -all- and press OK. In the next screen, unselect “filled” and press OK. To
change the contour divisions choose Plot → Modify, select Phi and change the number of
Divisions (20). The values should be similar to Figure 14.
Figure 15: Phi scalar surface plot form for Problem 5
21. Then select E Vector → Surface -all- and press OK. In the arrow options select Type 2D and
press OK. Since the pattern of arrows is not too consistent, go to Plot → Modify... and in Arrow
Options change Size and Spacing to approximately 4 and 2 respectively. Uncheck map size then
press OK. (The size and spacing values may be adjusted to your liking). The values should be
similar to Figure 15.
Figure 16: E vector surface plot for Problem 5
22. Answer the following questions for problem 5.
a) Plot the equipotential and electric field lines of your structure. 2 marks
b) *3 Compare results obtained here with those calculated in the pre-lab section. 2 marks
References
 http://en.wikipedia.org/wiki/Finite_element_analysis, accessed September 2008.
 Edminister, J.A., Schaums Outlines: Electromagnetics, second edition, 1993.
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