.MCAD 310000000 \  docDocument MmcObject[00 d2_graph_format graphData% axisFormat)L)Ltrace2D&&&&&&&&& & & & & &&& dim_formatTmasslengthtimecharge temperature luminosity substanceNumericalFormatQdii  shpRectVh5mcDocumentObjectState\ mcPageModelK????mcHeaderFooterI@I |P CHeaderFooterJ@{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fswiss\fprq15\fcharset0 Arial;}{\f1\fnil\fcharset0 Arial;}} \viewkind4\uc1\pard\f0\fs18\{d\}\f1 \par \par \par \par } @{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fswiss\fprq15\fcharset0 Arial;}{\f1\fswiss\fprq15 Arial;}{\f2\fnil\fcharset0 Arial;}} \viewkind4\uc1\pard\qc\f0\fs18 Electrostatics\f1 \par \f2 \par \par \par \par } @{\rtf1\ansi\ansicpg1252\deff0\deflang1033{\fonttbl{\f0\fswiss\fprq15\fcharset0 Arial;}{\f1\fswiss\fprq15 Arial;}{\f2\fnil\fcharset0 Arial;}} \viewkind4\uc1\pard\qr\f0\fs18 Laplace Solver \f1\{n\}\f2 \par \par \par \par } @J@{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss Arial;}{\f4\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\plain\f4\fs18 \par } @{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss Arial;}{\f4\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\qc\plain\f3\fs20 \par } @{\rtf1\ansi\deff0\deftab720{\fonttbl{\f0\fswiss MS Sans Serif;}{\f1\froman\fcharset2 Symbol;}{\f2\fnil\fprq15 Arial;}{\f3\fswiss Arial;}{\f4\fswiss\fprq15 Arial;}} {\colortbl\red0\green0\blue0;} \deflang1033\pard\qr\plain\f4\fs18 \par } @J@J MbP?MbP? 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This type of plot can be selected from the graph menu. The potential values of 5.1 and -5.1 were used so that the "5" contour and the "-5" contour would pass close to the capacitor plates. Otherwise, we could not tell exactly where they are. 79A}<A~:@W1A</A<A0@NormalArial A*@Uf-Plots of potential contours 79A<A:@W 1A</A<A0@NormalArial A@B@U0"RA@@ pA@@^AA@@@AA@@AA@@<@AA AAA< 2CVSOleClientItem ࡱ> Root EntryI>'r !@ContentsOlePres000   x:m@e?A&qFz?nK@ ; b@?333d   ?$@  ?`ff`ff@  ?$@     ??@???@??@ ???D# >@33@33 A A?69@@d8j:@@81;@@#<@@dNANIi?@@=@@@@d?j@A@@?1@B@@"2@C@@@D@@@@C@E@@d@DQ@F@@ @D@G@@d@Fi@H@@@Fj@I@@@CC@J@@@K@@d@Jj@L@@@J@M@@t@L1A@@APhiA@B@UQA@@ pA@@^AA@@@AA@@AA@@<@AA AAA< CVSOleClientItem ࡱ> Root EntryI>'r !@ContentsOlePres000   [A5?DJWƿ ; b@?333d   ?$@  ?`ff`ff@  ?$@     ??@???@??@ ???D >@33@33 A A?NANI+00#C:\1.1*XQDocuments and SettingsDOCUME~1'1+Scott RobertsonSCOTTR~11+RecentRECENTPO :i+00#C:\1.1*XQDocuments andA@@APhiA*@U#)C01! ! -@You can click on the contour plot at right and rotate it. The axes are labelled with the grid point number. You can change this by clicking on the graph, then on axes, and filling in the limits. 