{VERSION 2 3 "IBM INTEL NT" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "R3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "Helvetica" 1 12 255 0 0 1 2 1 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Fo nt 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 12 0 0 0 1 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "PHYSICS RESOURCE PACKETS P ROJECT File: edipole.MWS" }}{PARA 0 "" 0 "" {TEXT -1 94 "Rose-Hulman Institute of Technology \+ Author: mjm 2/26/96" }}{PARA 0 "" 0 "" {TEXT -1 97 " http://www.rose-hulman.edu/~moloney \+ Software: Maple V Release 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 375 "Along the axis of a finite dipole, the electri c field points one way in x=-a to +a and find out the the region betwe en the two charges, and it points the other way outside the charges. W hich field produces the greater effect? The one in between is stronger , but it doesn't last very long. To answer this question, we set up a \+ line of dipoles, and integrate along the line of " }}{PARA 0 "" 0 "" {TEXT -1 80 "the dipoles, although we do not run right over the top of any individual charge." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 137 "Set up a line of 4 dipoles in the y-direction and i ntegrate from x=-a to +a and find out the average electric field. Will it be + or - ? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 209 "A line of + charges at (0,1/8), (0,3/8), (0, -1/8), (0, \+ -3/8) and a line of - charges at x= +1/4. This gives 4 dipoles pointin g in -x direction, from y=-3/8 to y=+ 3/8 m. Distance is 1/4 in x and \+ y directions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 438 "Predict the sign of Ex before beginning. [It should be m ainly negative outside +/- charges and positive between the charges.] \+ Graph Ex from -2 to 2 and see Ex is negative (mostly) outside dipoles, and large positive between them. Who wins? Integrate Ex from -2 to 2 a nd see. \{Easy way to understand result is that V is positive for x<0 \+ since all + charges are closer, and V is negative for x> 1/4 because a ll negative charges are closer !\}" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "restart; with(plots): k:=9e9: q:=1 e-9:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Ex:=-k*q*(a-x)/((a- x)^2 + b^2)^(3/2); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#ExG,$*&,&%\" aG\"\"\"%\"xG!\"\"F),&*$F'\"\"#F)*$%\"bGF.F)#!\"$F.$!\"*\"\"!" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 55 " Ex at point (x,0) from a unit cha rge located at (a,b) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "Edipole:=s ubs(a=0,Ex)-subs(a=1/4,Ex);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(Edip oleG,&*&%\"xG\"\"\",&*$F'\"\"#F(*$%\"bGF+F(#!