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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;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The net charge " }
{TEXT 258 1 "q" }{TEXT -1 71 " on the rod is distributed uniformly, 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;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "Gauss:=Phi=q/epsilon;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "Eofr:=solve(Gauss,E);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 68 "We found the expected result that the electric field decr
eases 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 suggest
ed to use a cylinder as a Gaussian surface, and thus the surface integ
ral could be calculated in a trivial way." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT 265 60 "2) the electric field of a unifor
mly charged spherical 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 surrounds 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;" }}}{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;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "Eofr:=solve(
Gauss,E);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "Note that " }{TEXT 
266 1 "r" }{TEXT -1 73 " is the spherical radial distance and that thi
s result is valid only for " }{TEXT 270 2 "r " }{TEXT -1 2 "> " }
{TEXT 269 1 "R" }{TEXT -1 36 ", i.e., outside the spherical shell." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 71 "Inside th
e spherical shell no charge is enclosed and 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 understood as the cancel
lation of contributions from Coulomb's law pulling in all directions w
ith equal magnitude." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 
0 "" {TEXT -1 74 "3) a solid non-conducting (insulating) uniformly cha
rged sphere of radius " }{TEXT 273 1 "R" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 87 "Outside the sphere the result is ide
ntical to the previous one for the spherical shell." }}{PARA 0 "" 0 "
" {TEXT -1 53 "Inside the sphere we use a Gaussian 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 are held in place, since th
e sphere is not conducting." }}{PARA 0 "" 0 "" {TEXT -1 81 "The consta
nt charge density is given as total charge divided by the total volume
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "rho:=Q/(4/3*Pi*R^3);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "We equate the constant charge d
ensity 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);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Qpr:=solve(eq,Qpr);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Gauss:=Phi=Qpr/epsilon;" }}}{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 "Ein" }{TEXT -1 17 ") as g
iven below:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "Ein:=solve(G
auss,E);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "At the surface they d
o match:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "subs(r=R,Ein); \+
subs(r=R,Eofr);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "We graph the e
lectric field after a choice of constants:" }}}{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
\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "The electrostatic potenti
al is spherically 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);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Vofr:=-int(Eofr,r);" }}}
{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:=pl
ot(subs(epsilon=1,R=1,Q=1,Vofr),r=1..5):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "display(\{P3,P4\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 118 "The integration constant should be adjusted for the integratio
n of the inside part to match the potential function at " }{TEXT 282 
2 "r " }{TEXT -1 1 "=" }{TEXT 283 2 " R" }{TEXT -1 65 " (with continuo
us derivative). 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 315 11 "Exercise 1:" }}{PARA 0 "" 0 "" {TEXT -1 14 "Use the
 large-" }{TEXT 317 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 318 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 320 1 "r" }{TEXT -1 1 "=
" }{TEXT 319 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 "=" }
{TEXT 290 2 " a" }{TEXT -1 44 ", and then it decreases linearly to zer
o at " }{TEXT 293 1 "r" }{TEXT -1 2 " =" }{TEXT 292 2 " R" }{TEXT -1 
4 " = 4" }{TEXT 316 1 "a" }{TEXT -1 23 "/3. The region between " }
{TEXT 321 1 "a" }{TEXT -1 6 " and 4" }{TEXT 322 1 "a" }{TEXT -1 93 "/3
 can be considered as a surface skin region where the charge density d
iminishes gradually. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 40 "The charge distribution (up to a factor " }{TEXT 19 4 
"rho0" }{TEXT -1 22 ") is given as follows:" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 8 "rho0:=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
5 "a:=1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "c1:='c1': c2:='
c2':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "rho:=rho0*(c1+c2*r)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eq1:=subs(r=a,rho)=rho
0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eq2:=subs(r=4/3*a,rho
)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "sol:=solve(\{eq1,eq
2\},\{c1,c2\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(s
ol);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "rho;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "P1:=plot(rho0,r=0..a):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "P2:=plot(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 12 "with(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "display(\{P1,P2,P3\});" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 240 "It is of interest to draw the charge den
sity contribution in such a way that one understands how the amount of
 accumulated charge increases as one steps from the inside out. To und
erstand the amount contained in a spherical shell at radius " }{TEXT 
323 1 "r" }{TEXT -1 23 ", i.e., in the window [" }{TEXT 324 1 "r" }
{TEXT -1 1 "," }{TEXT 325 1 "r" }{TEXT -1 2 "+d" }{TEXT 326 1 "r" }
{TEXT -1 46 "] it is 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,P
3\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The flux for  " }{TEXT 
298 1 "r" }{TEXT -1 3 " < " }{TEXT 297 1 "a" }{TEXT -1 9 " and for " }
{TEXT 296 1 "a" }{TEXT -1 2 " <" }{TEXT 295 2 " r" }{TEXT -1 4 " < 4" 
}{TEXT 294 1 "a" }{TEXT -1 87 "/3 has to be determined by calculating \+
the amount of enclosed charge. The total charge " }{TEXT 302 1 "Q" }
{TEXT -1 13 " is split as " }{TEXT 301 1 "Q" }{TEXT -1 2 " =" }{TEXT 
300 2 " Q" }{TEXT -1 3 "1 +" }{TEXT 299 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 304 2 " r" }{TEXT -1 2 " <" }{TEXT 303 2 "
 a" }{TEXT -1 34 " we have a result that depends on " }{TEXT 327 1 "r
" }{TEXT -1 68 "-cubed (which is obvious after integrating the r^2 tim
es a constant)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Q1:=rho0*
(4/3*Pi*a^3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Q1p:=rho0*
(4/3*Pi*r^3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "Up to the facto
r epsilon (permittivity) this constitutes the flux in the innermost pa
rt, 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 linearly decreasing density we ca
lculate the integral:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Q2
:=4*Pi*int(r^2*rho,r=a..4/3*a);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
61 "The flux (from the decreasing density alone, i.e., excluding " }
{TEXT 306 1 "Q" }{TEXT -1 57 "1) is calculated by terminating the inte
gral at a radius " }{TEXT 305 1 "r" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Q2p:=4*Pi*s
ubs(rp=r,int(r^2*rho,r=a..rp));" }}}{EXCHG {PARA 0 "> " 0 "" {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 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "display(\{P4,P5,P6\});" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 43 "We see that the flux continues smoothly at " }{TEXT 
331 1 "r" }{TEXT -1 1 "=" }{TEXT 330 1 "a" }{TEXT -1 26 ", and becomes
 constant at " }{TEXT 329 1 "r" }{TEXT -1 2 "=4" }{TEXT 328 1 "a" }
{TEXT -1 136 "/3, the point from which on no additional charges are in
cluded when the probe sphere increases. Now we can calculate the elect
ric field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 16 "In region I (0 <" }{TEXT 308 2 " r" }{TEXT -1 2 " <" }{TEXT 
307 2 " a" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "Phi:=E*4*Pi*r^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Gau
ss:=Phi=Q1p/epsilon;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "E1:
=solve(Gauss,E);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "In region 2 (
" }{TEXT 309 1 "a" }{TEXT -1 3 " < " }{TEXT 310 1 "r" }{TEXT -1 4 " < \+
4" }{TEXT 311 1 "a" }{TEXT -1 4 "/3):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "Gauss:=Phi=(Q1+Q2p)/epsilon;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 19 "E2:=solve(Gauss,E);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 17 "and in region 3 (" }{TEXT 313 1 "r" }{TEXT -1 4 " > 4" }
{TEXT 312 1 "a" }{TEXT -1 4 "/3):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "Gauss:=Phi=(Q1+Q2)/epsilon;" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 19 "E3:=solve(Gauss,E);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 61 "For graphing we set epsilon=1, i.e., we choose our own un
its:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "epsilon:=1;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "P7:=plot(E1,r=0..a):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "P8:=plot(E2,r=a..4/3*a):" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "P9:=plot(E3,r=4/3*a..5*a):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "display(\{P7,P8,P9\});
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The electric field function i
s now differentiable at the matching points." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 42 "subs(r=a,diff(E1,r)),subs(r=a,diff(E2,r));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "subs(r=4/3*a,diff(E2,r)),sub
s(r=4/3*a,diff(E3,r));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 189 "It is \+
worth noting how the electric field grows for constant radial charge d
ensity and then turns around to decrease while the density is decreasi
ng linearly. This is followed by the usual " }{TEXT 19 5 "1/r^2" }
{TEXT -1 192 " dependence outside the charged sphere. The skin area of
 the charge distribution now contains a field that interpolates smooth
ly between the growing and decreasing parts of the electric field." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "The elect
rostatic potential can be obtained from a integration, which we now ca
rry out carefully:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "V1:=-
subs(rp=r,int(E1,r=0..rp));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
17 "c1:=subs(r=a,V1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "V2
:=c1-subs(rp=r,int(E2,r=a..rp));" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "c2:=subs(r=4/3*a,V2);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 38 "V3:=c2-subs(rp=r,int(E3,r=4/3*a..rp));" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "P10:=plot(V1,r=0..a):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "P11:=plot(V2,r=a..4/3*a):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "P12:=plot(V3,r=4/3*a..5*a):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "display(\{P10,P11,P12\}
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "The potential is continuou
s, but has an unusual constant: as the integral was started at the ori
gin, " }{TEXT 314 1 "V" }{TEXT -1 61 "(0)=0; usually one defines the p
otential as zero at infinity." }}{PARA 0 "" 0 "" {TEXT -1 71 "We now d
etermine the constant by which the potential should be shifted." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "c3:=limit(V3,r=infinity);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "P10:=plot(V1-c3,r=0..a): P
11:=plot(V2-c3,r=a..4/3*a): P12:=plot(V3-c3,r=4/3*a..5*a,0..-1.1*c3):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "display(\{P10,P11,P12\}
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 337 11 "Exercise 2:" }}{PARA 0 "" 
0 "" {TEXT -1 102 "Construct a radial density of your own choice with \+
regions where the density grows as a parabola from " }{TEXT 336 1 "r" 
}{TEXT -1 6 "=0 to " }{TEXT 334 1 "r" }{TEXT -1 1 "=" }{TEXT 335 1 "a
" }{TEXT -1 53 ", where it has a maximum and then falls to vanish at \+
" }{TEXT 333 1 "r" }{TEXT -1 1 "=" }{TEXT 332 2 "b " }{TEXT -1 6 "(e.g
.," }{TEXT 339 2 " b" }{TEXT -1 2 "=2" }{TEXT 338 1 "a" }{TEXT -1 55 "
). Calculate the electric field and electric potential." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 342 11 "Exercise 3:" }}
{PARA 0 "" 0 "" {TEXT -1 231 "Adjust the solution from Exercise 2 such
 that the same total amount of charge is included as in the solved pro
blem (by adjusting an overall factor to the density). Compare the pote
ntial from both problems, and comment on the large-" }{TEXT 341 1 "r" 
}{TEXT -1 18 " versus the small-" }{TEXT 340 1 "r" }{TEXT -1 11 " beha
viour." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "99" 0 }{VIEWOPTS 1 
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }