University:
Washington State University
Course:
Physics 104 | Problem Solving for Physics 102
Academic year:

2022
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406

Pages:

27
Author:

Janor

Electrostatic fieldTypes of charge Sources:-Q1point charge-Q2lLine charge ; Pe = Q = line charge density (4/m)3QSurface charge ; ls = Q = Surface charge densityArea=AA(4/m2)4Q1/4Volume charge, P=Q = = volume charge density ( %Volume VV# Law : coulombs force :- 1. F a o,2. F a o2F a Q,Q23. F x -d2d22F =KQ, O22wherek= = 9 x10"d24 totrtrN-w2)for free space ; Er=1Coulomb Force in Vector FormQ,R12Q2^->?,R12 =R12P2R12ooriginF12 = K (Q,)(Q2) ( R12)NewtonIR1212It represents Force on Q 2 due to Q,FILr12=KO.O2 R12Newton12,213 E = Total premitivity (F/m)E= Eotrtr = t. =Relative permitivityocunit less)to = Permitivity of Free space or airto = 8.854 x10th F/mto = + x159 F/m36 TT#Fa Q,=>F= K Q, or y=mx:Force is linear w.r.t Charge Source.0,=-1x154cO2=6x104cQP(2,0,5)P2(1,2,3)medium isair (tr=1)Then find force on charge 2 (O2) due to charge / (Q1)or F12 = ?Sol F12 = K Q,02 R12 where R12=(1-2) i +(2-o)j + (3-5) kIR1213R12=1923-28(Riz 1 = 3F12 = 9x109 x (-1x154)x (6x104) (it 25-2K)(3)3F12 = -2 x 10 (-it 2j -2 i ) = 20 i - 40 ] - + 40 k Newtons Image theory of chargeLord Kelvin"Image theory states that the Q charge near by aconducting surface is replaced with charge itself +Qits image charge -Q, so that conductor surface is replacewith a "o volt equipotential plane ":Qhconductorplaneh:Q= Q Find force on Q charge in fig.+Q (0,0,d)dz=orconducting1 (o,o,d)planeSolR12Z=O-> R12 = (0-0) ax+ (0-0)0y+(d-1-d))az2o(0,0,-d)R12 = 2d az i IR121 = 2d^R12 = a3ANow F12 = K Q, 02 R12 = -9x10 o2 az = 9x109(29)IR12/(2d)38d3Force on Q charge is 9x10'03 towards 1=0 plane8 d3 Electric field Intensity (E) It is defined as vector force on onthe charge (i.e unit positive test charge)E =F#E12 is Electric field intensity vector atqP2 due to Q charge at point P,AE12 =QR12kQ=K124TTtto 2( 1/m)tr (Riz) 32->E12 Pz (22,72,22)op2OR12 = (x2-x1) ant(y2-71) cig + (22-21)dz7yof,a+P, (x1,41,21)# i. E d Q (source charge) i.c E = const Q: E is Linear work Source charge.2. Field ; Is a region where we measure Same physicalquantity.3. Electric field:- Is a region where electric effect is present.#E in all region due to Q' point charge Locate at originR12 = r asQR12#P2 (r,o,p)E12=45t 1R1213R12R12/ = rTP1E12=Q rar(origin)^QE12=QUTE r31arV245 Er2sisin #1.Eatp(r) 0,4) due to Q point charge at originis given as1to theE Q arsurface+as(5,0,0)45 E j2Qradiallyoutwards2. E at p(r,0,0) due to -Q' point charge at origin isTH - - Q^aiaara(,0,4)4 K t r2Qradially7ITinwards= Q (-ai)456 y21rd to surface3P2->A+tPtE12=ls dsR12dsR12E12+at P24T E2#+R12/due tois Surface chargeSurface=Ps dsR12(m2) densityClement456 (R12)=)do = Psdsds4A+++ dr +R12E12E12 =Pv dv R12tPvg7P24T tP(c/33)volume charge density= 111CrdsR12:)do = Pvdv4rt (R12)^5E12 =/P, dlR12P2LdidP,E124THE 121212Ri2E12=/le dlR124nt Ri213line charge density #E in all regions due to infinite long line having P1 (4/m)along Z axis iszP^CleE=apP,P2(P,P,Z)yxe is Ind distance from line charge to field pointae is unit Vector along 1rd line drawn from line chargeto field point.For infinite long line along 2- axis (or) parallel to z-axisE is independent on Z- - axis.#E in all the regions due to infinite area surface havingls (c/m2) surface charge density is .where ae is unit vector inE = Its ae ^E direction2 €TH is Lrd outward to the +ve chargeSurfaceE is Ird towards to the -ve chargezSurfaceEx-(-a,b,c)Eat Point "P(-a,b,c) = les ac ^2 EE(ax)y= (-ax)2=axls& x=a; a>o2t Find the value of Fully in the below question:-2Eat (8,0,0)= ?ycx-Q.6(0,01-6)SolA+= Qsource R12GIT 1R121RACRAC = 892-602 ; Racl=10cRBCB -Q F2RBC -> = 8ai+693 ; (RBcl =10E,E,= = Q (8ai-693) -456 (10)3tate = E,+E2E2==0 (892 + 693 )= Q45t(10)34TH (10)Eare=4TE(10)Qx12 [-9zDue to22 Line charget at C ( 3,60,9) =?Se = lo n(c/m)es3Sol E = Pl ap ^yEy=125t e(Friespace)X10x109 ae=2kto (3)(infinite length)10 x 10 9ae=25 x10-9<(3)36/5= 60 ai Oll Given P1=20 n(c/m) is an x-axis from x = 5 to isfind E at origin?(G.1)Field Point2->originSourceSdbeR12 = (0,0,0) - (x,0,0)aeoy= -x an(5,0,0)IRI2 = xdx of (x,0,0)(15,0,0)dE =do R12 = ledx (-xax)at o4T Eo (Rizl3 41Tx10-9 x336/TSate =dx (-ax)1x10"x259E= 180SZ = 180E = 24 (-ax)l = 30 n c/m is at y=3; 2=4 find E at (4,5,10).SalR12= Field Point - Source point= (4,5,10) - (x,3,4) where x combe (-s to a)= (4-x)ax + + =(4-4)an + 2aj + 6G3->R12 = 2ay + 693 & 1R12 = 540R12 A = 20j+6a3R12 is Ind vector from x-axis540towards given point i.e e = IR12& ae = R12Eat(4,5,10) =Ceap =10x15 /29g + 69325to P2T x 10-9 ( / 40)36/5(T40)->E= gay + 27 az ^ Electric flux or Displacement flux (Y) :Fluxis flow ; Diclectric fluxflow of electric charge effect}Total Electric flux = Qwhere G is totaltocharge enclosedD Electric flux Density or Displacement then Density= Electric flux A^7an=anAreaArea1. Electric fleex crossing The open Surface is:-Y/open surface =2. Electric flux crossing the closed surface isY/closed surface = $ B.d's3.B= E Ehollow conducting spherehollow conductorytosphereQ#bbtQSolid4y^E= Q arconductingspherea456 82tooisQE= 0^Eout =Osolid( outs de)ar4/TE r2 Gauss Law66 states that the Electric flex crossing closed Surface [(or) gaussSurface ] is equal to the total charge enclosed by that closedSurface".Mathematically : T Eo = Qclosed surface$ Bods = Qenclosedlips1.Gauss Surface is closed SurfaceIn rectangular System Gauss Surface is cubeIn cyllindrical System Gauss Surface is cylinderIn Sph. systemisspherewhile finding E equation using gauss law, take GaussSurface at the location ofa=If D= xy a: + yz ay + Z x a then find electric flux densitycrossing y=2 plane for 0< x $ 4 & 052 V 2 ?Sol Y=11D.D's = 16B. dxdzaj)B(0,2,2)(4,2,21)>42A(0,2,0)dy2 4=yZ dxdz = 2 dx S 2 dz = 2 2Lx]DC4,2,0)002y=2 constant00= [4](4] - 16 coulombsO J= 2r2 ar + 2sino ao + Sinp aip Find