Problems set # 3 Physics 169 February 24, 2015

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1 Pof Anchodoqui Polms st # 3 Physics 169 Fuy 24, A point chg q is loctd t th cnt of unifom ing hving lin chg dnsity λ nd dius, s shown in Fig 1 Dtmin th totl lctic flu though sph cntd t th point chg nd hving dius R, wh R < Solution Only th chg insid dius R contiuts to th totl flu, hnc Φ E = q/ɛ 2 A point chg is loctd just ov th cnt of th flt fc of hmisph of dius R s shown in Fig 2 Wht is th lctic flu (i) though th cuvd sufc nd (ii) though th flt fc? Solution With δ vy smll, ll points on th hmisph nly t distnc R fom 1 th chg, so th fild vywh on th cuvd sufc is 4πɛ dilly outwd (noml to R 2 th sufc) Thfo, th flu is this fild stngth tims th of hlf sph Φ cuvd = 2πR R 2 = 2 2ɛ (ii) Th closd sufc ncloss zo chg so Guss lw givs E d A = 1 4πɛ Φ cuvd + Φ flt = o Φ flt = Φ cuvd = 2ɛ 3 Th lin g in Fig 3 is digonl of cu A point chg q is loctd on th tnsion of lin g, vy clos to vt of th cu Dtmin th lctic flu though ch of th sids of th cu which mt t th point Solution No chg is insid th cu Th nt flu though th cu is zo Positiv flu coms out though th th fcs mting t g Ths th fcs togth fill solid ngl qul to on-ighth of sph s sn fom q Th totl flu pssing though ths fcs is thn 1 q 8 ɛ Ech fc contining intcpts qul flu going into th cu: = Φ E,nt = 3Φ E,cd + q 8ɛ Thfo, Φ E,cd = q 24ɛ 4 A sph of dius R suounds point chg, loctd t its cnt (i) Show tht th lctic flu though cicul cp of hlf-ngl (s Fig 4) is Φ E = 2ɛ (1 cos θ) Wht is th flu fo (ii) θ = 9 nd (iii) θ = 18 Solution Th chg cts unifom E, pointing dilly outwd, so Φ E = EA Th c lngth is ds = Rdθ, nd th cicumfnc is 2π = 2πR sin θ Hnc, A = 2πds = θ (2πR sin θ)rdθ = 2πR2 θ sin θ dθ = 2πR2 cos θ θ = 2πR2 (1 cos θ) nd Φ E = 1 4πɛ R 2 2πR 2 (1 cos θ) = 2ɛ (1 cos θ), i, indpndnt of R! (ii) Fo θ = 9 (hmisph), Φ E = 2ɛ (1 cos 9 ) = 2ɛ (iii) Fo θ = 18 (nti sph), Φ E = 2ɛ (1 cos 18 ) = ɛ ; this is foml divtion of Guss lw 5 An insulting solid sph of dius hs unifom volum chg dnsity nd cis

2 totl positiv chg A sphicl gussin sufc of dius, which shs common cnt with th insulting sph, is infltd stting fom = (i) Find n pssion fo th lctic flu pssing though th sufc of th gussin sph s function of fo < (ii) Find n pssion fo th lctic flu fo > (iii) Plot th flu vsus Solution Th chg dnsity is dtmind y = 4 3 πρ ρ = = 4π3 ρ 3 4π 3 ctd y th nclosd chg within dius : Φ E = q in ɛ th nsws to pts (i) nd (ii) g t = (iii) This is shown in Fig 5 (i) Th flu is tht = 3 ɛ 3 (ii) Φ E = ɛ Not tht 6 A solid insulting sph of dius cis nt positiv chg 3, unifomly distiutd thoughout its volum Concntic with this sph is conducting sphicl shll with inn dius nd out dius c, nd hving nt chg, s shown in Fig 6 (i) Constuct sphicl gussin sufc of dius > c nd find th nt chg nclosd y this sufc (ii) Wht is th diction of th lctic fild t > c? (iii) Find th lctic fild t c (iv) Find th lctic fild in th gion with dius wh < < c (v) Constuct sphicl gussin sufc of dius, wh < < c, nd find th nt chg nclosd y this sufc (vi) Constuct sphicl gussin sufc of dius, wh < <, nd find th nt chg nclosd y this sufc (vii) Find th lctic fild in th gion < < (viii) Constuct sphicl gussin sufc of dius <, nd find n pssion fo th nt chg nclosd y this sufc, s function of Not tht th chg insid this sufc is lss thn 3 (i) Find th lctic fild in th gion < ( ) Dtmin th chg on th inn sufc of th conducting shll (i) Dtmin th chg on th out sufc of th conducting shll (ii) Mk plot of th mgnitud of th lctic fild vsus Solution (i) q in = 3 = 2 (ii) Th chg distiution is sphiclly symmtic nd q in > Thus, th fild is dictd dilly outwd (iii) Fo c, E = 1 q in 4πɛ = 2 2πɛ 2 (iv) Sinc ll points within this gion loctd insid conducting mtil, E =, fo < < c (v) Φ E = E da = qin = ɛ Φ E = (vi) q in = 3 (vii) Fo <, E = 1 q in 4πɛ = 3 2 4πɛ 2 (dilly outwd) (viii) q in = ρv = π3 3 π3 = 3 3 (i) Fo, E = q 3 in 4πɛ = 3 2 4πɛ 3 (dilly outwd) () Fom pt (iv), fo < < c, E = Thus, fo sphicl gussin sufc with < < c, q in = 3 + q inn = wh q inn is th chg on th inn sufc of th conducting shll This yilds q inn = 3 (i) Sinc th totl chg on th conducting shll is q nt = q out + q inn =, w hv q out = q inn = ( 3) = 2 (ii) This is shown in Fig 6 7Consid long cylindicl chg distiution of dius R with unifom chg dnsity ρ Find th lctic fild t distnc fom th is wh < R Solution If is positiv, th fild must dilly outwd Choos s th gussin sufc cylind of lngth L nd dius, contind insid th chgd od Its volum is π 2 L nd it ncloss chg ρπ 2 L, s Fig 7 Bcus th chg distiution is long, no lctic flu psss though th cicul nd cps; E d A = EdA cos(π/2) = Th cuvd sufc hs E d A = EdA cos,

3 nd E must th sm stngth vywh ov th cuvd sufc Guss lw, E d A = q coms E cuvd da = ρπ2 L sufc ɛ Now th ltl sufc of th cylind is 2πL, yilding E2πL = ρπ 2 L/ɛ Thus, E = ρ 2ɛ dilly wy fom th cylind is 8 A solid, insulting sph of dius hs unifom chg dnsity ρ nd totl chg Concntic with this sph is n unchgd, conducting hollow sph whos inn nd out dii nd c, s shown in Fig 8 (i) Find th mgnitud of th lctic fild in th gions <, < <, < < c, nd > c (ii) Dtmin th inducd chg p unit on th inn nd out sufcs of th hollow sph Solution (i) Fom Guss lw E da = E(4π 2 ) = q in ɛ Fo <, q in = ρ 4 3 π3, so E = ρ Fo < < nd c <, q in =, so E = 4πɛ Fo c, E =, sinc E = insid conducto 2 (ii) Lt q 1 th inducd chg on th inn sufc of th hollow sph Sinc E = insid th conducto, th totl chg nclosd y sphicl sufc of dius c must zo Thfo, q 1 + = nd σ 1 = q 1 = Lt q 4π 2 4π 2 2 th inducd chg on th outsid sufc of th hollow sph Sinc th hollow sph is unchgd, w qui q 1 +q 2 = nd σ 2 = q 1 = 4πc 2 4πc 2 9 An ly (incoct) modl of th hydogn tom, suggstd y J J Thomson, poposd tht positiv cloud of chg ws unifomly distiutd thoughout th volum of sph of dius R, with th lcton n qul-mgnitud ngtiv point chg t th cnt (i) Using Guss lw, show tht th lcton would in quiliium t th cnt nd, if displcd fom th cnt distnc < R, would pinc stoing foc of th fom F = k, wh k is constnt (ii) Show tht k = 2 4πɛ (iii) Find n pssion fo th fquncy f of simpl hmonic R 3 oscilltions tht n lcton of mss m would undgo if displcd smll distnc (< R) fom th cnt nd lsd (iv) Clcult numicl vlu fo R tht would sult in fquncy of Hz, th fquncy of th light ditd in th most intns lin in th hydogn spctum ɛ, Solution Fist, consid th fild t distnc < R fom th