2014-G12-Physics-E

All rights mztrttcd with Fut1iahTc:tthuok Board. Lahtwc.Appmvud by tht: Federal Minis-‘try t;:['E,|;1u|:;|Ii0r|_ [5I;g|1-;aI;t3d_ ‘Vida Letter I'~lt:|-I-'.lIi-4-I'Zl]'|]21' b-.5) dalmzl IT May, 2003 _ __ Guftarnmenl 0|\" Pttltislan. Islamabad- Sr. Ha. - Chgpter |=-Q9; 12. Electrctstattcs 1 13. C‘.-urre nt Electrictty 3,1 14. Electra-magnetism 55 15. . Electromagnetic lnductiflfl E1 15. Altemating Current 111 1?. Physics of Solids 135 13. Electmnics 154 19. Dawn uf Mun‘-arn Physics 1 T5 20. Atomic Spectra 3433 -21. Nuclear Physics 219 1. Glossary 360 ii. Btbliu-grap hy ' 25.4 iii. Index 255\"--- . __ .PM Br. llln. l\"dI1'.i| Saint-mi mt. Mnhdnmnd Mi-Silhid _ Eillrl-I-|l11lil-L|h'|..'\:-'t~FaI:ct:ht. I-:1:-i-F|¢p¢lg|Pm-I. Dr. M- I-lltrin Btttt mt. Sh. Allah nttma tum.) Ttl1,g]1I-1-lrn1|ll:Fmf. Nu-nr Aland QIu1'r.1hl{ll|d.| Pnul‘. 5. H.Altl1lnr[Rld'.jPmll IHIIII-tltrll-d Niflr Prof. Dr. ljn Mujubn Ghuril\"rnf- Dr. Muhammad Fllfiq Dr. Muhammad lhflq S:hl{RI:|.] PI'fl|'- Muhammad l\"itI\~\"lr KhanPmfl Yl'lttl1ltluIld.|l|.l_iSl1:hid |'*.ruI. 5l1_.I|.l't|i1-.~\l1m|-d {ll'.ld.j| l:u.1'.~|-F|.au:¢||| _ . Prqt Bf. Mn. I\"u-tlzil Sllulni ' S\";-ll‘!-i-|fl1|I-lLI|'iil'rir|:I.i‘.|2Ili'|I J’ - \-I. |'|.1ttr§-|tt.\. fllliscrfillccrnPutalllhnl by: Mum-II Brdl:l'IirlJlhIum. I; ,1 :1 - _ IPrin-tldhyrz Zahid BI:hlt'F'I'|l‘|lfl'l'! I_ahm- \" \"'\"flflqgflidrllng Euluu-n Imptld-1-|dII\ Nu-fllfirpill PI'lfl'“H mu, \" 151 1-1,. noon mm

12_- --_—Attlreehdofflrlc dreptertheshrdsnlswllhedale to:Understand and d-ee-clibe C~eu|r.trmb'e lew.' Deccrl-hetl1ert__achergehesal‘lelrle1tcroeeieundit.Uniderslendfietcsofllkeandunlllteclurges.-.‘\"£\"'!\"'-\":!\"‘ filroarteeflco-plehehnlhernenfi'pnm'lmeee'. f'Inn‘at D\"‘nlsrssncl prtctcstsl mm‘eras-ensppllcetlenot‘Etrplcinflteelectrtclnteneltylnalreeepeoeend In other media.. ‘Slate end prove Gs-|.|ss's law. _' 'Appreoletelheepplice1icnsctGsuss'slew. _‘F'l‘*F\"\"!-\"' mcrgeaphhehclemclnlfirntcpltE,r‘t.r;1tfl‘[|?leattpe?nf.°'m|'m rerrneotwcrkrioneInl:m' am' a eunllcoslllve9- Fteleteelectric field strerrgthencl potential gradient.1|}. Ftnd Expression [DI potentislste lil'fliI'll!d|.|e!.c1a pcinlcharge.11. Elsscrih-s and ccrivethe value o1eleo1ric_r':herge by Millil-tan's met:-ice,12. Cielrnalatc the cspscitanceof parallel plate capacitor.11 Reccgnlzclhie elfect eldielectriccnthecapccllsncecilpereltel plate capecilier.14, Llnderstartrt and cleectihe electric pelerizelion ofdielerzhic.' 15, gacw mergtrgaegfinefacgtaefigggralmd discharging c1 e cececltor through e resistance1B- Find energyerrereeeierrefecherged capacitor. '!~|e study of electric charges at rest under the ectien of electric forces is ltrrctrm asec1rcstalics.An electric leroe is the force which holds the peoilive end negative charges-lhal matte up atoms and molecules. The human body is composed entirety ol etcms sntlrrtolecutes.tl'tuswe owe cur ettislencclolhe electric force. '

We l-tnow that there are two l-tinds ot charges. namely, positive and negative charges. The charge on an electron is assumed to be negative and charge on a proton is positive. Moreover. we also learnt that lilte charges rep-al each other and urllilte charges attract each ether. Now we investigate the quantitative nature of these forces. The first measurement ol the loroe between electric charges was made in 1BTtl- AD by Charles Coulomb. El Frert-tat military engineer. Clrt the basis ol these measurements, he deduced slew ltnowrt as C‘-oulorrtt:|'s law. it statesthat.n teen-in-|; nun !'1_:‘-l.l|-.*.I_.'¢--f.1‘'FI_-.\";:.'_..._-_.-_. 1'.- t-._it..,|_-.-,—_|It'-Ir-*+--l--'-4i!.,'-:3-I -'..--,-.| -. . -I,‘.\"....--_. -.-t._ .1- '-it-_T---;\".i.r1:r_;'[\"_§_g-i\": ‘5?T;.=;.ii=-._-is :.'.\".'L.,\"rt-'.'.‘-.'-.'-',,-=-'{?.f!'1.-- -151 ;.j_-;._ L: its E7-;' ital. -._15.F.:,.e|_-:_i-r:_,'€.;:‘-rf_|_.__ _L,__-_F_-_._I-_f all-I-.-__—_--||t'd.|_i.t:-._---. .-_ ._ .i.- t;--.1-.'--:E.;:.‘-u-1_-.I.--c--l|-.,t--';''l,,{:.1.-‘--1r.;-_.-fl_ft-i.;_'1-§,t.F:-.:-1*-1:.._\";'.E1: \"\"1 ' I ' - - _ _7.-.t-_-_.__ ._ '- ...I_.'.-.j-'-t.rj,_ _F,‘ \"It. ‘II. F_ Fetiiigl er F.._t¢@ ........ -. 112.1;-l—-I ---- -~-3--l~ r\" Fi=-r- “ \"\"\" ' '1' Where F is the magnitude of the mutual loro-E that acts on r=,, |=,, each cri the two pcrnt charges qr, , q_. and r is the distance ‘It. j_§.----.q_-J-c. between them. The torce F alwatrs acts along the line joining -t------------> the two point charges (Fig. 12.11, lr is the constant til |t:. orooortionalittr. its value depends upon the nature oi medium betweorl the two charges and system oi units in which F, qF. 121 tlt lhoulutvtt hrou and r are rneasured. It the medium between the two pointbttilirltt llltl cltlriel Ind [hit charges is tree sea-so and the system of units is SI, then it isdllutmn-tohne lttteto ltehveert unlit: represented as it _ 5\"]; ........ .. [t2.2i where t:_ 15 an elactrtcal constant known as |;t-err'n|tt|~.'|t;,r of tree space in 5| units. its value is B B5 I 113\" hlt't't'C' Eubslttuttngtha value of: the constant it 4% =s=-=1o*r~.|m-t:- Thus Cctutornlzfs force in free space is nae; F H ____i__ 951 ........ .. em, '\ tu e'1r-

-he stated eainier. l1o|.tlomtis‘ force is mutual force. it meansIhel it ti. eiterts at foroe on o,. than q, also etterts an equal andopposite force on o.. lfvre denote the force cttertedoncghy e.arsF_.. and thaloncharge o. dueto e_.as F.,. thenF,--t=,, {tat} t.The magnitude ol both these two lioroes is the same ani:t is ‘Ligiven by Eg. 12.3- To represent the direction ol these forceswe introduce unit vector-s.'lf |'-,, isthe urtitv-actor directed from Jr.]I,toe,ar'|t:l|i,,istl1aunitve~ctori:Hreote'dfro|nq,tog,.flten I _.\" -'rF ti_ t=,, _-Ett1a,-r-gr._.., 12.51:} qt 1\". r J‘ -so t=,,-;—;;5f_§1ie taste} § If I ll-lit qt ,\"The tproes F,. and F._.are shown in Fig.12.2 l_a 8. D]. ltcan he J '||seen that i,. =-i.,,scEqe.12.sras cisnewit-lat tot ng :2 :r For = ' Fl: Iii E |:,.The sign cl the charges in Eqs. 12-5 ta 8- ti] determine vnctun ' 1whether the forces are attractive or repiulsive. iiI .tH'1I It'll] I ‘l.{IIfl- BllttilttrlWe shell now consider the ettect oi medium oetweenthe me 't'iilt'|'ltIit.|tt't I2-35charges upon the Coulomb's force- ll the medium l5 an 31: 5-1-Binsulator. it is usually retened as dielectric. It has been fflqmfi him 2.28-Ithat the presence or a dielectric always reduces the P-itfl paper 1!electrostatic loroe as compared with that tl'l tree space by e Flltllfll '.-‘Mucertain lactor which is a constant for the given dielectric. This Flttttiillt‘ 3-1.5constant is ttrtown as relative penhitlivity and is represemed Tilttti Ebye. The values ofrelalirve permittivityo-‘i difierent dietectncs Tfitfl‘ H 5.-ll}are given in Table 12.1 't\'ii|rllflIcl| 2.9! 2.1Thus the Coutcmtzrs loroe in a medium cl relative permittivity 2.1c, ls given by I'll F =‘fl¢L,EfErit ........ .. \"2.E}ltcan be seenin the tabtethel r._ forairls 1.tZiiJtZti;'i. This valuetsso close to one that with negligible error. the Eq. 12-3 givesthe electric toroeln air.!h'It1th11fl: t2l'ugecq,=1DlIl_pl3mdq,f5Dpt3at'elomldhtry-phtedpctilornr, =10] u'ii:lr,=-it.tItltmier:-firttly. 3

nlrerefliedialmseseremeaeumdinmelresfiaiuflateflwe\" ferceer\e,[Fig.12.3}. 3-fill-I’flen: q,é100|.1C. q,=50|.|G Feaiticrrvqrserefcirelefivelfler =r,,Ir,-r.I-4-l-Bl _ _ \,\ .r=n1egnltud|eh',.I J =.'5n1 .I. is - -- ME.‘ MEI i':||:|!—-“Ina: ~‘H 5 \" _ *- 4--=..-f -F n1q‘Q1 .-- »r~ Q.h|llil|,\"\" 3 _9_:r ie'_r~r_|ifc:'-’ 1_ rec I1- 11‘ cs ms 1e\"c 1 4 r- s 1.-\.DI-_--1*; r,l~l.0l \ ' rs in?’ s Fla 1:: F\" ' =1-rH.-i.-'l.lJBait \"ins-;,|i1r|u<issr|§,.=F=,,lr| .-14; ’ + r-r-es]»' = |.s er: DlreclicnefF,.= tan\" -3T\"wilh:r-axis _ I, 1.44; , I: - Han-ten's urivsrsal gs-uitaliensl law and C-eulen'|l:'s law enable us ie calculele the magniluee as well as the clireciisns eflhe grerritetierral and eleclricferces. respect?-sly. Hewaver one may q|.|es1ien. ta} W'ha1 are the eriglns cf lhese fercee? {bl Haw are these [areas lransmlttec frem ene mesa le- anelrisrnrlrern ens snargsle anether? The ansn-er‘ In [is] is slil unimmvn: lhe eirlslense of lhese forces is accepted as il is. Thal is wlw ll'ie{|' are called basic fereesefnslure. Te describe Ilia mecnanlsm by which electric ferce is iransmlliacl, Michael Faraday [1?91-186?} lntrerlucecl the eense|:tefanaisr:|risfis|d.Ai:sereinglehlslrierir1r.ltls1he iruiinaiserepertyelnelurelrrataneiscuicfisidsisstshflne space erer.mn:l err electric charge. This electric field la censiceredteb-eafereefieltllhelexeriaefereeenefliar charges plecercl In that lleld. Fer example. e charge e preeusee an electrlr: field in the space surreunting ii. This 4

field exists whether the ether charges are present in space er [Illncil. Huweuer, the presence el field cannpt be tested untilanelher charge q_, is l:irciug1'i| irtte the field. Thus the field ofcharge e interacts with qi, lie prccluce an electrical farce. Theinteractlen between g and g, is accomplished in twp stepe:{at the charge e produces a field and lb) the field interactswith charge g,_te preduce a leree Fen cu. These tiuesleps areillustrated in Flg. 12.4.In this figure the density cl clets is prep-nttienal te the strengthcrl the field at the --arieus paints. We may define electric fieldetrenglherelectric field intensity-Ealanp peintinthefield as E - q-|=p l12.1Iwhere F is the farce eitperlencecl by a peslllire test charge pi, _,'. ' --. F=I-c|_Eplaced at the pcilnt. The test charge -rt, has te be uergismalsetrial itrnay net dlslertlhefield which ilhaslc measure.Since electric field intensity is tcrce per unit charge, it ismee5t.tre-cl in newton perceutemlzi i1 SI units. lt is a vector-quantity and its directien is lhesameasthatefthe fierce F. llll tiltHThe fence BlqJBl'iEl'lCE'l1l by a lest charge c, placed in the field _ 1:I}miI_—'I.I'I*'fl-!cl‘ a charge g in vacuum is given by Eu. £12.33.Eq. 12.? nbe used tc evaluate electric intanslty due tci apaint charge g ata peint distant rt'rcijn IL Place a peeitivetestcharge qr, at this paint. The C.‘-eutnrnl:i's Terce that this chargewill experience due llci q is F =_‘m\"_u‘??'-l'P-r' .......... rthetwhereiis aunll -recterdlrected tremlne pelntchargeg te thetest point where qr, has been placed. l-e., the petnl where theelectric intensilyis teheevaluated- By Eq. 12.?EiqFaflin1c qFq3.9-r1lqiihztn Fgit.’ ........ .. H19}:_-I_.'-_-_'._._\"._~ _ --,3;-_'-_-._.-...._-'.—._'-:,. -. . --. .-. ._ -_ ,... ... '-:_..2-:11-. ,. '1];-__'. -|u='.-;=. -1 _ 5 -5--4 Iu _-|— I-.t'1-15--‘:!-2:‘ 1-; '.‘ .¢.Q. \"al F em..:!i\"-I=-‘,-l-‘lIl-1t|i'--l''-u,.\"-\"I'rT|_. ._I\".'_-.',i-;I.'_1,T5!§\",I‘. ,' -. - -_t \".P- - |1'i!'|t-‘T-_I.n'.I‘L._‘J_'*! i1--5I' 41\"---2]-.I}n-t----'ll' sures __l|t'~|E-_,l‘PI':I7I‘: -_ lIlI.-:-1.'5 |l L -1'! |';:t.-J:-TI|l':l l I:.?#. -J\":. 1.-:‘l Iii--=5-lei. .'.=.*.*l~'tl.'-rd{I1-.'l—''l'lI.\" -I I :-I- I ;'-l-|-f'I:\"' 1 -1|---ll-Ir 1-» “I I I 5

