000000000., +, +, +, +, +, +, +, +, +, +, +, +, +, +/111111111-AIEEEPHYS I CSS T U D Y M A T E R I A LNARAYANA INSTITUTE OF CORRESPONDENCE COURSESF N S H O U S E , 6 3 K A L U S A R A I M A R K E TS A R V A P R I Y A V I H A R , N E WD E L H I - 1 1 0 0 1 6PH.:(011)32001131/32/50FAX:(011)41828320Websi t e : w w w . n a r a y a n a i c c . c o mE- mai l : i n f o @ n a r a y a n a i c c . c o mELECTROSTATICS 2004 NARAYANA GROUPThis study material is a part ofNARAYANA INSTITUTE OF CORRESPONDENCE COURSES for AIEEE, 2008-09. This is meantforthepersonaluseofthosestudentswhoareenrolledwithNARAYANAINSTITUTEOFCORRESPONDENCECOURSES,FNSHouse,63,KaluSaraiMarket,NewDelhi-110016,Ph.:32001131/32/50.AllrightstothecontentsofthePackagerestwithNARAYANA INSTITUTE. No other Institute or individual is authorized to reproduce, translate or distribute this material in any form,withoutpriorinformationandwrittenpermissionoftheinstitute.PREFACEDear Student,Heartiest congratulations on making up your mind and deciding to be an engineer to serve the society.As you are planning to take various Engineering Entrance Examinations, we are sure that this STUDY PACKAGE isgoing to be of immense help to you.At NARAYANA we have taken special care to design this package according to the Latest Pattern of AIEEE, whichwill not only help but also guide you to compete for AIEEE & other State Level Engineering Entrance Examinations.The salient features of this package include:! Power packed division of units and chapters in a scientific way, with a correlation being there.! SufficientnumberofsolvedexamplesinPhysics,Chemistry&Mathematicsinallthechapterstomotivatethestudents attempt all the questions.! All the chapters are followed by various types of exercises (Level-I, Level-II, Level-III and Questions asked in AIEEEandotherEngineeringExams).These exercises are followed by answers in the last section of the chapter. This package will help you to know whatto study, how to study, time management, your weaknesses and improve your performance.We,atNARAYANA,stronglybelievethatqualityofourpackageissuchthatthestudentswhoarenotfortunateenough to attend to our Regular Classroom Programs, can still get the best of our quality through these packages.We feel that there is always a scope for improvement. We would welcome your suggestions & feedback.Wish you success in your future endeavours.THE NARAYANA TEAMACKNOWLEDGEMENTWhilepreparingthestudypackage,ithasbecomeawonderfulfeelingfortheNARAYANATEAMtogetthewholehearted support of our Staff Members including our Designers. They have made our job really easy throughtheir untiring efforts and constant help at every stage.We are thankful to all of them.THE NARAYANA TEAMELECTROSTATICSTheorySolvedExamplesExercisesLevel ILevel IILevel IIIQuestions asked in various ExamsAnswersCONTENTSCONTENTSCONTENTSCONTENTSCONTENTSPhysics:Electrostatics1FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAELECTROSTATICSAIEEES y l l a bu sElectric charge itsunit and cons ervation, Coulombslaw , dielectric cons tant, electric field,linesof force, field due to dipole and itsbehaviour in a uniform electric field, electric flux,Gaus s s theoremandits applications .Electricpotential,potentialduetoapointcharge.Conductorsand ins ulators , dis tribution of charge on conductors .CONTENTSINTRODUCTIONWhenaglas s rodis rubbedw iths ilk ,this acquirespow ertoattractlightbodies s uchas s mallpieces ofpaper. The bodiesw hich acquire thispow er are s aidto be charged. If thes e chargesdo not move theyarecalled s tatic chargesand the branch of phy s icsw hichdealsw ith s tatic chargesiscalled electros tatics ." Electric charge" Coulombslaw" Electros taticsfield" Ele ct ros t a t ics lin e s offorce;Properties" Electric field due to a pointcharge" Superpos ition of Electricfield" Electric field intens ityontheaxis ofauniformlycharged ring." Electric flux" Gaus s sLaw ." ApplicationofGaus s slaw" Ele ct ricpot e n t ia la n dpotential energy" Equipotential s urfacePhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 418283202INSTITUTE OF CORRESPONDENCE COURSESNARAYANAELECTRICCHARGECharge is a scalar quantity which is categorised into two types.a) Positive charge (anciently called Vitreous)b) Negative charge (anciently called Resinous)A body having no charge, is said to be neutral in nature i.e. on a neutral body the sum of positive charges isequal to the sum of negative charges.The positive charge means deficiency of electrons, whereas the negative charge on a body implies excess ofelectrons.The S.I. unit of charge is coulomb (C).I. UNITS OF CHARGEa) S.I. unit of charge is coulomb (C). One coulomb of charge is that charge which when placed atrest in vacuum at a distance of one metre from an equal and similar stationary charge repels itand is repelled by it with a force of 99 10N .b) CGS unit of charge is stat coulomb1 coulomb = 93 10esu of charge 110 emu of charge.i.e. 1 coulomb 93 10 stat coulomb110 abcoulombc) Practical units of charge are amp hr = 3600 coulomb and faraday( )96500 C d) Smallest unit of charge is stat coulomb and largest is faraday.II. METHODS OF CHARGINGa) By Friction : In friction when two bodies are rubbed together electrons are transferred from onebody to other. As a result one body becomes positively charged while the other negatively charged.Examples : When glass rod is rubbed with silk, the rod becomes positively chargedwhile the silk negatively charged.Clouds also become charged by frictionb) By Induction : If a charged body of same sign is brought near a neutral body charged body willattractcharge of opposite sign and repel charge of same sign present in neutral body. The natureas induced charge is always opposite to that of inducing charge. Charging of body by Induction isshown as : Charged body brought near uncharged bodyUncharged body connected to earthUncharged body disconnected from earth++ + ++ + +++++ + + + + ++ + + + ++ + + +++ + +++++++++ Physics:Electrostatics3FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAc) By Conduction : If a charged body is in direct contact with an uncharged body, charge flows fromformer to latter till both are at same potential. This flow of charges is due to mutual repulsionsbetween same kind of charges on charged body.Charging body is removedIII. PROPERTIES OF ELECTRIC CHARGEa) Like point charges repel each other and unlike charges attract each other.b) Charge ConservationThe algebraic sum of all the charges in an isolated system is a constant. In crude language wecan say that charge can neither be created nor be destroyed, however it can simply betransferred from one body to the other.c) Relativistic InvarianceCharge on a body is relativistically invariant. i.e. charge on the body at rest equals the chargeon the body at relativistic speeds. However charge density is not relativistically invariant.Mathematically( ) ( )at rest in motionq q or ( ) ( )at rest at relativistic speedsq q d) Charge QuantisationCharge on a body q must always exist as an integral multiple of some fundamental unit ofcharge (called electronic charge) e , where 191.6 10C e .Mathematically,q ne t ,1, 2, 3, ..... n From here we conclude that a neutral body can have1 C +of charge when it falls deficient of186.25 10 electrons.e) A charged body can attract light uncharged body. (Due to charging by induction)f) Charges are always added algebraically.For example, if a neutral body is first given a charge of5 C +and subsequently a charge of7 C , then it will finally have a charge of5 C 7 C 2 C + .EXAMPLESBASEDONELECTRICCHARGEExample - 1 Calculate protonic charge in 100 cc. of water.Solution: 1cc=1gmforwaterasdensity=1000kg/m3Now no of atoms in 18 gm (atm weight) = 6.023 1023 and each molecule of H2O contains10proton(8ofoxygen+2ofH2)soNo.ofprotonin100gmwater252310 3 . 3 10 1001810 023 . 6