cnt of unifom sph of positiv chg ( = +) with dius R Fom Guss lw, 4π 2 E = q in ɛ = ρv ɛ = π2 4 3 πr3 ɛ, so E = 4πɛ R 3 dictd outwd (i) Th foc td on point chg q = loctd t distnc fom th cnt is thn F = qe = 2 4πɛ = k (ii) k = 2 R 3 4πɛ (iii) F R 3 = m = 2 4πɛ, so R 3 = 2 4πɛ m R = ω 2 Thus, th motion is simpl hmonic with fquncy f = ω 3 2π = 1 2 2π 4πɛ m R 3 (iv) f = Hz = N m 2 /C 2 ( C) 2 2π, which yilds R kg R 3 = m 3, o 3 R = m = 12 pm 1 An infinitly long cylindicl insulting shll of inn dius nd out dius hs unifom volum chg dnsity ρ A lin of unifom lin chg dnsity λ is plcd long th is of th shll Dtmin th lctic fild vywh Solution Th fild diction is dilly outwd ppndicul to th is Th fild stngth dpnds on ut not on th oth cylindicl coodints θ o z Choos Gussin cylind of

4 dius nd lngth L If <, Φ E = q in ɛ nd E2πL = λl/ɛ, so E = λ 2πɛ ˆ If < <, E2πL = [λl + ρπ( 2 2 )L]/ɛ nd E = λ+ρπ(2 2 ) 2πɛ ˆ If >, E2πL = [λl + ρπ( 2 2 )L]/ɛ nd E = λ+ρπ(2 2 ) 2πɛ ˆ 11 A pticl of mss m nd chg q movs t high spd long th is It is initilly n =, nd it nds up n = + A scond chg is fid t th point =, y = d As th moving chg psss th sttiony chg, its componnt of vlocity dos not chng ppcily, ut it cquis smll vlocity in th y diction Dtmin th ngl though which th moving chg is dflctd [Hint: Th intgl you ncount in dtmining v y cn vlutd y pplying Guss lw to long cylind of dius d, cntd on th sttiony chg] Solution Th vticl vlocity componnt of th moving chg incss ccoding to m dvy dt = F y m dvy d d dt = qe y, s Fig 9 Now d dt = v, hs nly constnt vlu v Hnc dv y = q mv E yd, yilding v y = v y dv y = q + mv E yd Th dilly outwd compnnt of th lctic fild vis long th is; it is dscid y + E yda = + E y2πd d = /ɛ Putting ll this togth + E yd = 2πdɛ nd v y = θ = tn 1 q 2πɛ dmv 2 q mv2πdɛ Th ngl of dflction is dscid y tn θ = v y /v, so 12 Two infinit, nonconducting shts of chg plll to ch oth, s shown in Fig 1 Th sht on th lft hs unifom sufc chg dnsity σ, nd th on on th ight hs unifom chg dnsity σ Clcult th lctic fild t points (i) to th lft of, (ii) in twn, nd (iii) to th ight of th two shts (iv) Rpt th clcultions whn oth shts hv positiv unifom sufc chg dnsitis of vlu σ Solution Consid th fild du to singl sht nd lt E + nd E psnt th filds du to th positiv nd ngtiv shts, s Fig 1 Th fild t ny distnc fom ch sht hs mgnitud givn y E + = E = σ 2ɛ (i) To th lft of th positiv sht, E + is dictd towd th lft nd E towd th ight nd th nt fild ov this gion is E = (ii) In th gion twn th shts, E + nd E oth dictd towd th ight nd th nt fild is E = σ/ɛ to th ight (iii) To th ight of th ngtiv sht, E nd E + gin oppositly dictd nd E = (iv) If oth chgs positiv (s Fig 1), in th gion to th lft of th pi of shts, oth filds dictd towd th lft nd th nt fild is E = σ/ɛ to th lft; in th gion twn th shts, th filds du to th individul shts oppositly dictd nd th nt fild is E = ; in th gion to th ight of th pi of shts, oth filds dictd towd th ight nd th nt fild is E = σ/ɛ to th ight 13 A sph of dius 2 is md of nonconducting mtil tht hs unifom volum chg dnsity ρ (Assum tht th mtil dos not ffct th lctic fild) A sphicl cvity of dius is now movd fom th sph, s shown in Fig 11 Show tht th lctic fild within th cvity is unifom nd is givn y E = nd E y = ρ [Hint: Th fild within th cvity is th supposition of th fild du to th oiginl uncut sph, plus th fild du to sph th siz of th cvity with unifom ngtiv chg dnsity ρ]