79A<A:@W1A</A<A0@NormalArial A*@USs`1   -@Mathcad has a special functions "mulitgrid" and "relax" which will solve Laplace's equation. These solvers converge more quickly, but basically do the same thing we have done here. 79A<A:@W1A</A<A0@NormalArial A*@U9 11`1`-ANotes: 1. We initially loaded zeros into Phi as our intial guess. This is not a great guess. But the solver converges anyway. The values in the rows and columns at the boundaries are not averaged because this is a boundary value problem with fixed boundary conditions. 2. The averaging was done 40 times. Is that enough? One way to tell is to double the number of iterations. If they are within 1% of one another then the answer is probably good to 1%. 79A<A:@W1A</A<A0@NormalArial A*@U X  9PP-+Being more clever with boundary conditions:7+9+A<+A:@W1A</+A<+A0@NormalArial A*@U .+  1&P&P-AWe could have a program that allowed us to put different shaped central objects without having to put new lines in the programming block. Let's put the values for our new object in a matrix called BC for boundary condition. We will set the unused members of this matrix equal to 1001. Then we will put into the program an "if" statement: if the value is greater than 1000, it is not averaged because it is a boundary value. 79A<A:@W1A</A<A0@NormalArial A@B@U4 OL -@ A@@ pA@@ AA@@@AA@@dABCA@@ AA@@dAiA@@AjA@@A1001A*@Uh3 C q@ ??->This "works" for all i,j values. Test it with any i,j values: 7>9>A<>A:@W1A</>A<>A0@NormalArial A@B@U, .L @ A@@ pA@@AA@@@AA@@dABCA@@ AA@@tA3A@@A4A@@AA@@+@A@XA@@AA*@US ^c ` !VV-;Use the same capacitor plates, only turn them horizontally.7;9;A<;A:@W1A</;A<;A0@NormalArial A@B@U| K 2 A@@ pA@@ AA@@@AA@@dABCA@@ AA@@tA4A@@A4A@@A5.1A@B@U`| A@@ pA@@ AA@@@AA@@dABCA@@ AA@@tA5A@@A4A@@A5.1A@B@U|  A@@ pA@@ AA@@@AA@@dABCA@@ AA@@tA6A@@A4A@@A5.1A@B@U| b B A@@ pA@@ AA@@@AA@@dABCA@@ AA@@tA4A@@A6A@@KAA@@A5.1A@B@Ux|   A@@ pA@@ AB@@@AB@@dBBCB@@ BB@@tB5B@@B6B@@KAB@@B5.1B@B@U| "  B@@ pB @@ BB @@@B B @@dB BCB @@ B B @@tB 6B@@B 6B@@KB B@@B5.1B*@U@  H -We average 60 times. 79B<B:@W1B</B<B0@NormalArial B@B@U   B@@ pB@@ BB@@dBkmaxB@@B60B@B@U  ! B@@ pB@@ BB@@dBPhiB @@BB!@@@B B"@@@B!B#@@@B"B$@@dB#iB%@@B#0B&@@B"B'@@@B&B(@@@B'B)@@@B(B*@@dB)PhiB+@@ B)B,@@dB+iB-@@B+jB.@@B(B/@@dB.ifB0@@pB.B1@@ B0B2@@ @B1B3@@)@B2B4@@@B3B5@@dB4BCB6@@ B4B7@@dB6iB8@@B6jB9@@B31000B:@@B2B;@@dB:BCB<@@ B:B=@@dB<iB>@@B<jB?@@B10B@@@B'BA@@dB@iBB@@B@BC@@tBB0BD@@BBimaxBE@@B&BF@@dBEjBG@@BEBH@@tBG0BI@@BGjmaxBJ@@B!BK@@@BJBL@@@BKBM@@@BLBN@@@BMBO@@@BNBP@@dBOAvgBQ@@ BOBR@@dBQiBS@@BQjBT@@BNBU@@@BTBV@@@BUBW@@@BVBX@@@BWBY@@dBXPhiBZ@@ BXB[@@@BZB\@@dB[iB]@@B[1B^@@BZjB_@@BWB`@@dB_PhiBa@@ B_Bb@@@BaBc@@dBbiBd@@Bb1Be@@BajBf@@BVBg@@dBfPhiBh@@ BfBi@@dBhiBj@@BhBk@@dBjjBl@@Bj1Bm@@BUBn@@dBmPhiBo@@ BmBp@@dBoiBq@@BoBr@@dBqjBs@@Bq1Bt@@BT4Bu@@BMBv@@dBujBw@@BuBx@@tBw1By@@BwBz@@dByjmaxB{@@By1B|@@BLB}@@dB|iB~@@B|B@@tB~1B@@B~B@@dBimaxB@@B1B@@BKB@@@BB@@@BB@@@BB@@dBPhiB@@ BB@@dBiB@@BjB@@BB@@dBifB@@pBB@@ BB@@ @BB@@)@BB@@@BB@@dBBCB@@ BB@@dBiB@@BjB@@B1000B@@BB@@dBBCB@@ BB@@dBiB@@BjB@@BB@@dBAvgB@@ BB@@dBiB@@BjB@@BB@@dBjB@@BB@@tB1B@@BB@@dBjmaxB@@B1B@@BB@@dBiB@@BB@@tB1B@@BB@@dBimaxB@@B1B@@BJB@@dBkB@@BB@@tB0B@@BkmaxB@@B PhiB*@U '  / -,It is good practice to initialize a matrix. 