\"$F+$\"\"*\"\"!*&,&#F(\" \"%F(F'!\"\"F(,&*$F4F+F(F,F(F.F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "TotalDipoleEx:=0:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "for k from -3 to 3 by 2 do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "TotalDipoleEx:=TotalDipoleEx+subs (b=k/8,Edipole):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "plot(TotalDipoleEx,x=-1/2..1/2,titl e=`E_x of a finite dipole`);" }}{PARA 13 "" 1 "" {INLPLOT "6&-%'CURVES G6$7\\p7$$!1+++++++]!#;$!18J<)pd'**[!#97$$!1LLLe%G?y%F*$!1+gFH'*yG`F-7 $$!1mmT&esBf%F*$!1ss5qFEVdF-7$$!1LL$3s%3zVF*$!1k!*eQuwhiF-7$$!1LL$e/$Q kTF*$!1U)oPA8#[oF-7$$!1nmT5=q]RF*$!1],jKyi1vF-7$$!1LL3_>f_PF*$!1g]p#3^ Y>)F-7$$!1++vo1YZNF*$!1%p(HG['))**)F-7$$!1LL3-OJNLF*$!1P/d$HEc%**F-7$$ !1++v$*o%Q7$F*$!1gq&\\pTE5\"!#87$$!1nmm\"RFj!HF*$!1o164GnI7Fen7$$!1LL$ e4OZr#F*$!1['=.L'**f8Fen7$$!1+++v'\\!*\\#F*$!1-wa6g/F:Fen7$$!1+++DwZ#G #F*$!1!)*)pK/g?Fen7$$!1LL3-TC%)=F*$!1Fbn Ayo[@Fen7$$!1nmm\"4z)e;F*$!1?'4:n]ZU#Fen7$$!1nmmm`'zY\"F*$!1pJ\"*HVKgE Fen7$$!1++v=t)eC\"F*$!1O:?wGN&*GFen7$$!1M$3x'*)fZ6F*$!1gjJ#*p:mHFen7$$ !1nmm;1J\\5F*$!1&)oD+Kr,IFen7$$!1.](=n;R&**!#<$!1=8?)oG8+$Fen7$$!1NL3x rs9%*F`r$!1vm*)RBa$)HFen7$$!1m;H#oPb())F`r$!1lZQ/3[XHFen7$$!1)***\\(=[ jL)F`r$!1TnHD%zS)GFen7$$!1*****\\Pw%4tF`r$!11y6ad,\"p#Fen7$$!1****\\iX g#G'F`r$!1oJ96QTzBFen7$$!1KL3_:<6_F`r$!1>/V$)RS1>Fen7$$!1lmmT&Q(RTF`r$ !1,d#GlVuE\"Fen7$$!1mm\"HdGe:$F`r$!17AnO;*\\K&F-7$$!1nm;/'=><#F`r$\"1] -mC=8CLF-7$$!1,+D\"yQ16\"F`r$\"1?@Vd#R0Q\"Fen7$$!1EMLLe*e$\\!#>$\"1Fp[ pQ_*\\#Fen7$$\"1imT5D,`5F`r$\"1$Gihd@An$Fen7$$\"1em;zRQb@F`r$\"1e9m%Gf %)y%Fen7$$\"1FLL$e,]6$F`r$\"14`o^$p)pcFen7$$\"1&***\\(=>Y2%F`r$\"1JK2' Hz0W'Fen7$$\"1GLe*[K56&F`r$\"1\"R2KU!*\\8(Fen7$$\"1hmm\"zXu9'F`r$\"1GA Bv!fjo(Fen7$$\"1HL$e9i\"=sF`r$\"1y_To49=\")Fen7$$\"1'******\\y))G)F`r$ \"1d>L?6vH%)Fen7$$\"1$***\\ibOO$*F`r$\"1$fp\\$>iR')Fen7$$\"1****\\i_QQ 5F*$\"1#=(oi?qu()Fen7$$\"1*\\(=U,1*3\"F*$\"1fIIR?#y\"))Fen7$$\"1**\\(= -N(R6F*$\"1B,t1'Q#[))Fen7$$\"1*\\i:!*4/>\"F*$\"1iP\"e/oo'))Fen7$$\"1** *\\7y%3T7F*$\"1h:!Qf*Gu))Fen7$$\"1*\\7`f]tH\"F*$\"1_)3?]s'p))Fen7$$\"1 **\\P4kh`8F*$\"1:XF8(Q8&))Fen7$$\"1*\\PMA#))49F*$\"1!>rx!He=))Fen7$$\" 1****\\P![hY\"F*$\"1)fl(*)e=q()Fen7$$\"1mmT5FEn:F*$\"1^%=J@j\"Q')Fen7$ $\"1LLL$Qx$o;F*$\"1di/.PLO%)Fen7$$\"1+++v.I%)=F*$\"1e%)QqU!3p(Fen7$$\" 1L$ek`H@)>F*$\"1ju$[E\"GvrFen7$$\"1mm\"zpe*z?F*$\"1(\\_M)RiKlFen7$$\"1 L$e9\"=\"p=#F*$\"1W=)=:vNo&Fen7$$\"1+++D\\'QH#F*$\"1iC>[h'pp%Fen7$$\"1 m;zp%*\\%R#F*$\"14m6W(4Vn$Fen7$$\"1KLe9S8&\\#F*$\"1]d,xZO/EFen7$$\"1mm T5hK+EF*$\"1`iy'y%[\"\\\"Fen7$$\"1***\\i?