5 Solution Th sultnt fild within th cvity is th supposition of two filds, on E + du to unifom sph of positiv chg of dius 2, nd th oth E du to sph of ngtiv chg of dius cntd within th cvity Fom Guss lw w hv 4 π 3 ρ 3 ɛ = 4π 2 E +, so E + = ρ ˆ = ρ Using gin Guss lw, 4 π1 3ρ 3 ɛ = 4π1 2E, so E = ρ 1 ( ˆ 1 ) = ρ 1 It is sily sn in Fig 11 tht = + 1, so E = ρ( ), yilding E = E + + E = ρ ρ + ρ = ρ = î + ρ ĵ Thfo E = nd E y = ρ t ll points within th cvity 14 A solid insulting sph of dius R hs nonunifom chg dnsity tht vis with ccoding to th pssion ρ = A 2, wh A is constnt nd < R is msud fom th cnt of th sph (i) Show tht th mgnitud of th lctic fild outsid ( > R) th sph is E = AR5 5ɛ 2 (ii) Show tht th mgnitud of th lctic fild insid ( < R) th sph is E = A3 5ɛ [Hint: Th totl chg on th sph is qul to th intgl of ρdv, wh tnds fom to R; lso, th chg q within dius < R is lss thn To vlut th intgls, not tht th volum lmnt dv fo sphicl shll of dius nd thicknss d is qul to 4 2 d] Solution Fom Guss lw E da = E4π 2 = q in /ɛ (i) Fo > R, q in = R A2 4π 2 d = 4πAR 5 /5, nd E = AR5 5ɛ (ii) Fo < R, q 2 in = A2 4π 2 d = 4πA 5 /5, nd E = A3 5ɛ 15 A sl of insulting mtil (infinit in two of its th dimnsions) hs unifom positiv chg dnsity ρ An dg viw of th sl is shown in Fig12 (i) Show tht th mgnitud of th lctic fild distnc fom its cnt nd insid th sl is E = ρ/ɛ (ii) Suppos n lcton of chg nd mss m cn mov fly within th sl It is lsd fom st t distnc fom th cnt Show tht th lcton hiits simpl hmonic motion with fquncy s = 1 2π ρ m ɛ (iii) A sl of insulting mtil hs nonunifom positiv chg dnsity ρ = C 2, wh is msud fom th cnt of th sl s shown in Fig 12, nd C is constnt Th sl is infinit in th y nd z dictions Div pssions fo th lctic fild in th tio gions nd th intio gion of th sl ( d/2 < < d/2) Solution (i) Consid cylindicl shpd gussin sufc ppndicul to th yz pln with on nd in th yz pln nd th oth nd contining th point : Us Guss lw: E da = qin /ɛ By symmty, th lctic fild is zo in th yz pln nd is ppndicul to da ov th wll of th gussin cylind Thfo, th only contiution to th intgl is ov th nd cp contining th point Hnc EA = ρa/ɛ, so tht t distnc fom th mid-lin of th sl, E = ρ/ɛ (ii) Us Nwton s lw to otin = F m = ρ m ɛ Th ccltion of th lcton is of th fom = ω 2 with ω = ρ m ɛ Thus, th motion is simpl hmonic with fquncy f = ω 2π = 1 ρ 2π m ɛ

6 16 In th i ov pticul gion t n ltitud of 5 m ov th gound th lctic fild is 12 N/C dictd is pplid ugh cssuming th pln th y is, is N/C ists stom is y 3 m d t 1 of th c lctic fild ound Th 5 Nm 2 /C hoizon- 4 N/C s though slntd 6 A point chg q is loctd t th cnt of unifom ing hving lin chg dnsity nd dius, s shown in Figu P246 Dtmin th totl lctic flu though sph cntd t th point chg nd hving dius R, wh R λ q 7 A pymid with hoizontl squ s, 6 m on ch sid, nd hight of 4 m is plcd in vticl lctic fild of 52 N/C Clcult th totl lctic flu though th pymid s fou slntd sufcs 8 A con with s dius R nd hight h is loctd on hoizontl tl A hoizontl unifom fild E pntts th con, s shown in Figu P248 Dtmin th lctic flu tht nts th lft-hnd sid of th con R Figu P246 Figu 1: Polm 1 E δ h R R Figu P248 c of ufc lis y pln? Sction 242 Guss s Lw 9 Th following chgs loctd insid sumin: Figu 2: Polm P C, 9 C, 27 C, nd 84 C () Clcult