7,9,B<,B:@W1B</,B<,B0@NormalArial B*@Uk ; x 0-@The updated matrix value is set equal to the average of the surrounding values UNLESS it is a boundary value. If it is a boundary value, the original value is put into the matrix instead of the average.79B<B:@W1B</B<B0@NormalArial B*@U )3  !! ! -@How "if" works: When the condition following the "if" is true, the first value (BC) is used, and if the condition is not true, the second value (Avg) is used. 79B<B:@W1B</B<B0@NormalArial B*@U[ # h 1PP-A0Further reading: 1. Numerical Recipes, The art of scientific computing, by Press, Flannery, Teukolsky and Vetterling (Cambridge Press, New York, 1985) Chapter 17.5. 2. Solutions to Laplace's equation using a spreadsheet on a personal computer, T. T. Crow, American Journal of Physics 55, 817 (1987). 720B<B8BB<2B8BkArialBB Root EntryI>'r !@ContentsOlePres000   [A5?DJWƿ ; b@?333d   ?$@  ?`ff`ff@  ?$@     ??@???@??@ ???D >@33@33 A A?NANI+00#C:\1.1*XQDocuments and SettingsDOCUME~1'1+Scott RobertsonSCOTTR~11+RecentRECENTPO :i+00#C:\1.1*XQDocuments andB@@BPhiB@B@U@ 2( )B@@ pB@@^BB@@@BB@@BB@@<@BB BBB< B CVSOleClientItem ࡱ> Root EntryI>'r !@ContentsOlePres000   3x?y=[w@*ðs? ; b@?333d   ?$@  ?`ff`ff@  ?$@     ??@???@??@ ???D# >@33@33 A A?69@@d8j:@@81;@@#<@@dNANIi?@@=@@@@d?j@A@@?1@B@@"2@C@@@D@@@@C@E@@d@DQ@F@@ @D@G@@d@Fi@H@@@Fj@I@@@CC@J@@@K@@d@Jj@L@@@J@M@@t@L1B@@BPhiB*@U;1{ H*@@-A.We did the second problem with 60 iterations instead of 30. Are the answers the same as for 30? The two problems are the same except the capacitor plates are turned 90 degrees. Let's turn our second answer 90 degrees (a matrix transpose operation from the matrix menu) and compare to the first answer.A7.9.B<.B:@W1B</.B<.B0@NormalArial B@B@UOz'B@@ pB@@BB@@@BB@@dBPhiB@@BB@@@BC@@@BC@@BC@@<@CC BCC< 1CVSOleClientItem ࡱ> Root EntryaB.dK8!@Contents( :5  -di i`xArialp p Arial88lUw%w<T1c$}{\s8\f1\fs36 \qc \sbasedon0 Subtitle;}}\plain\fs20 \pard\plain \s0\f0\fs20 \pard We did the second problem with 401 iterations. Are the answers the same as for 400? The two problems are the same except the capaC@@Crad@XC@@B3J0`C*@U!  -@There is a difference in the third decimal place for some of the values. So our answer is good to at least two decimal places. 