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.|Henceprotoniccharge=3.310251.61019=5.4106CPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 418283204INSTITUTE OF CORRESPONDENCE COURSESNARAYANAExample - 2 A copper sphere of mass 2.0 g contains about 2 x 1022 atoms. Atomic weight of copper is64 and atomic number 29.(i) How many electorns must be removed form the sphere to give it a charge of +2C?(ii) Determine the fraction of electrons removed.(iii) Isthereanychangeinmassofspherewhenitisgivenpositivecharge?Solution: (i) Number of electrons to be removed 1319610 x 25 . 110 x 6 . 110 x 2eQn (ii) Totalnumberofelectronsinthesphere=29x2x1022=5.8x1023Fraction of electrons removed =11231310 16 . 210 8 . 510 25 . 1 Thus 2.16 x 10-9% of electrons are to be removed to give the sphere a charge of2C.(iii) Yesmassdecreases,whenbodyisgivenapositivecharge.Decreaseofmasskg 10 x 125 . 1 10 x 25 . 1 x 10 x 9 m17 13 31 .COULOMBSLAWThe magnitude of the force (F) of attraction or repulsion between two point charges 1qand 2qplaced invacuum at separation r isa) directly proportional to the product of the magnitude of the two charges.1 2F q q ...(i)b) inversely proportional to the square of the distance of separation between them.21 Fr...(ii)COULOMBS LAW IN VECTOR FORMConsider two point charges 1qand 2qseparated by a distance r. If>1 20 q qi.e. if both 1qand 2qare +veor both1qand 2qare negative then the charges repel each other otherwire they attract each other.If !12Fis the force exerted on charge 1qby charge 2q .1 212 21 20q q 1 F r4 r= 1 221 301. .4q qrr !If !21Fis the force exerted on 2qdue to 1qthen!1 221 12 2014q qF rr= 1 212 301. .4q qrr !Both 21 rand 12 rhave same magnitude i.e. unity and are oppositely directed( ) ! !1 221 21 12 2014q qF r Fr ! !21 12F FSo the forces exerted by charges on each other are equal in magnitude opposite in direction12 rF12F21q1q221 rPhysics:Electrostatics5FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAPROBLEM SOLVING TRICKS1. While calculating the force between the two charges from Coulombs Law, never takeinto account, the sign of the two charges. The sign just indicates the nature of the force.2."!12 F is the force on charge 1 due to 2. "!21 F is the force on charge 2 due to 1. !ijFis theforce on charge i due to j.3. When we are to calculate the force on charge 1 due to charge 2, then we assume charge2 to be fixed and vice versa, unless and otherwise stated."!121 1F ma"!212 2F m ai.e. 1 1 2 2ma m a(by Newtons Third Law)where 1mand 2mare the masses of charges 1qand 2qrespectively.4. If we are to calculate the force on charge 0q , due to assembly of charges 1q , 2q , . nq ,then we have + + + +"! "! "! "! "!0 01 02 03 0 ..... n F F F F Fwhere "!01 F is calculated as if only 0qand 1qare present and all others are absent and so onfor other combinations. This principle is called the Principle of Superposition.Note : When two identical bodies having charges 1qand 2qrespectively are brought incontact and separated, then the charge on each body is +1 22q q.EXAMPLESBASEDONELECTROSTATICSFORCEExample-3 Atomicnumberofcopperis29,itsatomicweightis63.5gm/mole.Iftwopiecesofcopper having weight 10 gm are taken and on one of the pieces 1 electron per 1000atom is transferred to the other piece and there after this these pieces are placed 10cmapart,thenwhatwillbethecoulombforcebetweenthem.Solution: In 1 mole copper (63.5 gm) there are 6 1023 atoms (NA = Avogadro number = 6 1023atoms)Numberofatoms= 5 . 6310 10 623 =9.451022For every 1000 atoms, 1 electron is transferred therefore total number of transferred electronis192210 45 . 9100010 45 . 9 Therefore charge on one piece is 9.45 1019 e and on the other will be (9.45 1019e)Forcewhentheyarekept10cmapart22 10rq q41F 2 22 2 190) 10 10 (e ) 10 45 . 9 (41 = 2 22 19 2 19 9) 10 10 () 10 6 . 1 ( ) 10 45 . 9 ( 10 9 =21014 NPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 418283206INSTITUTE OF CORRESPONDENCE COURSESNARAYANAExample-4 On an insulated rod having length L, Q charge is evenly distributed and a point chargeq is placed at a distance d from one of its ends as shown. What will be the total electricforceonq.Solution: Consideringasmallelementdxoftherodatadistancexfromq,forceonqduetothissmallelement20x 4qdQdF dxpqB A# dxwheredQischargeinsmallelementdxanddQdxLQ

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.|F=dF=20Lx 4qQdx LqQ410 2L ddxdx+ LqQ410 L ddx1+]]]

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+ L d1d1LqQ410 ) L d ( d 4qQ0+ Example-5 Three equal charges (Q, each) are placed on the vertices of an equilateral triangle ofsidea.Whatistheresultantforceonanyonechargeduetotheothertwo?Solution: The charges are shown in fig. The resultant force 60 cos F F 2 F F F2 12221+ + a aaQ Q30F1F2F60 QwithF1=F2=kQ2/a2223kQFaFromsymmetrythedirectionisasshownalongy-axis.Example-6 Five point charges, each of value +q are palced on five vertices of a regular hexagonof side L m. What is the magnitude of the force on a point charge of value -q coloumbplacedatthecentreofthehexagon?Solution: If there had been a sixth charge +q at the remaining vertexof hexagon, force due to all the six charges on -q at Owill be zero (as the forces due to individual charges willbalanceeachother),i.e., RF!=0.AFEDCBqqqqq-qoNow Iff!is the force due to sixth charge and F! due to remaining five charges,0 f F +! !i.e. f F! ! or,2020Lq41Lq x q41f F]]]

Physics:Electrostatics7FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAExample-7 TwochargedspheresofradiusRarekeptatadistanced(d>2R).Onehasacharge+qand the other -q. The force between them will be-[1] 220dq41[2]> 220dq41+++++++dR R[3]< 220dq41[4]NoneoftheseSolution: [2] Redistribution of charge will take place due to mutual attraction and hence effectivedistancewillbelessthand.ELECTROSTATICFIELD ( )!EThe region of space around a source charge( ) qin which it can exert a force on a test charge( )0q .Mathematically, electric field is the force experienced per unit test charge 0qplaced in the electrostaticinfluence of source chargeq ."!"!0FEqElectric field strength is a vector quantity directed away from a positive charge and towards the negativecharge. SI unit of electric field is newton/coulomb( )1NC or( )1volt metre Vm .The dimensional formula for E is 3 1MLT A.Since"!02014qqF rr "!"!20 04F qE rq ris the electric field due to a source point chargeqat a distance rfrom it.ELECTROSTATICLINESOFFORCE:PROPERTIESA line of force is an imaginary path straight or curved such that the tangent to it at any point gives the directionof electrostatic field at that point. A field line is an imaginary line along which a unit positive charge would movewhen set free. The lines of force are drawn such that the number of lines per unit area of cross-section, (areaheld normally to the field lines) is proportional to magnitude of "!E .I. PROPERTI ES OF FI ELD LI NESThe electric field lines for a point charge +2q and a second point charge q. Note that two lines leave the charge +2q for every one that terminates on q.- q ++2qa) Field lines always come out of positive chargeand enter the negative charge.b) Field lines never cross each other.c) Field lines never form closed loops.d) Fieldlinesarealwaysdirectedfromhigherpotential to lower potential.e) Field lines never exist inside a conductor.f) If 1Nis the number of field lines coming out of acharge 1qand 2Nis the number of field linesentering 2q ,then1 12 2q Nq NPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 418283208INSTITUTE OF CORRESPONDENCE COURSESNARAYANAg) If 1Nis the number of field lines coming out of a charge 1qand 2Nis the number of field linescoming out of charge 2q ,then1 12 2q Nq NThis relation also exists if field lines are entering both the chargesh) Tangent to field line at a point gives the direction of field at that point.i) Field lines exhibit longitudinal (length wise) contraction, thus indicating that unlike chargesattract each other. (See Figure 1)j) Field lines exhibit lateral (sideways) expansion, thus indicating that like charges repel eachother. (See Figure 2)Figure 1 : The electric field lines for two equaland opposite point charges an electric dipole.Note that the number of lines that leave thepositive charge equals the number thatterminate at the negative charge.+ -equal positive point charge.k) Field lines always enter or leave a conducting surface at right angles.l) Since electric field inside a conductor in electrostatics is zero, electric field lines do not existinside the conductor.EXAMPLES BASED ON ELECTRIC LINES OF FORCEExample-8 ASolidmetallicsphereisplacedinauniformelectricfield.WhichofthelinesA,B,CandDshowsthecorrectpathandwhy?Solution: Path (A) is wrong as lines of force do not start or end normallyon the surface of a conductor. Path (B) and (C) are wrong aslines of force should not exist inside a conductor. Also linesof froce are not normal to the surface of conductor. Path (D)represents the correct situation as here line of force does notexistinsidetheconductorandstartsandendsnormallyonitssurface.AEBCDExample-9 Ametallicslabisintroducedbetweenthetwoplatesofacharged parallel plate capacitor. Sketch the electric lines offorce between the plates.+Q -Q +Q -QMetal+++++++++E =0E0E0Solution: Keeping in mind the following properties :1. Lines of force start from positive charge and end on negativecharge.2. Linesofforcestartandendnormallyonthesurfaceofaconductor.3. Lines of force do not exist inside a conductor (as field inside aconductor is zero)The field lines between the plates are as shown in the figure.Physics:Electrostatics9FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAELECTRICFIELDDUE TOAPOINTCHARGErrq41E20 rrq41E30qXrPEY+(A)0XrPr r0Y(B)+Err r0If the position vector of the charge is 0r,thenelectric field at position r is3000| r r |) r r ( q41E SUPERPOSITION OF ELECTRIC FIELDSTheresultantelectricfieldatanypointisequaltothevectorsumofelectricfieldsatthatpointdueto various charges....... E E E E3 2 1+ + + Themagnitudeoftheresultantoftwoelectricfieldsisgivenby + + cos E E 2 E E E2 12221E2E1EE2EThedirectionisgivenby + cos E Esin Etan2 12ELECTRIC FIELD INTENSITY ON THE AXIS OF A UNIFORMLY CHARGED RINGdEEPrar + a22++++++++++++++++ +dqLet dE be the electric field at a point P on the axis, due to charge dq, while total charge on the ring is q.Then 2 20a rdq41dE+ As only axial components of electric field due to all elements remain, while perpendicular components arecancelled.Therefore E at Point P =dEcos $= ( )3/ 2 2 22 2 2 20 01 dq r 1 r. dq4 r a 4r a r a + + + ( )]]]

+ q dq asa rqr412 / 32 20Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832010INSTITUTE OF CORRESPONDENCE COURSESNARAYANASPECIALCASES(i) At r = 0 (i.e. at centre of the ring) E = 0(ii) At r >>a (i.e. at for off points on the axis) 20rq41E . It shows that the ring behaves like a pointcharge for far off points.The magnitude of E increases from zero as we move on the axis starting from centre of the ring. It attainsa maximum value at some point on the axis and thereafter again decreases. At the point, whose E ismaximum, 0drdE2 2 3/ 20d 1 qr0dr 4 (r a ) ] ] + ]++++++++++++2ar =2 2 3/ 2 2 2 1/ 23(r a ) r. (r a ) . 2r 02+ + ar2EXAMPLES BASED ON ELECTRIC FIELDExample-10 Charges with magnitude q are placed at 4 corners of a regular pentagon. These chargesare at distance a from the centre of the pentagon. Find electric field intensity at thecentre of the pentagon.Solution: ChargesareplacedatcornersA,B,CandDofthepentagon.Ifchargeqisplacedatthefifthcorneralsothenbysymmetrytheintensity EatcentreOiszero.Let 1E! be the resultant electric field due to charge at A, B C and D and let 2E! be theelecricfieldduetothefifthcharge.Now,1 2E E 0 + ! !1 2E E ! !ElectricfieldduetochargesatA,B,CandD.B AE CDqqq qOa2 20qE4 q!along EO"""!Example - 11 Three charges +q, +q, +2q are arranged as shown infigure. What is the field at point P (center of side AC)PaaC+q +2qBA+qSolution : The sum of fields at P due to charges at A and C add upto zero (because of equal magnitude and oppositedirection) . Thus the net field at P is that due to +2qcharge. Its direction is along the line BP and its magnitudeis041E2) BP (q 2Physics:Electrostatics11FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA2 2PC BC BP = 2 / a a2 2 = 2 / aThus 20qEaExample-12 An infinite plane of positive charge has a surface charge density . A metal ball B ofmass m and charge q is attached to a thread and tied to a point A on the sheet PQ.FindtheanglewhichABmakeswiththeplanePQ.Solution: Due to positive charge the ball will experience electrical force Fe = qE horizontally awayfrom the sheet while the weight of the ball will act vertically downwards and hence ifTisthetensioninthestring,forequilibriumofball:Alonghorizontal,Tsin=qEAndalongvertical,Tcos=mgSo, mgqEtan andT=[(mg)2+(qE)2]1/2++++++++qqEmgBPAQTThefieldEproducedbythesheetofchargePQhavingchargedensity is 02ESo, mg 2qtan0 i.e., ]]]

mg 2qtan01ELECTRICFLUXThe mathematical quantity related to number of lines passing through a surface is called the electric flux .The electric flux through a surface which is perpendicular to a uniform electric field E is defined as the productof electric field E and surface area A : EA Since the electric field is proportional to density of lines of force, the electric flux is proportional to number oflines of force passing through the surface area :N .If the surface area is not perpendicular to the electric field, then the electric flux is given byA E A cos E A n . En !AEnwhere n is a unit vector perpendicular to the surface and nE is the componentof electric fieldperpendicular to the surface (normal component).The electric flux over a curved surface over which electric field may vary in direction and magnitude can becomputed by dividing the surface into large number of very small area elements. Let inbe the unit vectorperpendicular to such an area element and iA be its area.The flux of the electric field through such an areaelement is i i i iA n . E !!Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832012INSTITUTE OF CORRESPONDENCE COURSESNARAYANAThe total flux through the surface area is found by adding flux through each area element. In the limit ,whennumber of area elements approaches infinity and area of each element approaches zero, the sum becomes anintegral. ii i indA n E A n E . . lim! !.Quite often we are interested in finding out the fluxthrough a closed surface. The unit vector n in such a caseis defined to be directed outward from each point. Note that when an electric line comes out of the closedsurface, then n E .!is positive and if it enters the surface n E .!is negative.The net flux through a closed surface is written as !$ $ net n E . n dA EdA. .Please note that netis proportional to the net lines of force going out of the surface, ie, number of lines goingout of the surface minus the number of lines going into the surface.GAUSSLAWI. STATEMENTenclosedneto surface Sq E .n dA E dA "! "! !$ $where qenclosed is the total charge enclosed inside the surface. This important result is called Gausss Lawand can be stated as follows:For a system of charges, the net flux through any closed surface S is equal to o/ 1 times the netcharge inside the surface.Example If a dipole is enclosed by a closed surface S, then total flux due to this chargedistribution is zero.( ) ( ) + "! "!$010SEdA q qAPPLICATIONOFGAUSSSLAWELECTRIC FIELD DUE SOME CHARGE DISTRIBUTION DERIVED BY USING GAUSSLAW.(i) Electricfieldduetolinecharge(infinitelength)Theelectricfieldatadistancerfromalinecharge(density)isr 2E0=k. r2rE(a) (b)ErFig.1The direction is outward perpendicular to the line charge. The E (1/r) dependence is showninfig. 1(b)Physics:Electrostatics13FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(ii) ElectricfieldduetocylinderIfthelinechargeisacylinderofradiusR,then(a) Electric field outside+ +Rr(a)E0rEr E r1(b) Fig . 1 Fig . 2r 2E0(r>R)(b) Electric field on the surfaceR 2E0(c) Electricfieldinside(atadistancerfromtheaxis)20R 2rE(ifr R )The direction of the field is outwards (normal to the axis). The dependence of the field on risshown in figure.2 (b). Inside the charged cylinder,Eroutsider1E (iii) ELECTRIC FIELD DUE TO A PLANE SHEET (INFINITE DIMENSIONS)(a) SinglesheetofchargeE++++++++++++For the surface charge density (coulomb/metre2) the fieldatadistancerfromthesheetis 02E directedtowardsoutward normal (from a positively charged sheet)Theelectricfielddoesnotdependondistance.(b) ChargedmetalplateInside charged metal plateE=0Outside charged metal plateEE++++++E=0++++++ 0EThefieldisnormallyoutwardsforpositivelychargedplateand inwards for negatively charged plates.(iv) ElectricfieldduetoachargedsphericalshellThechargeQresidesonthesurfaceofthesphericalshell(radiusR) .EEBOr ARCQ++ ++++ +++Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832014INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(a) FieldatoutsidepointArrQ41E20201 QE4 r(b) FieldatsurfacepointBrRQ41E20201 QE4 R(c) FieldatinsidepointCE 0 O RE=EEkQ1Rr22rE=0The variation of field with distance r from the centre O of the shell in shown in fig. The field atthesurfaceismaximum.AndoutsidetheshellfieldvariesasE1/r2 .(v) ELECTRIC FIELD DUE TO CHARGED CONDUCTING SPHEREThe entire charge Q resides on the surface of a charged conductor.Any charge given to the interior, flows to the surface in less thana nanosecond.Soachargedconductingspherebehaveslikesphericalshell.Thus,+++++++ +++++EAEr0RCB(a) Fieldoutside 201 QE4 r(b) Fieldonsurface 201 QE4 R(c) Fieldinside E 0 Specialnote:Thesurfacechargedensityintheabovecaseis=Q/4R2.Intermsofthefieldsareoutside

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.|220rREon surface0Einside E=0Physics:Electrostatics15FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(vi) ELECTRIC FIELD DUE TO A CHARGED SPHERE (NON CONDUCTING)In case of a charged non conducting (plastic etc.) sphere, chargesdo not flow. As a result, charges exist inside the sphere as wellas on the surface. Assuming uniform charge distribution, the electricfield,outside(pointA),onthesurface(pointB)andinside(pointC)areasfollows.RC++ ++++ ++++++Ro+++++++++CEAErEB(a) Fieldoutside20rQ41Edirected radially outwards (for positive Q)(b) Fieldonsurface201 QE4 R(c) Fieldinside301 QE r4 Rdirected radially outwards for positive Q.O RE=EEkQ1Rr22rThe variation of the electric field with distance r from the centre of the charged nonconductingsphere is as shown in fig. The field outside varies inversely as square of the distance. The fieldatthesurfaceismaximum.Thefieldinsideisdirectlyproportionaltothedistance.Specialnote:Thevolumechargedensityinabovecase,is ) R34/( Q3 .Intermsof,thefield areoutside

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.|230rR3Eon surfaceR3E0insider3E0EXAMPLEBASED ON ELECTRIC FLUXExample - 13 If a point charge Q is located at the centre of a cube then find (i) flux through the totalsurface, of the cube (ii) flux through one surface.Solution : (i) According to Gausss law for closed surface = Q/0 , (ii) Since cube is a symmetricalfigure thus by symmetry the flux through each surface is = Q/60Example-14 A point charge +q is located L/2 above the centre of asquare having side L. Find the flux through this square.Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832016INSTITUTE OF CORRESPONDENCE COURSESNARAYANASolution: The charge q can be supposed to be situated at the centreof a cube having side L with outward flux . In this casethesquareisoneofitsfacehavingflux/6fluxthroughthesquare 06qOL/2qLExample-15 WhatisthevalueofelectricfluxinSIunitinY-Zplaneofarea2m2,ifintensityofelectricfieldis ) j2 i5 ( E + N/C.Solution: dA . Ei2 ). j2 i5 ( + =10V-mExample-16 2C charge is in some Gaussion surface, given outward flux , what additional chargeisneededifwewantthat6fluxentersintotheGaussiansurface.Solution: Accordingtoquestion +6C ) Q 2 (0or 6) Q 2 (0 +