7 f ic u- t s of, t t ld d s is f t- n- d t st n- n, n point sufc Find chg th lctic qfild is t points loctd () just outsid on th tnsion of lin g, vy shll nd () insid th shll clos A thin to squ vt conducting plt of 5 th cm on cu sid lisdtmin th lctic flu in th y pln A totl chg of 4 1 though ch of th sids 8 C is plcd on th plt Find () th chg dnsity on th of plt, th cu which mt t th () th lctic fild just ov th plt, nd (c) th point Sction 243 Figu 3: Polm P Appliction of Guss s Lw to Vious Chg Distiutions θ 23 Dtmin Rth mgnitud of th lctic fild t th sufc of ld-28 nuclus, which contins 82 potons nd 126 nutons Assum th ld nuclus hs volum 28 tims tht of on poton, nd consid poton to sph of dius m 24 A solid sph of dius 4 cm FIG hs totl P2453 positiv chg Additionl of 26 Polms C unifomly distiutd thoughout its volum 54 A nonunifom lctic fild is givn y th pssion indpndnt Clcult E yî zĵ ckˆ, th wh of mgnitud, R!], nd c constnts of th lctic fild () cm, Dtmin th lctic flu though ctngul sufc () in th 1 y pln, cm, tnding (c) fom 4 to cm, w nd fom(d) 6 cm fom th cnt 49 lctic fild just low th plt You my ssum tht th chg dnsity is unifom 5 A conducting sphicl shll of inn dius nd out dius cis nt chg A point chg q is plcd t th cnt of this shll Dtmin th sufc d chg dnsity on () th inn sufc of th shll nd () th out sufc of th shll q ctic fild lins tsid 51 A hollow conducting sph is suoundd y lg concntic sphicl conducting shll Th inn sph hs chg, nd th out shll hs nt chg 3 Th chgs in lctosttic quiliium Using Guss s lw, find th chgs nd th lctic filds vywh 52 A positiv point chg is t distnc R/2 fom th cnt of n unchgd thin conducting sphicl shll of dius R Sktch th lctic fild lins st up y this ngmnt h oth insid nd outsid th shll Sction 245 Foml Divtion of Guss s Lw 53 A sph of dius R suounds point chg, loctd t its cnt () Show tht th lctic flu though cicul cp of hlf-ngl (Fig P2453) is 2 (1 cos ) Wht is th flu fo () 9 nd (c) 18? Th c lngth is ds cos 2Figu RP24531cos Figu 4: Polm 4 FIG P2452 Rd, f g f c g

8 Guss s Lw () Constuct sphicl gussin sufc () E Not tht th nsws to pts () nd h ditionl Polms, wh c, nd find th nt chg 59 A 54 In gnl, E y this sufc (f) Constuct yi zj ck z sphicl gussin In th y pln, z nd Eyic k y = y = h l f dius, wh, nd find = th nt (c) E zedaz yick k y j da closd y this sufc E u w w (g) Find th lctic fild E ch d ch ion (h) chw = w 2 2 da = hd 2 2 z Constuct sphicl gussin f dius, FIG P2454 ch nd find n pssion fo th nt 55 () q ci in closd y 3 this 2 sufc, s function of Not D () Th chg distiution is sphiclly symmtic nd q hg insid this sufc is lss thn in Thus, th fild is dictd 3 (i) Find dilly outwd d ic fild in th gion ( j) Dtmin th m (c) E n th inn kq 2k fo sufc c 2 2 of th conducting shll lo min (d) th Sinc ll chg points within this on gion th loctd out insid conducting sufc mtil, of E th fo ch g shll (l) c Mk plot of th mgnitud of