.1 sta-'l -u rFelgl.fit2e.5v.nieettrrteedleIsctetneceeldittP-etreorcnlgc,bieeort.i-a- elltthIitilh=ifi1-t;e; I~\.f-.-‘HQ! ii,’ .-55 ME E_nEh |||.qj{|;lil|'|-'!_|;l_|gg1j'|qr| .:§5 ;'§:t' tiittJ 'l as cl-rt\" est’ ; ;l=.;\" t q‘, =1‘ ‘it -q '-F‘ '==‘.=:i-TE - .. u i --It - -r *¢| \" \"lib I 1'‘it’. ll, ll‘ W 1B.fltt“tlll\"C 4.'D:r'..’llIl\"G - E: I;_..§ ‘qr We or e+e -\"ea\"\"'i1.i=i;—d — - . .. ....._.\"l...!-__. i;\"\"-.H.__1£|\"F_.iI\"'li‘ irir-an-;a-o. hilicltglvea trlgl-_1;i_ii_i;=s'ii-.f_l-ll There are htiio peeslttle _ir_a1ues of cl, oor1'es|:iondetoeiocati:inot'ItotlierlghtoIhottt;trie_et1u1gpe \\’/.. urheren1egniti.ideeofE.and E.ere eorrltfl-hut\"dlmcIp|@s~are sante.li1tt\iseeeeE,endE_,iitonolc=eneeletfliea|:_irtt‘l§lie positive velieccrroet:icnde'lici\"tIte.-iioatton shoinrr'1n=flipu'I'iB. ertdlstherterofieldlocetlor|,henee.d¢=+1.tlrrt,.;_==-5:~15!-\"T r a +\\*~. rlwisual represantaliion cl‘ lheelectric field can be obtained in terms of electric field lines; an idea proposed by liilichaelrs 1' ' l Faraday. Electric field lines can he thought cl a ‘map’ that provides iitlorrnalicirii ah-out the diection and 51.l‘Bt’lgl.l't of ll'ierg.t:_ag1!ltlI=iII'rtrHIdlltI*rl.pHildl'i1ililrl'iII|ll'|IilliclfllItill electric field at various pleoea- As electric field linea providedefile:-:i'ld'in'ip|le.i|i-iiterllrieei information about the electric force etrerted on a charge. theI tqtrtfltil tense tilcld H11- lines are commonly called ‘lines oliorce\". To introduce electric field lines-. we place posi'li'ire lesticnl-lruwsed.n-.u_-|iIy'rI=d-emtuhtt'i:I%nnl'llrluIIt charges each ofmagnitude q,,at ditterenlptaces out at equalneiilhIircI1l=I1IaI*~c- distances froth a positive charge +q as shrnvn in the figure. Each test change will experience a repulsive force, as\. l/3-\- ?” ll\"l2ll.t,‘sfllfEd by enztrws in Fig- 12.E[a]. Therefore, the electric; field created by the charge +o is directed radially outward. Fig. 12.6 {bl shows corresponding field lines which showlhe field direction. Fig. 12.? shows the electric field lines in the vicinity ol a negative charge -e In this case the tines are directed radially ‘inward’, because the tdrce on a positive test charge is new or attraction, indicating the electric field pointsin-ward. _i“T_njtH‘\ ll/t\"'-i: Figures 12.6 and 12.? represent two dimensional picture-sot the field lines. However, electric field lines ernarge from the charges in three dimensions. and an infinite ni.irnb-er of tines couldbedrawri- The electricfield lines ‘map’ also provides information about ti

the strength oi‘ the electric field. As we notice in ‘|rii|i|-lie;-llFigs. 12-5 and 12.? that field lines are closer to each othernear the charges where the field is strong vthie they ‘I. .1continuously spread outindicating a continuous decrease inthefieldstrength. -_- --Il'I I!|,-l_t-_-l\"'_|_- __1 . i\"i \" .p ;\" -§*‘\"'lr \" \"is\" 1 -I -I --.. .’r'itii; l| \" L .- U‘-.The electric field ices are curved ln case oi hvo identical _1 _.-rl .. ‘-separated charges. Flg.12.B shows the partem ot linesassociated virith two identical positive point charges of equalmagnitude. lt reveals that the linesin the region brB'l;1i'tl$Bt'lt'r'r\l;tlilte charges seem to repel each other. ‘I1-ie behaviour i;i‘itwpidentical negatively charges irril be eiteclly the same, Themidcle region shows the presence ot a zero field spot orneutrelzorie.The Fig.12.9 sltovrs the electric field pattern cl hvo opposite .4‘ W-1; trcharges oi same magnitudes. The ileld lines start from 't¥i-atlxlufi I.-'..|r~_;—I-I_I_I5'_ gs, ‘--.positive charge and end on a negative charge. The electric '-.~-_ Itfield at points suoh as 1, 2, 2 is the resultant otfietde created |I \"t,i..-' _. ‘ \"by the two charges at these points. The directions or the | _ yresultant intensities is given by the tangents drevim to the fieldlinesat these points. i i”ITl il2>iIn the regions where the field lines are parallel and equallyspaced. the same number or lines pass per unit area andtherefore, field is uniform on all points. Fig. 12-11] shows l'.l1efield lines between the plate-sot aparallel plate capacitor. Thefield is unil‘orm in the middle region where field lines areequally spaced.We are new in a position to summarize the propertiesoteleciricfieldlinee.ti Electric field lines originate lrom positive chargesand end on negative charges. -2] The is-irigettt to a field line at any point gives the diireclionotthe elactricfieldatthat point.3] The fines are closer vritere the field is strong and the linesaretanherapehnheretrtefialdisvieahttlltilfglr<-c§l lgl,lasttilNo two lines crosseach other. This is be-muse E hasonly onedireclionatanygiven point. tithe lines cross,Ecould have more than one direction. T

i[|| Jtorography[Photo-copier] Fig.12.11 illustrates a photocopy machine. The copying pa-grain-bocopled-la-oodcrirrrn larrMip _ __ lens lortorooifirrdge :,lr—I\"'I ocrritl-nrrtg bleolt lonerrhrst u—rdr-rial i hqggigji-gigy, papero-n'rIr11lcl'it1~rI irrtngets-prritrl-d t'.'*..\"-...t'...’“s'f.‘;.\"L'l.t'...\"°“\"\"\"’.....,..l 2.'.tt'...\"’.......\"\"\"\"...':‘..'tt.t'.';'.,“s%'t iII'lIHII'.TIlIII'fl'-Irililfir-W. _ ' fl '_\" ,2 process is celled rrerography. irom the Grreelr word 'xeros' and ‘graphosf, theaning \"dry writing\", The heart of machine is a drum which is an atuminrum cylinder coated with a layer , ol selenium. Aluminium is an excellent conductor. CI-n the other hand, selenium is an insulator iri the clarlr and becomes a conductor when exposed to light; it is a photoconductor. As a result. if a positive charge is sprinltlad over the selenium it will remain there as tong as it remains in darir. if the drur'n is exposed to light. the electrons from aluminium pass through the conducting selenium and neutralize the positive charge.fry; rig ||-q “yaw pg it It the drum is exposed to an image oi the document to beltmhi-ehusinlHl-latr,rl ehapniioi-rM’tIit'lmrtlh|ieidmlo.IrlIi'rr‘Ili'rIInroor.irr,go,h,r copied, the clarlt and light areas of the docuntent produce corresponding areas on the drum. The darlt areas retain their positive charge. but light areas become conducting, lcse their positive charge and become neutral.“'°\"\"““”\"'“““\" In this way, a positive chargelmage of the document remains on the selenium surlace. Then a special dry. blaclr. powder called ‘toner’ is given a negative charge and spread overthe dn,im,1.rrl'1ere itsticlrs to the positive charged areas. The loner [rem the i;ln,.im is translerred on to a sheet-bl paper on which the document is to be copied. Heated pressure S

rollers than nri-air the toner into the paper which is also givenarr e’-tlcess positive charge to produce the permanentimpresrsion oftha document.Ill] lrl!-Ijcl Pr‘ r'rl.r2-I\";An inltjet printer {Fig 12.12 a] is a hrpe of printer wfiripit usesetaotnc charge in its operation, while shuttling tiaclt and forthacroae the paper. the initjet printer ‘ejects’ a thin stream orinlr. The inlt is forced outcria small nozzle and breaks up intoetttrarrtel-;.r smalldroplets. Dunng their flight. the droplets passthrough two elaotrioal components. a ‘charging electrode‘and the \"deflection plates‘ [a parallel plate capacitor). Whenthe printhead moves over regions or the paper which are notto he inlted, the charging electrode is tart on and gives the inkdroplets a net charge. The deflection plates ditrert suottcharged tlrepsintoa gutter and in this way such drops are netable to aaadt the paper. Whenever ink ire to he placed on thepaper, the charging control. responding to con1ooter.turnsol‘fthe charging electrode- The uncharged droplets fly straightthrough the deflection plates and sirilte the paper. Schematicdiagram ofsu-ch a printeris shown by-Fig.12.12th}. rflnarprr-5 Ilr'l-|I.l-:1 to-I‘-I1 Ht-II-HJI i\"t ...;'\"-HII-IJIIIIIIIFIIF-ItitEl53$ii? iiilit‘atmoan-glddrnshllannimlernei-nnooonlnltjel printers can ate-D produce coloured copies.when we place an element oi area in art electric field. someortheiirtes olforca pass throughitt_Fig. 12 t3al.The number of the field lines passingthrough ecertain elementi i?/'or area ts ltnoir-n as electric ltutr through thatarea. It is usuallydenoted try Graelt letter ¢'. For tell-Bmplre the flt.|:-I otthrough theareartin Fig. 12.13 ratisd wlhletlteflutr Irlrrough Ei is2.In order to gitra a quantitative meaning to flux. the field linesare drawn such that the number or field lines passing through ‘ti

l ll—r- s unil area held perpendicular lo field Ines at a point represrerttthelrttenslty Eolthe llald aluial paint. Suppose at—-+ a given paint lha tralue crl E is -rill’-ill‘-\". This means that ll 1m‘ area is hsldporporidlculartolleldllheeatli-tiepolrit,4fiaii! inaswilipessutrsugltlt.ihordertaeslahiishrolstion bohrrean elactriciluttrlli. electrlcinlensit-;rE and area rt. wel—-n —+ rnrtsidar the Flg-12.13 r[lfll,¢.d] triiltieit shows the three tllrhertslrllrtfll representation of the electric field lites due to ai-:__L_i-J.. 1--|__,.,_,_._-I u'iilor'melech1cfieldoli1lensil1rE-.\._-.- .- -.|»._ - . qr. _.r In Fig.12.13 (bl. also is hold perpendicular to the field lines. lhenErt_. lines pass fltrrougit ll. Tl'te1'lu:rtlll,ln ihlscasais l—-ii -“-.-s t-1==+*=-t»_ir,- Eli?- . _. \"--' ll -' ' -—'-it ——Ir trrhere A_t:lttr'id1iee that the area is held perpendicular to field*4 lino-s. in Flg.12.13lc},a.rea A is hel-dparallaltofiaid lines and. aslrs-ch»v'it:iushelinerscroesllisaraa, aatl-|atrt1nrrp_ir1 thisii -\"'—Ir caseis in .Lflfli vrhera A, indicates that A is held paralal to the field lites. Fig.12.13{dj|shorrrsthec=aserrrhttrrtr1ls-neither perpenrzlioular Fl|.12.1'! up nor parallel to field lites but ls inclined at angle El with lzha lines. in this case we have-to find the projection ol the area tahiclt is perpendicular tie use field lines. The area at this prcrlectlon_{Flg.12-13d}isAeee£l.Ti-ie fluxtbin lhiscaqnls 1iI_=EAoorsEl Usualtrlltoalamantolaraoisrepresantsdhyavectorareah ruhosomsrgniltrdeisequeitodtesurtacearea.-tiorll'vs slomsntt_tind1t1'rosedlrectlcnlsdirectianefnorntaltothe area. The electric llutt ttr, through a patch ot‘ flsl surface in lotrrtsolEandill.istl'tangirrenl:iy tmerareisutaartgiaoettvaonlltefleldllnesandtlterramtalte theema. Eleotric11r.ttr.traing ascaiarproduct. is a scalar quantity. its 5| urtitiel~lrn’C'*. L-etuscelculatell'ieelec1|icfluttU1r'ougltsdrosedsurfsoe.irt shape ole sphereerradiusrdua loa point charger; Ill