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.|02Q = 14CExample-17 A cylinder of length L and radius b has its axis coincident with the x axis. The electricfieldinthisregion i200 E .Findthefluxthrough(a)theleftendofcylinder(b)theright end of cylinder (c) the cylinder curved surface, (d) the closed surface area of thecylinder.Solution: from fig. then(a) cos EA A . EaEdASadAESbdASc=200b2cos=200b2(b) b=EAcos0=200b2(c) c=EAcos90=0(d) =a+b+c=200b2 +200b2+0=0Physics:Electrostatics17FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAELECTRICPOTENTIALANDPOTENTIALENERGYLINE INTEGRAL OF EThelineintegralofelectricfieldsisdefinedastheintegral # d . EBAEABd#The value of line integral depends only on the position of points A and B, and is independent of thepath between A and B # # d . E d . EABBALineintegralforaclosedpathiszero . 0 Ed #$(such electric fields are called conservative fields).Note : There exists one more type of electric field, called induced electric field.It is non-conservativeforwhich . 0 Ed #$ELECTRICPOTENTIALElectric potential and potential energy are defined only for conservative fields.Definitionintermsofworkdone:Potential at any point A is equal to the amount of work done (by external agent against electric field)inbringingaunitpositivechargefrominfinitytothatpoint.qWVAAUnitofpotential(V)=J/CorvoltPotential at a point is said to be one volt if the amount of work done in bringing one coulomb of positivechargefrominfinitytothatpointisonejoule.Sinceworkandcharge,botharescalars,theelectricpotentialisascalarquantity.Thedimensionsofelectricpotentialare TAT ML2 2 or [V] = ML2T3A1Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832018INSTITUTE OF CORRESPONDENCE COURSESNARAYANAPOTENTIALDIFFERENCEPotential difference between two points f (final) and i (initial) is defined to be equal to the amount ofworkdone(byexternalagent)inmovingaunitpositivechargefrompointi(initial)tof(final)qWV Vifi f If work done in carrying a unit positive charge from point 1 to point 2 is one joule then the potentialdifferenceV2V1issaidtobeonevolt.Potential difference may be positive or negative.The work done against electrical forces in transporting a charge q from point i (potential Vi) to pointf(potentialVf)isW =qVwhereV=VfVi.RELATION BETWEEN E AND ELECTRIC POTENTIAL VV=VfVi= # . EIn one dimensionsdrdVE ......(1)dr E V ......(2)[If electric potential is known, electric field can be determined from eq. (1) and if E is known, V canbe determined from (2)]In generalE V AA(a)(b)q-qAABBV >VV r) suchthatthesurfacedensitiesareequal.Findthepotentialatthecommoncentre.Solution: Ifq1andq2arethechargesonspheresofradiirandRrespectively,then q1+q2=QAccordingtogivenproblem1=2or 2221R 4qr 4qor 2221RrqqRq1roq2Physics:Electrostatics25FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANASo ) R r (Qrq2 221+and ) R r (QRq2 222+Now as potential inside a conducting sphere is equal to its surface, so potential at thecommon centreV=V1+V2= 041]]]

+Rqrq2 1= 041 ]]]]

+++ ) r R (QR) r R (Qr2 2 2 2= 041 ) r R () r R ( Q2 2++EXAMPLES BASED ON ELECTRIC POTENTIAL ENERGYExample-29 Threecharges,q,Q,qareplacedatequaldistancesonastraightline.Ifthetotalpotentialenergyofthesystemofthethreechargesiszero,thenQ:q=.....Solution: Letdbetheequaldistance.Thetotalpotentialenergyofthesystemis,U=U12+U23+U31]]]

+ + d 2q qdq qdq q41U1 3 3 2 2 10 ]]]]

+ d 2qdQqdqQ4120or,

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.|+ 2qQ Qdq41U0SinceU=02Q+ 2q=0 or,4Q+q=0or,4Q=qor, qQ=1:4Example-30 Three charges are arranged as shown in fig. Find the potential energy of the system.Solution: ThepotentialenergyofthesystemisU=U12+U23+U31]]]

+ + aq qaq ) q (a) q ( q411 3 3 2 10]]]]

+ 1 . 010 21 . 010 81 . 010 44114 14 140+q= 4 1027C+q= + 1037 Cq= + 1017Ca = 10 cm a = 10 cmd3]]]]

+ ) 2 8 4 (1 . 01041140orU=91091013 (10)=910910129103joulesPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832026INSTITUTE OF CORRESPONDENCE COURSESNARAYANAExample - 31 An electron (charge, e) is placed at each of the eight corners of a cube of side a and -particle (charge, +2e) at the centre of the cube. Compute the P.E of the system.Solution: Fig., shows an electron placed at each to the eight corners of the cube and particleatthecentre.Thetotalenergyofthesystemisthesumofenergiesofeachpairofcharges. There are 12 pairs like A and B (separation a), 12 paris like A and C (separation,a 2),4pairslikeAandG(separation a 3)and8parislikeAand0(separation 23 a).Hencee eeeeeeFEHGDABCe+2eo041U ]]]]]

+ + + a) e 2 ( ) e ( 8a 3) e ( ) e (4a 2) e ( ) e (12a) e ( ) e (1223=(9109) ae2]]]

+ +3323421212=(9109) ae2(4.32)=3.91010

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.|ae2J.Example-32 Thevalueofq1andq2are2108 coulomband0.4108coulombrespectivelyasshowninfig.Athirdchargeq3=0.2108coulombismovedfrompointCtopointDalongthearcofacircle.Thechangeinthepotentialenergyofchargewillbe-Solution: Potentialenergyofq3atpointC]]]

+ 1q q8 . 0q qk U2 3 3 1C .....(1)Potentialenergyofq3atpointD0.2m0.6mDq1q3C0.8 mq2]]]

+ 2 . 0q q8 . 0q qk U3 2 3 1D .....(2)UDUC=kq2q3]]]

112 . 01=91090.41080.21084=2.88107jouleNote : Note Charge potential energy charge will ne same even if we had takenanyotherpathaselectricfieldisconservative.)ELECTRICDIPOLE+dp q-qAn arrangement of two equal and opposite charges separated by a smalldistanceiscalledanelectricdipole.Physics:Electrostatics27FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAThe dipole moment pis a vector quantity whose magnitude is equal to the product of magnitude ofone charge and the distance between the two charges. It is directed from negative charge to positivecharge.Unitofdipolemoment(p)=coulombmetre =C.mDimensions of p = M0L1T1A1ELECTRIC FIELD AT AN AXIAL POINT2 2 2] ) 2 / d ( r [pr 2 kE-q +qodPE+Thedirectionisalongtheaxis,parallelto p.Themagnitudeis041E2 2 2] ) 2 / d ( r [pr 2Forr>>d,30rp 241EELECTRIC FIELD AT AN EQUATORIAL POINTTheelectricfieldatanequatorialpointPis2 / 3222drp kE

