ELE366 LAB1

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ELE366 EE08U155 ARSAL JAVID

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ELE 366 Lab Experiment 1: Numerical Solution OfLaplaces Equation

Introduction

This experiment consists of lab simulation which provides a clear understanding of Laplaces

Equation. Geometry is usually used to describe the solutions to the Laplace equation,

however, real world problems are solved numerically. Solving problems numerically

provides a solution to a greater accuracy. Solving problems hypothetically may not be able

to provide an ideal solution. In order to deal with numerical solutions, computers can be

used as they are powerful enough to perform iterations.

The experiments consists of the derivation and application of Laplace equation. Thisequation is a partial differential equation named after Pierre-Simon Laplace and is used in

the field of electromagnetism and fluid mechanics. This is used due to involvement of

gravitational, electric and fluid potentials.

In order to gain a deeper understanding of Laplaces Equation, an experiment was

performed using Microsoft Excel. This software provided us with a numerical of the

geometric shapes that turn up in the course.

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Experiment 1

Assuming that = :

22

+22

= 0

The first term represents the rate of change of Potential V with respect to x, whilst the

second term represents the rata of change of V with respect to y.

The rate of change of potential points 1 and 0 is given by:

10

1 0

The rate of change between points 0 and 2 is:

02

0 2

The second order differential equation can be found by similar reasoning to be given by:

2

2

10

02

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The second order differential equation can be given with respect to y in a similar manner

using points 3 and 4:

10

02

+

30

04

This can be rearranged to give:

0 = 0.25(1 + 2 + 3 + 4)

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Experiment 2

The diagram above was implemented using a predefined 29 by 29 grid based on the five point

average formula. The convergence of this formula was then examined by performing iterations of

this formula. It will be assumed that the potential is zero at every location except the electrodes.

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2.2)

After 20 iterations:

Computation Region after 20 Iterations

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Magnitude of Electric Field after 20 iterations

Laplace Representation after 20 iterations

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After 100 iterations:

Computation Region after 100 Iterations


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Laplacian Representation after 100 iterations

The converged solution:

Computation Region after 101 iterations

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Laplacian Representation after 101 iterations

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As convergence is performed, the voltage pattern around these boundaries (layout of electrodes)

stabilises. Hence, the pattern stabilises as the number of iterations increases. As the number of

iterations come closer to that of number of iterations required for convergence, the grid becomes

similar to the problems faced that require the use of the five point average equation. Therefore, it

can be seen that there is no charge stored inside the conductor as any existing charges are attached

at the electrodes that were used in the spreadsheet. It can be seen that the charge propagates

quickly in certain directions and slowly in opposite direction. Any cell not represented by an

electrode is a vacuum cell. Each of these store a five point average formula. If the solution to the

Laplacian is equal to the average of the four surrounding cells. Each Vaccum cell is recalculated

repeatedly as iterations.

Free charges exist outside the boundaries drawn in Figure 1 due to the fact that two positive and

negative charges are repelling each other and moving the sets towards the boundaries around the

edge of the spreadsheets represented by the purple area on the diagram showing the potential

distribution. At 20 iterations, it can be seen that the charge stored inside the electrodes merge with

the charge initially on the electrodes which is why it can be seen that the values represented in the

red and black change significantly between 20 and 100 iterations. This also applies to the charges

contained outside the boundaries, as the iterations take place, these charges combine with the

charge already existing on the boundary around the edge of the screen and increases the number of

positive charges. The change in charge at a point on a boundary is proportional to the charge density

in the area.

In this particular instance, 101 iterations were performed for convergence which is slightly more

than the previous number of 100; therefore this value could be used if required. Despite this, it is

important to note that the value required for convergence varies due to the sizes of the electrodes

in cells. Any zeros in this computation region can be be derived using Gauss s law. The boundary

conditions guarantee that any field line that leave the top of the universe re-enter at the bottom of

the universe. Consequently, there is not net flux flowing into the universe (this can be seen as the

edges are displayed with the value of zero). Therefore, the accuracy of Gauss s law depends wholly

on the structure of the operator (a + b + c + d -4w).

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2.3) Using the converged solution, The equi-potential lines between 0 Volts and 100 Volts in

intervals of 10 Volts is given as:

The electric field lines for the given configuration can be given as:

By looking at both graphs, it can be seen that Electric Potential is known to be perpendicular to the

electric field which is poof of one of the basic electromagnet concepts in physics.

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Experiment 3

3.1) The 5 point average formula will be implanted with the boundary conditions specified in part 2

using the same 29 by 29 grid. The diagram shows the initial implementation of the problem with the

potential of the electrodes set at 100 Volts.

Computation Region with Cross-shaped electrode defined

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Computation Region with Cross-shaped electrode defined once convergence is performed

Equipotential Surface Line of cross shaped Electrode

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Electric Field Lines of Cross Shaped Electrode

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3.3) In order to calculate the capacitance per unit length, the Gaussian surface needs to be found.

This can be done by calculating Gaussian Surface area for one side of the surface:

= 3 + 5 + 8 + 10 + 13 + (2 16) + (2 19) + (2 21) +(2 23) + (2 24) + (5 25) + 14 + 11 + 9 + 7 + 4 + 2 = 456

456 10 3 = 0.456

= 456 10 3 8.85 10 12 = 4.0356 10 13 = 4.0356 10 12 4 = 1.614 10 11 C

We know that : = and that V = 100 Volts

= 1.614 1011

100 = 1.614 10 13 1

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ELE366 LAB1