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. * 90 volts NWO006 O volts 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 AV-4 =Vs-V4. The separation along the z axis between these points is Anh. By definition the first derivative of the potential with respect to the z axis is: av lim Vs+h)-VE) (P-5) If at the moment we ignore the lim as h we see that V(z+h)-V(:) is the difference in voltage between adjacent grid points separated by A=h. Thus an approximation to the first derivative can be obtained AV V . So now we have a way to calculate the first derivative by examining voltage values of by 2 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 points. OV(: +h)a(3) @ @_. The derivative of the derivative is also offset by h/2 in grid 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: OV, V, and Vas - V7V (P-6) Let's get the derivative of the derivative between points 4, 5 and 6: (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 =0 which when using (P-7) gives: Vs+V-2V5 = 0 after rearranging VTVA =V (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 = 4 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-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. * 90 volts NWO006 O volts 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 AV-4 =Vs-V4. The separation along the z axis between these points is Anh. By definition the first derivative of the potential with respect to the z axis is: av lim Vs+h)-VE) (P-5) If at the moment we ignore the lim as h we see that V(z+h)-V(:) is the difference in voltage between adjacent grid points separated by A=h. Thus an approximation to the first derivative can be obtained AV V . So now we have a way to calculate the first derivative by examining voltage values of by 2 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 points. OV(: +h)a(3) @ @_. The derivative of the derivative is also offset by h/2 in grid 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: OV, V, and Vas - V7V (P-6) Let's get the derivative of the derivative between points 4, 5 and 6: (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 =0 which when using (P-7) gives: Vs+V-2V5 = 0 after rearranging VTVA =V (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 = 4 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