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.|

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.|+ +pq-qdpErThefieldisdirectedoppositeto pandthemagnitudeis041E2 / 3 2 2) ) 2 / d ( r (p+Forr>>d30rp41EPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832028INSTITUTE OF CORRESPONDENCE COURSESNARAYANAELECTRIC FIELD AT (R, ) POINTForr>>d,theradialcomponentErandangularcomponentEofelectricfieldduetoadipole,are3rrcos p 2k E3rsin pk E+pPErEE0ZXrTheresultantfieldis2 2rE E E+ + 23cos 3 1rkpTheangleinfigureissuchthattan= tan21EErFORCE AND TORQUE ON A DIPOLE PLACED IN AN ELECTRIC FIELD(a) A positive charge +q experiences a force parallel to the electricfield E q FF = -qEF = qE++(b) A negative charge q experiences a force in a direction oppositetothatoftheelectricfield E q F(c) Thetotalforceonadipoleplacedinanuniformelectricfieldiszero) E q ( E q F + =0(d) Thetorqueonadipoleplacedinuniformelectricfieldis E p sin pEELECTRIC POTENTIAL DUE TO A DIPOLEThepotentialatanaxialpoint(P1)isqd-q0132PPP+-2 20) 2 / d ( rp41VIfdd)201 pcosV .4 rPotentialatapointwhichisequidistantfrom+qandqchargeiszero.Thuspotentialatallpointslyingontheequatorialplaneiszeroforadipole.POTENTIAL ENERGY OF A DIPOLEThe work done in rotating a dipole placed in a uniform electric field E, from initial angle 1 to final angle 2 is d W21=pE(cos1cos2)whereU2,U1arethepotentialenergyofthedipoleinthetwoorientations.Potential energyU2U1=pEcos2(pEcos1)=pE(cos2cos1)Thezeroofthepotentialistakenat=90.Thus,potentialenergyofthedipoleis E. p U=pEcosUmin=pE,Umax=+pEWork done in rotating a dipole from = 0 (aligned parallel to E), to = 180 (aligned antiparallel toE) isW 2pE EXAMPLES BASED ON DIPOLE SYSTEMExample-33 The work required to turn an electric dipole end for end in a uniform electric field whentheinitialanglebetween pand Eis0is-Solution: W=pE(cos1cos2),inthiscase1=0and2=+0.ThusW=pE{cos0cos(+0)}=2pEcos0Example-34 Calculatetheelectricintensityduetoadipoleoflength10cmandhavingachargeof500Catapointontheaxisdistance20cmfromoneofthechargesinair.Solution: Theelectricintensityontheaxiallineofthedipole2 2 20) d (pd 241E# 2l=10cm#=5102mPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832030INSTITUTE OF CORRESPONDENCE COURSESNARAYANAd=20+5=25cm=25102mp=2q#=250010651025103102=5105C-m 2 2 2 82 5 9] 5 25 [ 1010 25 10 5 2 10 9E 6.25107N.C.Example-35 Calculate the electric intensity due to an electric dipole of length 10 cm having chargesof100Catapoint20cmfromeachcharge.Solution: The electric intensity on the equatorial line of an electric dipole is 2 / 3 2 20) d (p4E# + 1p=2#qC-m=10102100106=105 C-md2+#2 =(20102)2=4102 2 / 3 25 9) 10 4 (10 10 9E 735 910898 1010 10 9 =1.125107N/CExample-36 Find out the torque on dipole in N-m given : Electric dipole moment) k2 ji5 ( 10 P7 + coulombmetreandelectricfield) kji( 10 E7+ + Vm1 is-Solution: E P= 1 1 12 1 5kji) 1 5 ( k) 5 2 ( j) 2 1 ( i + + + k4 j7 i3 + 6 . 8 | | N-mCHARGEDLIQUIDDROPIfnsmalldropseachofradiusrcoalescetoformabigdropofradiusR,then(i)3 3r34n R34 R=rn1/3(ii) Ifeachsmalldrophasachargeq,thenthechargeonthebigdropnq ' q Physics:Electrostatics31FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(iii) IfVisthepotentialofthesmalldrop,thenthepotentialofthebigdropwillbe2/ 31/ 3Kq' KnqV' Vnr rn (iv) If E is the electric field intensity at the surface of the small drop, then the electric field intensityatthesurfaceofthebigdropwillbeE nrKqnRKnqR' Kq' E3 / 123 / 12 2 EXAMPLE BASED ON CHARGED LIQUID DROPExample-37 1000equaldropsofradius1cm,andcharge1106Carefusedtoformonebiggerdrop.Theratioofpotentialofbiggerdroptoonesmallerdrop,andtheelectricfieldintensityonthesurfaceofbiggerdropwillberespectively-Solution: LetthepotentialofonesmallerdropbeBthenpotentialofbiggerdrop,isV=n2/3V3 / 2nV' V =(1000)2/3 =100V:V=100:1AlsolettheelectricfieldonthesurfaceofsmallerdropbeEthenelectricfieldonbigger drop isE=n1/3E=n1/3 2rkq =(1000)1/3 2 26 9) 10 1 (10 1 10 9 =9 108V/mFORCEONACHARGEDSURFACE(i) If we consider an element of the charged surface, then the charge on the element experiencesa repulsive force due to the charge on the remaining part. As a result, a resultant force acts onitperpendiculartothesurfaceintheoutwarddirection.(ii) The charged surface behaves as a stretched membrane.(iii) If is the surface charge density, then the electric field intensity at external points close to thesurface is 0 and at internal points close to the surface the field is zero. Thus the average intensityatthesurface.02EPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832032INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(iv) Therepulsiveforceactingonaunitareaofthesurfacewillbe.022E F N/m2(v) The repulsive force acts in the outwards direction. The force acting on a unit area of the surfaceiselectricalpressure.02elec2P=2K2N/m2EQUILIBRIUMOFACHARGEDSOAPBUBBLE(i) In equilibrium the air pressure inside a soap bubble is greaterthan the atmospheric pressure. This excess pressure is producedduetosurfacetensionofthesoapsolution.Ifristheradiusof the bubble and T is its surface tension, then for the unchargedbubble in equilibriumP + PexP Forceduetoexcesspressure=forceproducedduetosurfacetension2exr p =T(22r)rT 4pex(ii) If a bubble is charged, then electrical pressure due to charge acts in outwards direction on thebubble.02elec2pwhereisthesurfacechargedensity.(iii) Inequilibrium,theforceproducedduetosurfacetensionisequaltothesumofforcesduetoexcessairpressureinsidethebubbleandtheelectricalpressureduetocharge,i.e.(pex+pelec)r2=T(22r)or pex+pelec= rT 4orrT 42p02ex+(iv) For charged bubble,02ex2 rT 4p (v) Iftheairpressureinsidethebubbleisequaltotheatmosphericpressureoutside,i.e.,pex=0thenrT 4202orKrT 2rT 80 Physics:Electrostatics33FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(vi) Ifchargegiventothesoapbubbleisq,then2r 4q The charge on the bubble 2r 4 q= 300 2Tr 2 8rT 8r 4 K / Tr 323 (vii) Theintensityofelectricfieldatthesurfaceofthebubbler / T 8 r / TK 32 E0 (viii) Theelectricpotentialatthesurfaceofbubble0/ Tr 8 rTK 32 V (ix) Onchargingabubbletheairpressureinsideitdecreasesbecausetheradiusofthebubbleincreased due to charging.(x) A soap bubble always expands on giving any kind of charge (positive or negative)MOTIONOFACHARGEDPARTICLEINELECTRICFIELD(i) A charged particle at rest or moving experiences a force E q F in the presence of electric field.The acceleration, velocity and displacement are given bymE qaLqDScreenYVVyVxu0EFig. (A) + at u v + 2at21ut s(ii) ForachargedparticlewithinitialvelocityperpendiculartotheelectricfieldNotethat Fx=0,ax=0,Vx=uatalltimesFy=qE,ay= mqE,Vy= mqEtThedisplacementcomponentsarex=ut2tmqE21y Eliminatingt, 221 qEy x2 muwhichistheequationofaparabola.Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832034INSTITUTE OF CORRESPONDENCE COURSESNARAYANA(iii) The path of a charged particle entering a region of electric field with initial velocity perpendiculartothefieldfollowsaparabolictrajectory.The time spent in electric field is uLt (see fig. A) The y component of velocity when it emergesoutofthefieldregionismuqELVyThe resultant velocity2y2xV V V + 22muqELu ,`

.|+ TheangleatwhichtheparticleemergesxyVVtan 2muqELTheheightYatwhichtheparticlehitsthescreen(seefig.A)Y=Dtan2muqELDY K 2qELDY (whereKisinitialkineticenergy)EXAMPLES BASED ON MOTION OF A CHARGED PARTICLE IN ELECTRIC FIELDExample-38 A positively charged oil droplet remains stationaryin the electric field between two horizontal platesseparatedbyadistanceof1cm.Ifthechargeon the drop is 9.6 1010 esu and the mass ofdroplet is 1011 g, what is the potential differencebetween two plates ? Now if the polarity of theplatesisreversedwhatistheinstantaneousaccelerationofthedroplet?[g=9.8m/s2]+qE mg+++++ +F0mg+++++ qE Fe==Solution: As the droplet is at rest, its weight W = mg will be balanced by electric force F = qEi.e.,qE=mgor qmgdV ) 10 3 / 10 6 . 9 () 10 1 )( 8 . 9 )( 10 (V9 102 14 =3062.5voltNow if the polarity of the plates is reversed, both electrical and gravitational force will actdownwardso,F=mg+qE=2mg[asmg=qE]Andhenceinstantaneousaccelerationofdrop: mmg 2mFa =2g=19.6m/s2Physics:Electrostatics35FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAELECTROSTATI CBEHAVI OUROFCONDUCTORSMaterials which allow flow of current through them are called conductors. They contain free electrons whichare donated by the atoms to the material as a whole and are able to move around freely inside the material.These electrons come from the outermost orbit of atoms as they are very loosely bound to the atom.The electrostatic behaviour of an insulated charged conductor or a conductor placed in an electric field can besummarized as follows.a) The electric field in the interior of the conductor is zero everywhere.b) The electric field immediately outside the surface of the conductor is perpendicular to the surface.c) Any excess charge given to a conductor remains on the outer surface.d) All points on and within the surface of the conductor are at the same potential.It follows that there is no difference between the electrostatic behaviour of a solid conductor and ahollow conductor of the same shape and size.ELECTROSTATICSHIELDINGIf there is a cavity of any shape inside a conductor, the field there will be strictly zero. This property is calledelectrostatic shielding because any thing placed inside the cavity will be completely shielded from externalfields.Examples : In a thunderstorm accompanied by lightning, it is safer to be inside a car, rather than near a treeor on the open ground.CONDUCTORHAVINGSHARPPOINTS(ACTIONOFPOINTS)Another interesting result is that the charge density and the electric field tend to be relatively high on sharppoints and low on plane regions of a conducting surface. This can be attributed to the fact that charge densityis proportional to radius of curvature.Glow discharges from sharp points during thunderstorms are due to this reason. The lightning rod, which hassharp points, is thus able to neutralize charged clouds and prevent lightning strokes.INSULATORS(DIELECTRICS)Materials which do not allow current to flow through them are called insulators or dielectrics. While theelectrons in such materials remain bound within their individual molecules, thus preserving the overall neutralityof each molecule, they are affected by external electric fields because the positive and negative charges tendto shift in opposite directions.Dielectric substances are of two types1. Non-polar dielectrics : In such dielectrics the centre of mass of all the positive charges (protons) in amolecule coincides with the centre of mass of all the negative charges (electrons). Therefore, they arenot only electrically neutral but also possess zero dipole moments.In the presence of an external field, the two centres of charge get slightly separated and each moleculebecomes a dipole, having a small dipole moment. This is because the protons experience a force in thedirection of the applied field while the electrons experience a force in the opposite direction. Thus in thepresence of a field the dielectric gets polarized.2. Polar Dielectrics : In polar dielectrics the centre of mass of the protons in a molecule do not coincidewith the centre of mass of the electrons. This is due to the asymmetric shape of the molecule. Thuseach molecule behaves as a dipole having a permanent dipole moment.Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832036INSTITUTE OF CORRESPONDENCE COURSESNARAYANAIn the presence of an external applied field, these dipoles tend to align themselves along the field andtheir dipole moments may also increase.a) DielectricConstant:Supposeadielectricslabisplacedinauniformelectricfield 0E .Theelectric field will polarize the slab, i.e., the positive charges of the molecules will be shifted slightlytowards the right and the negative charges towards the left. The right surface of the slab getspositivelycharged and the left surface negatively charged. This sets up a field pEin the oppositedirection which is less than 0E . The net field inside the slab becomes 0 pE E E . The ratio0EKEis called the dielectric constant of the material.EpE0E + + + + + + + +E0EPb) Dielectric Strength : The dielectric strength of a dielectric is the maximum value of the electricfield that can be applied to the dielectric without its electric breakdown, i.e., without liberatingelectrons from its atoms (or molecules).ATMOSPHERICELECTRICITYThe atmosphere surrounds the earth like a blanket. The atmosphere is a system with lots of variations.Therefore only its average properties may be described. The radius of earth is about 6400 km and theatmosphere extends to about 300 km above the surface of earth which is about 1th20 radius of earth.When we go up, the temperature and density vary. At about 300 km, its density falls to 1010 times itsground level value.ELECTRICALPROPERTIESOFATMOSPHEREa) At low altitudes the atmosphere is a poor conductor. The conductivity is only due to presence of ions,small nuclei of dirt, water vapour carrying static charge etc. The conductivity in low atmosphere variesa lot. It even varies from day to day.b) At the top of stratosphere (i.e. about 50 km) the atmosphere is fairly conducting. The conductivityincreases from earths surface towards the top of stratosphere.c) At ground level there is a vertical electric field of about 100 Vm1 all over the earth. The field weakens athigher altitudes and becomes negligible a 50 km.d) The potential drop from 50 km to earths surface is nearly 400 kV. Most of the potential drop occurs atlow altitudes.e) The surface density of earth is 9 210Cm the total charge being 60.5 10C.f) The number of protons entering the earths surface per sec is 72 10 per 2m. This is equivalent topositive charge of +1800 C.Physics:Electrostatics37FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAg) To maintain the constancy of negative charge of earth and potential difference between earth andstratosphere, there are about 44 10 thunder storms per day world-wise. This means a storm starts offsomewhere in every two second. The duration of each storm is about 1 hour.h) Within each thunder cloud (or storm) positive charge is carried upward to a height of about6 km while negative charges collect at about 2 km to 3 km above ground, the bottom of the cloud.i) The total amount of negative charge may be 20 C to 30 C. At the end of storm, the negative chargebursts along narrow path from cloud to earth to maintain earth at negative potential.j) In the last stages of a storm, there are about 200 flashes or bolts, each lasting about 32 10sec. Thepeak current in each bolt is about 410A in the downward direction. The calculation shows that eachbolt deposits 20 C of charge on earth. After each bolt the thunder cloud gets charged again and getsready for next bolt.NAME ALTITUDE DENSITY TEMPERATURERANGE IN KM BEHAVIOUR BEHAVIOURTrophosphere 0 to 12 km falls 1 to 110Falls uniformlyfrom 290 K to 200 KStratosphere 12 to 50 km falls from 310to rises uniformlyfrom 200 K to 280 KMesosphere 50 to 80 km falls from 310 ot 510falls uniformlyfrom 280 K to 180 KIonosphere 80 to 300 km falls from 510 to 1010to rises uniformlyfrom 180 K to 700 KPOINTS TOREMEMBER! Fundamental forces of nature :(i) Gravitation, (ii) Electromagnetic, (iii) Nuclear, (iv) Weak! Relative strength 1 : 1036 : 1039 : 1034! Charge is quantised, the quantum of charge is e = 1.6 1019 C.! The charge on the electron or proton denoted by eis the smallest charge that exists in nature.All othercharges are integral multiple of this charge.(q = ne)! When two bodies are rubbed against each other there is generally a transfer of electrons from one bodyto the other.A body which loses electrons is positively charged and the other negatively charged.If thebody is an insulator, electric charge remains confined to the rubbed portion but in case of conductor itspreads out on the surface of the conductor.! Electric field inside a charged conductor is zero. However there is electric field on the surface of conduc-tor which is normal to the surface.In case of irregularly shaped conductor, the electric field E andsurface charge density vary from point to point (0E ).At sharp points surface charge density andelectric field have greater value.However potential at each point of the same conductor have equal value.Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832038INSTITUTE OF CORRESPONDENCE COURSESNARAYANA! Electricfieldlineswhicharetangenttoelectricfield,originatefrompositivechargesandendonanegative charges.They are perpendicular to the surface of conductor (or any equipotential surface) butdo not pass through conductor.! Work done in a moving a point charge q from point A to point Bindependent of the path taken and isequal to) V V ( qA B .! A free stationary positive (negative) charge moves from higher (lower) to a lower (higher) potential.! The value of dielectric constant k is infinite for metals and one for vacuum.! Work done in rotating a dipole is given bypE U U Wi f ) cos (cosf i ! An electric dipole always tries to align itself with the electric field because its potential energy in thisconfiguration is minimum.! For a charge moving in an electric field, gain (less) in kinetic energy is always equal to loss (gain) inpotential energy.Hence,( )i f2i2fV V q mv21mv21