79C<C :@W1C </C <C 0@NormalArial C *@U A88-ATry it: Show that a smoother plot is obtained with a larger number of grid points. Try it: Put a circular object in the center with a potential of +10. You could define a circular boundary of radius r using something like BCi,j := 10 if (i-5)2 + (j-5)2 < r2 and 1001 otherwise.7C<C8C mArialC@@ CiC@@@C>jCA@@C;CB@@dCAifCC@@pCACD@@ CCCE@@ @CDCF@@(@CECG@@@CFCH@@dCGBCCI@@ CGCJ@@dCIiCK@@CIjCL@@CF1000CM@@CECN@@K@CMCO@@CN1CP@@CM4CQ@@CD0CR@B@U 'CS@@ pCT@@ CSCU@@@CTCV@@dCUbCW@@ CUCX@@dCWiCY@@CWjCZ@@CTC[@@dCZaC\@@ CZC]@@dC\iC^@@C\jC_@B@U X;(C`@@ pCa@@ C`Cb@@@CaCc@@dCbcCd@@ CbCe@@dCdiCf@@CdjCg@@CaCh@@dCgaCi@@ CgCj@@dCiiCk@@CijCl@B@U)Cm@@ pCn@@ CmCo@@@CnCp@@dCodCq@@ CoCr@@dCqiCs@@CqjCt@@CnCu@@dCtaCv@@ CtCw@@dCviCx@@CvjCy@B@U 3$#"Cz@@ pC{@@ CzC|@@@C{C}@@dC|eC~@@ C|C@@dC~iC@@C~jC@@C{1C@B@Uh $#C@@ pC@@ CC@@@CC@@dCfC@@ CC@@dCiC@@CjC@@CC@@dCifC@@pCC@@ CC@@ @CC@@)@CC@@@CC@@dCBCC@@ CC@@dCiC@@CjC@@C1000C@@CC@@dCBCC@@ CC@@dCiC@@CjC@@C0C@B@U8 VG$C@@ pC@@ CC@@dCuC@@CfC@B@U %C@@ pC@@ CC@@dCrjC@@C0.99C@B@U4E@ C@@ pC@@ CC@@dCSC@@CC@@dCrelaxC@@pCC@@ CC@@ @CC@@ @CC@@ @CC@@ @CC@@ @CC@@ @CC@@dCaC@@CbC@@CcC@@CdC@@CeC@@CfC@@CuC@@CrjC*@U3C@!Y--Same result as using relax(a,a,a,a,e,f,f,rj).7-9-C<-C:@W1C</-C<-C0@NormalArial C@B@UPMB%C@@ pC@@CC@@@CC@@dCSC@@CC@@@CC@@@CC@@CC@@<@CC CCC< /QCVSOleClientItem ࡱ> Root EntryaB.dK8@!@Contents( :5  -di ii xArialp p Arial88lUw%w<T1c$- -  - -!,{ - -!0 - -!,C@@Crad@XC@@CC*@UK1X))`)`-AAbove the matrix transpose of S has been listed so it can be compared with the other listings. The transpose takes the horizontally oriented plates and makes them vertically oritented. Note that it agrees with the 60 iteration version and differs a little from the 30 iteration version. We did not have to tell the relax function how many iterations to use. The number of iterations is determined by the tolerance variable TOL which Mathcad sets to 0.001 if you do not specify a different value. 79""C<"C:@WC<C:@WCCC1C</C<C0@NormalArial C*@U1)) ) -@The equation below from the definition of the relax funtion will help you see why the values a,b,c,d,e and f have the values we have assigned to them:79C<C:@W1C</C<C0@NormalArial C@B@UHC@@ pC@@ CC@@@CC@@@CC@@dCeC@@ CC@@dCiC@@CjC@@CC@@dCAvgC@@ CC@@dCiC@@CjC@@CC@@@CC@@dCfC@@ CC@@dCiC@@CjC@@pCC@@CC@@@CC@@@CC@@@CC@@@CC@@dCaC@@ CC@@dCiC@@CjC@@CC@@dCPhiC@@ CC@@@CD@@dCiD@@C1D@@CjD@@CD@@@DD@@dDbD@@ DD@@dDiD@@DjD @@DD @@dD PhiD @@ D D @@@D D @@dD iD@@D 1D@@D jD@@CD@@@DD@@dDcD@@ DD@@dDiD@@DjD@@DD@@dDPhiD@@ DD@@dDiD@@DD@@dDjD@@D1D@@CD@@@DD@@dDdD @@ DD!@@dD iD"@@D jD#@@DD$@@dD#PhiD%@@ D#D&@@dD%iD'@@D%D(@@dD'jD)@@D'1D**@U 3K1+@+@-ADOur assignments make Avg equal to the average of the four adjacent values of Phi unless we are on an interior boundary point. In that case a,b,c and d are zero so the adjacent values are not averaged. f at an interior boundary is nonzero and is equal to boundary values. At these points the boundary value is placed in Avg. 7D9DD+