th z () E EdAqin E ld vsus Figu 5: Polm 5 (f) qin 3 FIG P2434(c) (g) E kq 3k in 2 2 F (h) qin V F I KJ F H G (dilly outwd) fo 3I 3 K J 3I KJ 3 (i) E kq in k 3 3 k (dilly outwd) fo (j) Fom pt (d), E fo c Thus, fo sphicl gussin sufc with c c, qin 3qinn wh q inn is th chg on th inn sufc of th conducting shll This yilds q 3 (k) (l) inn Sinc th totl chg on th conducting shll is qnt qout qinn, w hv Figu inn P g 2 qout q Figu 6: Polm 6 This is shown in th figu to th ight 3 3 two idnticl conducting sphs whos sufcs E c FIG P2455(l) 6 R g it th l th w f ( f l d 2

9 th cnt of th wll to th fild point is Figu P2455 th d insid 2 L 57 s cuvd gth 56 Consid two idnticl conducting sphs whos sufcs sptd y smll distnc On sph is givn lg nt positiv chg whil th oth is givn smll nt positiv chg It is found tht th foc twn thm is ttctiv vn though oth sphs hv nt chgs of th sm sign Eplin how this is possil A solid, insulting sph of dius hs unifom chg dnsity nd totl chg Concntic with this sph is n unchgd, conducting hollow sph whos inn nd out dii nd c, s shown in Figu Figu 7: Polm 7 P2457 () Find th mgnitud of th lctic fild in th gions, FIG, P2429 c, nd c () Dtmin th inducd chg p unit on th inn nd out sufcs of th hollow sph Insulto y wy fom th cylind is c Conducto h t 1 5 cm w hv Figu P2457 Polms 57 nd 58 Figu 8: Polm 8

10 chg j 6 Rviw polm An ly (incoct) modl of th hydo- tom, Lsuggstd y J J Thomson, poposd tht pos- 2 2 gn itiv cloud of chg ws unifomly distiutd thoughout th volum of sph of dius R, with th lcton n qul-mgnitud ngtiv point chg t th cnt () Using Guss s lw, show tht th lcton 2 j would in quiliium t th cnt nd, if displcd fom th cnt distnc R, would pinc stoing foc of th f fom F K, wh K is constnt () Show tht K k 2 /R 3 (c) Find n pssion fo th fquncy f of simpl hmonic oscilltions tht n fild du to th ch sht givn y g chg v y dictd v v lcton of ppndicul mss m would undgo to if th displcd sht smll y distnc (R) fom th cnt nd lsd (d) Clcult t nd lt E numicl nd E vlu fo R tht would sult in fquncy of 6nd 247 ngtiv CHAPTER 1 q 15 Hz, 24 shts th Guss s fquncy Th of Lw th light ditd in th most intns lin in th hydogn spctum d o s th 61 An mgnitud lft infinitly of th long pi givn cylindicl of y shts, insulting oth shll of filds inn d th dius lft nd out th dius nt hs fild unifom is volum chg on dnsity th ight A lin hs of unifom unifom lin chg dnsity dnsity is Clcult FIG 67 P2463 A solid insulting sph of So plcd long th is of th shll Dtmin th lctic th fild lctic vywh fild t points () to th lft of, () in twn, chg dnsity tht vis with to nd 62 th Two (c) lft infinit, to th nonconducting ight of shts th two of chg shts plll A 2, wh A is constnt to ch oth, s shown in Figu P2462 Th sht on Figu th FIG 9: Polm P lft hs Wht unifom If? sufc Rpt chg th dnsity clcultions, nd th on fo Polm 62 th cnt of th sph () Show t, whn E is oth dictd shts towd hv positiv th unifom sufc chg dnsitis of th vlu shts, th filds du to th individul shts () oppositly Show tht th mgnitud o lctic fild outsid ( R) th twn h nd th nt fild ov this 4 itud A nt sph fild of th of is ( R) th sph is E A 3 / fild dius du 2to is th md ch of sht nonconducting givn y mtil chg on th sph is qul to 248 fild tht is hs vis unifom long volum th chg is, dnsity ut is (Assum dscid tht y tnds fom to R; lso, th th mtil dos not ffct th lctic fild) A sphicl R is lss thn To vlut cvity E of dius dictd is ppndicul now movd to fom th sht th sph, s ts, E nd 2 E oth volum lmnt dv fo sph o th shown ight in Figu of th P2464 pi of Show shts, tht oth th lctic filds fild within dictd towd th ight thicknss d is qul to 4 th nt fild 2 d) ld th is cvity is unifom nd is givn y E σ nd E y n th /3 gion (Suggstion: to th lft Th of th fild pi within of shts, th cvity oth is filds th supposition th towd ight of th th fild Figu lft du nd P2462to th th nt oiginl fild isuncut sph, plus R t distnc fom th pln 68 A point chg is loctd on σ ictd to FIG P2462 FIG P2463 th fild du to sph th siz of th Figu cvity 1: Polm with 12 unifom E ngtiv to th chg lft dnsity ) chg psss though th disk, t Show tht if on fouth of th ngl of dflction is dscid y hin t, th E cvity nd Eis th gin supposition oppositly of dictd two nd E n q unifom sph of positiv chg of dius th gion twn th shts, y th filds du to th individul shts oppositly u ictd to 2 dmvnd sph th nt of ngtiv fild is chg of dius vity E R n th so gion cnt Eto th of ight unifom of th pi sph of shts, oth of positiv filds chg dictd towd th ight nd th nt fild is E I to th ight so E KJ g 3 dictd Figu P2464 outwd tnt 3 Figu 11: Polm 13 4 fild Rwithin th cvity is th supposition of two FIG P2464 E du to f unifom sph of positiv chg of dius Figu P24 oth 3 E du to sph of ngtiv chg of dius 5 A unifomly chgd sphicl shll with sufc chg within th loctd cvity t distnc fom th cnt is thn 69 A sphiclly symmtic chg

11 z F I K J his coms E2 g dv Th lmnt of volum is cylindicl, lngth, nd thicknss d so tht dv 2 d 2 IF KJ 2 3 I K J so insid th cylind, E 2 F 2 3 I K J uss s lw coms F g I d K J 2 o outsid th cylind, E 2 R 2 F 2R 3 I K J indicl shpd gussin sufc ppndicul with on nd in th yz pln nd th oth nd point : y A z w: EdA q in th lctic fild is zo in th yz pln nd is to da ov th wll of th gussin cylind only contiution to th intgl is ov th nd th point : EA Af nc fom th mid-lin of dth sl, E O y 73 () Using th mthmticl simility twn Coulom s lw nd Nwton s lw of univsl gvittion, show tht Guss s lw fo gvittion cn wittn s th gussin sufc () D t distnc gussin fom th cnt ssuming sufctht th Eth s m Answs to uick uizzs 241 () Th sm num o sph of ny siz Bcus sph clos to th ch 242 (d) All fild lins tht nt contin so tht th totl ntu of th fild o th co 243 () nd (d) Sttmnt () F I n qul num of positiv Figu P2471 Polms 71 nd 72 Figu 12: Polm 15 FIG P2471 psnt insid th suf m KJ sily tu, s cn sn 72 A sl of insulting mtil hs nonunifom positiv lctic fild ists vyw chg dnsity C 2, wh is msud fom th n of th lcton is of th fom 2 with chg is not nclosd wit cnt of th sl s shown in Figu P2471, nd C is m flu is zo constnt Th sl is infinit in th y nd z dictions 244 (c) Th chgs q 1 nd q Div pssions fo th lctic fild in () th tio 1 contiut zo nt flu th on is simpl gions hmonic nd () with th fquncy intio gion f of th sl (d/2 d/2) 2 2 m 245 (d) W don t nd th su point in spc will pin locl souc chgs z 246 () Chgs ddd to th m will sid on th out su