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.|! Electric field intensity due to a monopole (point charge) varies inverselyas a square of the distancewhereas in case of dipole it varies inversely as the cube of the distance.! Potential of an isolated conductor is inversely proportional the dielectric constant of the medium.! Potential energy of system having two point charges :rq q41U2 10Physics:Electrostatics39FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAExample - 1 A polythene piece is rubbed with wool. It developes a (-) ve charge of 710 2 . 3 C.(i) Calculate the no. of electrons transfered?(ii) What is the charge on wool after rubbing?(iii) Calculate transfer of mass from wool to polythene. 10 1 . 9 10 6 . 131 19kg m C ee Solution : Here,710 2 3 . q(i)1219710 210 6 . 110 2 . 3 eqn(ii) Charge on wool after rubbingC710 2 . 3 + (iii) Transfer of mass from wool to polythene em n kg19 31 1210 2 . 18 10 1 . 9 10 2 Example -2 Two small charged spheres repel each other with a forceN310 2 . The charge on onesphere is twice that on the other. When taken 10 cm further apart, the force isN410 5 .What are the charges and what was their original distance ?Solution Let q and 2q be the charges and x the distance between them.Then2232.4110 2xq Again( )2241 . 02.4110 5+ xq( )1 . 0 21 . 021 . 052022+ ++ x x orxxorxxor m x 1 . 0 Substituting the value of x so obtained

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.| 9229 310 9411 . 0q 210 9 10 2 or C q or q97 14210 33 . 33310910 SOLVEDEXAMPLESPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832040INSTITUTE OF CORRESPONDENCE COURSESNARAYANAandC q910 66 . 66 2 So, the charges on the spheres are C and C9 910 66 . 66 10 33 . 33 Examples - 3 Two free charges, each +Q, are placed at a distance r form each other. A third charge qisplaced on the line joining the above two charges such that all the charges are in equilib-rium. What is the magnitude, sign and position of the charge q ?Solution Let the two free charges be placed at points A and B and the third charge q be placed at apoint C onAB such that x AC and x r BC .Consider the equilibrium of the charge Q at A. The forces acting on it due to the charges atB and C must be equal and opposite. This is possible only if the charge q is negative, andAB ACF F or 222rQ41xQq41 orQrxq22.............................(1)Now considering the equilibrium of charge q at C, we haveCB CAF F or( )2 2x rQq41xQq41 or x r x or 2 r x Substituting in Eq. (1)4 / Q q With proper sign:4 / Q q Example - 4 Three charges, each of value Q, are placed at the corners of an equilateral triangle. Afourth charge q is placed at the centroid of the triangle.a) IfQ q ,will the corner charges move towards the centroid or away from it?b) What should be the relation between Q the q so that the charges remainstationary ?Solution: Let a be the length of each side of the triangle.Distance( ) 60 sin32a OA> 60OCQQFF1FAB 23. .32a 3 / a Physics:Electrostatics41FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAa) The forces on the charge at A due to those at B and C have magnitude.2241aQFThe resultant of these forces is 30 cos 21F F

23.aQ41. 222

OA alongaQ4322 The force on the charge Q at A due to chargeQ q at O is222aQ 341F along AOSince 1 2F F > , the charge at A will move towards O.By symmetry, the charges at Band C will also move towards O.(b) For equilibrium of the charges2 1F For2 2233aQqaQor 3 / Q q with proper sign 3 Q q Example - 5 Four chargesq q q + + , ,and q are placed respectively at the corners A,B,C and D of asquare of side a , arranged in the given order. Calculate the electric potential and intensity atO, the centre of the square. If E and F are the midpoints of sides BC and CD respectively,what will be the work done in carrying a charge e from O to E and from O to F.Solution Potential at O

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.| +OD OC OB OAqV1 1 1 14 ( ) OD OC OB OA as 0Intensity at O : Let C B AE E E, , and DEbe the intensities at O due to thecharges at A,B,C, and D respectively, Their directions areas shown. We have2 2241) 2 / (41aqaqE E E Eo oD C B A OE =E +E1 B DE =E +E2 A CFD -q -qEA+q +qBa 21441aqE E EoD B + Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832042INSTITUTE OF CORRESPONDENCE COURSESNARAYANAand22441aqE E EoC A + 2o2221aq 2 441E E E + Work done in carrying charge e from O to EPotential at E,]]]

+ CE1DE1BE1AE14qVoEBut AE = DE and BE = CE0 VE [ ] 0 V V e ) E 0 ( W0 E Work done in carrying charge e from 0 to FPotential at F, ,`

.| +CF DF BF AFqVoF1 1 1 14

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.| 151 441aeqV V e F O Woo FExample - 6 A charge of 60nC(nanocoulomb) is placed at the corner A of a square ABCD of side 10cm.Another charge of -40nC is located at the centre of the square.Find the work done incarrying a charge of +5nC from the corner C to the cornerr B of square.Solution Work done per unit charge is always equal to potential difference. Hence we will calculatethe potential at C, and the potential at B and take the potential difference between them.This will give the work done per unit chargeFrom geometry of the figurem cm OD OC OB OA210 2 5 2 5 Potential at C292910 2 5 410 4010 2 10 410 60 + +o o ABCDOnC 40 nC 60 +29 959 910 2 510 40 10 910 2 1010 60 10 9

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.| 910 941oV 6 . 1272 2 900 2 3600 2 2700 Potential at 292910 2 5 410 4010 10 410 60 + o oB

V 6 . 309 4 . 5090 5400 2 3600 5400 Physics:Electrostatics43FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAPotential difference between B and C C BV V Work done on unit charge from C to B = 1582.2JWork done in taking 5nC from C to B J6 910 91 . 7 10 5 2 . 1582 Example - 7 Three point charges, each of 0.1C, are placed at the corners of an equilateral triangleofside 1m. If this system is supplied energy at the rate of 1kW, how much time will be requiredto move one of the charges on to the midpoint of the line joining the other two ?Solution Initially the three charges are placed at the three corners A,B,C of the triangle. Initial poten-tial energy]]]]

