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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 265 "" 0 "" 
{TEXT 257 114 "1999 Oregon State University              Capstones in \+
Physics: Electromagnetism                  Philip J. Siemen" }}{PARA 
4 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {TEXT -1 17 "METHOD OF IMA
GES\n" }}{PARA 0 "" 0 "" {TEXT -1 378 "\nThe uniqueness theorem for el
ectrostatics guarantees that there is only one potential function V(x,
y,z) that satisfies both Poisson's equation and a given set of boundar
y conditions.  Thus, if we discover some clever trick that allows us t
o find a function that satisfies these requirements, we have solved th
e problem.  One such trick is referred to as \"The Method of Images\"
\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "restart: with (linalg): with \+
(plots):" }}}{EXCHG {PARA 260 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 47 "A POINT CHARGE NEAR A G
ROUNDED CONDUCTING PLATE" }}{PARA 258 "" 0 "" {TEXT -1 214 " \n(ref. L
CL Example page 121)\n\n\nConsider a point charge +Q located a distanc
e z=+d from a grounded conducting plate located at z=0 in the x-y plan
e.  The conducting plate is oriented as shown in the following plot." 
}}{PARA 258 "" 0 "" {TEXT -1 2 "  " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
70 "plot3d(0,x=-2..2,y=-2..2,orientation=[-6,83],axes=boxed,shading=zh
ue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 189 "\nIf we want to find the
 potential above the plate, we need to solve Poisson's equation in the
 region z>0, with the single point charge Q at (x=0,y=0,z=1) subject t
o the boundary conditions:" }}{PARA 0 "" 0 "" {TEXT -1 116 "\n     (1)
  V=0 when z=0 since the conducting plate is grounded.\n     (2)  V go
es to zero far from the point charge  " }{TEXT 256 6 "( i.e." }{TEXT 
-1 97 " for x^2+y^2+z^2 >> d^2)\n\nFor comparison, let us examine the \+
following configuration of charges:\n" }}{PARA 261 "" 0 "" {TEXT -1 8 
"A DIPOLE" }}{PARA 0 "" 0 "" {TEXT -1 198 "\nRemove the conducting pla
te, and locate a charge +Q at z=+d and a charge -Q is at z=-d.  The el
ectric field of this dipole configuration (omitting the constant facto
r  Q/4piEpsilon0 ) is given by :" }{MPLTEXT 1 0 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "r1:=sqrt((x-0)^2+
(y-0)^2+(z-d)^2):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "E1:=Q*[x/(r1^3
),y/(r1^3),(z-d)/(r1^3)]:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "r2:=sqrt((x-
0)^2+(y-0)^2+(z+d)^2):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "E2:=-Q*[x
/(r2^3),y/(r2^3),(z+d)/(r2^3)]:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E12
:=E1+E2;" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "d:=1;" }}{PARA 0 "" 0 "" 
{TEXT -1 40 "\nPlot the electric field of the dipole: " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "fieldplot3d(s
ubs(Q=1,E12),x=-1.5..1.5,y=-1.5..1.5,z=-1.5..1.5,grid=[5,5,5],axes=box
ed,orientation=[-16,70],shading=zhue); " }}}{EXCHG {PARA 262 "" 0 "" 
{TEXT -1 0 "" }}{PARA 262 "" 0 "" {TEXT -1 18 "\nE-FIELD DIRECTION" }}
{PARA 262 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 291 "What
 is the direction of the electric field at the z=0 plane? (ie the plan
e formerly occupied by the plate).  Recall, the boundary condition on \+
the electric field requires that it be normal to the surface of the co
nductor. Changing the plot range and grid size will make it easier to \+
see.  \n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "fieldplot3d(subs(Q=1,E
12),x=-0.5...0.5,y=-0.5..0.5,z=-0.5..0.5,grid=[5,5,5],axes=boxed,orien
tation=[-16,70],shading=zhue); " }}}{EXCHG {PARA 263 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 9 "P
OTENTIAL" }}{PARA 0 "" 0 "" {TEXT -1 41 "\nWhat is the potential in th
e z=0 plane? " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 237 "             Hint, the plane formerly occupied by the pl
ate is equadistant from the two charges.\n\nRecall, the potential of a
 point charge is a scalar...so the potential of the dipole is just (om
itting the constant factor 1/4PiEpsilon0 ):" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "V12:=Q/sqrt(x^2+y^2+(z-d)
^2)-Q/sqrt(x^2+y^2+(z+d)^2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 196 "
\nSince the potential is zero in the z=0 plane, condition (1) is satis
fied.  Also note, the potential goes to zero at distances far from the
 point charge, thus, condition (2) is satisfied as well. " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 " From the uniquene
ss theorem, this solution must be THE correct solution.  " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "Therefore, the po
tential and electric field in the upper half plane are just those prod
uced by the dipole.  (Keep in mind the solution is only valid in the u
pper half plane.)\n \n" }}{PARA 264 "" 0 "" {TEXT -1 22 "INDUCED SURFA
CE CHARGE" }}{PARA 0 "" 0 "" {TEXT -1 384 "\nThe point charge Q causes
 a surface charge sigma to be induced on the conductor.  \n\nTo calcul
ate the surface charge density on the grounded plate, recall that E = \+
sigma/Epsilon0 at the surface of a conductor, and is directed outward.