placietl at this oantre of sphere as sriorri-n in Fig. 12.14. Toapplrrthelonhttla lil,= Ejlilorttiaootrqsutalioitoleleotitcflita.tlteaurlacia qreasltoultlb-all-sl. Forthi.eroeaonl1'ielr:i'tolstrfaceareaofthesphoroisdiviiadirrionsrriollpahttltssurihtareas olntagrtlttrdaa M..M,M, ......i|iliil..,-lfn is vary large.each palchtrrr:|r.|ld lsa'aIoteiiernentolorere-'l'l1ieoorreaponding tractor areas are dull... fliig,respiectivel1r.Tha dlractionoleactttroctorareaiaparpenticular drawn outward to the ooirrespondi3iiiTliaater.ttt1cin'r-ansitiesatti-ieoentraeolrractor£|.illi..£irli.,. E,. E, ---.----E_respecti-reigr.According to Eq.12.12. the total 'l’I.t:tr. passing through the I--.-“-l .IT._.'-l.,1_. _.|.:..'..;-. .‘|i_J-,,.-_.-_I.-._-.cioisadattrfaoais _- _ __ ‘_J_T.,.T_.L .1-i.i_ -_--J__-I‘_-:. -. -T .. ._1t'1;.; ;~ -' -- -_ _- _ - .Thedi|'ectiono1alectriclntenailyarid\recloraroaisaatrtoateecltpatr:h.liloreovar.bacatraeolspltar1cda';immatr3r.atlltosurfaosolspriere. ->t-.~\"-=- fr-:'1-- -- ‘-'r-__ \"-lbs-, tnJ-1' .i.ti.._ 'r_i ..- 5\"t;-'-_--5.2- .-.-:-_7__r-;.|:2 J-‘U.__3r=:. |-T. -£1'-5.\"3-j:ir1-_. . \"1\"\"..':I'l.'_ ll‘.ll- tli,IEo.-i.+Eo.A,+E.oA,+ .......+Eil!iA,. =IEtt[.ilL-li,+.ilii.ll,+iM,+ .......+a.it.t =Ett {total spherlcelsurlacia area} i--‘-fi-—1t,—:H3- ! 4IF 3 . _ \"r\" \ .-,4 __ t=tq,t:_ta_1. _ I‘. '.:.-.F‘. is _._| ‘I _. ._f_..- -1\"I-i '-1\'- .!Ir .-_:_1.i.. 1.._\" ,;;:E1-.“-.._4_. E-1|! 3‘-\"'I“f-'-lrlowlniaginetltataclossdsurlaioafiisarioiositgltiasphare.llcanbeaeer1lrtFig.12.1EithatthallutrtlIooghlhoolosorr:lsurlacefiietlteeameasthatthroughthospltartetsottrooartooricludelzhattortalfluttdt I'll\"-ill haclosadsurlaroadoasnotdEflrEl'tdU[II1|hBShE|3EDfQflflfl1D'U'§'Dffl1IdflGId5lJffIOl.lldepends uponlhiemeditrn and lhiecriargo enclosed.Suppose poiint charges qt, rt, qr, . rt, are arbitrarilydistdhutad it an arhltt'a|1,rahaped closed aurla aaahorirrrt inFig. 12.15. Using idea giren it previous section. tlte electricltuiipassinglliit:tug|h1ltac:iorseidsurlaioe_ls HI H 1-

(Q-=gL |. q .| HI\". ;, 1'1 Ea E: E-U ED at-_= —1 .'r[q, +13, +41, -........+q,,) c,=__—1 at flclat c5h“arge enclosed byclcsec surface] re ~it,=—1 I'D ........ .. (12.15) EU where D = q, + q_. +q-_ + .... .. +q,1, is the tctalcharge ertdcsed by closed surface. Er:|-12.'1E is mathematical expression of Gauss's law which can bestatetl as. - .. ._._. ....::.. ‘Iii____ . F -r-*'+ Gauss's law is acplleo to calculate the electric Intensity clue to rtitlereht charge confrguraucns. in all such cases, an Q fir imaginary clc-sect surface is considered which passes -l- -l- through the paint at '|'|‘l‘ltCl'l the electric intensity is to be1- GilLt1-=__lluIt_'l,_S._ltiiflfi ‘I evaluated. This clcsed surface is itnown as -l\".-iatlssian surface. Its choice is such lhatthe flu: through itc-an be easily evaluated. Nest the charge enclosed by Gaussian surface is calculated anrt finally the electric intensity is computed by applying -t'5auss's law E-q.12.1E. We will iilustrate this prcceclure lsyrccnsirlering scrne examples. 1I1] Ir_iteris._lty._u_‘ .F.ield irws_icl.c-a_rlcllcw Ch_ar_g:iirt F Supposelhat a hollow ccncluclihg sphere cf radius Ens given a positive charge c. We wish to calculate the iielcl intensity first atapcrnt inside the sphere. .- -I‘ New imagine a sphere cf radius Fits R In be inscribed within Q the i1r:l|hew charged sphere as shcum in Fig. 12-17’. The s l. um’-Ir'3- surface cl‘ this sphere is the Gaussian surface- Let ti? he flux ‘.1-.1‘ J 1- through this closed surface. It can be seen in the figure that '\_ rt‘ _‘_ the charge enclcseu by the Gaussian surfaces is zero. .r r\" Applying Gaussian law. we have ‘\. |'\-i R. i Since o,= 5 = c 1 1- ,|, 4- 1' E9 n; 1111 ¢t,= EA = D as Ar Cl. therefore. E = D I2

Thus the int-erler cf a hollow charged metal sphere is e field ._ ,.|tree regicn.A.s a consequence. any apparatus pieced withinemelalenclesureis'st|ielded_'fromelecai_cfieli:ie._ __it:-| Electric lnti:-risity ljtiie to an ln*iuitr- Sheet -ct 1,4 -;i-__ ____;_____ Eli:-irge _| _Suppcse we heire B plane sheetcfinfinite estenten yiihich ' ',.__, . __._In H,_ .-pcisltiug dflrges. are urifclnnly l:li=et1ibu'bed_ The ul'til'tJi|‘l'l'l '' \"‘“surface charge density ts. say. cr.Atinite pertctlhie sheet isshown in Fig. 12.18. TocalculatellteelectricintensityEetepeint P, clcse to the sheet. imagine a closed Gaussiansurfacei-i thetermctecyiindarpessingdtrtmgltttie sheet.whceecnetlatfcce ccnleins poi-nt PL Frcm symmetry we canconclude that E pcinisat right angletetheend faces andainrayfrcrhtl'le plane. Since Eisperalteltethecuruedsurfaceciftl1ecytlnder.setharelsrrpccrrlIihulienlcfltnrhcn1tltecurved wall of the cylinder. White it will he,EA +EA=2EA.ihrcughtl1ehucllatandfacesofd1ec:lceedcylindrical surface, where A is the surface area ct the fiatfeces (Fig- 12.13]. As the charge enclosed by the closedsuriaceia tr.-lt.1;l1erefcre. eccardingtc Gauss's law. c_=£l itciiargeenc-lcsedhycdcseclsurfece Q_. _. _- e==§-intuit‘ -__-=\"-1;_ - . -fi- _. ..., - ._... .-_1s '1,Therefore. 2E-'l=£lxeA5, - , E..L\",1--=1-=3:s..::.;..--7'1=?lfesis-. .__| .- s .I_. _ . _-.-_:-_-,-_..-s__ -_ 1- -- .i...-:.---'If1I\"'_\"u _-. _!.;' '.\"T‘?'t\"_' 'r_-'----_--i.-.--'.!1\"-\"“'““‘“\"\"-\"'7\"'- -' - E ‘is’?-.111:-s'='é-1-'.ij_i'i‘;ii'i_”t3jfififiwfwhereFisaunitvectnrnornuitcti'Iesheetttirectedawayfi'cmiL1L| Eiijiijttric ||‘itij!r\"i:5.it'y E-r-tween Turci Clpp-:s|.'.Lly Cliurguct F'cr._il|e-- Plates upp-use I et | para = a I i-\"l\"-.\"' \":\"- ~-:- :- --i Iesinfinite extent separated hy yacuurn are griren cppcslteduarges, Under these eenditiens the charges are essentiallycchcentratecl en the inner surfaces cf the plates. The fieldlines which originate on pesitiue charges cnthe inner facepi‘ one plate. terminate en negalnre charges on the innerface pf the ether plate {Fig 12.191 Thus the charges- I3

ereurilcnnlytllstrlhutedcnttielmersurfacecftheptatetna icrmcfsheetefrcttargesofsurfapedensllyesqt.-\t_tthere.it isthaareecfptateendqlstheamountofchergecnetthercf Itepietes.tin knsglneno\v_aGet:sslensu'facelntl'iefcn'nelehet+cwihcs wlthltstcplnsicelheupper|netai1:tetear\"iditshottomlnti1e specehe'trneentlIepletasessltcnni1F|g;12.2tl.Asthefield lines are parallel tc the sides ct the cos. therefore, the tttltt thrcughthas-idesis-aerc.Thettetdtlnasare urtltonnty -_=-_.- -..-..-_--._ clistlitiiuted ch the tower bettcrn face and are directed'' ' '-'-.;. '-_‘_~'§'-1. tnnotermheaRllyy'tBottllt.BI£ttAtisl.s1hthseflLahrettat1ctctutghhisittvarpcetldattnedEE.i\tt.hTeheefleecistnricc‘- fltptfltrcughtheupp-ereridctlriehoxbecaLmetltet\"e.isriefield — \"' ihsidelhemelslplstsi.111L|$tt'ietetxIfl1|:ttIl_tl1rei,|ghtt1e GaussianaurfacaksEA.Thecd1argeer1cdceedhyti'ie Gaussian surfeceiaefl- Applying Gauss-‘slaw c,-l ltfllilt pr E.-\t=—:tcut an-hf‘ - .|_.- .1-u \" ; - - -_ __._s-_i ':-']--'!,'-t.'\"-\".j-__!- -1.;-1-:33. - @.- - . s, c-_-.=._-,;r.-; -. _. ___q_...|_ - . 1 __ ,|. :__:_. , _. __ .| _ . -._l' --.- -:I,-..:,;. -- .?\".'--., ' \"\",__|;|..|.--.1;-' ;'_1.-.--r.-\"'1-s_14-.H - ' .--rt_-.1'.:--.- _ . -',- -'.- , - '-:.e\" \"\"\'#\"+'~1c“\"-\"I\"II: 1--\"=,J1.’ it ttlefialdittetisityisthesameataltpcittsbeheeantheptates. thelzliecticrtctfietdisfrcrtipcstthletortegaltseptatebecause amltpcstthecttageenyvrherebettueenttteptatesueutdhe repelledtrcmpestliueanctattractedtcnegativeplaternd tl1esefercesarei1lhesameclirecticn.lnvectcrlcrmEI- wtrereiisaunttvectprdiecaedtmmpoetfiuetpnegettyeptele. LetLiscorrsider*apesiliuecd'iargeg,wt1ictiisallcwedtcn1cy-elt- lnamelechtcttetdpreducechetneentvieeppoeilelydterged Fl. 1121 tit pal-eletphtesessticv-ninFig.12.21[e}.Ttrepcsitiuec=l1arge ttriltrricuefmrncilateBtc.handwitlgelnK.E-1titlstobemeved ficrnA.tciEl.aI1e1dsmalfcrceisneeideidtomattett1er:herge meueagaii1sttheetectrtcttetdar|d\ttltgaitFtE.Letirstrnpese accntltltprithatesthec:t1arIgetsn1cuedfrcm.iflitcB_itlsrricued loeeplngetectrestatlceqiilitiriuh, i.e.,llmcueswiI1 uniferm veicciry.1hse:rie|scnccueeesciseyecbysPnlri'learcice Fe-quelandoppoeitetcq,Eateverypcirrtalchgitspati1 H

as shown in Fig. 12.21 (ti). The wort done by the etrternal iq1—.|I.n.lorceegairisttheelectlicfietrzltrtcreasesetectrtcdpotenttaleriergyefthec=l1ai'gelttattsmgued.Let W.“ theworkctene tiylheforcein cenying tliepoeitilrecharge tL'tl't1tIl'l'l.Pt|t1lBti\l‘|\"lt|fl keeping the charge in equilibrium.Triediangeinitspo-tenfialenercry cl.-'=litr‘.,it ' - - '\"-==1 .i-_&\"‘.-I-lit-\"~1'l'li\"i\"‘='.=l:-'.*-I‘l'-so.-'.~. =1whereU..and Li,.arede1tnedt-otielheprrlieritialenervgiieset F‘.1L21-fitpolntsiltand B. respectively.Tcoescrioeelech-tcfieldw-einuodi.icelrieideeofeiectr1cpotential The potential clttererice between twopctntsfliandfl nanetectrtcfteld istleflned as theworttdonein carrying a unit positive charge trornnto Bwhile keeping thechargeinequiitirium. thatis.=--=- - - ~*-- -5:\" -1.\". 1. .r .- rr|.ur-nn-o|rruuInn-:llrruelnln-.sI-uneuII.nI.I-IIuIIIsJeI-I-In1uuIsI'neII-nIalcnIuInmIiIsennI:aliIlrghiffi-iifs,-einoiewtti':,i?tcdie§ll>m!i‘c~esle’c\"ttr%icpotentials sipt-ofiiinnionieeieif IIIIItI-IiIailIcIlIIlniIIteIlilIsIiIiIaiIIlnIlialrsII-nIIiIIniI|Ia-I-|IiI|iI.IlIniIIlaIl-tI.Iln1mInIIL|IialII-nIIrlaIdII|I|I-.IInBII'nIIlI|aIiI.luiJr1| |taI-.nII ,n‘iIP|l.Iuir1I.n4Il1\lIn.IlI-Iul4I'lli1-nl.1|lI..lI|“'I.Il|-.I.r.-uIH‘lIInl.II_llI.‘PI.n|I_I|II1I'.II4I1I;'-nI‘PI'ncIIFIaIrI-I'|II-InI-.III_n'I-1I-n-''-III|,IPnI.II-'a1lI-'InI.J'I-I. respectively. lectric potential energy ciiference and IIIr--I|lrI- -HIIsIIu-IIIan'IIl-n\"Inl|I-ul1I1l-I-'u--In-rPuIu-n'naII1'|nIlIIl:IIrIII.III.\"II':rIII-III4lI-II1IlIFIIII:-II1I.IIII.-I'IIltlaIltIl't.IIiIIItIl]aI1-'tII.aegct-ac potential difference between the points-hand E are '1 1:“: .' . 'F:1':' l:H,':-': l:::'i.\"..' _ as ..I-:l.F:::l:ll; -:7 :3 ::Il:-::'::- iii I -.iiss4i t:;i..t:t::.ttitE|.t §ir'.'::i'. §:'t:t§i§ II‘|'III'I|'I'I1lFiI‘|I'II' .'IrII1IIII'I'II'Il'ITl'mstl'iepoteritialdiffer*encebetweenlhetwopoi1ts-car1tia II-IHILII-I-J'1I--rI-II-IsIiIlIl-:rnluIlrIln-IIJll1lillllIl-llI|-III E -‘II 1.defined as the difference of the potential energy P91’ unltcharge.PlSl'BLIi'lClP.E. lsiouie.Eq.1?_2s iiisis itstiitttpi_;iterilielcliflsrsriceisior.teperccr.licIr't:i.-is s-='-..i-'-\".--1'--:ti-‘=I\":‘-i''! ‘rac'|Ee_-.-E.*1'-::7?11. r'f.ii\".-;.:''i.t?\"‘?t'*- 1-That is. a potential difference cf 1 volt exists between twopoints ifworlr clone in moi-ri_ng a unit poeitiye chem frornonepoint toother. iteepiing equilibrium. isonepute.In order to ghre a concept ofelectdc potential ate poii-it in anelectric field. we must have a reference to which we assignzero electric potential. This point ls ursuatly talten at infinity.Thus in Ec|.12.2T3. lfwetal-tehtp beat lnflnlty and chooseVac. the electric potential at Bwllt he v,=w_..e.orcmppiigthe subscripts.\"s-____ .i_f.'-—--|- :-\".c',I__ ._ : ‘\"1 r 15

