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hi [DD—FIFE: The ﬁeld is uniform and the surface is ﬂat, so use Iii; = EAcos p. SEI L'P: at is the angle between the normal to the surface and the direction of E . so g3 = TD”.
EXECI‘IE: Clay. = {T5 .ﬂ N.'C}I:ll.4flll mjuﬁflo'ﬂfl m] cos T0” = 5.16 N m:."C
ETALI‘AIE: If the field were perpendicular to the surface the ﬂux would be CPR = EA =1 3.0 N  mil{1. The ﬂue: in this problem is much less than this because only the component of E perpendicular to the surface contributes to the
ﬂute. [DD—HIE The ﬂux through the curved upper half of the hemisphere is the same as the ﬂux through the ﬂat circle
deﬁned by the hottan of the hemisphere because every electric ﬁeld line that passes through the ﬂat circle also must
pass through the curved surface of the hemisphere. SET UP: The electric fieldis perpendicular to the ﬂat circle, so the ﬂux is simply the product of E and the area of
the ﬂat circle of radius 3'. EXECI‘II: $5 = 5.4 = Et .sr‘] = .rr‘E
ETALI'AII: The ﬂux would be the same if the hemisphere were replaced by any other surface bounded by the ﬂat
circle. (a) IDEHEFY: L'se Eq.{22.5} to calculate the ﬂux through the surface of the cylinder. SET UP: The line of charge and the cylinder are sketched in Figure 22.?
II L‘:I'II.1l.'!'.'r It.
11—b
E
Figure 22.?
ELECI‘II: The area of the curs‘ed part of the cylinder is A = 2e? .?
The electric field is parallel to the end caps of the cylinder. so If  .71 = 'D for the ends and the ﬂux through the
cylinder end caps is zero.
The electric field is normal to the curs‘ed surface of the cylinder and has the same magnitude E = Stare; at all points on this surface. Thus c3 = 0': and I y a; {ﬁﬂﬂlxlﬂ'ﬁ c:.1u‘H'o.4oo my . _ .
sit=ﬂ4cosp=ﬂi=j£.'23re..rt[2mf]=4=%=2.?1slfJ'hm‘IC
" “ ' 5; 3.3343110" 031% m' {h} In the calculation in part {a} the radius t' of the cylinder diyided out. so the ﬂux remains the same.
lur=._.=2.t1:<ioi ttinisc. 13 first: 3: 1ft" c:.m‘H'o.soo mt _ i . _ _ (c) tith = — = = 3.42 x 10' N m' ."C {twrce the ﬂux calculated in parts {b} and {c}}.
e; 3.3:4xld " C‘ .'I\ m‘ ETALI'AII: The ﬂux depends on the number of field lines that pass through the surface of the cylinder. H
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'45 [DIE—FIFE: Apply the results in Example 21.1CI for the field of a spherical shell of charge. SEI L'P: Example 2210 shows that E = [J inside a unifonn spherical shell and that E = 2E outside the shell.
F Exact—rt: (a) E =D  —s
{s} r = asst: m and a = {3.992105 2st— ruse: = sasxis? 315C
{01360 mjt'
 —s
(c) }'=C.11C m and E=(B.99><ltjt"Ifini."CiJW=l.llxlﬂi515C
[0.110 m]' ETALI'AIE: Outside the shell the electric ﬁeld is the same as if all the charge were concentrated at the center of
the shell. But inside the shell the field is not the same as for a point charge at the center of the shell, inside the shell
the electric field is zero. 22.19. IDENTIFY and SET UP: Example 22.5 derived that the electric ﬁeld just outside the snrfaoe of a spherical contractor that has net charge g is E‘ = Calculate q and from this the container of exoess electrons.
are: ' 2 2
Enema: 9': RE = “mm “mm: =3.2?5x1ﬂ"'C. lf42r 3.933 3: ll?!9 N  mitt?
q] Each electronhasa chargeofniagnitude e=l.dﬂ2xlﬂ'” C, solhe nurnherofexcesselectrons noededis 3.225 x 1'0"" C rattanzip"? c
E‘t'ALI'ATE‘. The result we obtained for q is a typical value for the charge of an object. Such net charges
correspond to a large number of excess electrons since the charge of each electron is very small. = 2.114 x101”. 22.21. [DD—MY: Add the vector electric ﬁelds due to each line of charge. E (3') for a line of charge is given by
Example 22.6 and is directed toward a negative line of chage and away from a positive line. SET UP: The two lina of charge are shown in Figure 22.21. _‘I h
loam.“ Jr: 3tupt_'.'m 1 A “' [taIr": m j" = —:l.lll'lftrl'lttl IIJCII m Figure 22.21 EXECUTE: (a) At point a, E: and E2 are in the +y—direction (Durward negative charge, awa},r from positive charge}.
at = (1222:“ [(4.sz 1E!" Chn]f[ﬂ.2lm m)] = 4.314212 we a. ={l.t2rrer_,1[[2.4ﬂslﬂ'“ ctmtttsaos mt] = 2.1 st x lﬂ’ Nat: E = E1 + E1 = 6.42 2 1t]1 NE; in the ydirection. (h)Atpointb,E1 is inthe ﬁll—direction was; isinthe —y—direclion.
