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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 10 "Gauss' law" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 275 "In electrostatics w
e deal with positive and negative charges that serve as origins and te
rminators of electric field lines. If we consider closed surfaces in s
pace and count the number of (imaginary) field lines that enter and le
ave the surface, one of three cases can occur:" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "(i) more field lines leav
e than enter the surface" }}{PARA 0 "" 0 "" {TEXT -1 65 "(ii) the same
 number of field lines enters and leaves the surface" }}{PARA 0 "" 0 "
" {TEXT -1 52 "(iii) more field lines enter than leave the surface." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "Clearly \+
the cases correspond to the inclusion of" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "(i) positive net charge" }}{PARA 
0 "" 0 "" {TEXT -1 18 "(ii) no net charge" }}{PARA 0 "" 0 "" {TEXT -1 
26 "(iii) negative net charge." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 144 "Note that case 2 does not mean that no c
harge is included, it could be an equal number of positive and negativ
e charges enclosed by the surface." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 370 "In Gauss' law this observation is quanti
fied to determine the electric field generated by some charge distribu
tion. The total flux is given as a directed surface integral of the el
ectric field vector. Using Coulomb's law for the electric field of a p
oint charge and the superposition principle to generalize to an arbitr
ary charge configuration one arrives at the result" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "int(E dA, closed surface)
 = Q/epsilon" }}{PARA 0 "" 0 "" {TEXT -1 47 "(epsilon=permittivity=8.8
54E-12 C^2 N^-1 m^-2 )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 141 "Gauss' law is of practical use to calculate the ele
ctric field if symmetries guarantee that the field is constant on the \+
surface of interest." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 257 8 "Examples" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 264 16 "1) a charged rod" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 588 "This problem has axial (cylindrical) sy
mmetry. The electric field lines extend radially in a plane perpendicu
lar to the rod. Thus, the field has zero component along the axis defi
ned by the rod. We can make use of a cylinder as a Gaussian surface an
d will obtain zero contributions to the flux from the top and the bott
om (surface vector perpendicular to the electric field vector). The cy
linder is centered on the rod, thus, the radially emanating field line
s are aligned with the surface vector on the side surface of the cylin
der, and the field vector has a constant magnitude E there." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Given the  radi
us " }{TEXT 262 1 "r" }{TEXT -1 12 " and height " }{TEXT 261 1 "h" }
{TEXT -1 40 " of the cylinder we obtain for the flux:" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Phi:=2*P
i*h*r*E;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$PhiG,$**%#PiG\"\"\"%\"h
GF(%\"rGF(%\"EGF(\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The net
 charge " }{TEXT 258 1 "q" }{TEXT -1 71 " on the rod is distributed un
iformly, i.e., the linear density lambda =" }{TEXT 260 2 " q" }{TEXT 
-1 1 "/" }{TEXT 259 1 "h" }{TEXT -1 13 " is constant." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "q:=lambda*h;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"qG*&%'lambdaG\"\"\"%\"hGF'" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 21 "Gauss:=Phi=q/epsilon;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&GaussG/,$**%#PiG\"\"\"%\"hGF)%\"rGF)%\"EGF)\"\"#*&*&
%'lambdaGF)F*\"\"\"F1%(epsilonG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "Eofr:=solve(Gauss,E);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%EofrG,$*&%'lambdaG\"\"\"*(%#PiG\"\"\"%\"rG\"\"\"%(epsilonG\"
\"\"!\"\"#\"\"\"\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "We found
 the expected result that the electric field decreases as 1/" }{TEXT 
263 1 "r" }{TEXT -1 206 " with the radial distance from the rod. Gauss
' law was useful, since the rod's symmetry suggested to use a cylinder
 as a Gaussian surface, and thus the surface integral could be calcula
ted in a trivial way." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT 265 60 "2) the electric field of a uniformly charged spherica
l shell" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
197 "To obtain the field outside the shell we use a sphere that surrou
nds the charged shell. In analogy with the previous example we can use
 the constancy of the field on the Gaussian surface with area " }
{TEXT 19 8 "4*Pi*r^2" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "Phi:=E*4*Pi*r^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%$PhiG,$*(%\"EG\"\"\"%#PiGF()%\"rG\"\"#\"\"\"\"\"%" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 18 "The entire charge " }{TEXT 267 1 "Q" }{TEXT -1 
49 " is contained on the spherical shell with radius " }{TEXT 268 1 "R
" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Gauss:=
Phi=Q/epsilon;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&GaussG/,$*(%\"EG
\"\"\"%#PiGF))%\"rG\"\"#\"\"\"\"\"%*&%\"QGF.%(epsilonG!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Eofr:=solve(Gauss,E);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%EofrG,$*&%\"QG\"\"\"*(%#PiG\"\"\")%
\"rG\"\"#F(%(epsilonG\"\"\"!\"\"#\"\"\"\"\"%" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 10 "Note that " }{TEXT 266 1 "r" }{TEXT -1 73 " is the sphe
rical radial distance and that this result is valid only for " }{TEXT 
270 2 "r " }{TEXT -1 2 "> " }{TEXT 269 1 "R" }{TEXT -1 36 ", i.e., out
side the spherical shell." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 71 "Inside the spherical shell no charge is enclosed a
nd we must find that " }{TEXT 271 3 "Q' " }{TEXT -1 10 "= 0, i.e. " }
{TEXT 272 1 "E" }{TEXT -1 152 "=0. Based on forces the result can be u
nderstood as the cancellation of contributions from Coulomb's law pull
ing in all directions with equal magnitude." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 74 "3) a solid non-conducting (in
sulating) uniformly charged sphere of radius " }{TEXT 273 1 "R" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "Outside t
he sphere the result is identical to the previous one for the spherica
l shell." }}{PARA 0 "" 0 "" {TEXT -1 53 "Inside the sphere we use a Ga
ussian sphere of radius " }{TEXT 274 1 "r" }{TEXT -1 30 " and realize \+
that it encloses " }{TEXT 275 2 "Q'" }{TEXT -1 16 ", a fraction of " }
{TEXT 276 1 "Q" }{TEXT -1 93 " that changes with radius. The charges a
re held in place, since the sphere is not conducting." }}{PARA 0 "" 0 
"" {TEXT -1 81 "The constant charge density is given as total charge d
ivided by the total volume:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "rho:=Q/(4/3*Pi*R^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$rhoG,$*
&%\"QG\"\"\"*&%#PiG\"\"\")%\"RG\"\"$F(!\"\"#\"\"$\"\"%" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 88 "We equate the constant charge density to \+
an expression that leads to the partial charge " }{TEXT 278 2 "Q'" }
{TEXT -1 2 " (" }{TEXT 19 3 "Qpr" }{TEXT -1 33 ") enclosed by a sphere
 of radius " }{TEXT 277 1 "r" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 25 "eq:=rho=Qpr/(4/3*Pi*r^3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#eqG/,$*&%\"QG\"\"\"*&%#PiG\"\"\")%\"RG\"\"$F)!\"\"#
\"\"$\"\"%,$*&%$QprGF)*&F+\"\"\")%\"rG\"\"$F)F0F1" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 19 "Qpr:=solve(eq,Qpr);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$QprG*&*&%\"QG\"\"\")%\"rG\"\"$\"\"\"F,*$)%\"RG\"\"$F
,!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Gauss:=Phi=Qpr/ep
silon;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&GaussG/,$*(%\"EG\"\"\"%#P
iGF))%\"rG\"\"#\"\"\"\"\"%*&*&%\"QGF))F,\"\"$F.F.*&)%\"RG\"\"$F.%(epsi
lonG\"\"\"!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Thus, we have \+
the " }{TEXT 279 1 "E" }{TEXT -1 37 " field outside the charged sphere
 as " }{TEXT 19 4 "Eofr" }{TEXT -1 14 ", and inside (" }{TEXT 19 3 "Ei
n" }{TEXT -1 17 ") as given below:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "Ein:=solve(Gauss,E);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%$EinG,$*&*&%\"QG\"\"\"%\"rGF)\"\"\"*(%#PiG\"\"\")%\"RG\"\"$F+%(ep
silonG\"\"\"!\"\"#F)\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "At t
he surface they do match:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
30 "subs(r=R,Ein); subs(r=R,Eofr);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
,$*&%\"QG\"\"\"*()%\"RG\"\"#F&%#PiG\"\"\"%(epsilonG\"\"\"!\"\"#\"\"\"
\"\"%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%\"QG\"\"\"*()%\"RG\"\"#F
&%#PiG\"\"\"%(epsilonG\"\"\"!\"\"#\"\"\"\"\"%" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 56 "We graph the electric field after a choice of constant
s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "P1:=plot(subs(epsilon
=1,R=1,Q=1,Ein),r=0..1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 
"P2:=plot(subs(epsilon=1,R=1,Q=1,Eofr),r=1..5):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "display(\{P1,P2\});" }}{PARA 13 "" 1 "" {GLPLOT2D 
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "The electrostatic potential is sph
erically symmetric (depends on " }{TEXT 280 1 "r" }{TEXT -1 47 " only)
 and can be obtained by integration over " }{TEXT 281 2 "r " }{TEXT 
-1 47 "(in general a line integral would be required):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Vin:=-int(Ein,r);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%$VinG,$*&*&%\"QG\"\"\")%\"rG\"\"#\"\"\"F-*(%#PiG\"
\"\")%\"RG\"\"$F-%(epsilonG\"\"\"!\"\"#!\"\"\"\")" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 19 "Vofr:=-int(Eofr,r);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%VofrG,$*&%\"QG\"\"\"*(%#PiG\"\"\"%(epsilonG\"\"\"%\"
rG\"\"\"!\"\"#\"\"\"\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
45 "P3:=plot(subs(epsilon=1,R=1,Q=1,Vin),r=0..1):" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 46 "P4:=plot(subs(epsilon=1,R=1,Q=1,Vofr),r=1..5
):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(\{P3,P4\});" 
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118 "The integration constant should be adjusted for the integration o
f the inside part to match the potential function at " }{TEXT 282 2 "r
 " }{TEXT -1 1 "=" }{TEXT 283 2 " R" }{TEXT -1 65 " (with continuous d
erivative). However, the second derivative of " }{TEXT 284 1 "V" }
{TEXT -1 1 "(" }{TEXT 285 1 "r" }{TEXT -1 27 ") remains discontinuous \+
at " }{TEXT 286 2 "r " }{TEXT -1 2 "= " }{TEXT 287 1 "R" }{TEXT -1 8 "
, since " }{TEXT 288 2 "E'" }{TEXT -1 1 "(" }{TEXT 289 1 "R" }{TEXT 
-1 85 ") is not defined. An example where the constants have been calc
ulated is given below." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 316 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 14 "Use the
 large-" }{TEXT 318 1 "r" }{TEXT -1 44 " part obtained from definite i
ntegration of " }{TEXT 19 4 "Eofr" }{TEXT -1 47 " as a start, and use \+
definite integration from " }{TEXT 319 1 "R" }{TEXT -1 186 " inwards t
o find the potential as a continuous function which vanishes at infini
ty. Note that the inside part of the potential will have to be shifted
 by the potential value obtained at " }{TEXT 321 1 "r" }{TEXT -1 1 "=
" }{TEXT 320 1 "R" }{TEXT -1 6 " from " }{TEXT 19 4 "Vofr" }{TEXT -1 
2 ". " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1082 "Gauss' law permits us \+
to determine electric fields for simple charge distributions with a hi
gh degree of symmetry so that the choice of Gaussian surfaces becomes \+
obvious. It has also a great practical use for the understanding of ch
arge distributions in conductors. In particular it follows directly th
at excess charges in a conductor have to reside on the outside surface
. This follows as the presence of field lines inside the conductor is \+
impossible: a conductor contains many freely moving charges that would
 move along the field lines until the field would be compensated. Thus
, field lines can originate only from the surface of the conductor. A \+
Gaussian surface located just underneath the surface of the object wou
ld contain zero flux, i.e., zero net charge. Only as the Gaussian surf
ace moves infinitesimally outside the object's surface does the situat
ion change. Thus, one introduces a surface charge density to describe \+
the distribution of charge over the surface. It is possible to derive \+
that the electric field near the surface is proportional to the surfac
e density." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 127 "We conclude with a slightly less trivial example of an electri
c field associated with a nonuniformly charged insulating sphere:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "The charg
e is distributed uniformly up to " }{TEXT 291 2 "r " }{TEXT -1 1 "=" }
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o at " }{TEXT 294 1 "r" }{TEXT -1 2 " =" }{TEXT 293 2 " R" }{TEXT -1 
4 " = 4" }{TEXT 317 1 "a" }{TEXT -1 23 "/3. The region between " }
{TEXT 322 1 "a" }{TEXT -1 6 " and 4" }{TEXT 323 1 "a" }{TEXT -1 93 "/3
 can be considered as a surface skin region where the charge density d
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=FbuF(7$$\"1AAAZrug=FbuF(7$$\"1++voH5v=FbuF(7$$\"1666J$H*))=FbuF(7$$\"
166h$yoI!>FbuF(7$$\"1cbIa64<>FbuF(7$$\"1nmmci(*H>FbuF(7$$\"1cb0QSuW>Fb
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F(F(Fet-%+AXESLABELSG6$Q\"r6\"%!G-%%VIEWG6$;F(F_]m%(DEFAULTG" 1 2 0 1 
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{TEXT -1 240 "It is of interest to draw the charge density contributio
n in such a way that one understands how the amount of accumulated cha
rge increases as one steps from the inside out. To understand the amou
nt contained in a spherical shell at radius " }{TEXT 324 1 "r" }{TEXT 
-1 23 ", i.e., in the window [" }{TEXT 325 1 "r" }{TEXT -1 1 "," }
{TEXT 326 1 "r" }{TEXT -1 2 "+d" }{TEXT 327 1 "r" }{TEXT -1 46 "] it i
s helpful to include the volume element:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "P1:=plot(r^2*rho0,r=0..a):" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 29 "P2:=plot(r^2*rho,r=a..4/3*a):" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 25 "P3:=plot(0,r=4/3*a..2*a):" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 20 "display(\{P1,P2,P3\});" }}{PARA 13 "" 1 "
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LABELSG6$Q\"r6\"%!G-%%VIEWG6$;F(F_cm%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 
"The flux for  " }{TEXT 299 1 "r" }{TEXT -1 3 " < " }{TEXT 298 1 "a" }
{TEXT -1 9 " and for " }{TEXT 297 1 "a" }{TEXT -1 2 " <" }{TEXT 296 2 
" r" }{TEXT -1 4 " < 4" }{TEXT 295 1 "a" }{TEXT -1 87 "/3 has to be de
termined by calculating the amount of enclosed charge. The total charg
e " }{TEXT 303 1 "Q" }{TEXT -1 13 " is split as " }{TEXT 302 1 "Q" }
{TEXT -1 2 " =" }{TEXT 301 2 " Q" }{TEXT -1 3 "1 +" }{TEXT 300 2 " Q" 
}{TEXT -1 26 "2 between the two regions:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "For 0 <" }{TEXT 305 2 " r" }{TEXT 
-1 2 " <" }{TEXT 304 2 " a" }{TEXT -1 34 " we have a result that depen
ds on " }{TEXT 328 1 "r" }{TEXT -1 68 "-cubed (which is obvious after \+
integrating the r^2 times a constant)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "Q1:=rho0*(4/3*Pi*a^3);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#Q1G,$%#PiG#\"\"%\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 23 "Q1p:=rho0*(4/3*Pi*r^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$
Q1pG,$*&%#PiG\"\"\")%\"rG\"\"$\"\"\"#\"\"%F+" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 136 "Up to the factor epsilon (permittivity) this constitut
es the flux in the innermost part, i.e., the flux grows as the cube of
 the radius." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 113 "To determine the total charge contained in the region with lin
early decreasing density we calculate the integral:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 33 "Q2:=4*Pi*int(r^2*rho,r=a..4/3*a);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#Q2G,$%#PiG#\"#n\"#\")" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 61 "The flux (from the decreasing density alo
ne, i.e., excluding " }{TEXT 307 1 "Q" }{TEXT -1 57 "1) is calculated \+
by terminating the integral at a radius " }{TEXT 306 1 "r" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "Q2p:=4*Pi*subs(rp=r,int(r^2*rho,r=a..rp));" }}{PARA 
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{MPLTEXT 1 0 21 "P4:=plot(Q1p,r=0..a):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "P5:=plot(Q1+Q2p,r=a..4/3*a):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 29 "P6:=plot(Q1+Q2,r=4/3*a..2*a):" }}}{EXCHG {PARA 
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$\"166')z0L0=F^uFjjl7$$\"1+++&*4f>=F^uFjjl7$$\"1cbIM*3I$=F^uFjjl7$$\"1
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))=F^uFjjl7$$\"166h$yoI!>F^uFjjl7$$\"1cbIa64<>F^uFjjl7$$\"1nmmci(*H>F^
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45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "We see that the fl
ux continues smoothly at " }{TEXT 332 1 "r" }{TEXT -1 1 "=" }{TEXT 
331 1 "a" }{TEXT -1 26 ", and becomes constant at " }{TEXT 330 1 "r" }
{TEXT -1 2 "=4" }{TEXT 329 1 "a" }{TEXT -1 136 "/3, the point from whi
ch on no additional charges are included when the probe sphere increas
es. Now we can calculate the electric field." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "In region I (0 <" }{TEXT 309 2 
" r" }{TEXT -1 2 " <" }{TEXT 308 2 " a" }{TEXT -1 2 "):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Phi:=E*4*Pi*r^2;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$PhiG,$*(%\"EG\"\"\"%#PiGF()%\"rG\"\"#\"\"\"\"\"%" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Gauss:=Phi=Q1p/epsilon;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&GaussG/,$*(%\"EG\"\"\"%#PiGF))%\"
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "E1:=solve(Gauss,E);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E1G,$*&%\"rG\"\"\"%(epsilonG!\"\"#
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{TEXT 310 1 "a" }{TEXT -1 3 " < " }{TEXT 311 1 "r" }{TEXT -1 4 " < 4" 
}{TEXT 312 1 "a" }{TEXT -1 4 "/3):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "Gauss:=Phi=(Q1+Q2p)/epsilon;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&GaussG/,$*(%\"EG\"\"\"%#PiGF))%\"rG\"\"#\"\"\"\"\"%*
&,&F*#F/\"\"$*&F*F.,(*$)F,F/F.#!\"$F/#!\"(\"#7F)*$)F,F3F.F2F)F/F.%(eps
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\"rG\"\"%\"\"\"\"\"**$)F,F(F.!#;F.*&)F,\"\"#F.%(epsilonG\"\"\"!\"\"#!
\"\"\"#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "and in region 3 (" }
{TEXT 314 1 "r" }{TEXT -1 4 " > 4" }{TEXT 313 1 "a" }{TEXT -1 4 "/3):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Gauss:=Phi=(Q1+Q2)/epsi
lon;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&GaussG/,$*(%\"EG\"\"\"%#PiG
F))%\"rG\"\"#\"\"\"\"\"%,$*&F*F.%(epsilonG!\"\"#\"$v\"\"#\")" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "E3:=solve(Gauss,E);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#E3G,$*&\"\"\"F'*&)%\"rG\"\"#F'%(eps
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or graphing we set epsilon=1, i.e., we choose our own units:" }}}
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1 "" {XPPMATH 20 "6#>%(epsilonG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
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d then turns around to decrease while the density is decreasing linear
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" dependence outside the charged sphere. The skin area of the charge d
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338 11 "Exercise 2:" }}{PARA 0 "" 0 "" {TEXT -1 102 "Construct a radia
l density of your own choice with regions where the density grows as a
 parabola from " }{TEXT 337 1 "r" }{TEXT -1 6 "=0 to " }{TEXT 335 1 "r
" }{TEXT -1 1 "=" }{TEXT 336 1 "a" }{TEXT -1 53 ", where it has a maxi
mum and then falls to vanish at " }{TEXT 334 1 "r" }{TEXT -1 1 "=" }
{TEXT 333 2 "b " }{TEXT -1 6 "(e.g.," }{TEXT 340 2 " b" }{TEXT -1 2 "=
2" }{TEXT 339 1 "a" }{TEXT -1 55 "). Calculate the electric field and \+
electric potential." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 343 11 "Exercise 3:" }}{PARA 0 "" 0 "" {TEXT -1 231 "Adjust th
e solution from Exercise 2 such that the same total amount of charge i
s included as in the solved problem (by adjusting an overall factor to
 the density). Compare the potential from both problems, and comment o
n the large-" }{TEXT 342 1 "r" }{TEXT -1 18 " versus the small-" }
{TEXT 341 1 "r" }{TEXT -1 11 " behaviour." }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{MARK "97 3 0" 11 }{VIEWOPTS 1 1 0 1 1 1803 }