+ +lqlqlqUoi2 2 241lllAB MC lqo 432Suppose the charge at A is moved to the midpoint M ofBC. The final potential energy of the system]]]]

+ +lqlqlqUof2 2 22 / 2 / 41

lqo 452Increase in energy( ) 3 542 lqU U Uoi f

lqo22.41

( )11 . 0 2 10 92 9

J710 18 Energy supplied per second = 1000 JTime taken. 50100010 187hrs Example - 8 Three point charges 1C, 2C and 3C are placed at the corners of an equilateral triangle of side1m. Calculate the work required to move these charges to the corners of a smaller equilat-eral triangle of side 0.5m, as shown in the figure.Solution Work required = (P.E of the charges on the innertriangle) (P.E of the charges on the outer triangle)]]]

++13 213 112 15 . 03 25 . 03 15 . 02 141oC 1m 1m 5 . 0C 2 C 3] 6 3 2 12 6 4 [ 10 99 + + J1010 9 . 9 Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832044INSTITUTE OF CORRESPONDENCE COURSESNARAYANAExample - 9 A particle having a charge of C .1910 6 1enters between the plates of a parallel platecapacitor.Theinitialvelocityofthe particle is parallel to the plates.A potential differenceof300Visapplied to the capacitor plates.If the length of the capacitor plates is 10cm andthey are separated by 2cm, calculate the greatest initial velocity for which the particle willnot be able to come out of the plates.The mass of the particle is . kg2410 12Solution Here,V=300V, d=2cmcm 2y cm 1+ + +cm x 10 MVdVE 1500010 23002 Now,221

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.|uxmqEymyqExu2221 m . cm x 1 0 10 m . cm y 01 0 1 ( )( ) 01 0 10 12 21 0 15000 10 6 1242 192.. .u s / m u410 Example-10 Two similar small balls having mass m and charge q are suspended by silk strings havinglength " , according to the figure. Ifin the figure is an acute angle then for equilibriumwhatwillthedistancebetweenthecentreofthetwoballs.Solution: Theforce&actingonthesystemareasfollows.Tistensioninstring,Fiscoulombforceandmgisweight.For equilibrium,Tcos=mgandTsin=F mgqx 41mgFtan220 xT T " "OF Fmg mgIfissmalltan ~sin" 2x mgqx 412x220 "3 / 102mg 2qx'' "Example-11 Two similar negative charges q are situated at point (0, a) & (0, a) along Y-axis. ApositivechargeQisleftfrompoint(2a,0).AnalysethemotionofQ.Solution: Due to symmetry the y components of forces acting on Q due to charges q at A andBwillbalanceeachotherandthexcomponentswilladdupalongdirectionO.IfatanyinstantQisatadistancexfromOF=F1cos+F2cos=2F1cos[F1=F2] 0412 ]]]]

+) x a (qQ2 2]]]]

+2 / 1 2 2) x a (x-qB+qOxQaaF2F1C2aA2 / 3 2 20) x a (qQx 241F+ i.e.FxPhysics:Electrostatics45FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANAFx,themotionisoscillatoryhavingamplitude2abutitwillnotbeS.H.M.Ifx EC > ED(2) EA < EB < EC < ED(3) EA = EB = EC = ED(4) EB = EC and EA > ED2. If the above question, the potential has correct relations as given -(1) VA> VB > VC > VD(2) VA> VB VC > VD(3) VD = VC = VB = VA(4) VC< VB > VA > VD3. In the above question, the surface charge densities have the correct relation is -(1) A > B > C > D(2) A = B = C = D(3) D > C > B > A(4) C < B > A > D4. Van-dee Graph generator is used in -(1) nuclear power establishments(2) in accelerating charged particles to very high potential(3) lighting and heating(4) lighting only5. Two identical pith-balls of mass m and having charge q are suspended from a point by weight-less stringsof length ". If both the strings make an angle of with the vertical, then the distance between the ballswill be (tanking to be small) -(1) 3 / 102) mg /2 q ( " (2) 3 / 102) mg /4 q ( "(3) 3 / 102) mg /4 q ( " (4) 3 / 102) mg /2 q ( "6. For the isolated charged conductor shown in fig. the potential at points A, B, Cand D are VA, VB, VC and VD respectively. Then -(1) VA = VB > VC > VD(2) VD > VC > VB = VA++ ++ +++++++A+D BC(3) VD > VC > VB > VA(4) VD = VC = VB = VA7. A non conducting sheet S is given a uniform charge density s. Two unchargedthin and small metal rods X and Y are placed near the sheet as shown.Then, the correct statement is -(1) S attracts both X and Y(2) X attracts both S and Y++++++++++++X YS(3) Y attracts both S and X(4) all of the above8. The variation of potential with distance R from fixed point is shown in fig.The electric field at R = 5m is -(1) 2.5 V/m (2) 2.5 V/m 23451 234 5 6 0Potential in VDistance R in M(3) 2/5 V/m (4) 2/5 V/m9. A charged spherical conductor of radius R carries a chargeA point test charge 0qis placed at adistance x from the surface of the conductor . The force experienced by the test charge will be proportionalto(a)2) ( x R +(b) 2) (1x R +(c)2) ( x R (d) 2) (1x R Physics:Electrostatics55FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA10. Two charges 9e and 3e are placed at a distance r. The distance of the point where the electric fieldintensity will be zero is -(1)) 3 1 /( r +from 3e charge (2)) 3 1 /( r +from 9e charge(3)) 3 1 /( r from 3e charge (4)) 3 / 1 1 /( r +from 3e charge11. The electric field in a region surrounding the origin is uniform and along thex-axis. A small circle is drawn with the centre at the origin cutting the axes atpoints A, B, C, D having coordinates (a, 0); (0, a); (a, 0); (0, a) respectively asshown in fig. Then the potential is minimum at -(1) A (2) BBADCE(3) C (4) D12. If an electron has an initial velocity in a direction different from that of an electric field the path of theelectron is-(1) a straight line (2) a circle(3) an ellipse (4) a parabola13. Consider equal and oppositely charged oppositely charge large parallelplates, with charge density t . A small charge q0ismovedalongtherectangular path ABCDA where side AB = x and side BC = y. Then correctstatement(s) is (are) -(1) work done by electric field along path AB is positive and equal to 0 0/ x q .(2) work done by electric field along path BC is zeroD CBA+ (3) work done by electric field along the path ABCDA is zero(4) all of the above14. Charge on an originally uncharged conductor is separated by holding apositivelychargedrodverycloselynearby,asinFig.Assumethattheinduced negative charge on the conductor is equal to the positive charge qon the rod then, flux through surface S1 is -(1) zero (2) 0 0/ q (3) 0 0/ q (4) none of the above15. Some equipotential surfaces are shown in the figure. The magnitude anddirection of the electric field is-(1) 100 V/m making angle 1200 with the x-axis(2) 100 V/m making angle 600 with the x-axis10 20 30 20V 30V 40V(cm)0 = 30 (3) 200 V/m making angle 1200 with the x-axis(4) none of the above16. Themetalplateontheleftinfig.carriesasurfacechargeof+sper unit area. The metal plate on the right has a surface charge of -2sper unit area. It is assumed that the plates are large and the central plate isconnected to zero. Then the charge densities on the left and right surfaceofthe central plate are, respectively -(1) - , + (2) - 2, + 2++++++2 (3) - , + 2 (4) none of the abovePhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832056INSTITUTE OF CORRESPONDENCE COURSESNARAYANA17 Point charge q moves from point P to point S along the path PQRS (asshown in Fig.) in a uniform electric field E pointing co-parallel to thepositive direction of X-axis. The coordinates of the points P, Q, R, andS are (a, b, 0), (2a, 0, 0), (a, -b. 0) and (0, 0, 0) respectively. The workdone by the field in the above process is given by the expression-(1) qEa (2) qEaS QPRYE(3)2 qEa (4) ( )2 2b a 2 qE +18 An electron is projected as in fig. with kinetic energy K, at an angle q = 450between two charged plates. The magnitude of the electric field so that theelectron just fails to strike the upper plate, should be greater than -(1) K/qd (2) 2K/qd45KEd+ + + + + + (3) K/2qd (4) infinite19 Two plastic rods of equal lengths (L = R) one of charge q and other of charge -q, form a circle of radiusR in an xy plane. The charge is distributed uniformly on both rods. Then the electric field at the centre ofcircle is-(1) zero (2) 20R 4 / q (3) 202R 2 / q (4) 202R / q 20 Two identical thin rings, each of radiusR metre are coaxially placed at distance R metre apart. If Q1 andQ2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving acharge q from the centre of one ring to that of the other is-(1) zero (2)( ) ( ) R 4 2 / 1 2 ) Q Q ( q0 2 1 (3)R 4 / ) Q Q ( 2 q0 2 1 (4)( ) R 4 2 / 1 2 ) Q Q ( q0 2 1 + +21 A and B are concentric conducting spherical shells. A is given a positive charge while B is earthed. Then-(1) A and B both will have the same charge densities(2) the potential inside A and outside B will zero++++ +++AB(3) the electric field between A and B is none zero(4) the electric field inside A and outside B is non zero.22 A semicircular ring of radius R is given a uniform chargeThen the electric field and electric potential atits centre willbe -(1) R 4Q,R 4Q020 (2)R 4Q,R 2Q02 20 +++++++++++ ROQ(3) R 2Q,R 4Q0 0 (4) zero, zero23 Mark the wrong statement -(1) Equipotential surface never cross and other(2) For a uniformly charged nonconducting sphere, the electric potential at the centre of the sphere is1.5 times that at the surface(3) If potential in a certain region in non zero constant, then the electric field in that region will also be nonzero constant(4) Inside a spherical charged shell, the electric field is zero but the electric potential is the same as thatat the surface.Physics:Electrostatics57FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA24 A charge +q is fixed at each of the points 0x x , 0 0x 3x ,x 5x ... ad inf. of the x-axis and a charge -q is fixed at each of the points x= 2x0, x=4x0 , x=6x0..... an inf. Here x0 is a positive constant. Take theelectric potential at a point due to a charge Q at a distance r from it be Q/40r. Then, the potential atte origin due to the above system of charge is(1) 0 (2) 0 0q8 x log2 (3) (4) 0 0glog24 x 25 In the follwoing fig. where the change q must be kept so that the potential energy of the system will beminimum?q2q 8q9 cmx (9-x)(1) 3 cm (2) 2 cm(3) 4 cm (4) none of thesePhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832058INSTITUTE OF CORRESPONDENCE COURSESNARAYANAPROBLEMS ASKED IN VARIOUS EXAMS1 Two identical conductors of copper and aluminum are placed in an identical electric field. The magnitudeof induced charge in the aluminum will be - [AIIMS 1999](1) zero (2) greater than in copper(3) less than in copper (4) equal to that in copper2 The study of the effects associated with electric charge at rest is known as -[AIIMS 1999](1) electrostatics (2) magnetostatic(3) electromagnetism (4) none of these3 Two electric charges 12C and 6C are placed 20 cm apart in air. There will be a point P at whichelectric potential is zero on the line joining these two charges and outside, excluding the region betweenthem. The distance of P from 6C charge is - [EAMCET ENGG. 2000](1) 0.10 m (2) 0.15m(3) 0.20m (4) 0.25m4 The displacement r of a charge Q in an electric fieldke je ie E3 2 1+ + is jb ia r + . The work doneis - [EAMCET ENGG. 2000](1) Q(ae1 + be2) (2)2221) be ( ) ae ( Q +(3)2 22 1b a ) e e ( Q + +(4)) e e (2221 + (a + b)5 AnelectronofmassmandchargeeisacceleratedfromrestthroughapotentialdifferenceVinvacuum . The final speed of the electron will be - [MP PMT 2000](1)m / e V(2)m / eV(3)m / eV 2(4) 2eV/m6 An insulated charge conducting sphere of radius 5 cms has potential of 10 V at the surface. What is thepotential at centre - [AIIMS 2000](1) 10 V (2) zero(3) same as that at 5 cms from the surface (4) same as that at 25 cms from the surface7 Let Ea be the electric field due to a dipole in its axial plane distant 1 and let Eq be the field in the equatorialplane distant ". The relation between Ea and Eq is - [AIIMS 2000](1) Ea = Eq(2) Ea = 2Eq(3) Eq = 2Ea(4) Ea = 3Eq8 A point positive charge of Q unit is moved round another positive charge of Q unit on a circular path. If theradius of the circle is r, the work done on the charge Q in making one complete revolution is -[HARAYANA PMT 2000, BHU MED. 2000](1) r 4Q0(2)r 4' QQ0(3) r 4' Q0(4) zero9 Three point charges Q1 , Q2 and Q3 in that order are placed equally spaced along k straight line. Q2 andQ3 are equal in magnitude but opposite in sign. If the net force on Q3 is zero, the value of Q1 is -[UPSEAT 2000](1) Q1 = |Q3| (2) Q1 = 2|Q3|(3) Q1 = 2 |Q3| (4) Q1 = 4 |Q3|Physics:Electrostatics59FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA10 As shown in the figure, charge +q and q are placed at thevertices B and C of an isosceles triangle. The potential atthe vertex A is- [MP PET 2000](1)0412 2b aq 2+(2) zeroAb b B C+qa-q(3)0412 2b aq+(4)0412 2b a) q (+11 Three charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown. Thenet electrostatic energy of the configuration is zero if Q is equal to - [IIT SCREENING 2000](1) 2 1q+(2) 2 2q 2+2 aQ+q+qaa(3) 2q (4) +q12 Two point charges (+Q) and (2Q) are fixed on the X-axis at positions a and 2afrom origin respectively. At what positions on the axis, the resultant electricfield is zero - [MP PET 2001](1) only x = 2a (2) only x = 2a(3) both x = 2a (4) x = 2a 3 only13 Three positive charges of equal value q are placed at the vertices of an equilateral triangle. The resultinglines of force should be sketched as in - [IIT SCREENING 2001](1)(2) (3)(4) 14 A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the pointon the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. Then the potentials at thepoints A, B and C satisfy - [IIT SCREENING 2001](1) VA < VB(2) VA > VB(3) VA < VC(4) VA > VC15 Charge density on upper half is and in lower half charge density is . Direction of electric field at O is[UPSEAT 2001](1) along OA (2) along OBBODCA+++(3) along OC (4) along OD16 The work done in placing four charges at the corners of a square asshown in the figure, will be -[RPET 2002](1)aKq) 2 4 (2 (2)aKq) 2 4 (2+1 23 4+q -q+q -q(3)22aKq) 2 4 ( (4)22aKq) 2 4 ( +Physics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832060INSTITUTE OF CORRESPONDENCE COURSESNARAYANA17 Two spherical conductors B and C having equal radii and carrying equal charges on them repel eachother with a force F when kept apart at some distance. A third spherical conductor having same radius asthat of B but uncharged is brought in contact with B, then brought in contact with C and finally removedaway from both. The new force of repulsion between B and C is - [AIEEE 2004](1) 3F/8 (2) 3F/4(3) F/8 (4)F/418 A charged particle q is shot towards another charged particle Q, which is fixed, with a speed v. Itapproaches Q upto a closest distance r and then returns. If q were given a speed of 2v, the closestdistances of approach would be - [AIEEE 2004](1) r/4 (2)2 rqQrv(3) r/2 (4)r19 Four charged equal to Q are placed at the four corners of a square and a charge q is at its centre. If thesystem is in equilibrium the value of q is - [AIEEE 2004](1) ) 2 2 1 (2Q+ (2)) 2 2 1 (4Q+(3) ) 2 2 1 (2Q+ (4)) 2 2 1 (4Q+ 20 An electric-dipole is placed at an angle of 30 to a non-uniform electric field. The dipole will experience(1) a torque only [AIEEE 2006](2) a translational force only in the direction of the field(3) a translational force only in a direction normal to the direction of the field(4) a torque as well as a translational force21 Two insulating plates are both uniformly charged in such a way that the potential difference betweenthem is V2 V1 = 20 V. (i.e. plate 2 is at a higher potential). The plates are separated by d = 0.1 m andcan be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. Whatis its speed when it hits plate 2 ? (e = 1.6 10-19C, me = 9.11 10-31 kg)[AIEEE 2006]XY0.1m1 2(1)1932 10 m/ s (2)62.65 10 m/ s (3)127.02 10 m/ s (4)61.87 10 m/ s 22 Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and areuniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, theratio of the magnitude of the electric fields at the surfaces of spheres A and B is[AIEEE 2006](1) 1 : 4 (2) 4 : 1(3) 1 : 2 (4) 2 : 123 An electric charge 103C is placed at the origin (0,0) of X-Y co-ordinate system. Two points A and Bare situated at( )2, 2 and ( ) 2,0respectively. The potential difference between the points and A andB will be[AIEEE 2007](1) 9 V (2) zero(3) 2 V (4) 4.5 VPhysics:Electrostatics61FNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 41828320INSTITUTE OF CORRESPONDENCE COURSESNARAYANA24 Charge are placed on the vertices of a square as shown in Fig. Let $E is along (2). Potential at centreremains same[AIEEE 2007](1)$E remains unchanged, V changes (2) Both $E and V change(3)$E and V remain unchanged (4)$E changes, V remain changed25 The potential at a poit x (measured inm) due to some chargfes situated on the x-axis is given by( ) ( ) 2V x 20/ x 4 volts. The electric field E at x = 4mis given by[AIEEE 2007](1)53volt/ mand in the -ve x direction (2)53volt/ mand in the +ve x direction(3)109volt/ m and in the -ve x direction (4)109volt/ m and in the +ve x directionPhysics:ElectrostaticsFNS House, 63, Kalu Sarai Market, Sarvapriya Vihar, New Delhi-110016! Ph.: (011) 32001131/32 Fax : (011) 4182832062INSTITUTE OF CORRESPONDENCE COURSESNARAYANAEXERCISELEVEL - I1. (3) 2. (3) 3. (1) 4. (1) 5. (2)6. (1) 7. (3) 8. (1) 9. (2) 10. (3)11. (3) 12. (2) 13. (3) 14. (3) 15. (1)16. (2) 17. (3) 18. (4) 19. (2) 20. (1)21. (4) 22. (4) 23. (3) 24. (4) 25. (3)LEVEL - II1. (1) 2. (4) 3. (3) 4. (3) 5. (2)6. (3) 7. (4) 8. (3) 9. (4) 10. (4)11. (2) 12. (3) 13. (1) 14. (2) 15. (3)16. (3) 17. (2) 18. (1) 19. (3) 20. (2)21. (1) 22. (1) 23. (2) 24. (2) 25. (3)LEVEL - III1. (1) 2. (3) 3. (1) 4. (2) 5. (1)6. (4) 7. (4) 8. (1) 9. (2) 10. (1)11. (1) 12. (4) 13. (4) 14. (2) 15. (3)16. (3) 17. (2) 18. (3) 19. (4) 20. (2)21. (3) 22. (2) 23. (3) 24. (4) 25. (1)PROBLEMS ASKED IN VARIOUS EXAMS1. (4) 2. (1) 3. (3) 4. (1) 5. (3)6. (1) 7. (2) 8. (4) 9. (4) 10. (2)11. (2) 12. (2) 13. (3) 14. (2) 15. (4)16. (1) 17. (1) 18. (3) 19. (4) 20. (4)21. (2) 22. (2) 23. (2) 24. (4) 25. (4)ANSWERS