E1={1f2rrq}[(4.EU:tlﬂ" on}t(o.sco m)]=i.4asxio5 we 22 = [122aeu}[[2.4ﬂ KID—ﬂ otm}r{n.2on 111)] = 2.152 x 1115 NE: a = a2 —21 = 2.221s4 MC, in the —ydirection.
EVALUATION: At point a the two ﬁelds are in the same direction and the magnitudes add. At point it the two ﬁelds
are in opposite directions and the magnitudes subtract 22.23. [DD—MY: The electric field inside the conductor is zero. and all of its initial charge lies on its cutter surface. The
introduction of charge into the cavity induces charge onto the surface of the cavity, which induces an equal but opposite charge on the outer surface of the conductor. The net charge on the outer surface of the conductor is the
sum of the positive charge initially there and the additional negative charge due to the mttoduction of the negative charge into the cavity.
(a) SET "UP: First find the initial positive charge on the outer stuface of the conductor using Q; = owl. whereA is
the area of its outer surface. Then find the net charge on the surface after the negative charge has been introduced into the cavity. Finally use the deﬁnition of surface charge density.
EECI‘II: The original positive charge on the outer surface is qr; = 5:! = Emmy!) = :63? a 1ft"6 Cs’anrtlﬁﬁlﬁﬂ m3} = Sill] rt 1t)"5 Con:
After the introduction of 431.5130 ,riC into the cavity. the outer charge is now
ifJCI INC —O.SDO ,nC = 4.5[1 INC  e
The surface charge density is now c: = i — q — 4'33 x 1 D C. = 5.23 x 19—5 Con2 A _ 4m: _ raroaso m) {h} SET UP: L'sing Gauss's law, the electric ﬁeld is E = I?" =i4= q ._ A €911. £04m"
ESECI‘II: Substituting numbers gives  —s
g as? as m.
[3.321 :3 lll' ' (IiTs m‘}(43tj[ﬂ.2afl 1111‘
{c} SEI L'P: We use Gauss‘ s law again to find the flux. Cb}: = i .
Es
EECI‘II: Substituting numbers gives
= —G.50Et>< 1D" [3 = _..‘ ﬁg X194}: _m:_.C:
* ass all)": Gotm: ”' ” ' ' ETALI'AIE: The excess charge on the conductor is still +5.0ﬂ tall, as it originally was. The introduction of the 43.5% INC inside the cavity merely induced equal but opposite charges {for a net of zero] on the surfaces of the
conductor. [DD—rift Use Eq.[22.3} to calculate the ﬂux through each surface and use Gauss’s law to relate the net ﬂux to
the enclosed charge.
SEI L'P: Flint into the enclosed volume is negative and ﬂux out of the volume is positive. EEECI‘IE: (a) sb = 5.4 = (125 Iffﬂijlﬁdﬂ m2} = 25'!) N  111120
{b} Since the field is parallel to the surface. II! = D.
(c) IChoose the Gaussian surface to equal the volumes surface. Then Till Ifinlt’C—EA = {piss and 1
till] m" net ﬂux ﬂowing in so the ﬂux is — ETALI'AIE: [{1} gr :: CI but we have E pointing away from face I. This is due to an external ﬁeld that does not
affect the flux but aﬁects the value of E. E: {2.40 x 1 ll"3 (2.55.; + 2'50 N  mist} = 5]"? NC . in the positive .tdirection. Since :3: c [l we must have some EAl on second face. 22.49. {DIXIE}? Use Gauss's law to find the electric field .E' produced by the shell for r c R and r :2 R and then use
F = gE to ﬁnd the force the shell exerts on the point charge. {:1} SET 'L'P: .'‘Lpp1‘j.P Gauss’s law to a spherical Gaussian surface that has radius r o R and that is concentric with
the shell. as sketched in Figure 23.4%. r ' EXECI‘I'E: o! = name} ''I. I. Quecl = _'2 Figure 2149's in. = Q“ gives E [43nd } = 6:. ' Eu:
The magnitude of the ﬁeld is E = Q . and it is directed toward the center of the shell. Then F = qE = [IQ 4mg: r‘ 4350?": ’ directed toward the center of the shell. {Since gr is positive, E" and F are in the same direction} {h} SET UP: Apply Gauss: s law to a spherical Gaussian surface that has radius r < R and that is concentric with
the shell. as sketched in Figure 22.4%. Exact—tr: o! = spar!)
GILcl = '3 figure 21.4911 of$ gims(4m”)=o
ﬁn
Then E=ﬂsoF=ﬂ
EVALUATE: Outside theshell the electric ﬁeld andthefotce it exerts is the sameas forapoint charge —Q located atthecenter ofthe shell. Imidettleshell E=EI andthereisnoforce. ...
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This note was uploaded on 10/17/2009 for the course PHYS 2306 at Virginia Tech.
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