=bq#F*$\"1>4v**)>dU%F-7$$\"1m ;H2FO3GF*$!1B:!)o09KZF-7$$\"1LLL3s?6HF*$!16h\"4e#p[7Fen7$$\"1m;zpe()=I F*$!12acA5x%*=Fen7$$\"1++DJXaEJF*$!1WEh<'zIP#Fen7$$\"1L$e*[ACIKF*$!1?J sdXI*o#Fen7$$\"1nmmm*RRL$F*$!1F.cL&)\\%)GFen7$$\"1n;a8H'pQ$F*$!1tb&==S \\%HFen7$$\"1nmTge)*RMF*$!1=!)p(GyF)HFen7$$\"1n;H2)3I\\$F*$!1J6)y_I4+$ Fen7$$\"1mm;a<.YNF*$!1[<$*4n8-IFen7$$\"1**\\PM&*>^OF*$!1n/1V(3T'HFen7$ $\"1LLe9tOcPF*$!1frG]O@')GFen7$$\"1+++]Qk\\RF*$!1Fka/*f?o#Fen7$$\"1LL$ 3dg6<%F*$!1;)[y)QS4CFen7$$\"1mmmmxGpVF*$!1-t?hZXm@Fen7$$\"1++D\"oK0e%F *$!1$4h&Hy>E>Fen7$$\"1++v=5s#y%F*$!1i,o]uO? " 0 "" {MPLTEXT 1 0 36 "evalf(int(TotalDi poleEx,x=-20..20));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"*g%Q*\\%!#5 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Integral of Ex is positive; \+ stronger positive part in middle wins; " }}{PARA 0 "" 0 "" {TEXT -1 204 "Dipoles like these would be created (or aligned)by an electric fi eld pointing in the -x direction, so the field of the dipoles opposes \+ the field that created it. This means in capacitors when dipoles are \+ " }}{PARA 0 "" 0 "" {TEXT -1 100 "present in the dielectric, the elect ric field is smaller in magnitude than without the dielectric. " }} {PARA 0 "" 0 "" {TEXT -1 102 "Smaller electric field means smaller pot ential difference, which (for the same charges on the plates) " }} {PARA 0 "" 0 "" {TEXT -1 21 "a larger capacitance." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "Yes, but what if we make \+ the dipole distances tiny? (substitute a=1/100 instead of 1/4)" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "DipoleEx:=0:" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 26 "for k from -3 to 3 by 2 do" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 61 "DipoleEx:=DipoleEx+subs(a=0,b=k/8,Ex)-subs(a=1/100, b=k/8,Ex):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "od:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(DipoleEx,x=-1/2..1/2);" }}{PARA 13 " " 1 "" {INLPLOT "6%-%'CURVESG6$7gp7$$!1+++++++]!#;$!1\\*4?0%*4,$!#:7$$ !1LLLe%G?y%F*$!1I&)=PZe4LF-7$$!1mmT&esBf%F*$!1!*oFs$z?g$F-7$$!1LL$3s%3 zVF*$!1etmYJ^tRF-7$$!1LL$e/$QkTF*$!1iwz))ow+WF-7$$!1nmT5=q]RF*$!1#o!\\ DA3*)[F-7$$!1LL3_>f_PF*$!1X>+GL^3aF-7$$!1++vo1YZNF*$!1%\\*ykO_EgF-7$$! 1LL3-OJNLF*$!1baqpnDnnF-7$$!1++v$*o%Q7$F*$!1Z@M&ofqi(F-7$$!1nmm\"RFj!H F*$!1&R!4R8Ff')F-7$$!1LL$e4OZr#F*$!1C>([vD)3(*F-7$$!1+++v'\\!*\\#F*$!1 C#e7\"Q)e5\"!#97$$!1+++DwZ#G#F*$!1;`y&Hb%e7Fdo7$$!1+++D.xt?F*$!1P-Oh$o gT\"Fdo7$$!1LL3-TC%)=F*$!1bm^+S.c:Fdo7$$!1+](ofh:x\"F*$!1Nf68_16kA#pbeFdo7$$!1mm\"HdGe:$Fis$\"1aCY4r0W lFdo7$$!1n;a)etQm#Fis$\"1!\\Z]%)=>?