I crossing r=3 SurfaceI=||3.ds = SS J. (r2sino dods )ar = 284 / Sinodo do = 162X2X2508:30r=3 constat= 6487 QFind E in the region due to infinite long coaxial cable havingPsi (c/m2) an P=a cylindrical Surface Pso (c/m2) and P=bcylindrical Surface as Shown in fig.LettotalchargeonP=acylindricalSurfaceforhheightisQ thenlsi = QQ = (2 Tah Psi - I25 ah#Total chargeonl=bsurfaceforhheightiscylindrical- Q thenlso = - QQ = -(2TT bh) lso -225bhBy compair4/&2eqlsi = ((6) lsowhere lsi = inner surface (l==) charge densityPso = outer in (p=b)y13nCSd Regions - oPvr a:0<02RasD (4582) = 4TR3 evD =33Pv R 3^D = Pv R3=DararD3 62: ePvrasas:0282R3 82E =36er R3A36 82ar ; P.RNon-conducting sphere3 €Elat1E188RrRIn the Last question find The at r=R & r=2R.Sd ocreR : E = er as A = PvR as3f3 +6 €Rereoo :- E = e R3 ar = Rv R3 ar = CR ar36823 ( ( 2 R ) 212 t+++-ls2= 4 n c/m2Find D at r= 3m?Tris/rs2,\+Sal Oenc in 3m radius Surfacett/Ps, = 20 n c/m2Oenc = es, (45(2)2)8=4+++go ds = Oeuc = Ps, (4/(2)2)D(45(3)2) = Ps, (4F(2)2)=^: if = Dar = 4Ps, a a = 4x20 9 x10 ar ^ (c/m2)-9Note: 1. Gauss law isHD.D = Q - I2. By divergence theorem@B.ds =SSS(v.B)dv -2By definitionQ= III Pr dv -3By 1, / 2 & 3 eq'V.B = Pvi.e Electric field hasDivergence. # Work done1. Energy stored in Capacitor L 1cv2 2 = 2 / Q2 C = 1/2 QVJouleWork done simply = QV -Work done : By External Surface in moving Q charge frompoint B to point A in the presence of Electricfield (E) is(trud) AW-Q - / E.dl=QVAB=Q=(VA-VB)JouleBiintial)IfEis uniformthen W = -O[E.LBA . JouleLBA - A-Bi.e length Vector from B to ADifferential work done in moving Q charge fordldistancedw = -o[E.difieldQB initialPointdlQs/finalPoint#Energy stored in the system of Q, , O2 / ONpoint chargesNW = 1 2 E Q Vi Joulei=1Where V, is potential at O, location due to other chargesVNis"inOninis-"#EnergydensityofElectricfieldis2dWE = 2 1 t 1212 = 1[3.2] 2 D E(Joule/m3)dv Potential DifferencePotential Difference b/w Point A and Point B 1/6 It is defined aswork done in moving "+1 c charge from point "B" to point AFind(A)VAI AB = E.dl = VA -VBinitial(B)VAR AB is potential difference between A and BorPotential A w.r.t 3 where VA Potential at A w.r.t OVVB potential at B wirstABAABE. I = - - S E.dl + SEED = E.dl -wB00as00AE.dl = VA- VBBVBBVBB = Vg(t=t) - V3 (t==)Battoo& t=t,=O [only for static field]=>ie-E.dl=0BFO [for time varying field]For Static field;PE.DD=0from Stokes theorem6 E.dl = [j(DXE) ids ->P E.dl = J|(DXE) ds=oTXE =0Static Electic field is irrotational or conservative QFind potential field in all region due to Q charge located atorigin.Case1:-Ovoltisatr=8distancePrp(r,0,0)-E.dl=Q ai.