whicitstatestltatiheelactricpotartlieiatartypeiritinat etectricttetd is equal to worltdone in bn'ng'|1g a unit positive charge from infinityto that ppintjsaep-irtct it in equilibrium. It is tobenoted that potential eta point is stilt potentielditlerence between the potential at that point and potential at infinity. Both potential and potential differences are scalar quantities because both Wand q.are scalars. Elli?-ittflll--tliltfllifii In this section we will establish a relation between electric to-ttinlclclr-Ind iitensity and potential diftenenoe. As a special case. let us racer-tigonice consider the situation shown in Fig. 12.21 tb).The electric ttetd between the twocherged plates is unilionn. let tlsvatue - beE.Ti1apotentielditferencebetweeniitendBtsgivenbytt1e flIl{IlItIllflltlIij'l|ll'II | equation where til-',., = Fe’ = - o_,Eo' {the negative sign is needed i-1- becauseFmustbeappliedopposi'tetoq,Esoastolieepitin equitibnunli-ttnthtnis.Eq.12.2Tbecon1es . V,-‘t-\",.=-E! =-Ed We 5._=' - I _ 'L--_:..i_-_\"-'\":-.'-=i1__1_5»_;..-_,..'..\"=;~:--.=-§-.-r-.oc.-=-c--it'i'i;-'\"-'; _'.i|';f.-..,t-#;¢..._\"I- I=.' 5._i=i:?~s.;i_-'1_._'--;:fi_l_-rl_ .. r . --._1....|. .-.- _ .l Ii’ the plates It is B are separated by inlinitesirnsly entail Poterfl clistarrcad.r.tlisE.12.2tiisrncdifiedas _; *_ _ dllrilri-ct _ -- -'-~ \" -'- -= .- -- .-\";;t;;..-n-.t'i'- 1-,... _...1|. cThhaengqeuoefnptoilt-eoynrgti‘avl ewsiththdeismtaan:ictiembuemcavuasleuethoeftchheargraetshaosf been rnovedslong alietdtinezaongwhich the distanced:his between the two plates is minimum. tt is itnown as potential gradient. Thus the electric intensityis equal tothe negative of; l‘I-- r I ' I -H-..s-‘II-s'-\"‘I-,5I-?--Ir't it--t.-.-.I|.tJ--..I'-:: '- the gradient of potential. Thenegative sign indicates trial the; '\"%t.='l r=t~i-.sl1..-':*_:-J'-.r:n-.-- .|5 ‘ i.\"'.-\" ' directional E is along the decreasing potential-_ -. .'.' -I \".y-1-F|_.-'‘I.L? -'.i.r'---- '\"-._..\_.-.:l_;q_.. \" - \"1\"\": -'- -1- Theurrit ofeiectric intensity from Eq.12.29ls voithnetre wriiei iseoualtotitti ‘as shown below. 1i=1h.\"'°l\"l.E1‘E ,1 =1h metre metre I rnetretieoutonlt cotlomb lit

E||;h,\";1_1'iC F'{HF1|‘|HEJiflt a Point due tn a Point - -.f'. --- Cher 9 tr__t,q¢m;.derb4eane:p|'esstd|1terltrepete|1tia1a1aeertainpnimin ..-.... .3 d1efialdulapdeimrepeinld1argeq.Ttisem1beau:e|11pHhed byhringingaunfl pueirive magefmmmflnltyleflmtpcint F|q, 1Ll'.E Ivtflep-I fl1e|;:hargein equi|ib|iurn.Tt1elargeteanbeaet'tieved uelngngq. 12.29 in the form av = -E hr, provided electric ir|ta11.sityEren1airte:I:m-tentHdwe\Ieri1tl'ieeaseEvarie-e inversely as square crf cietenee from the p-uinl I:-harge. fl 11¢ mereren'|ai1s::dnetanleuweus=ebasiepflrlciJIe-alndemptlte maateuhidpdlantielalapei1t.111efieldiera1:Iia1esshu\m1in_ Fig.12.22_Lel us lake two points A and B. infimteelmally eleee to eachdthemlethat Eremains alt-ndsteenetanl between them. Thedistance et pqinteh and B from q are r, and r,_ respdelivatyand distanced!‘ I‘l'lid|JIDinh‘;ll5-FELDB interval behueennand B isr [rem q-Then ac::erdlngtuFig.1l'-22._;- _. ' r,_Ir,, 1-Eur £12.30] -._m-'=r,-r, (12.31;Asrra-|;|re5B|'|lE|'|‘|id pDi|1lDHfllBN3|bBMfBflflP\3Ild Ben -_ ,--&;T\":_ ' ' 112.521Themagnitudedieleclrieinterlsityatthispninlifi. £12.33]-, _ El du1n 1r,A3 the paints A and B are very cldee lhen. as a first HllfifiiiflifieB|3|Jl'I;HItiIIlBllr3l1.\\'B' can lake the arithmetic mean to be equal ihtlitlflilhliliflrflwlh.tO\"gB='UII'IBlfii:l'l'iEBI1\\'hi|::hg'vvB5 H'fl'kil'I'I'liiI'§I,i- I10-'l|1'|I-bl di L-erI|'ntH. Int In L_5 ~ vI|IlI1'||IijnIniudI'i-|:|\iI_-Ind-:i|'l -dllpd pdhrnll i:fllrI|'l:\u cl thi- r, r‘ mwdnumulflmbuunn'-|'hlrl=Iure, r.‘ =r, r, - (12.34; il\"!\"I\"\"l'YP\"l'°*llfi'-Trtus.Eq.12.33c-‘an be-written as '._ 1q __ ' El --—-— £12.35}_ I .I I I ‘Ian far: _Mmv, if a unit peeitive charge ls moved frem B teflt. the werhdune is equal tufl1a putenlialdifferenee belweenhand B. '|.|\"_-L-\"_,= -E[ri- 1\",} ' 'lT

Suhelfluth ' aqtzas '' _ Teeelcalateahedlut-updlu|1llddrpetunflulatA.p0ir|tBls.~ fi'_,_;‘§,\",,,\"§',\",§\"\"\"\" assunedl:dbohir|ltypdi1ledfl1atV,=flendhenee '“H*u*|# I lll-1-D '.. - _ri ‘fl In. 111e5|enera|exp|e::idnfdre¥e:htep|dle|1l1a|\F,aladislinue rfrdmqie.;Pu||-u|ln'I'CI RBI?-P._% HUI-ll-lulfllllllfll- §i“‘ .».. FlBfBlTi'lQFig.12.21.WBhrlWtfl\BlIhBl1BIlGI'tlflBflHi'lBfQE qmweefmm pelntAwitl1pdtenl;ie| V,,tdepelntBwifl1 '-. .. pdtert-fie! V, keeping eiedt|1delnfidequilbr1um.tl1eehar1gein_'_ -.' -.,.._.-,=-=.. ., .n|..-r_'--.\"\\":... potanlt=eienetgy.fl.Udfperdetele.' IIr|dautle|ndinrdua»e|een'I1ed'iu'ge'td'n1}i11ilhu:|L|I:rh|n. lhtsd1ungui1FtEnppaarsi1l1afdn'ndlet'nr|gei1K.E. Euppde-echargecarfiadbyfl1eputlcialeqIeI1.Bx1IJ”'¢.

‘hue. in this case. ttteenergy acquirer! tnytliac-t1a|=;a willbe else =qsv=ssv-l1.a I 1o\"‘t:t[eu1»ioreo1rar.asaurnett1atM-‘=1 1-roit.harioe oi-LE - qov={1.es1o\"‘c):{1l-on; £it{.E =(1.s=1c\"'}:{cxv}=1.es1o\"‘-t‘he amount cl‘ energy ec|uatto1.E~ t-:10\" J is celed cnerlectron-volt and is denoted by 1eV. It is defined as ‘thelmoont ol energy acquired or teat by an electron as itraveraeaa potantiald_itieren-oeolonevoit'.Tt1ue. it} . '-t_‘_|'1itg:.1.e==1a\"J ' ........ (12421Eiitarxqoamumgrhspuleeb1p_1erlt2e. .r1le:ssAdtpliuerfla- dnosecoalnarimnq. aCaclctuialartgeathoelieenelargttey q=ile , .lllll'l3.lt‘ll'|'ltt!tH'lt2lfflt_|'l=ttlt]ltlt'Ittlltl[|t'llte|:let|1;it‘;laIi$ ' . sic: =qev-las]ra.ow=eoav I=$.litr1.t5::1lZl\"'..i =tl.fl-:r1fl-\"'Jn chapter 4, we pointed out that gravitational force is amnaervative loroa. that is. wort: done in a-ooh a field isndependent ol path. It can also he proved that C‘-oulomlataelectrostatic force is also conservative force. The electric'oroabet\\-aentwo charges F = £211; ,issirnitarirl1‘orrn tothe gravitational force between tlle two |:l-oint rnaaaea.= =t3-mr-\"':-31.Bofl1foroesvaryinverselyvdfl1lheaouareolfl1edistance between the two charges or the two masses.l-tow-ever, the value of gravitational oionstant G is very smallea oompared to-electrical oonalant e4115. It is because oi thisi-act that the gravitational tome is a\"u-ery weal: foroa asooropare-d to electrostatic force- As regards their qualitativeaspect. trreelectrcataticiioroe could beattractiveor repulsivelvhie. on theother hand. gravitational loroe is only attractive.Anolherdiflareooetobeno-ted iethatttiealactmstatictoroeismedium dependant and can be shielded while gravitational 19

force laclts this property. In 190$. FLA tutill'lr.ar1 devised a technique that resulted i precise rneaeuremenrtotthe charge onanatectron. C . - \"..\":|;':.Ii:.':.££'d:i.'-;rr,J 5ri-if.- a _..__ __ fig. lump F. - ii! A schematic diagram of the htilliltan oil drop experiment l. shown in Fig. 12.23 la}. Two parallel plates PF‘ are piece: Di mp inside a container C. to avoid disturbances due to ai cmranls. The separation between the plates is d. Tris uppe F.*\"\"'l plaleF' hasa snlallholeH.asst1osinintttefigore.Au-oltagel - - -sq -- - is applledtlothe plates dueto-nhichtiieel-acizricfield Eit_ .._____l'f'f'§§1§-'_'-;§j=_j-fit;-_.r:\" _ setup between the plates. The magnitude of its value it E = ea. An atomizereis used for spraying oil dropslnto the container through a nozzle. The oil drop gets charger because of friction between walls ol atomizer and oil drops These oildrops are verysrnall. and are actuallyln the form o n1iat.$oma ofthase drops happerlto pass throughtheholeir the upper plate. Thespaoe between the pistes is ilurninater bythe ligil coming from the souroe S through the lens L ant Wlndthtl W',. The path oF|'|1|olion oi’ i;hes.e drops can bl carefully observed byamiorosoope M. -it given droplet between the two plates could he suspendec in airiflha gravitational loroe F, I mgacilingonthedropit aquatic the eieolricalt‘oroe F, - us. asshotrrn in Fig-12.23{t:|} The F_c-anbeediusted equal tofibyradjtietitgttievolage. tr this case. we can write. _ or ' qE=mg ltlfisthevalt1eo1'p.d-hehveenli1eplalealorti'|is ae‘tt:ing.1her H Ii] '

= E=Eif ' wemaywrita 'il'1!1'' =\"\"Q humarludatalmir-H\emanamuifl'|a¢l'cplat.maalac=1‘n=|ldbfln'aanl1apl|hu:siu'nitd'~adufl=Tlrad|'unlaIlaIIs-undurmacliunolql-niylvaughai.nau.uln:mrninal:und\¢a1moslallheifllnltI'lIlil1:lrl¢fiHdill'hi:hB5nfl'.ltilIm'iIrBl:pfildv,i$uhmirodhyfinirlglwialnlhawnflflwaranuasumdb§'tlnUl.$iMiIPItI‘I§iUI'¢lFdulhliri¢l.iflQ\|nMU'|idI'DQlHfiB11itilIfi\"|Qwflh¢¢|'|IiInltGflfllflI|&nild|5B£l|fll0fbiVIiQht.' FIEnn1rv.Imgrhuarlsfluradlusnilheurup-latand :1 lsmacnalfldqmuiisnusilyfurai.lfpis-lhadansityu'Ilr|udr=n|:lat,H10hHfltl, ‘E §1|:r‘pg=BII:\1rv,r I‘? 2 9 “H-I I-’|-W _ .irumiwgihavalua olr. the mas-5 mcan tyac-abculalad byslng Eq.{1.'?,-15). This value of mis s|.|b:|:i'oulad in E_q.12.4-I:gel1i1uvaIuunId1a|'=g|q0nll'\udmplat. 'Iilkmmaasuradmachargann marqdrbpsaawdlbund mman-I1 U1-'iI'EB was ail integral mullipla crfa rrinimurn value ofhargenqualtu1.Bx1D'\"G.Ha.li1enaIum.umc|L|1uflIh=atlhlainhummkmnlmedurgablhadwargammulecrun. 1|

Ferfltedlapietelflledtflfflflq-weheve _. _ _ . fa _ .-R;-e_ I: _q..e_e_1g_!F_|g eea me\" ire.e.rr1e'l' m_- ' - v , \" : -raw = - r _qI3.2:r1D\"'G \" '__';- _,0 . _.;, Acepaciler la a |:le\-'i|:_e that can store charge- It ceneiete -r twe eenduetera piaeecl neer one another separated l: _’_*|:“‘“\"- g uaeuum. air er any ether ineuleter. ltnewn as rlleleclrii Usually lheeencluciere are it thetermetperelet ptelee. an1- - ptltaerceaepuericeiuheernieelegiaepuemeateupreermeleleeipnlnaeteelaepuamclalmnweetteenrymr I12 voltage V|[Flg.12.24}|, iteelabli-ehee e potential clitlereaeer Fla-11-H Vault: between lhehllrepleles enrllltebalileryplaeee el-large +C.len the plate eennectecl vi-ilh ite peeiti-we termim ende charge--Den the otherplate. eennectedte its negalhr tenninal. let Cl-aethe megnitudeelthe charge en either-elth pletee. It iefeund that Darin\" of .flI_GV DI‘ ‘C-\"lg \"['I2.'lifl The 1!lt‘t)|DIDfliOl'kB|H‘_|l' cenetanl C la celled tl1e eeperzitarrce 1 the capacitor. he we e-hall see leler. it e'ep-encle upen th gee-rneu'3r'eIlne plates endthe medium between lhern. It is rneeeure ef the eniity el capacller lle atere charge. Th r,-apaeitanee er a capacitor can he clellner! as the arneunti charge en ene plele nepeeeery tap raise the petential at rm plate by ene --elt with respect te the ether. The SI unit i3 fcraepqaI1ceitnatunscee.islacc+emeumipemnhl'y'pcearlleveelli‘le. rewr:hl{iFch]|.eblteecI’etuheeeIeal1‘tltni.'W5‘-aw,“ .-,-,-5 Englieh S-ClBfl|fl5I..FH|'H'dH!|'. II. “:“\"~“m\"\"m.m. .\".__\"-.H|.iR.F“.2_F\",'F._.F..W._l“w\".hl...lh.'.fl-u\"\"\"\" d\"e1iue'lr'pta|Hne-:alflplflelcfltlwlallt'cl'_:e'Bi:nfiltacle|n'm1:=hnqh-nhr.-uureeaudruueuc=:eaeltaIlr1pq'eip.hnla|!ullIehIe_I; Censiclar a parallel plate capacieer canslellng cl twe plan metelpietee. eeehetareaa. eepalatedby edietance in shown in Fig. 12.24. The dielenee cl ls small ee lnal lh electric field E between the plelee is unilerrn and eenfine 22

atrneat entirely in the raglan between the plates. Let initiallythe medium between the plat-ee be air er vacuum. ThenaccerdingteEq.12.4ti. _ __ __ _ I G_- TD; (12171where Ole the charge an the capaciter and 1-its the p-ctentialdiflerence between the parallel plates. The magnitude E el’etectrtcinlensltyle related wilttthetlietaneedbv Eq.12.2-Bee' _ Er; pa.-tat “I.AeQiethechergeenei1heref1tteplateselarea.ll,ttteeurfeaedensitvetchergeenthepteteeieae - I =1?-Ae aieedtr ah-awn in sectren 12.3. the electric intensity pnnlll-elllatltlilvbetween tvra dppttraile-Iv ctterged plates is given by E = ' Httrtrtfi ttrleillSaba-ltt|.li'|glhevaltreefc.weheve ’\" EI_r® _ 1'.._'?_ ' t:t:t.4a; I.-‘<1.-'_- d kg I Ft|,i 1I.I5Efldl2IlkI'.ilSI!l1I'I %i'HI'flIi'- Ittpt-in - c,,-g-4’? 112.:-atIf an insulating metettal. celled dielectric. cl relatlvep-etmittivil-tr r, is introduced between the plates, thecapacitance at cepeciter ls enhanced by the teeter 1,.ti?-epacitela ccmrnentv have sernedielectric medium. therebyc,leeteeceltedaadielect|'iccenetant.Fellcrwlng experiment give-stheettect cl insertion at dielectricbetween the plates eta capaciter.Cencider a charged cepaclterwheee plates are cennected ta-a vettmeter (Fig. 12.25 e). The detteclicn ct‘ the meter ls etneaeure at the potential difference between the plates.When a dielepttie material is inserted between the plates,reeelng drape a decrease In the petentialdillerenee between the plates (Fig. 12.25 bt. Frcm thedefinitien, C = O.-‘V. eince V decreases whlte O remaineccnetentthevalueeftilncreaeee.ThenEq.t2-5-flbectcmea. _ _ ' c_-it? 112.51} 23

Ee.12.5tt shows the dependence oi a capacitor upon the area oi plates. the separation between the plates and medium betweanthem. Dividing Ec|.12.51 by Eq.'i2.5t) we get expression tor diatarztriccenslantias, . I - s.I%5- i[t2.ti2}_ From Eq.12.52 dielectric co-eiticient or dielectric constant is defined as !'-\"'-I\"'l.t_ei_. i \"-\"_1-'- -1 .ole --I. -._ .it- |i- Int.'I_ ---.---i -l\"..-_I 2- t ' -. '-'-' i- \" mt.“ _ 1:;--“\"\".\"5'T]i'.'!\"-<~i.-.=.-ii'_i‘-‘= ;.. .. -. . ‘riiiieerirliiriirreiriiiriiriiiiii-r_iiinirt-i The increasein the capacity eta capacitor due to presence oi diatectricisduetoeiectric polarization oidielectric.1'-'I‘KII“'lit The dielectric consists ot atoms and molecules which are electrically neutral on the average, i.e, they contain equal amountsotnegative and positive charges. Tlmedisrnbutionot these charges in the atcme and molecules is such that the centre oi the positive charge coincides with the centre of negative charge. Wltien the molecules oi dielectric are subjected to an electric iield between the plates oi a capacitor. the negative charges [electrons] are attracted towards the positively charged plate cl the capacitor and the positive charges (nut:-lat} towards the negetiiieiy charged plate. The electrons in the dielectric iinsulatcrji are not tree to lttcive but it is possible that the electrons and nuclei can undergo slight displacement when subjected lib an electric; field. As a result of this displacement the centre of positive and negative charges nevrnoionger coincide with each other and one end oi molecule-s shows a negative charge and the bitter end. an equal amount oi positive charge but the molecule as a whole is still neutral. Two equal arid opposite charges separated by a srnalt distance are said to constitute adipola- Thus the molecules otihe dielectric under the action at elechricfield become dipoles and the dielectric is said to be p-alitrildtil. ‘i'heeltectetthepclari2a'lieneti:tielactncissnonni1 14

sias tsitE?asget;tiiitithe dielectric in contact with the nege _ —' _.. -. _‘ _.__|__ 'pieceealtiryerofpoei1iverharges.ltelactivetydeo'easeethe ,.flllTB'EBt3i'Bfi5i|}t'UlihEd1EIQ'BfltI1‘i|lBDiBlB'§.tl5i5li']B electricinteneityEbettveenthepia1iesis §'.soEdect1aaseedueto |[C'j_e_iitir-I;i\"s.I\".L\"_C':'_.i-'-...iI;_I- 1,__]|polaitzationottliedieiectrlcttisresuits-ittoedecaaseet Fig. 12-Hpotential riflerencebetweentheplatesdue to presencec'i_dielectitcasdei1'ior'rslraledbytl'ieeicpedntentdesc:i:adinthaAcapacitoris adevtceto stare charge. Fiitemattveiy. itiepeeeibtetcthlnitctecapacttcrasadevlcaiorstorlngelectrical energy. iiittereli. the chargeon the plate possesseselectrical potential energywhich arises becausewertt tstobedonetodepositchargaen theplates- This isduelotltetactthat with each small increment at charge heiig depositedduring the tifialjiifla process. the Potential differencebetweenthe plates lncreeeae.and a larger BI'l'ttJl.ll'tiD‘iW0l'|'l isneededtd bring upnerttlnuement ofcharge.initially when the capacitor is uncharged. the potentialditterence between plates is zero and tlnetiy it becomes ifvmengctiaigeisdepcettedoneactt plate. Thus, theaveragaeeiencereisere-ieee!3%‘i=%u .riiereeire as. =Energy = %e vusiiigiiviriitauiiiii;-=cvirireipei;itiiriiiegiitr\" i'_:|:-_.-r\"'-- !-.--.;j...e.:...eit,p;qg, . - ..it is also possible to tegard the energy as being stored instectricfield bet'iveentheplates.ratherthenthepo'nenltalenergy ct‘ the charges on the plates. Such e view point isusetutvihen electric Field strength behveenthe plates itstaadotcharges on the plates causing field is to he considered.This relation can be obtained by Etlsatiiuting iv’ = Ed andC= Ara-.Jd in Ei:|.t2.53.' - 25

R +1; ~¢ Emm,,E[1_..Aés._sq.)5d]F—iec 5 =1I=eE1mm A Fl-E[Ad]iiu'G|umBbHWBflfl'|L|iBplEliBBBfl _ 'i\". '--' '_-1''- '__. E-'1.'|-_. ' 'l.-v:'' ' -.‘:''-._';t:''-|,__\"|'\".' II.-._F'r,\" |-II. -.1-‘F.-.3JJ\"‘ r‘-!_;1-!,_\"-:'...IL-.:|._3|I.'''_'.-.:E- -.._,2_ . .~:_-:-'-. -'.I .L'_.,'_ _.¢ .- -..1-L 1|5.I- .,._ -_.. r‘-1.. .-l_ !-!:|I --..,.\"_..'=r ‘l'nisequa1inn isuaiidferan1_reieeb'iefieide'b'eflQfl1-. TJ. -.3 ll\" Br.-n. Many eieetlie tireiilis sensisl ei hem eeeesilbls and nesislaers. Fig.12.2? ehe-as e resieiibr-sapacilel einmit eeiied F!-G-c=irv::ui1;. ‘Mien the ewil1.':h S is set al tenninai A. ihe R-E\" \"\" ; Ll“ enirbineliginisebnneclied ieeheiteryeflelteee 'i-\",Wi'I|-ei1 starts charging lheca|:iesitbrtl1reughlheresistnr R. : 5'i The sepecih:-r is nnl charged immedieiehr. reihel‘ eflefeee -build up gredualy le ihe equilibrium velue ef q. = CV, The 1-: L_ gmwtll ef ¢l'|el'ge\uflh lime ier diiierent-resistenees is ehbum F1 11-H in Fig.12.2B. Aseerclirig in ihis graph q = D at i = D and increases gradually witi'| time lil ll reaches its equilibrium R .||¢.q vaiue q_, I Cifl. Thevelege Vecmss pacilnarlarvyirisiant{ C canbenbminedbydiifidingqby G, as i-'= qifi. \"' 5 I-lcrnleslerhnwsiewtl1esepaeitce'issharginger BA dieehergirig. depends upen the predusi of ihe resisienee R and ihe uapasiiame G used in Ihe circuit. A-e lihe uflii el pmdu-|:!RCiBH1HfC||'fil'fl0, sethis pruduetislinuumeeiinie censtenl end is defined as lhe lime required by the GBDB-flier + A tedepesit 0.53 timeelhe eqiiilbfium charge q,,. Thegraphsei‘ [Ii Fig.1:-_*.1'.B ghewthatthaeiiarge reaeries ils equilibrium value1? snenerwhenliietime eeneieniisemell.CV. Fig. 12.291111} iiiusirales ihe dieel1ergi1g of a capacitor through a resislnr. in this iigure. the ewiieh 5 '5 eel ei peinl B, ' 5InlI'?¢ seihe::i1arge+qnnt|1eieitpieiessI1flewerifi~eie-cinidse through ihe reeielenee and neulraiize lhe charge -q en the right piale. . The grqziesin Fig. 12.29ibj|-eheurs ihelc|is|:i'ierging begins at i= Human q = CV, and decrees-ee graduelyhcizere. E-male! in '- values uilime eensiamfl-Cieed teemererepiddiseharge.@ 26-

EJ:ampie1i.5: 'I'I'|eti'l1e00I‘IIlfl1lU|l series HG clrcuilisI=R‘G.Vu1fyil1ateneiuI:ti|ieefe1'ediseqiivaie|'uteeeound. : ohm‘: E~iv'iF1 iem*e'uFpnmrIE\" BIIereneo\"_\"\"\" ‘IIcurreriuandreeisimeeflcanbeiirriitenes. V=IR rpqum ;-qr; um gqiimni-|' unmiuu-nu h'i:eihe equelien. nu M n eleenll-llilifl-iililflill‘ v=%e _ wiper: ei eerl he he-use-d“I HIE ' l'hn'IllIIj'fl'l'!§-IQ!-li'hD.||\ lillindeelqlimlieubln Q Mifliflfilflfllfllfli I.Il'i1II llilfi'.ThII'iiIII'lIll'i-'flqI:llIi\fi0fll'li'lflTUBqilBlifl1lI'CV.CI¢\"V Khfifi HID HI hi-if I ifiur-a|iI:||'urIliflen.|IlLIliIiFi'lflil1ieequeien\ii'd1abeueec||1aiion-fives. _ -1 1 'i-ienoe 1eh1n-x1feredI‘ieeeend_wi'|ereoiI'nisfl1eu|it-alresietaricefi.'. Tiiefiwhniflsieueieieeflietfliefnrbehememnmpeirldulgeehdheeiiy pmpuflenalmmepudueteifliemegniudeabidwgesmdhwuueiyplupuflemih lieequme-aili1edi:lanuebehueentherr1-1 Eieebiefieidhmepermfidiergeetepehliscdhdehebiefilfiflrengmersledrh flddhllnlljralfllalpditit.e Tl1en|||'nbe:di'1heiieidiir|eepeeehgiiu1:I.|gl1e eensinelm-'rmlde|Ie ;ilme'e'flBe eiaic:rii:fl1ncli'Inughli'ielaree.derlbi|db'yfli.. ¢flrIEee.Ale=eEhdieenfias-sEfi.fluiflm1ieqrheaBuieelcritenarangmieehhehtdwmeedie1sflbnifleefiiddidheieiksdaer|ldleflliyeinieeg|'inu1flaihiiby eurfaee area.- Gain-s'siawisetawdes'fl1eflusdIeughenyc:nsedu|fieeehiIs,flmeel1eumi diageensiibsldhil.e Tl1einlariornIai'ueiiewci1lgedmelaienl1e|1eieefieldfieeIegi1e. U1 fl1eeieehsH¢eneIyheheu1hneeppnsinehrei'iergedpu'eleipieieeleE= t— .. flwarmumefiwlkduiehbnrighigeuritpudfluedulgehumhfldhrhepdm ageinsleieclri<:filidisiheiiec:i:iepei:e|1iielelihetpein1. 2'7

‘I .111 fl1epeierflielieemslenimmu%rnmeghenmgienefspene.lefl1eehehicHfieHzereernen-zeminilfleregien? spleln- _.11.2‘ Suppnsefl1etyeuhIuuw1Hedriefieidinedueieepesifiveeei|'flfl\e:‘ge.De eiec.-briei'ieid andlhe peteniieiirisreaeeerdeiieeee?‘IE-3 Huiinyeu_idenfifyjthatudflc#ipiaieefacepec#ieIispeeiflveiyd1a1;ed?12.4 gigs“-ginmmefermmbrmemapeflflmpbimdmrgenhmpheedbemewiperflifl iai wiihsintiisuandeqiiaidaerges (bi wilhnppnsileandequdelnarges12.5 Eleei1'ieii1eeeffeteeneveru'nss.\|ilhy'?12.6 Hepdrfldwlqeeelmessmismleesedinenerwfifiermeieehiefiaflwfiiifiehiries p0|l'Iii'iQli‘ifl'ifl'B-Bi'I'iB'iii'BlZi'i;iliiIl, Wi||i;ITiB|i'H'~BrBGi|il_BflrI'flU‘liUl'l‘?12.? isEnenes-eerily zeru-insides-ei1arged rubberbeiieun ifbeleenie spherieel? Resume ihat charge isdisb'ibuieduniiem1l1;eveIihesurfeee.'I18 is_ilhnflulGmss‘sieusHmsd\mdBlmaimn|hueiHrp§_:jfe[eesu'eesfl1gany eieeeflsurieeeiniheeuhuarddieelinnisprnpnninneite nelpeslthrechatge enzsnsednimhsufleee?123 Deeiedrenemndlegehunegimeirighpelmflaiereiiewpetemiai? ''111 Cenuuermgnflxndesefeieeuleflsneg-ewmennuiermsuenenunnnebiuc: ~[m-ass-=1fl.Ug,el1e|'ge=2fl._iiu'C}b3re|1ifleflfieelebieI21fl'ialiepleeed 11'Ii.i]i;:n'|fmm thefiral-{G=B.H?x1fl\"‘Hm‘hg\"'} ' (Ann: fl I 5.4 :1 10\"} FI IS

12.2 Calculate uectcrieity the net eieeireslsetic fierce en q as sheum in the iigiure. flilflpr _I -1 i Ii ii 1 -I I I Fl --- iI--I 1 I in -1,_ __M I “\"2 _,_ ___m 1 (Am: |= = 0.050 mIy12.3 .i.peir|tchargie e= -Ei.0:|r10“Cis|:|ieeed ettheerigin.t';eltui_ateeiett1ricfieidetepuint 2.0 tn from the eriginen the 2-esie. |A.ns: [-1.8 :10’12.4 Deterrrtinedteeiectficiieidettitepeeiiienr=0i-i+3]]mc=eusedbyapdint_dtergee=5.0x10\"Cpie0edaterigin. [A.rll:-[1-I11i0i +10e0jJN 0\";12.5 T-we paint charges. e, = -1.0 s10‘C end qr, = + 4.0 it 10‘ C. are separated by e distance ef 3.0 m. Find and iuslity the zere-feid iecalien. tens: 3.0 mfi12.0 Fihdtheeleetriclieid strength requiredte hcidsuspendede |:\eri:ieleefrness 1-0 ::10'kg end charge 1.0|.ic between tyi-e_pietes 10.0 cm apart. {amt 0.0 'ut't1'f|12.? Apelticiehevirig acttargeef20etecb'ensenitiatieltu1:ughapetentialditerer1eebI 100 ydtts. Gaiwlate theenergy acquired by itin B1001-l'1;\n\Pt:||i3l{tt\"r\"}I. fem: [2 0 1: 10‘ em;12.5 in Mi|iiitan's experiment. nil drepiets ere introduced |t1i0 the space between twe flat herizentel pieles. 5.00 mrn apart. The plate vdttage is adjusted te exactly T0011’ se that the dmpiet isheidstetienery. The piateuelegeis switched eflendtheseleeted drepiet ieebsenred tefeii a measured dislaneeet 1-50 mm in 11,2 5. Gi‘uen1',i1a‘|;fl'Ua density elihe eilused is 900irg m“, end the viscbsitycfeirattaberatery temperature is1.00:r10\"i*-im\"'s,caicuia|e5; Themess.end a}_ Tirechergeenuiedrepiettflussume-g=0.ttms’i [Am-: {e} 5.14 it 10'” HQ, ib} 3.20 at 10'\" C]12.5 Apretcn pieced inaunflermeiectiie iieidet‘5000 NC‘ direciedterighi is eiicmled tc geedisiance cI'10.0cr'r|fremAte B. Deteulelere; Pct-entiai difference between thetwe pointsq 1; 5| were dene bythe iieldic] Theehengein P.E.eiprc1enidi Thechangein i{.E.ei'1heprnl.en .re} tisueie-city {mes-a0I|:r010fli51.l5?'.'t10'\"lItg} _ (Ana: ~5D0 V, 500 e‘i|\". -500 e\"-I. 500 e'u'. 3.00? 110’ ms\")12.10 Usingzere reiarenee peini ei iniiiity. determine the ameunt by which a painlcharge 01'-1.0::10\"(.1eiterstheeiectricpetentia|atape-int1.2me1Ire'y.1Irl1en re; Chargeispesdiye (bi _ Ctiergeisnegsttve’ (Ans: +3.0 rt 10‘ V, -3.0 at 10' Vi . 19

12.11 in ears atemic mcdeiethytregen Bl0i11,0‘lBB|Bi.'!0\"0l'i15ifi3fl{I'|ii0iBi'lZlI|FIii11I|'lB nuctear prctan at e distance cl 5.29 iii 10'\" m with a speed -at‘ 2.10 it 10‘ nee\". -[eI1.00:\t 10'“t3.maeeeteiect11:I1 =0.10ir10\"'kg]-. Findta} Titeatectricpeterttiattttateprtatcarteatertsettiitedietaneeib} Tetaienergyefthe ateminetttc} Theicriizaiierl ertergytermealcmin a\"u' _ [Ann +2120 ti. -12.0 av, +t3.0D av}12.12 The eiedrenic itash attachment tar a camera contains a capecitcr fer aicring theenergy uaed ta preducethetteah. In ene such unit, ihapetentiai diiierence betweenti-ueptetaeete?50uFpaciturieseettfleterrninethaenergythetisusedtepredueethatteah. (lu1s:40.aJ}12.13 Asap-aeiterhasalcep-ecitaneeet2.5:r10' F. _inthecharging precess.eiecl.ransarerernu-iredtremenepiatear1dpieee~de-ntbeethercne. Whenittepetentieidittierericebete'eenthepiatesis450V.hawmenyalecatat1ahevabeenharts1erred‘?{e =1.00:-1 10\"'tI‘.} ' {il|.ne:7.0:r10\"eiaclrens) 3-0

13 CURRENT ELECTRICITYLearning ObjectivesAt the end of this chapter the students w * be able to: Understand the concept of steady current. Describe some sources of current. Recognize effects of current. Understand and describe Ohm's law. Sketch and explain the current-voltage characteristics of a metallic conductor at constant temperature, diode and filament lamp. Understand resistivity and explain its dependence upon temperature. Understand and elaborate conductance and conductivity of conductor. Solve problems relating the variation of resistance with temperature for one dimension current flow. Know the value of resistance by reading colourcode on it. Know the working and use of rheostat in the potential divider circuit. Describe the characteristics of thermistor. Use the energy considerations to distinguish between emf and p.d. Understand the internal resistance of sources and its consequences for external circuits. Describe the conditions for maximum power transfer. Know and use the application, of Kirchboffs first law as conservation of chargev Know and use the application of Kirchhoff s second law as conservation of energy. Describe the function of Wheatstone Bridge to measure the unknown resistance. Describe the function of potentiometer to measure and compare potentials without drawing any current from the circuit.M ost practical applications of electricity involve charges in motion or the electric current.A light bulb glows due to the flow of electric current. The current that flows through the coil ofa motor causes its shaft to rotate. Most of the devices in the industry and in our homes 31

operate with current. The electric current has become a necessity o f o u r life 13.1 ELECTRIC CURRENT An electric current is caused by the motion o f electric charge. If a net charge AO passes through any cross section o f a conductor in tim e At, we say that an electric current / has beon established through the conductor where At (13.1) The SI unit o f current is ampere and it is a current due to flow o f charge at tho rate o f one coulomb per second. Motion o f electric charge w hich causes an e le ctric current is due to the flow o f charge carriers. In case of metallic conductors, the charge carriers are electrons. The charge earners in electrolyte are positive and negative ions e.g.. in a CuSO. solution the charge carriers are C u \" and SO . ions. In gases, the charge carriers are electrons and ions. Current DirectionWhan c«< « i» o » dangor. a x /n * Early scientists regarded an electric current as a flow ofinto a »W10b a O tfy Any onawt>0 positive charge from positive to negative terminal o f thoaracM a Mw*y k> j M a battery through an external circuit. Later on. it was found thatshock Th o po:«nt-*l a current in metallic conductors is actually due to the flow ofM M « n 9 * h M d on<3 L-vi oT jn nogative charge carriers called electrons m oving in the«Jo cti>C M l«nb *u plo 6C O V opposite direction i.e.. from negative to positive term inal of the battery, but it is a convention to take tho direction of current as the direction in which positive charges flow. This current is referred as conventional current. The reason is that it has been found experimentally that positive charge moving m one direction is equivalent in all external effects to a negative charge moving in the opposite direction. A s the curront is m easured by its external offects so a current due to motion o f nogative charges, after reversing its direction of flow can be substituted by an equivalent current due to flow of positive charges. Thus The conventional curront In a circuit is defined as that equivalent current which passes from a point at higher potential to a point at a lowor potential as if it represented a movement o f positivo charges. W hile analyzing the electric circuit, we use the direction o f the current according to the above mentioned convention. 32

If we wish to refer to the motion o f electrons, we use the term Electron Itowelectronic current (Fig. 13.1). Fig. 13.1 C u rre n t T h ro u g h a M eta llic C o n d u c to r <«>)In a metal, the valence electrons are not attached to F i g - 13.2individual atom s but are free to move about within the body.These electrons are known as free electrons. The freeelectrons are in random motion just like the molecules of agas in a container and they act as charge carriers in metals.The speed o f randomly moving electrons depends uponte m p e ra tu re .If we consider any section o f metallic wire, the rate at whichthe free electrons pass through it from right to left is the sameas the rate at which they pass from left to right (Fig. 13.2 a).As a result the current through the wire is zero. If the ends ofthe wire are connected to a battery, an electric field E will beset up at every point within the wire (Fig. 13.2 b). The freeelectrons will now experience a force in the direction oppositeto E. A s a result o f this force the free electrons acquire amotion in the direction o f -E. It m ay be noted that the forceexperienced by the free electrons does not produce a netacceleration because the electrons keep on colliding with theatoms of the conductor. The overall effect of these collisionsis to transfer the energy of accelerating electrons to the latticewith the result that the electrons acquire an average velocity,called the drift velocity in the direction of -E (Fig. 13.2 b). Thedrift velocity is o f the order o f 10 ’m s'1, w hereas th e v o lo d ty offree electrons a t room temperature due to their thermalm otion is several hundred kilometres per second.Thus, when an electric field is established in a conductor, thefree electrons modify their random motion in such a way thatthey drift slow ly in a direction opposito to the field. In otherw ords the electrons, in addition to their violent thermalmotion, acquire a constant drift velocity due to which a netdirected motion o f charges takes place along the wire and acurrent begins to flow through it. A steady current isestablished in a wire when a constant potential difference ismaintained across it which generates the requisite electricfield E along the wire.E x a m p le 1 3 .1 : 1.0 x 10’ electrons pass through aconductor in 1.0 ps. Find the current in am pere flowingthrough the conductor. Electronic charge is 1.6 x 10 ” C.S o lu tio n :- Num ber Of electrons = n = 1.0 x 10' 33

Charge on an electron = e = 1.6 x 10 '*C Time =Af= 1.0ns Current / through the conductor is given by 1 to to I « 1:0x- 9.7 x1 -6- x1()19P = 1 .6 x 10 6Cs‘‘ = 1.6 x 10'®A 1 .0 x 1 0 '6s 13.2 SO U R C E O F C U R R E N TF i g . 13.3 C o n vo n s o o a l current flows When two conductors at different potentials are joined by afrom h ig h * to lo w * poW rftal IN o o / t metallic wire, current will flow through the wire. The current continues to flow from higher potential to the lower potential until both are at the same potential (Fig. 13.3). After this tho current ceases to flow. Thus the current through the wire decreases from a maximum value to zero. In order to have a constant current tho potential difference across the conductors or the ends of the wire should be maintained constant. This is achieved by connecting the ends of the wire to the terminals of a device called a source ofcurrent (Fig. 13.4). Every source of current converts some non electrical energy such as chemical, mechanical, heat or solar energy into electrical energy. There are many types of sources of current. A few examples are mentioned below: (i) Cells (primary as well as secondary) which convert chemical energy into electrical energy.fbiaytlo1ryj imAamsouuurrc*e»ofnceuarrrlayr*otuocmfiu«n*i (ii)potennaf <M««rencebetween end*o<• (Iii) Electric generators which convert mechanical energy into electrical energy.C O rr\K ttx Thermo-couples which convert heat energy intoFo r Y o u r Info rm ation (iv) electrical energy. Solar cells which convert sunlight directly into electrical energy. 13.3 E FFE C TS O F C U R R E N T The presence of electric current can be detected by the various effects which it produces. The obvious effects of the current are: (i) Heating effectHeebng affect of currant is used In (ii) Magnetic effectotectnckettle (iii) Chemical effect 34