(Fdo7$$!1nm;/'=><#Fis$\"1*)\\IT1$3 \"yFdo7$$!1,+D\"yQ16\"Fis$\"1'>v`\"fEr))Fdo7$$!1EMLLe*e$\\!#>$\"1'ze!y ^eb%*Fdo7$$\"1H:/EvuV))Fex$\"1;iZ')QU!\\*Fdo7$$\"1[;a)3RBE#!#=$\"1bM]t _M:&*Fdo7$$\"1W\"zWU..k$F`y$\"1PHBaYFI&*Fdo7$$\"1RmTgxE=]F`y$\"1p!3A?n ^`*Fdo7$$\"1NTN'4KiR'F`y$\"1<(\\K@3+`*Fdo7$$\"1I;HKk>uxF`y$\"18[^ *Fdo7$$\"1E\"H#o2;_\"*F`y$\"1@+4hxi*[*Fdo7$$\"1imT5D,`5Fis$\"1QQd\"RFX X*Fdo7$$\"1g;zW#)>/;Fis$\"1#eFf9,v@*Fdo7$$\"1em;zRQb@Fis$\"1h')oM1>N)) Fdo7$$\"1$**\\7y#>NEFis$\"1wSJSR](R)Fdo7$$\"1FLL$e,]6$Fis$\"1C`&\\Lps( yFdo7$$\"1hmT&Q5[f$Fis$\"1!Q)y8Py!H(Fdo7$$\"1&***\\(=>Y2%Fis$\"1c.jg6C bmFdo7$$\"1h;aQe#Gf%Fis$\"1,s\"[hZN$fFdo7$$\"1GLe*[K56&Fis$\"1\"*e\\=& o_>&Fdo7$$\"1&*\\iS\"R#HcFis$\"1?5Fvs,fWFdo7$$\"1hmm\"zXu9'Fis$\"1&*oP @fzSPFdo7$$\"1&**\\(oR!Go'Fis$\"1pe(>Gi:.$Fdo7$$\"1HL$e9i\"=sFis$\"1k9 '4^YkO#Fdo7$$\"1jm\"HK?Nv(Fis$\"1x!\\&3w3`\"Fdo7$$\"1$***\\ibOO$*Fis$\"1UIO(H?Bv#F-7$$\"1****\\i_QQ5F *$!1x%eo$RD'H%F-7$$\"1**\\(=-N(R6F*$!1]*pr]nWG*F-7$$\"1***\\7y%3T7F*$! 1D)RV)3Ew7Fdo7$$\"1**\\P4kh`8F*$!1$*\\K\\Ry@:Fdo7$$\"1****\\P![hY\"F*$ !10yo8*>`l\"Fdo7$$\"1K$eRP0n^\"F*$!1PSJ$[Auo\"Fdo7$$\"1mmT5FEn:F*$!1%z 44RMdq\"Fdo7$$\"1**\\(o/?yh\"F*$!16'[iT%R7rJNx;Fdo7$$\"1+++v.I%)=F*$!1owM[&H7i\"F do7$$\"1mm\"zpe*z?F*$!1>*[J$3(o[\"Fdo7$$\"1+++D\\'QH#F*$!1Rx^eroC8Fdo7 $$\"1KLe9S8&\\#F*$!1S0.&*[;x6Fdo7$$\"1***\\i?=bq#F*$!1*=sr$))*p.\"Fdo7 $$\"1LLL3s?6HF*$!1w?urY5j\"*F-7$$\"1++DJXaEJF*$!1`*[ec$)*o!)F-7$$\"1nm mm*RRL$F*$!1$H%HY$pI;(F-7$$\"1mm;a<.YNF*$!1!ob47'pmjF-7$$\"1LLe9tOcPF* $!1!*=ASsT(o&F-7$$\"1+++]Qk\\RF*$!1T[#*\\1KX^F-7$$\"1LL$3dg6<%F*$!1>3D \"*fS0YF-7$$\"1mmmmxGpVF*$!1$[35%ft%=%F-7$$\"1++D\"oK0e%F*$!1F!QPE#z!z $F-7$$\"1++v=5s#y%F*$!1#y`,*4keMF-7$$\"1+++++++]F*$!1=)fr%>OVJF--%'COL OURG6&%$RGBG$\"#5!\"\"\"\"!Fehl-%+AXESLABELSG6$%\"xG%!G-%%VIEWG6$;$!++ +++]!#5$\"+++++]Fail%(DEFAULTG" 2 437 437 437 2 0 1 0 2 9 0 4 2 1.000000 45.000000 45.000000 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 74 32187 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " evalf(int(DipoleEx,x=-20..20));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\" )JZ*z\"!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "Nope, smaller, but \+ still positive. Dipole field always opposes applied field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "Final note: - \+ integral of Ex dx = V(2) - V(1). Since \{integral Ex dx\} is positiv e, V(1) > V(2). The potential" }}{PARA 0 "" 0 "" {TEXT -1 113 "at the \+ start is higher than the potential at the end. This is because all + c harges are closer to the start point" }}{PARA 0 "" 0 "" {TEXT -1 19 "t han the - charges." }}}}{MARK "10 0 0" 23 }{VIEWOPTS 1 1 0 1 1 1803 }