^dr asr= 00Vp =45E 826-00OvaltasrVpQdrQ(origin==456r2456r000=00V :=4THE Q ( \ r 8V(r,o,p) = QUTECase2:Potential at Pin) wrst M(r)p(r,0,0)rVp =OAMa,drar456 825Qr,00QVp ==+ (-QYITErr,4ktr yourUp =Q+CyoterQAQisatoriginaroundthisatA(2,0,0)potential'scharge15volts/o)potentialis30valt.atBo,FindPotentialatC(1,o,o)SofV= Q +c4rtVAO=+C=15Q = 10(4/T E):Vc=Q+ C45t22+0 +0245t1240402)C= 10QVB =+ (= 30= 10 + 104TT J= 20 volts a inrolthepc2a1 Equipotential surface of Q point charge are concentric Spheresthose center is at the location of a point charge) E lines and Equipotential surfaces meets at 90°OF is normal vector to equipotential Surface / so normalunit vector to Equipotential Surface isforAve charean = + THEIEve charge 3E = ya + x of : find V at (4,1,0) with (0,0,0)of(4,1,0)V -J E.di = - - I ya + aig) (dx a + dy by +de d3)(0,0,0)(0,0,0)V=+judy]Since (who) (90)y duL=OFq of liney=-4-0V=--da+Jtydy=)x-4y 2 dx - ddyV= 8 o + 4, 2 itV=- 2 + 2 ] = Electic DipoleElectric Dipole has o & - Q charge Separated by d distance[where d is very small compared to ratwhichwefindV,Efields ]todis distance vector from -Q to +Q(0,0,d)d =(0,0, d) - (0,0,- )1=d- 2 - ( I)] azd = daz- Q (0,0,-d).17Electric Dipolemomentpis defined ascoulomb - meterorp = QdDebye" PolarizationPis The total electric dipole momentper unit volume..NPSi.e=pii=o; cm or c/mVolumem3thIV at? r, 0,0) due to electric dipole is+ O-A(000,0)Vcr,0, =p.arQ d az.ar=^"daz04TH r 2456 82yor d coso-QV(r,o,p) =Dve Electric Dibole^t= - V V = - dv ar A - 1 dv ao#Vat&Eatdrr00r2g3Due to Qpoint chargeVal & t x-E = Qd822COSO as ^ + Sino ao45 t r3VLE decreases faster forElectric dipole then o Points#IIn air Capacitor there are no electric diputes:If we applyExternalElectric field thenIcD = to EIplates-VairIsJYatIsIn diectric capacitor electric dipoles arepresent When external E is not applied then all dipolesare randomly orientedSo,P=O# If External E is applied then dipoles Rotates and align/p 70ICSoi.eD = Co E + 12-2IDID DYdtLinearity In Dielectric ; ?& EIc F IDP = constant EWhere ; constant 1. Physical Process variable (Xe)2. Term to balance units ( to)->P = to Xe E- 3from3&2D = EOE +to XeTHEif = to (1+xe)E4LetEr= It Xe-5D = ford EWheret=total PermitivityD=EEEr= Relative Permitivity=1 for airApplicable for Linear matterial Note:#1 Ic = dq i conduction current (Arp)dt2 Jc = Ic ie conduction current density (Amp/m2)A3Id = dy i.e Displacement curent (Amp)dt4Jd= = Id i.e Displacement Current density (Amp/m2)A= 04/01 = X(4/A) = OD = JdAItatSome observationsQQ -YQ =CVC ===CabVVabVabCab = SS DidsSSEE.ds== Cad-VabE.dlbwhere i Cad is capacitor b/w platesVasis voltage b/w platesQ is charge on platesNote : condition to neglect fringing Effect in Capacitors A00ie After neglecting fringing Effects of finite area parallel platecapacitorC = EAd2. = Energy Stored in Capacitor ; We = 1/cv2 2 2 = QV = Q 2 Joules22CEnergy stored in Electric field , we i.e2WE= dv = = ff 1/2 D.E dv = SS 1 2 IDI2 Edv Some standard capacitances & their Dexi:Find capacitance b/w P=a and P=b surface of a Co-axial cablefor h heightE = Psi a apfor ourE PSol Cab = Q= SEEDS = Help E a P a.) (edg d3)aVas- SE.dl6-lsi a de aE Pb271hCab =dodzt(2 TT)()2/T E h=F==de- [ lan e]aln(b/a)PbbCab =2 T E h Ffor Coaxial cableln (b/a)FindCapacitanceofsphericalCapacitorShownbelow:-SolE=Qaoforacreb-Q45Er2Cab =QQ=a!QE.dl-do dr arYOTE6Cab=Q=4stQdrr2- (-1)Cab= 45E it b>a4KE bba1 -#If bas then r=a only present & r=a is called asisolated spherical conducting sphere of radius a / whoseCapacitance is "YITE a "Q Find the charge on isolated sphereif r=lm & t=to & 4V is appliedsol = =) Q=CV= (4Ttor)4 C Capacitor ConceptwaIfc, C2 are in Series then total C=C1C2i.e=)CAB=C, 22AC,C2BC,TC2If distance d between plates divided then V, E are devidedBut area ,P,P,I are not dividedId will be the Same in both diectric materialsTVdiTE,d,E4 EAd,V2 d2:IdE2,1d2E2=)C2 = EAd2#D=YQCVDisindependent of mediumAAAE, = D&E2 = DV=V1TV2E,E2# d = dit d2#V, D=&V2 = Dd, E1d2E2 Capacitor Conceptwa4It two capacitor connected parallel Then9C2C=GtczoIf the area of The plate is divided Q / Y Id is dividedId is divided but distance "d" is not divided, Sov, t are Same in parallel pathsY2,AzorTd-dxEO2!G42ddId,Yd2td2 - d 1/2dtADD9 GA, ; C2 = E2A2A2,02dd;FQ,- ++ C2D==-C,Y,& D2 - C2V2A, A,A1A2E, D&E2 = D2& E, = E2=L&V = V2=VE,E2QFind C, or E, : E2Fig:-E2E,E2ieSoDistances are divided area is not dividedlovSo charges Q, Q2Q are sameQ=OC,V, = C2V2E, 3EA ) V, EA E.AV2E2 - 2dE,2 = E2.6dd42 Some Imp. Equations:# Homogeneous mediumE, u, a values of the medium areconstants..Non- Homogeneous mediumE, u, to Values of the medium changefrom one location to other locationE, ce , 00 are functions of x,y, ZPoisson Equation ( P.E)V.D = ev ; D= EED E = - DVJ.EE = evV.EE= ev/-D. (-VV) = PvV.(DV) = PvFor homogeneousEE-2E constantorD. EVV] = -3non homogeneousEquation "3" is known as P.E of non- homogeneousmedium.For homogeneous medium E is constanti.e eq4'2' D. (DV)= - Pvt2 V = evt-4Equation "4" is known as P.E for homogeneous medium Laplace Equation (L.E)# If Pv=0 [By Q, le, ls can be presentThen Equation "3"V. (EVV)= - = 0i.e5-# Equation '5" is known as L.E of non - homogeneousmediumIf Pr=0inEqu4" i.eV'v - e==EViv=o-6Eq" 8 gives LOE in homogeneous medium# 4 in ques a,le,l's ask Then go for L.E# If ev ask then go for P.E QFind V, E, D equation between ria and r=bSurface of a spherical capacitor Shown in fig.diclectricFind Cas? Given i==0 and home mogeneousas+mediumV=Vo at r=aVoV=0 at r=bb+-H Given Pr=0 Select L.L E&V2v=02fig.In spherical. Co. Sys. :- v'v=)driveI+a Sinoav+r2drr2"2vorsino todor'sino002=0Since;V is the function of as we ad know foreviously2v =dr 2dvd=0p2 oror2dv do ) = odr,-Integrate eg" i both sides-r2 dv = O+KdrovK=dv= K ordr2r22Take integral both sides eg"V =K,[1/6+K2-3& given that V =0 at r=bO= + K2-k,K2=KI-46bBy 3 " & 4 "eg4V=-KI+K1-5rbNow; V=Vo at r=a : Vo= - K, +K,K1= Voab6a)"Put in eq" - " "5""V=Vo+bto - a)6- Eq" 6 :- V= Vo(1-1) [ 'b] & We know that E=-VVonly component of"r"E= av ar Vo- or = - d or (% ar( toE =Vo ar- 7D=EE DE VE =ar-8(a-b) 82Cab = II D.d'sEVo#ai. (r' sinodods) ar=(1-6) r2VabVo-02/EVo(1-1) 0=0sino do I doCab =E(2)(21)=Vo(----) a#Cab 4561-1a s

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