Assignment5 - H ya H H H u: hi hi [DD—FIFE: The field is...

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Unformatted text preview: H ya H H H u: hi 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'h-m‘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 tt-inisc. 13 first: 3-: 1ft" c:.-m‘H'o.soo mt _ i . _ _ (c) tit-h = — = = 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 ya '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"If-ini.-"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 rattan-zip"? 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: -3-tupt_'.|'m 1 A “' [ta-Ir": 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 y-direction. (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 —y-direction. 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' ' (Ii-Ts -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)": Got-m: ”' ” ' ' 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 If-inlt’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 .t-direction. 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. Que-cl = _'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!) GIL-cl = '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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Assignment5 - H ya H H H u: hi hi [DD—FIFE: The field is...

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