H e a tin g E ffe ct <*»>Current flows through a metallic wire due to motion o f freeelectrons. During the course o f their motion, they collidefrequently with the atom s o f the metal. At each collision, theylose some o f their kinetic energy and give it to atoms withwhich they collide. Thus as the current flows through the wire,it increases the kinetic energy o f the vibrations o f the metalatom s, i.e.. it generatos heat in the wire. It is found that theheat H produced by a current / in the wire of resistance Rduring a time interval f is given by H * l ‘RtThe heating effect of current is utilized m electric heaters,kettles, toasters and electric irons etc. M a g n e tic E ffectThe passage o f current is always accompanied by amagnetic field in the surrounding space. The strength o f thefield depends upon the value o f current and the distance fromthe current element. The pattern o f the field produced by acurrent carrying straight wire, a coil and a.solenoid is shownin F»g. 13.5 (a. b & c). M agnetic effect is utilized in thedetection and measurement o f current. All the machinesinvolving electric motors also use the magnetic effect ofcurrent. C h e m ical E ffectCertain liquids such as dilute sulphuric acid or coppersulphate solution conduct electricity due to some chemicalreactions that take place witnm them The study o f this processts known as electrolysis. The chemical changes producedduring the electrolysis o f a liquid are due to chemical effects ofthe current. It depends upon the nature o f tho liquid and thequantity o f electricity passed through the liquid.The liquid which conducts current is known as electrolyto. Them aterial in the form o f w iro or rod or plate which leads thecurrent into o r out of the electrolyte is known as electrode . Theelectrode connected w ith the positive terminal o f the currentsource is called anode and that connected with negativeterminal is known as cathode. The vessel containing the twoelectrodes and the liquid is known as voltameter. As anexample we w il consider the electrolysis of copper sulphatesolution. The voltameter contains drfute solution of coppersulphate. The anode and cathode are both copper plates 35

(Fig. 13.6). W hen copper sulphate is dissolved in water, it dissociates into C u \" and SO ', ions. On passing current through the voltameter. C u \" moves towards the cathode and the following reaction takes place. Cu“ ♦ 2 e ► Curi9 ,,< Tho copper atoms thus formed are deposited at cathode plate. W hile copper is being deposited at the cathode, the SO , ions move towards the anode. Copper atoms from the anode go into the solution as copper ions which combino with sulphate ions to form copper sulphate. C u \" +SO« --------- ► CuSO. As the electrolysis proceeds, copper is continuously deposited on the cathode-whiie an equal amount of copper from the anode is dissolved into the solution and the density o f copper sulphate solution remains unaltered This example also illustrates the basic principle of electroplating - a process of coating a thin layer o f some expensive metal (gold, silver etc.) on an article o f some ^ je a p m e ta l^ 13.4 OHM’S LAW We havo seen that when a battery is connectod across a conductor, an electric current begins to flow through it. How much current flows through tho conductor when a certain potential difference is set up across its ends? The answer to this question was given by a German Physicist George Simon Ohm. He showed by elaborate experiments that the current through a metallic conductor is directly proportional to the potontial difference across its ends. This fact is known as Ohms' law which states that \"The current flo w in g through a conductor is directly p ro p o rtio n a l to th e p o ten tia l d ifference across its ends provided the physical state such as tem perature etc. o f the c o n d u cto r rem ains constant\". Symbolically Ohm's law can be written as / oc V It im plies that V=Rl ............. (13.2) where R. the constant o f proportionality is called the resistance o f the conductor. The value o f the resistance depends upon the nature, dim ensions and the physical state o f the conductor. In fact the resistance is a measuro o f the 36

opposition to the motion of electrons due to their continuousbumping w th the atoms o f the lattico. The unit of resistance isohm. A conductor has a resistance o f 1 ohm if a current of 1ampere flows through it when a potential difference of 1 volt isapplied across its ends. The symbol of ohm is 0 . If / is measuredin amperes. I/in volts, then R is measured in ohms i.e.. V (volts) <13'3 >R (ohm s) ■ 7 (a m p e re s j .............A sam ple of a conductor is said to obey O hm 's law if its 0 7 Curronl - voK*g» gropft oTresistance R rem ains constant that is. the graph o f itsV versus / is exactly a straight line (Fig. 13.7). A conductorwhich strictly obeys Ohm's law is called ohmic. However,there are devices, w hich d o not obey O hms' law i.e.. they arenon ohmic. The exam ples o f non ohmic devices are filamentbulbs and semiconductor diodes.Let us apply a certain potential difference across theterminals o f a filam ent lam p and measure the resultingcurrent passing through it. If we repeat the measurement fordifferent values o f potential difference and draw a graph ofvoltage V versus current /. it w ill be seen that the graph is not astraight line (Fig. 13.8). It means that a filament is a nonohm ic device. This deviation o f / - V graph from straight line isdue to the increase in the resistance o f the filament withtcm porature - a topic w hich is discussed in the next section.As the current passing through the filament is increased fromzero, the graph is a straight line in the initial stage becausethe change in the resistance o f the filam ent with temperaturedue to small current is not appreciable. As the current isfurther increased, the resistance o f the filament continues toincrease due to rise in its temperature.Another example o f non ohm ic device is a semiconductordiode. The current - voltage plot o f such a diode is shown inFig. 13.9. The graph is not a straight line so semi conductor isalso a non ohmic device.Review o f Series and Parallel C om binations o f Resistors / R. r, r, i - > - V W “— A A A —/V \A -> -In an electrical circuit, usually, a num ber o f resistors areconnected together. There are two arrangements in which F tg .l) I 0 («]resistors can be connected w ith each o th e r.. one is known asseries arrangement and other one as parallel arrangement.If the resistors are connected end to end such that the samecurrent passes through all o f them, they are said to beconnected in series as shown in Fig. 13.10(a). There37

equivalent resistance R. is given by- e — A v v *w. v w — 4 - (13.4) \"VM VW W - lllh In parallel arrangement a number of resistors are connected side by side with their ends joined together at n»txwM two common points as shown m Fig 13 10(b) The eqvivalent resistance R, of this arrangement is given by R, 1 + —R12 + —1 + .......... (135) R, 13.5 RESISTIVITY AND ITS DEPENDENCE UPON TEMPERATURE It has been experimentally seen that the resistance R of a wire is directly proportional to its length L and inversely proportional to its cross sectional area A Expressing mathematically we have / ? «A- (13.6) R = l> where p is a constant of proportionality known as resistivity or specific resistance of the material of the wire. It may be noted that resistance is the characteristic of a particular wire whereas the resistivity is the property of the material of which the wire is made. From Eq. 13.6 we have P= V .......... (13-7) The above equation gives the definition of resistivity as the resistance of a metre cube of a material The SI unit of resistivity is ohm-metre (Q m) Conductance is another quantity used to describe the electncal properties of materials. In fact conductance is the reciprocal of resistance i.e.. Conductance resistance (R) The SI unit of conductance is mho or siemen Likewise conductivity, o is the reciprocal of resistivity i.e.. (13.8) 38

The SI unit of conducfcvtfy is o h m 'm ' or mh m \ Resistivity of Table 13.1various materials are given in Tabte 13.1. Subslarv* p(On» -<K'>It may be noted from Table 13.1 that silver and copper are twobest conductors. That is the reason that m ost electric wires Saver 1 52 * to* 000380are made of copper. 0 00390 C «w 1 S 4 - 10* 000340The resistivity o f a substance depends upon the temperature Odd 2 27 * 10* 000390also. It can be explained by recalling that the resistance ooo«eooffered by a conductor to tho flow o f electric current is due to AJumnum 2 63 • 10* 000020collisions, which the free electrons encounter with atoms of 000520the lattice. A s the temperature o f the conductor rises, the Tunpssn 5 00 ■ 10* 00001amplitude o f vibration o f the atoms in the lattice increases 000091and hence, the probability of their collision with free electrons ken 110 0 » 1 0 * 000020also increases. One may say that the atoms then offer a -0 0005bigger target, that is. the collision cross-section o f the atoms Ptaarum 1100* 10* -005increases w ith temperature. This makes the collisions Ccnstanun 49 00 * 10* ■007between free electrons and the atom s in the lattice morefrequent and hence, the resistance o f the conductor ..Msrtxry M 0 0 - 10*increases. 1000 - 0* rmyr&T'm. 3.5 * 10*Exporimontally the change in resistance of a metallic C trtxn 05conductor w ith temperature is found to be nearly linear over a Gsrmanumconsiderable range of tem perature above and below 0 °C(Fig. 13.11). O ve r such a range the fractional change in Sacoi 20-2301resistance per ketvin is known as the temperature coefficiento f resistance i.e.. « = (13.9) ow here R cand R,are resistances a t temperature 0 \"C and f *C.As resistivity p depends upon the temperaturo. Eq. 13.6 gives R,=pJJA and R,=pJUASubstituting th e valuos o f R, and R in E q . 13.9. w o get P. -P (13.10) M*C>-as a ® — \"• S t j. 11VenationOinuMYtyofCu Po* Iw here p. is the resistivity o f a conductor a t 0 °C and p, is theresistivity at f *C. Values o f tem peraturo co-efficients of Interesting Inform ationresistivity o f som e substances are also listed in Table 13.1 in sp e cto rs c a n e a t* y cnecfc theThere are som e substances like germanium, silicon etc.. rd d t-M y of a concrete t n J j c madewhose resistance decreases with increase in temperature, with carbon tb e rs . T h e ftbers conducti.e.. these substances have negative temperature coefficients. e lo c tn o ty ff s e n s o r s s n o w lh a l electrical resistance is s tc ra a s rgE x a m p le 1 3 .2: 0.75 A current flows through an iron wire over tm e the *bers a re le p a ra trv}when a battery o f 1.5 V is connected across its ends. The because ofcra A * 39

length o f the w ire is 5.0 m and its cross sectional area is 2 .5 * 10 W . Compute the resistivity of iron.Table 13.2 The C olour Code S o lu tio n :Colour Value The resistance R o f the w ire can be calculated by Eq. 13.2 i.e..Blec* 0Brown 1 yjv 7 “ oI s a \" 20 Va~' * 20 fi Red 2 T he resistivity p of iron o f which the w ire is m ade is g iven byOrange 3Y«eo*r 4 rA 2.0 0 x 2 .5 x1 0 rm* _ 41 n „ t n .y. PL .0 x 10 Om 5.0 m E x a m p le 13.3: A platinum wire has resistance o f 10 Q atGroen 5 0 ‘C and 20 O at 273 ’ C. Find the value of temperature Blue 6 coefficient o f resistance o f platinum.v«*rt 7 S o lu tio n :Gray 8 R, = 1 0 O . R, = 2 0 0 . I = 5 4 6 K -2 7 3 K = 2 7 3 Kwr*e 9 Temperature coefficient o f resistance can be found by a =^ = i o o l l oi L . _ L _ , R0 t 1 0 O x 2 7 3 K 273 K 13.6 COLOUR CODE FOR CARBON RESISTANCES R otittof Colour CcxJo Carbon resistors are most common in electronic equipment. They Fig 13.12 consist of a high-grade ceramic rod or cone (called tho substrate) on which is deposited a thin rosistrvo film of carbon. The numericalFor Your Intormatic value of their resistance is indicated by a colour code which consists of bands of different colours printed on the body of theRod Vk M Oango STrer resistor Tho colour used in this code and the digits represented by them are given in Table 13.2.27 000 t 10% Usually the code consists o f four bands (Fig. 13.12). StartingR « 27000 0 ( 1 10%) from left to right, the colour bands are interpreted as follows: The first band indicates the first digit in the numerical value o f the resistance. The second band gives the second digit. The third band is decim al m ultiplier i.e.. it g ive s the number o f zeros after the first two digits. The fourth band gives resistance tolerance. Its colour is either silver o r gold. Silver band indicates a tolerance o f ± 10%. a gold band shows a tolerance of 40

+ 5 % It there is t>o fourth band, tolerance is (at A R hociU t understood to be ♦ 20% By tolerance, we mean the possible variation from the marked value. For C example, a 1000 O resistor with a tolerance o f ± 10% will have an actual resistance anywhere between ;b Its u w H n r l t b l t rvsWtor 9000 and 1100Q. Fig. U . 1JR heostatIt is a w ire w ound variable resistance. It consists of a baremanganin w ire wound over an insulating cylinder. The endso f the wire are connected to two fixed terminals A and B(Fig. 13.13 a). A third terminal C is attached to a slidingcontact which can also be moved over the wire.A rheostat can be used as a variable resistor as well as apotential divider. To use it as a variable resistor one o f thefixed term inal say A and the sliding terminal C are inserted inthe circuit (Fig. 13.13 b). In this way the resistance of the wirebetween A and the sliding contact C is used. If the slidingcontact is shifted aw ay from the term inal A , the length andhence the resistance included in the circuit increases and ifthe sliding contact is moved towards A . the resistancedecreases. A rheostat can also be used as a potential divider.This is illustrated in Fig. 13.14. A potential difference V isapplied across the ends A and B o f the rheostat with the helpo f a battery. If R is the resistance of wire AB. the current /passing through it is given by / = V/R.The potential difference between the portion BC o f the wireAB is given bycurrent x resistance = (13.11)RRwhere r is the resistance o f the portion BC o f the wire. Thecircuit show n in Fig. 13.14 is known as potential divider.Eq.13.11 shows that this circuit can provide a t its outputterminals a potential difference varying from zero to the fullpotential difference of the battery depending on the position ofthe sliding contact As the sliding contact C is m oved towardsthe end B. the length and hence the resistance r o f the portionBC of the wire decreases which according to Eq. 13.11.decreases V* . On the other hand if the sliding contact C ismoved towards the end A. the output voltage increases. F ig .1S.14 41

f t ®|°| T herm istorsFig 1JM n * m to to n o l< X l9 r* rt A thermistor is a heat sensitive resistor. Most thermistors tfUpM have negative temperature coefficient of resistance i.e.. the resistance of such thermistors decreases when their F o r Y o u r In fo rm a tio n temperature is increased. Thermistors with positiveAioro-ofxn rcaiucr t .-.scaled try a temperature coefficient are also available.tingle bt*c* colour band around thebody of tb« r**«*tof. Thermistors are made by heating under high pressure semiconductor ceramic made from mixtures of metallic oxides of manganese, nickel, cobalt, copper, iron etc. These are pressed into desired shapes and then baked at high temperature. Different types of thermistors are shown in Fig .13.15. They may be in the form of beads, rods or washers. Thermistors with high negative temperature coefficient are very accurate for measuring low temperatures especially near 10 K. The higher resistance at low temperature enables more accurate measurement possible. Thermistors have wide applications as temperature sensors i.e.. they convert changes of temperature into electrical voltage which is duly processed. 13.7 ELECTRICAL POWER AND POWER DISSIPATION IN RESISTORS Consider a circuit consisting of a battery E connected in series with a resistance R (Fig. 13.16). A steady current / flows through the circuit and a steady potential difference V exists between the terminals A and B of the resistor R. Terminal A. connected to the positive pole of the battery, is at a higher potential than the terminal B. In this arcuit the battery is continuously lifting charge uphill through the potential difference V. Using the meaning of potential difference, the work done in moving a charge AO up through the potential difference Vis given by Work done = A W = V x AO (13 12).Fig. 13.16 The power of a bMtry This is the energy supplied by the battery. The rate at whichappears as P » poew *svpaled in the battery is supplying electrical energy is the power outputthoresistwft or electrical power of the battery. Using the definition of power we have Electrical power ■ =y ^ Time taken AT 42

Since l^ .s o AtElectrical power = Vx / (13 12a)Eg 13.12a is a general relation for power delivered from a source of current / operating on a voltage V. In the circuit shown in Fig. 13.16 the power suppliod by tho battery is oxpended or dissipated in the resistor R. The principle of conservation of energy tells us that the power dissipated in the resistor is also given by Eq. 13.12. aPower dissipated ( P ) * V x l ........... (13.13)Alternative equation for calculating power can be found bysubstituting V = IR . l = V/R in turn in Eq. 13.13 P= V x l* f R x l* l’R P= V x l = Vx i/ - i/* -R —RThus we have three equations for calculating the powerdissipatod m a resistor. P *V x l, P -t'R . P= s - (13.14)If Vis expressed in volts and /in amperes, the power isexpressed in watts.13.8 ELECTROMOTIVE FORCE (EMF) AND POTENTIAL DIFFERENCEWe know that a source of electrical energy, say a cell or a Fig. 13.17 EJ»c«rom»v« lore* erf •battery, when connected across a resistance maintains asteady current through it (Fig. 13.17). The cell continuouslysupplies energy which is dissipated in the resistance of thecircuit. Suppose when a steady current has been establishedin the circuit, a charge AO passes through any cross sectionof the drcuit in time At. During the course of motion, thischarge enters the cell at its low potential end and leaves at itshigh potential end. The source must supply energy AW to thepositive charge to force it to go to the point of high potential.The emf E of the source is defined as the energy supplied tounit charge by the cell. 43

It may bo noted that electromotive force is not a force and we do not measure it in newtons. The unit of emf is joule/coulomb which is volt (V). The energy supplied by the cell to the charge carriers is derived from the conversion of chemical energy into electrical energy inside the cell. Like other components in a circuit a cell also offors some resistance. This resistance is due to the electrolyte present between the two electrodes of the cell and is called the internal resistance r of the cell. Thus a cell of emf £ having an internal resistance r is equivalent to a source of pure emf E with a resistance r in series as shown in Fig. 13.18.F ig . S3 18 A n squvaiont orcuil of a Let us consider the performance of a cell of emf E andcad c * o m f £ m l internal roaatanoo r. internal resistance r as shown in Fig. 13.19. A voltmeter ofFig l i l t Tha MrirMtal pofen*al infinite resistance measures the potential difference across(3'Hp-fncc V c t i C H r i E / i the external resistance R or the potential difference V across the terminals of the cell. The current / flowing through the circuit is given by I E R +r or E *IR +Ir ............ (13.16) Here / R = V is the terminal potential difference of the cell in the presence of current /. When the switch S is open, no current passes through the resistance. In this case the voltmeter reads the emf E as terminal voltage. Thus terminal voltage in the presence of the current (switch on) would be less than the em f £ by / r. Let us interpret the Eq. 13.16 on energy considerations. The left side of this equation is the emf £ of the cell which is equal to energy gained by unit charge as it passes through the cell from its negative to positive terminal. The right side of the equation gives an account of the utilization of this energy as the current passes the circuit. It states that, as a unit charge passes through the circuit, a part of this energy equal to / r is dissipated into the cell and the rest of the energy is dissipated into the external resistance R. It is given by potential drop IR . Thus the emf gives the energy supplied to unit charge by the cell and the potential drop across the various elements account for the dissipation of this energy into other forms as the unit charge passes through these elements. The em f is the 'cause\" and potential difference is its 'effect*. The emf is always present even when no current is drawn 44

through the battery or the cell, but the potential differenceacross the conductor is zero w hen no current flows through it.E x a m p le 1 3 .4 : The potential difference between theterm inals o f a battory in open circuit is 2.2 V. W hen it Isconnected across a resistance o f 5.0 0 , the potential falls to1.8 V. Calculate the current and the internal resistance o f thebattery.S o lu tio n :G iven E = 2 .2 V, fl= 5 .0 fi. V = 1 .8 VW e are to calculate / and r.We have V = IR -1--.-8--V-- =0.36 Aor /,= —V 5 .0 0 R Do You Know?Internal resistance rc a n be calculated by using E -V */ro r 2.2 V = 1.8 V + 0.36 A x ro r r = 1.11 V A ’ = 1 .1 1 0 M axim um P o w e rO u tp u t (•)In the circuit o f Fig. 13.19. as the current / flows through the -Q h -resistance R. the charges flow from a point of higher potentialto a point o f lower potential and as such, they looso potontial -w w w -energy. If V is the potential difference across R. the loss of wpotential energy per second is V /. This loss o f energy persecond appears in other form s of energy and is known aspower delivered to R by current /. Powor delivered \o R = P ^ = V I - l’R (7 V =IR )As A vcftmtfor connected accost III Mminals <A a co * moasoros (a) E 2R e 2r om f of lha c « l o n open t i r o A (6 ) a (R + r f ( R - r j 2 + 4 R r ............. (13.17) terminal poumn dWarance o n dOMdcircuAwhen R = r, the denom inator o f the expression o f Pw is leastand s o Pw is then a m axim um . Thus w e see that m aximumpower is delivered to a resistance (load), when the internalresistance o f the source equals the load resistance. The 45

value o f this m aximum output power as given b y Eq. 13.17 is E2 4R . 13.9 KIRCHHOFF'S RULES Chm 's taw and rules o f series and parallel combination of asistance are quite useful to analyze simpte electrical circuits consisting o f more than one resistance. However such a method fails in the case of com plex networks consisting o f a numbor o f resistors, and a number o f voltage sources. Problems o f such networks can be solved by a system o f analysis, which is based upon two rules, known as Kirchhoffs rules. K irc h h o ffs F irst Rulo It states that the sum o f all the currents m eeting a t a point in the circuit iszeroi.e.. Z fO (13.18) It is a convention that a current flowing towards a point is taken as positive and that flowing away from a point is taken as negative. Consider a situation where four wiros meet a t a point A (Fig. 13.20). The currents flowing into the point A are /, and I, and currents flowing aw ay from the point are /, and /,. According to the convention currents /, and /, are positive and currents /, and /, are negative. Applying Eq. 13.18 we haven tF i g 1 1 j o A e c o f 4 n ( j l o K i r t f i h o » « 1* or /,♦ /,= /,♦/, (13.19) Using Eq. 13.19 Kirchhoffs first rule can be stated in other words as The sum o f all the cu rrents flow ing to w a rd s a p o in t is equal to th e sum o f all the currents flo w in g away from the point. Kirchhoffs first rule which is also known as Kirchhoffs point rule is a manifestation of law o f conservation o f charge. If there is no sink or source o f charge at a point, the total charge flowing towards the point m ust be equal to the total charge flowing aw ay from it. 46

K irch h o ff's S e co n d Rule It states that the algebraic sum of voltage changes in a closed rfc3 13 2«Accon*ng»K«N>o<r*2'‘ circuit or a loop must be equal to zero. Consider a closed nSi£../R,-£,•«, >o circuit shown in Fig. 13.21. The direction of the current / flowing through the circuit depends on the cell having the greater emf. Suppose £, is greater than Et . so the current flows m counter clockwise direction (Fig. 13.21). We know that a steady current is equivalent to a continuous flow of positive charges through the circuit. We also know that a voltage change or potential difference is equal to the work done on a unit positive charge or energy gained or lost by it in moving from one point to the other. Thus when a positive charge AO due to the current / in the dosed circuit (Fig. 13.21), passes through the cell E. from low (-ve) to high potential (♦ve). it gains energy because work is done on it.Usmg Eq 1312 the energy gam is E, AO When the current passes through the cell £ it loses energy equai to - £ . AO because here the charge passes from high to low potential. Ingoing through the resistor R „ the charge \Q loses energyequal to - / R, AQ where / R. is potential difference across R, The minus sign shows that the charge is passing from high to low potential. Similarly the loss of energy while passingthrough the resistor R, is - / R, AO. Finally the charge reachesthe negative terminal of the cell £, from where we started.According to the law of conservation of energy the totalchange in energy of our system is zero. Therefore, we canwrite £,A Q -/R ,A Q -£,A Q -/R ,A Q =0 (13.20)or E ,-/R ,-E ,-/R , =0which is Kirchhoff's second rule and it states that Tho algebraic sum o f potential changes In a closed c ircu it is zero.We have seen that this rule is simply a particular way ofstating the law o f conservation of energy in electricalproblems.Before applying this rule for the analysis of complex networkit is worthwhile to thoroughly understand the rules for findingthe potential changes. 47

(i) If a source of emf is traversed from negative to positive terminal, the potential change is positive, it is negative in the opposite direction.(ii) If a rosistor is traversed in the direction of curront. the change in potential is negative, it is positive in the opposite direction.E xam ple 13.6: Calculate the currents in the threeresistances of the circuit shown in Fig. 13.22.S o lu tio n :First we select two loops abcda and ebcfe The choice ofloops is quite arbitrary, but it should bo such that eachresistance tsincluded at least once in the selected loops.After selecting the loops, suppose a current /. is flowing in thefirst loop and /, in the second loop, all flowing in the samesense. These currents are called loop currents. The actualcurrents will be calculated with their help. It should be notedthat the sense of the current flowing in all loops shouldessentially be the same. It may be clockwise or anticlockwise.Here we have assumed it to be clockwise (F*g. 13.22).We now apply Kirchhoffs second rule to obta.m the equationsrequired to calculate the currents through the resistances. Wefirst consider the loop abcda Starling at a point 'a' wo followthe loop clockwise. The voltage change while crossing thebattery E, is * E. because tho current flows through it frompositive to negative. The voltage change across R, is - !JRrThe resistance R. is common to both the loops /, and /,therefore, the currents /, and /, simultaneously flow through it.The directions of currents /. and /, as flowing through R, areopposite, so we have to decide that which of these currents isto be assigned a positive sign. The convention rogardmg thesign of the current is that if we are applying tho Kirchhoffssecond rule in the first loop, then the current of this loop I.e. /,will be assigned a positive sign and all currents, flowingopposite to /. have a negativo sign. Similarly, while applyingKirchhoffs second rule in the second loop, the current /, will beconsidered as positive and /, as negative. Using thisconvention the current flowing through R} is (/, - /,) and thevoltage change across is - (/, - /,) R^ The voltage changeacross tho battery E, is E ,. Thus the Kirchhoffs second rule asapplied to the loop abcda gives -E ,-/,/?,-(/,-/,)« ,+ £ , =0 48

Substituting the values, we have - 40 V - /, x 10 O - ( / , - / , ) x 30 0 + 60 V = 0 20 V - 1 0 O x [ /, ♦ 3 (/, - /,)) » 0Of 4 /, - 3 /, = 2 V O ' = 2 A (13.21)Similarly applying Kirchhoffs second rule to the loop ebcfe.we getSubstituting the values, we have •60V - (/,- /,)x 3 0 0 -/,x 1 5 0 + 50V = 0 -10 V - 15 O x(/, + 2 (/,-/,)] = 0 (13.22)or 6 /,-9 /, =2 V fi’ = 2ASolving Eq. 13.21 and Eq. 13.22 for /. and l }. we get 22 / , = - A and / , = - AKnowing the values of loop currents/, and /, the actualcurrent flowing through each resistance of the circuit can bedetermined. Fig. 13.22 shows that /, and /, are the actualcurrents through the resistances R, and Ry The actualcurrent through R} is the difference of /, and /, and its directionis along the larger current. ThusThe current through R, = /. 2 = 0.66 A flowing in the =r Adirection of /. i.e.. from a to d.The current through = / , - / , = -2 A - -2 A = 0.44 A flowing inthe direction o f/.i.e .. from c to b.The current through R, = /} 2 = 0.22 A flowing in the =-Ad irectionof/, i.e.. from f toe.Procedure o f S o lu tio n o f C ircu it P roblem sAfter solving the above problem we are in a position to applythe same procedure to analyse other direct current complexnetworks. While using Kirchhoffs rules in other problems, it isworthwhile to follow the approach given below:(i) Draw the circuit diagram. The choice of loops should be such that each resistance is mdudod at least once in the selected loops. 49