{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "R3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "Helv etica" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 } {PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 18 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 37 "SCALAR POTENTIAL OF A CON TINUOUS RING" }}{PARA 0 "" 0 "" {TEXT 257 18 "by Corinne Manogue" }} {PARA 0 "" 0 "" {TEXT -1 30 "Copyright 2006 Corinne Manogue" }}{PARA 0 "" 0 "" {TEXT -1 44 "(Demo version by TD in 2005 and CAM in 2006)" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "In this worksheet you will examine the scalar potential due to a continuous r ing of charge with radius " }{XPPEDIT 18 0 "R = 1;" "6#/%\"RG\"\"\"" } {TEXT -1 41 " and with linear charge density given by " }{XPPEDIT 18 0 "lambda/(4*Pi*epsilon) = 1;" "6#/*&%'lambdaG\"\"\"*(\"\"%F&%#PiGF&%( epsilonGF&!\"\",$F&F&" }{TEXT -1 3 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "restart:with(plots):re adlib(unassign):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "The scalar po tential is given by an elliptic integral, which Maple cannot do in gen eral." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "V=Int(1/sqrt(r^2+1 -2*r*cos(phi-phi2)+z^2),phiw=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "V=int(1/sqrt(r^2+1-2*r*cos(phi-phiw)+z^2),phiw=0..2*P i);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "But Maple can \"do\" the i ntegral for particular values of the coordinates:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 17 "r:=4;phi:=0;z:=2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "V=int(1/sqrt(r^2+1-2*r*cos(phi-phi2)+z^2),phi2=0.. 2*Pi);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "which it can evaluate n umerically:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 223 "It can also use this numerical e valuation to plot the potential, but this is a slow process. How can \+ we speed it up? First, let's exploit the symmetries of the problem. \+ Since the value of the potential is independent of " }{XPPEDIT 18 0 "p hi;" "6#%$phiG" }{TEXT -1 46 ", we do not need to plot it for all valu es of " }{XPPEDIT 18 0 "phi;" "6#%$phiG" }{TEXT -1 26 ". Instead, we \+ can choose " }{XPPEDIT 18 0 "phi = 0;" "6#/%$phiG\"\"!" }{TEXT -1 10 " and plot " }{XPPEDIT 18 0 "V" "6#%\"VG" }{TEXT -1 19 " as a color on \+ the " }{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "z;" "6#%\"zG" }{TEXT -1 7 " plane." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "unassign('r');unassign('phi');unassign('z');" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "V:=Int(1/sqrt(r^2-2*r*cos(ph i2)+1+z^2),\nphi2=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "plot3d(V,r=0..3,z=-1..1,\nnumpoints=10000,orientation=[-90,0],\ns hading=zhue,axes=boxed,style=patchcontour);" }}}{EXCHG }{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "or we can plot horizontal cross-sections (const ant values of " }{XPPEDIT 18 0 "z;" "6#%\"zG" }{TEXT -1 20 ") for the \+ potential:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "addcoords(zcy l,[z,r,phi],[r*cos(phi),r*sin(phi),z]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "z0:=0.1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "V:=Int(1/sqrt(r^2-2*r*cos(phi-phi2)+1+z0^2),\nphi2=0..2*Pi);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "plot3d(V,r=0..1.2,phi=0..2* Pi,\nnumpoints=1000,orientation=[90,0],\nshading=zhue,axes=boxed,style =patchcontour, coords=zcyl);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 3 0" 43 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }