electric charges and fields - wordpress.com...[topic1] coulomb's law, electrostatic field and...

- CHAPTER 01

Electric ChargesAnd Fields

Chapter Analysis w.r.t. Lost 3 Yeor's Boord ExamsThe analysis given here gives you an analytical picture of this chapter and will help you toidentify the concepts of the chapter that are to be focussed more from exam point of view.

Number of Questions asked in last 3 years

2015 2016 2017Delhi All India Delhi All India Delhi All India

Very Short An~wer (1 ~~L __ IQ 1 Q IQ.-~-----------_._ .._- -~~~~-.---.--. .__ ._-_._._ ...._-_ ....-Short Type I Answer (2 marks)

___ • - 0- _. -.... -----------Short Type II Answer (3 marks) J Q lQ

Long Answer (5 marks) lQ 1 Q I IQ IQ:

i I-----j

Value Based Questions (4 marks) J• In 2015, one question of 1mark based on Electric Flux and one question of 5 marks

based on Application of Gauss Theorem were asked in both sets.• In 2016, in Delhi set, only one question of 1marks base on Gaussian Surface was asked.• In 2017, in Delhi set, one question of 3 marks based on Electric Field at a Point on

Equatorial Line of an Electric Dipole and ',' ql' . -f 'I1S of 5 marks based onApplication of Gauss Theorem were asked.In All India set, only one question of 5 marks based on Electric Field due to Dipole atAxial Point was asked.

On the basis of above analysis, it can be said that from exam point of view Electric Flux,Application of Gauss Theorem, Electric Dipole and Electric Field due to Dipole at AxialPoint are the most important concepts of the chapter.

[TOPIC 1] Coulomb's Law, Electrostatic Fieldand Electric Dipole

1.1Electric ChargeCharge is an intrinsic property associated withelementary particles of matter due to which itproduces and experiences electric and magneticeffect. Benjamin Franklin introduced two types ofcharges namely positive charge and negativecharge based on frictional (statical) electricityproduced by rubbing two unlike objects likeamber and wood.Transference of electrons is the cause offrictional electricity.

Basic Properties of Electric Charge(i) Additivity of Electric Charges Charges

are scalars and they add up like realnumbers. It means if a system consists of ncharges ql' q2' q3' ... ,qn' then total chargeof the system will be ql + q2 + ...+s:

(ii) Conservation of Electric Charges Thetotal charge of an isolated system is alwaysremains conserved i.e. initial and finalcharge of the system will be same.

(iii) Quantisation of Electric ChargesCharge exists in discrete amount ratherthan continuous value ani hence, they aresaid to be quantised.Mathematically, charge on an object,

q = ± ne.where, n is an integer and e = electroniccharge = 1.6x 1O-19C.

Units of Charge(i) SI unit coulomb (C)(ii) CGS system

(a) electrostatic unit (esu) or stat-coulomb(stat-C)

(b) electromagnetic unit (emu) or ab-C '>J • ,b ab-coulombj

1 ab-C = 10 C, 1 C = 3 xl O" stat-C

Conductors and InsulatorsThose substances which readily allow the passage ofelectricity through them are called conductors, e.g.metals, the earth whereas those substances whichoffer high resistance to the passage of electricity arecalled insulators, e.g. plastic rod and nylon.

1.2 Coulomb's LawIt states that the electrostatic force of attraction orrepulsion acting between two stationary pointcharges is given by F = _1_. qlq2

41tEo ,-2

where, ql and q2 are the stationary point charges andr is the separation between them in air or vacuum.

Also, _1_ = 9 x 109 N_m2C-2

41tEo

where, Eo = permittivity of free space= 8.85419 x 10-12 C 2/N_m2

In vectorform,F = _1_. qlq2 r or IFI= _1_. qlq241tEo I r f 41tEo r2

Absolute Permittivity of Medium[Dielectric Constant)The force between two charges ql and q2 located ata distance r in a medium other than free space maybe expressed as F =_1_. qlq2

41tE r2

where, E is absolute permittivity of the medium.1 qlq2

4,;~·7=~=E_1_. qlq2 Eo T

41tE r2

The ratio of force between two changes in vacuumand force acting between when they are shifted ina medium is called relative permittivity (ET) of themedium also called dielectric constant of themedium.

Fvacuum

Fmedium

4

1.4 Electric FieldThe space arround a charge in which its effect canbe felt significantly i.e. the area which appearsattraction or repulsion force on another charge iscalled electric field.

Electric Field IntensityThe electric field intensity at any point due tosource charge is defined as the force experiencedper unit positive test charge placed at that pointwithout disturbing the source charge.

. d I· FIt ISexpresse as E = lffi-qo~oqo

Here, qo -? 0, i.e. the test charge qo must be small,so that it does not produces its own electric field.S1 unit of electric field intensity (E) is xc' and itis a vector quantity.

Electric Field due to a Point ChargeElectric field intensity at point Pis, E = _1_. -; r

41tEo [r]

q P.-----------------------~EI. r >I

The magnitude of the electric field at a point P is--. 1 1 1 qgiven by E =-_.-

41tEo r2

Electric Field due to a Systemof ChargesElectric field due to a system of charges at a pointis equal to the vector sum of electric fieldsproduced by individual charges.

E;

F;

o ehopterwise eBSE Solved Popers : PHYSICS

n

E = El+E2+E3+ .... +En = LE;i=l

=> E=_I_~!lLr£.J 12 I

41tEo i = I IT;

Electric Field LinesElectric field lines are a way of pictorially mappingthe electric field intensity around a configurationof charge( s). These lines start from positive chargeand end on negative charge. The tangent on theselines at any point gives the direction of field atthat point. Electric field lines due to positive andnegative point charges and their combinations areshown as below:

Field lines of two Field lines of anequal positivecharges electric dipole

Different electric field lines

1.3 Electric DipoleTwo point charges of equal magnitude andopposite in sign separated by a small distancealtogether form an electric dipole.

Electric Dipole MomentThe strength of an electric dipole is measured by avector quantity known as electric dipole moment(p) which is the product of the charge (q) andseparation between the charges (20.

P = q x 21 => Ip 1= q (2/)

• •-q +q~1t-------21------~.'

. Direction Its direction is from negativecharge (-q) to positive charge (+q) .

. SI unit Its S1 unit is C-m.

CHAPTER 1 : Electric Charges and Fields

Electric Field at a Point on the AxialLine due to Electric DipoleThe electric field intensity at a point on axial line ofthe dipole at a distance r from the centre of thedipole is given by the formula.

E . =_1_. 2praxial 41tEo (r _ 12)2

P I- -I• • I _ • ------------_

-q +q p/+- 1 ---+l

1---21~

When 1« r,1 2p 1 21plEaxial= -_.- => IEaxiall=--·-

41tEo r3 41tEo r3

Electric Field at a Point on theEquatorial Line due to Electric DipoleThe electric field at a point on equatorial line of thedipole at a distance r from the centre of the dipole isgiven by the formula.

1 - PEequatorial= 41tEo· (r + p)3/2

1 I piIf 1« r, IEequatorialI= -4-.-31tEo r

E,

+qI- 21-----+\

Torque on An Electric Dipole Placed ina Uniform Electric FieldConsider an electric dipole consisting of two charges+q and -q placed in a uniform external electric fieldintensity E. The dipole makes an angtle 9 with the

5

direction of electric field, then torque acting onthe dipole is given by the formula.

"t = pEsin9

In vector form, "t = P x E

+~F=qE I J

pb :zF = qE=---4 .------------+·E

· Minimum torque is experienced by electricdipole in electric field, when 9 = 00 or 1t

"t = "tmin = 0

· Maximum torque t = "tmax'

when sin S = 1 => 9 = 1t/2"tmax = pE

• Dipole is in stable equilibrium in uniformelectric field when angle between pand Eis00 and in unstable equilibrium when angleis 1800

•

• There exists a net force and torque onelectric dipole when placed in non-uniformelectric field.

Work Done and Potential Energyof Electric Dipole

• When an electric dispole is placed in electricfield then work is done in rotating it. Inrotating the electric dipole from 91 to 92 is

W =pE (cos 91 - cos 92).

• Potential energy of electric dipole when itrotates from 91 = 900 to 92 = 9

W = pE (cos90° - cosS) = -pE cos 9= - p·E• Work done in rotating the dipole from the

position of stable equilibrium to unstableequilibrium, i.e. when 91 = 00 and 92 = 1t.

W=2pE

• Work done in rotating the dipole from theposition of stable equilibrium to the positionin which dipole experiences maximumtorque, i.e. when 91 = 00 and 92 = 90°.

W=pE


1.3 Electrostatic ForcesElectrostatic forces (Coulombian forces) areconservative forces. i.e. the work doneagainst these forces does not depend uponthe path followed.

Principle of Superposition ofElectrostatic ForcesThis principle states that the net electricforce experienced by a given charge particleqo due to a system of charged particles isequal to the vector sum of the forces exertedon it due to all the other charged particles ofthe system. The force between two charges isnot affected by the presence of other charges.

FCXJF02 F03

qoF

Superposition ofelectrostatic forces

where, fO' = fO - f" Fo, = force on qo due toq,.

Similarly, fOn = fO - fn; FOn = force on qo dueto qn

3

Electrostatic Force due to ContinuousCharge DistributionThe region in which charges are closely spaced in acontinuous manner is said to have continuousdistribution of charge. It is of three types (i) linear chargedistribution, (ii) Surface charge distribution and(iii) Volume change distribution.

(i) Force on a charge dueto linear chargedistribution (chargedistributed along aline) is give byF - qo J J... dl •

- 41tEo I ~ r

where J... is called linear charge density (chargeper unit length) and dl is a short length elementof linear charge distribution.

(ii) Force due to surface charge distribution (chargedistribution over a plane surface) is given by

dq = o dsF - qo J ods ,

- 41tEo s ~r

where a is called surface charge density (chargeper unit area) and dS is a small surface element.

+ + + ++ ++dS+ + + +

qo F

(Hi) Force due to volume charge distribution (chargedistributed over a volume) is given by

dq =. 1,..=r ~ F = ~ J pdV r. 41tEo v 1 fl2

+++++++ + r.+--+~+:-t'I:...+~ F+ ++++++ + + + +

dq= pdV

where a is called volume charge density and dV isa small volume element.

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions

1. Why do the electrostatic field lines notform closed loop? All India 2014, Delhi 2012

2. Two equal balls having equal positivecharge q coulombs are suspended by twoinsulating strings of equal length. Whatwould be the effect on the force when aplastic sheet is inserted between the two?All India 2014

3. Why do the electric field lines never crosseach other? All India 2014

4. Why must electrostatic field at the surfaceof a charged conductor be perpendicularto every point on it?Foreign 2014, Deihl 2012

5. Two point charges ql and q2 are placed ata distance d apart as shown in the figure.The electric field intensity is zero at thepoint P on the line joining them as shown.Write two conclusions that you can drawfrom this. Delhl2014C

6. Define dipole moment of an electricdipole. Is it a scalar quantity or a vectorquantity? Farelgn 2012; All India 2011

7. Draw a plot showing the variation ofelectric field (E) with distance r due to apoint charge q. Deihl 2012

8. A proton is placed in a uniform electric fielddirected along the positions X-axis. Inwhich direction will it tend to move?Deihl2011C

9. In which orientation, a dipole placed in auniform electric field is in (i) stableequilibrium (U) unstable equilibrium?Deihl 2011; All India 2008

10. Two point charges having equal chargesseparated by 1m distance experience aforce of 8 N. What will be the forceexperienced by them, if they are held inwater at the same distance?(Given, Kwater = 80). All India 2010C

11. A metallic sphere is placed in a uniformelectric field as shown in the figure.Which path is followed by electric fieldlines and why? HOTS; Foreign 2010

12. Point out right or wrong for the followingstatement. The mutual forcesbetween twocharges donot get affectedby the presenceofother charges. All India 2010

13. A dipole of dipole moment p is present ina uniform electric field E. Write the valueof the angle between p and E for whichthe torque experienced by the dipole, isminimum. Delhl2D09C

o 2 Marks Questions14. An electric dipole of lengthd em when

placed with its axis making an angle of60° with a uniform electric field,experiences a torque of 4../3 N-m.Calculate the potential energy of thedipole if it has charge ± 8 nC. Deihl 2014

15. An electric dipole of length 2 em whenplaced with its axis making an angle of60° with a uniform electric field,experiences a torque of 8../3 N-m.Calculate the potential energy of thedipole if it has charge of ± 4 nC.Deihl 2014

16. An electric dipole of length 1cm whenplaced with its axis making an angle of60° with a uniform electric field,experiences a torque of 6.J3 N-m.Calculate the potential energy of thedipole, if it has charge ± 2 nC. Deihl 2014


17. An electric dipole is placed in a uniformelectric field E with its dipole moment pparallel to the field. Find

(i) the work done in turning the dipoletill its dipole moment points in thedirection opposite to E.

(ii) the orientation of the dipole forwhich the torque acting on it becomesmaximum. All India 2014C

18. A small metal sphere carrying a charge+ Q is located at the centre of a sphericalcavity in a large uncharged metallicspherical shell. Write the charges on theinner and outer surfaces of the shell.Write the expression for the electric fieldat the point Pl' Oelhl20141!

Meta

19. Point charge (+ Q) is kept in the vicinity ofand uncharged conducting plate. Sketchelectric field lines between the charge andthe plate. Farelgn 2014

20. Calculate the amount of work done inturning an electric dipole of dipolemoment 3 x 10-8 C - m from its position ofunstable equilibrium to the position ofstable equilibrium in a uniform electricfield of intensity loB NC-l. Faralgn 2011

21. Plot a graph showing the variation ofCoulomb force (F) versus lIr2, where r isthe distance between the two charges ofeach pair of charges (lIlC, 2 IlC) and(lIlC, - 3 IlC). Interpret the graphsobtained. All India 2011C

22. Two identical metallic spherical shells Aand B having charges + 4 Q and - 10Q arekept a certain distance apart. A thirdidentical uncharged sphere C is firstplaced in contact with sphere A and thenwith sphere B, then spheres A and Bare

7

brought in contact and then separated.Find the charge on the spheres A and B.All India 2011C

23. A dipole with a dipole moment of magnitudep is in stable equilibrium in an electrostaticfield of magnitude E. Find the work done inrotating this dipole to its position ofunstable equilibrium. All India 2010C

24. A dipole is present in an electrostatic fieldof magnitude 106NC-1. If the work done inrotating it from its position of stableequilibrium to its position of unstableequilibrium is 2x 10-23 J, then find themagnitude of the dipole moment of thisdipole. All India 2010C

25. Deduce the expression for the electric fieldE due to a system of two charges ql and q2with position vectors r1 and r2 at a point rwith respect to common origin. Deihl 2010C

26. An infinite number of charges, each of qcoulomb, are placed along X-axis atx = 1m, 3m, 9m, and so on. Calculate theelectric field at the point x = 0, due tothese charges if all the charges are of thesame sign. Deihl 2009

27. The sum of two point charges is 71lC. Theyrepel each other with a force of 1N whenkept 30 em apart in free space. Calculatethe value of each charge. Forelg" 2009

28. Figure shows two large metal P1 P2plates Iiand P2 tightly heldagainst each other and placedbetween two equal and unlikepoint charges perpendicular Q Qto the line joining them. +

(i) What will happen tothe plates when theyare released?

(ii) Draw the pattern of the electric fieldlines for the system. HOTS; Farllgn 2009

29. Two charges + Q and - Q are kept atpoints (-X2' 0)and (~, 0)respectively, inthe XY-plane. Find the magnitude anddirection of the net electric field at theorigin (0,0). All India 2009C

8

o 3 Marks Questions30. (i) Derive the expression for electric field

at a point on the equatorial line of anelectric dipole.

(ii) Depict the orientation of the dipole in(a) stable, (b) unstable equilibrium ina uniform electric field. Delhi 2017

31. (i) Obtain the expression for the torque 't

experienced by an electric dipole ofdipole moment p in a uniform electricfield, E.

(ii) What will happen, if the field werenot uniform? Delhi 2017

32. A charge is distributed uniformly over aring of radius a. Obtain an expression forthe electric field intensity E at a point onthe axis of the ring. Hence, show that forpoints at large distances from the ring, itbehaves like a point charge. Delhi 2016

33. An electric dipole of dipole moment p isplaced in a uniform electric field E. Obtainthe expression for the torque 't

experienced by the dipole. Identify twopairs of perpendicular vectors in theexpression. Delhi 2015C

34. Two point charges + q Aand - 2q are placed at Dthe vertices Band C ofan equilateral MEC of a aside a as given in thefigure. Obtain the +q -2qexpression for B a C

(i) the magnitude and(ii) the direction of the resultant electric

field at the vertex A due to these twocharges. All India 2D14C

35. Define the term electric dipole moment. Isit a scalar or vector? Deduce an expressionfor the electric field at a point on theequatorial plane of an electric dipole oflength 2a. All India 2013; Foreign 2009

36. An electric dipole is kept in a uniformelectric field. Derive an expression for thenet torque acting on it and write itsdirection. State the conditions under whichthe dipole is in (i) stable equilibrium(ii) unstable equilibrium. Deihl 2012C

o Chopterwise eBSE Solved Papers: PHYSICS

37. Sketch the pattern of electric field linesdue to

(i) a conducting sphere having negativecharge on it.

