t2 CHAPTER 1. VECTOR ANALYSIS AND MAXWELL'S EQUATIONS IN INTEGRAL FORM
En, =
=
, Qo ^, ,"'^
ao .*x lo-e 'lA4' -n'
i :;=l:!-i; :;;;= t(o.soz., +0.535a, +0.267a.){x l0-'x 14x l0- \
5155.7a, +3439.3a.. + 1716.4a,
1.18
+Q
Eu' = 4tt.E# 1rq'a",)ulo-e t . --- \
=, o"'",.[O.gOSa +0.302a. +O.3OZa-l!X44xl0-4 \-- ' Y zl9
= 3702.3a "
+ 1235.5a, + 1235.5a,
E,o,ot = 8858'oa' + 467 4'8a, + 295 l'9a'
Assume at p(-x,O,O),Eo= 0, as shown in the figure'
-2Q
pA0B
s^-= *9- u.(-u,)AP 4ne,(x - a)"
E,o,ot=Enn*Eun =#.tA# * *fu)"r=o-1 2
(a- x)' (a+ x)"
2(a- x)2 -(a+ x)z =g
2a2 +2x2 - 4ax - a' - x' -Zax =0
x'-6ax+a2 =o
6t"lT64x=-.42
*= 4offi'(".)= 4rfr:v ",
IJ
1.19
xt = 5'83a
xz = 0.17 a where _r, is invalid
.'. There is a point (-5.83a,0,0) where E = 0.
After two balls are separated, they have the same charges:
Q,+42
anx{xl}-e xd2Q,..' ltr'l=
1.20
Qe
En = Err*Er4
l0-e
Q,=i;;7 ^'
Unit vector:
= T-------.9ix l0 '.1,
4 x 10-e
'-Txlv'z;'r' =o
Al a-+2aunr, =
m= l# = 0.4472a, + 0.8944a,
Bq AT, _AB=
--:=lnr,l lur,l
a,+2a.,-3a_ -2a +2a""r,
t , '' '
lBTrl= -ffif = -0J07a, + 0.707a,
o;Mu*'
10-e /
5', r' \0.++tza,+ o.seaaa,)
0.805a + 1.6Ia
l1
t.26 (a)
This integral can be broken up into five different parts:
ly az = fi' -r(,' + r)sin@ pdQ + pp'z,o,Qa,
*f " ,o1r' +r)cosQdp * Jo ,,-o(r'
+t)psinld| * l\p'r"orva,
=-fl's,rnvaE* prrar* J'"
ze pap- ft:r:sinQdQ* l\pi,ra,
(b)
(c)
=-25(-co'dll"*"(t)11,.^(t)|i,,-26p.(-cosa)1i,,*"'(t)lu,
= _25+ p2
f - f5 , l'5 "r:
I,: o' = )1, "'
01" + r) cos @dp = J ;ro,
- p'l'o "
= 2s - p:,
lr az * [t . a-t = (p3 - zs) + (zs - p'") = oJ. Jcl
.'. The iield F is conservative.
r.27
ro rufu ryrA^wE,LLJ tsQUA,l,loNS IN INTEGRAL FORM
(2,8)
--y-x3
(1,1)
w = I: ^"on
= nl,u'nd( = rll,is, - 6x'z)ax * l,?r, - oaor]
fc2 p8= nll,it,' - 6x'z)dx * f,irr - 4y,\dy]I
= n[t ' -z*')l' + r' -tr','li]- 3sq =3s x 10-6(,/)
. f- fr 1o51.28 A = J,On
= J,ur* * J,urdr = 4u,
1.29f ' \ - rt/2
J"(tin@, + pcosfar+tanpa,).av = Ji*t*, pd@,
= fi r"rwo= p2 =e
| .30 w = n I", . aV . fnrsintegral can be broken up inro rwo parts.
I.3l Flux =$Il1r,e,6.a,| ' "
F(r,O,Q) = rsin0a. +ao+aQ
ds, = 1'2 sinqd@Oa,
.'. Flux t, = [:'^l'" -{, siney, sin 0d0dQ = 2nr3 l"'' ,rn, ede = t- 13 = nz
.t e=o.t o=o Jo 2 2
w = ,Lf:ucosgsinda, 'u,ar+ flr'sin@a,
.arrsinqdp]
= ,lf;" o' - [i '^ *r]= qrs + 4 - 8t = +q111 J
t9
dsz = rsin9dMrae
.'. Flux t,= f*rf,=rt.rsintdQdr = frorf{a, =,
"' Total nux 6F 'd5=I-"ola
1.32 B=xax+yay+zaz
ds= dxdya,
fn a, = | 7d*dy = | +a*ay = 4.2tt. 22 = t6rJ, J, J-
1.33 B = ZlN"+ xar+ z2xa,
(a)
(3,2,4)
v(3,-l,0) (3,2,0)
X
ds = dzdya,
f fa fz '14 'r2
f na'=l"l zydydz=+l +l =t2J .rz=orr) - -r 2 lo 2 l_,
(b)
20 CHAPTER l. VECTOR ANALYSIS AND MAXWELL'S EQUATTONS IN INTEGRAL FORM
ds= dxdya,
.. Jn
. as = [ z'*a*dt = tzsra*at
{x = OcosQ
ldxdy = pd@p
. . Jn ^ = f,=,f;,25
p2 cos@oo, = n+li,sinpll" = s
1.34 Select as a Gaussian surface a cylinder which has radius = p and length = /.
(a) p<a
Since all the charge is on the surface of the cylinder, no charges are enclosed by theGaussian surface.
$"t," ds= Q=o
.'. E-0atPca
(b) a<p<b
In this region, only the charge from the inner cylinder will be enclosed by the Gaussiansurface.
f lI 12'Total charge Q= | pds = | arl p,. pdO =2tc(.ap,
J"" Jo Jo"
lfPtl2o{r,E. ds= le,,EopdOar= | | e"ErpdQdz=2n!.e,EopJ"" J'"' Jz=oJa=0"'
2nho ao.'.8 - =5, a<o<bo 2n!.e.p e,p
(c) p> b
There will be charge contributed by both cylinders.
o=lp"ar,+lp.ar,J"- "l ' Js^ -z -
= , ,f,=,f*fdwz - , ,f ,=,f*,uo*, = (a - b) p "
' 2ttl