  So, defining the unit vector to be in the +z direction, and reincorp
orating the constant Q/4PiEpsilon0, we find the surface charge density
 to be:  \n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "n:=[0,0,1];" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "sigma:=1/(4*Pi*epsilon0)*eps
ilon0*dotprod(subs(z=0,E12),n);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
161 "\nAs expected, the induced charge is negative (since the point ch
arge Q is positive) and is greatest near x=y=0.\n\nThe surface charge \+
is plotted below for Q = +1.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "p
lot3d(subs(Q=1,sigma),x=-2..2,y=-2..2,axes=boxed,shading=zhue);\n" }}}
{EXCHG {PARA 269 "" 0 "" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {TEXT -1 
21 "TOTAL INDUCED CHARGE\n" }}{PARA 0 "" 0 "" {TEXT -1 113 "To calcula
te the total induced charge, we simply integrate the induced surface c
harge sigma over the total area:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
70 "Qinduced:=int(int(sigma,x=-infinity..infinity),y=-infinity..infini
ty);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 268 "" 0 "" {TEXT -1 
31 "POINT CHARGE NEAR A DIELECTRIC\n" }}{PARA 272 "" 0 "" {TEXT -1 66 
"(References: Griffiths Sect. 4.4 Example 8;  LCL example pg. 213)\n" 
}}{PARA 0 "" 0 "" {TEXT -1 252 "\nNow, let us consider a somewhat more
 complicated situation.  Assume that the point charge Q now lies in ai
r close to a large block dielectric (ie make  the lower half (z<0) a d
ielectric).  Determine the electric field on both sides of the interfa
ce. \n" }}{PARA 270 "" 0 "" {TEXT -1 73 "\n***reference discussion in \+
class of Griffiths pg 183  or LCL pg 213***\n\n" }}{PARA 0 "" 0 "" 
{TEXT -1 44 "We will plot the resulting electric fields.\n" }}{PARA 
271 "" 0 "" {TEXT -1 30 "\nFIELD OUTSIDE THE DIELECTRIC\n" }}{PARA 0 "
" 0 "" {TEXT -1 75 "\nThe field outside the dielectric (i.e. above the
 dielectric) is given by:\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "E3:=
[Qprime*x/(r2^3),Qprime*y/(r2^3),Qprime*(z+d)/(r2^3)]:" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "E13:=E1+E3;" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "
\nwhere Q is the point charge and Q' is:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "Qprime:=-(epsilon[r]-1)/(eps
ilon[r]+1)*Q;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 128 "\nExamine the plot of this field (choosing Q = 1 and \+
epsilon-r = 5).  Note:  This solution is only valid outside the dielec
tric.\n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "fieldplot3d(subs(Q=1,ep
silon[r]=2,E13),x=-1.5..1.5,y=-1.5..1.5,z=-1.5..1.5,grid=[10,10,10],ax
es=boxed,orientation=[0,90],shading=zhue); " }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 41 "\nZoom in to take a look at the interface." }}{PARA 0 "
" 0 "" {TEXT -1 2 "  " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 133 "fieldplot
3d(subs(Q=1,epsilon[r]=2,E13),x=-0.5..0.5,y=-0.5..0.5,z=-0.5..0.5,grid
=[6,6,6],axes=boxed,orientation=[0,90],shading=zhue); " }}}{EXCHG 
{PARA 279 "" 0 "" {TEXT -1 0 "" }}{PARA 279 "" 0 "" {TEXT -1 31 "THE F
IELD INSIDE THE DIELECTRIC" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 137 "Inside the dielectric, there is no charge densit
y, so the field has no divergence.  It looks like the sources are outs
ide the dielectric." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 206 "In fact, the field inside the dielectric looks like th
e field there would be if there were no dielectric, but instead a poin
t charge, at the position of the actual, original point charge, but wi
th magnitude" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "Qdblprime:=2/(epsilon[r]+1)*Q;" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 21 "E4:=(Qdblprime/Q)*E1;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 67 "\nExamine the plot of this field (choose Q = 1 and epsi
lon-r = 5).  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 273 "" 0 "" 
{TEXT -1 57 "Note:  This solution is only valid inside the dielectric.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
135 "fieldplot3d(subs(Q=1,epsilon[r]=2,E4),x=-1.5..1.5,y=-1.5..1.5,z=-
1.5..1.5,grid=[10,10,10],axes=boxed,orientation=[0,90],shading=zhue); \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "\nZooming in on the interface
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
132 "fieldplot3d(subs(Q=1,epsilon[r]=2,E4),x=-0.5..0.5,y=-0.5..0.5,z=-
0.5..0.5,grid=[5,5,5],axes=boxed,orientation=[0,90],shading=zhue); " }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 274 "" 0 "" {TEXT -1 
62 "COMBINING THE CORRECT SOLUTIONS ABOVE AND BELOW THE INTERFACE\n" }
}{PARA 0 "" 0 "" {TEXT -1 31 "\nThe resulting total field is:\n" }}
{PARA 276 "" 0 "" {TEXT -1 60 "\nThe outside-the-dielectric solution i
n the upper half plane" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 91 "Eout:=subs(Q=1,epsilon[r]=5,[E13[1]*Heaviside(
z),E13[2]*Heaviside(z),E13[3]*Heaviside(z)]):" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 275 "" 0 "" {TEXT -1 58 "The inside-the-diele
ctric solution in the lower half plane" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "Ein:=subs(Q=1,epsilon[r]=5,[E4[
1]*Heaviside(-z),E4[2]*Heaviside(-z),E4[3]*Heaviside(-z)]):" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 277 "" 0 "" {TEXT -1 17 "add them together" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "Eactual:=Eout+Ein:
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 117 "fieldplot3d(Eactual,x=-1.5..1.5,y=-1.5..1.5,z=-1.5..
1.5,grid=[10,10,10],axes=boxed,orientation=[0,90],shading=zhue); " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 24 "Zo
om in on the interface" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 114 "fieldplot3d(Eactual,x=-1.5..1.5,y=-1.5..1.5,z
=-0.5..0.5,grid=[6,6,6],axes=boxed,orientation=[0,90],shading=zhue); \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "\nDON'T CLOSE THE WINDOW FOR \+
THE ABOVE FINAL RESULT.  INSTEAD," }}{PARA 0 "" 0 "" {TEXT -1 66 "\nCo
mpare this solution with that of the point charge in free space" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 119 "fi
eldplot3d(subs(Q=1,E1),x=-1.5..1.5,y=-1.5..1.5,z=-0.5..0.5,grid=[6,6,6
],axes=boxed,orientation=[0,90],shading=zhue); " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 278 "" 0 "
" {TEXT -1 23 "SURFACE CHARGE DENSITY\n" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 59 "The surface charge density is propor
tional to  Eout - Ein. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Ediff:=subs(z=0,E13)-subs(z=0,E4);
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 " Ediff:=simplify(Ediff);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 69 "A
s required by the boundary condition, there is only a z-component.  " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Can you
 see this result in Maple's expression?  HOMEWORK: SHOW THIS!" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "sig
ma:=dotprod(subs(Q=1,epsilon[r]=5,Ediff),n);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 54 "plot3d(sigma,x=-2..2,y=-2..2,axes=boxed,shading=
zhue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 44 "HOMEWORK:  What is the total induced charge?" }}{PARA 0 "
" 0 "" {TEXT -1 52 "Find the answer for an arbitrary value of epsilon-
r." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 267 "" 0 "" {TEXT -1 23 "v
ersion 12 January 1999" }}}}{MARK "28 2 0" 52 }{VIEWOPTS 1 1 0 1 1 
1803 }