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 V54 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 h0 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: V54 V5 V4 z h V65 V6 V5 z h and (P-6) Let’s get the derivative of the derivative between points 4, 5 and 6: 2V z 2 V65 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 averaging correctly linked. 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. Write your name in the sign in sheet when you arrive for the lab. 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 [1]. 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) [1] 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). [2] (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. 2. Extract the downloaded zip ﬁle Lab1 MaxwellFiles.zip to a folder in your H directory 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), [2] (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 [1] http://en.wikipedia.org/wiki/Finite_element_analysis, accessed September 2008. [2] Edminister, J.A., Schaums Outlines: Electromagnetics, second edition, 1993.

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Download PDF

advertising