(ii) an electric dipole. All India 2011C

38. A positive point charge (+q) is kept inthe vicinity of an uncharged conductionplate. Sketch electric field line originatedfrom the point on to the surface of theplate. All India 2D14; Delhi 2D08

39. (i) Define torque acting on a dipole ofdipole moment p placed in a uniformelectric field E. Express it in thevector form and point out thedirection along which it acts.

(ii) What happens if the field isnon-uniform?

(iii) What would happen if the externalfield E is increasing (a) parallel to pand (b) anti-parallel to p? Foreign 2016

o 5 Marks Questions40. (i) Derive an expression for the electric

field E due to a dipole of length 21 ata point distant r from the centre ofthe dipole on the axial line.

(ii) Draw a graph of E versus r for r » a.(ii) If this dipole is kept in a uniform

external electric field Eo,diagrammatically represent theposition of the dipole in stable andunstable equilibrium and write theexpressions for the torque acting onthe dipole in both the cases.All India 2017

o Explanations1. The electrostatic field lines do not form closed

loop because no electric field lines exist insidethe charged body. (1)

2. According to the question, let both the ballshave same charge q. Let the balls are separatedby a distance r, Hence, according to Coulomb'slaw, if F and F' are the force of attractionbetween balls in air and in medium respectively.

.. 1 q2Asmau,F=---

41tEo ,2

CHAPTER 1 Electric Chorges and Fields

1 2In medium, F' = -- q2' where

41tEo rF'= !..

Kwhere, K is dielectric constant of material andK > 1 for insulators. (1)

Hence, the force is reduced, when a plastic sheetis inserted.

_1_=K41tEo

3. At the intersection point, if electric field linescross each other, then there would be twodirections of electric field which is not possible,so lines of forces never cross each other. (1)

4. As, electric field inside a conductor is always zero.The electric lines of forces exert lateral pressure oneach other leads to explain of repulsion betweenlike charges. Thus, in order to stable spacing, theelectric field lines are normal to the surface. (1)

5. As per the condition given in question, twoconclusions that can be drawn are as follows:(i) The two point charges (ql and q2) should be of

opposite nature. (1/2J

(ii) Magnitude of charge ql must be greater thanmagnitude of charge q2' (1/2)

6. Electric dipole moment of an electric dipole isequal to the product of its either charge and thelength of the electric dipole.It is denoted by p. Its unit is coulomb-metre.

• 21I

Pp=qx2!

It is a vector quantity and its direction is fromnegative charge to positive charge. (1)

7. The plot showing the variation of electric fieldarid electric potential with distance r due to apoint charge q is shown as below:

5.-~~~~--~~~~~~4.5

4

t 3.5E 3

2.52

1.51

0.5 LL-L-L-2c:=:c::=;::=:;::=:c::..Jo 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

r ---+

•-q •+q

+-- E = _1_ ....!L41tEo r2

8. Force on positive charge is always in the directionof electric field. So, proton being positive willtend to move along the X-axis in the direction ofa uniform electric field. (1)

(1)

9

9. (i) For stable equilibrium, the angle between pand E is 00 i.e. it should be placed parallel toelectric field. (112)

---------------------.1 E

• •-q +qp

(ii) For unstable equilibrium, the angle between pand E is 1800 i.e. it should be placedantiparallel to electric field.--------------------+1 E

•+q -q(1/2)

10. Two point charges system is taken from air towater keeping other variable (e.g. distance,magnitude of charge) unchanged. So, only factorwhich may affect the interacting force is dielectricconstant of medium.Force acting between two point charges

F = 1 qlq2 or F oc.!...41t Eo K r2 K

~=K ~ _8_=80Fmedium Fwater

8 1Fwater = -- ~ Fwater = - N

80 10 (1)

11. Path d is followed by electric field lines. Electricfield intensity inside the metallic sphere will bezero, therefore, no electric lines of force existinside the sphere. Also electric field lines arealways perpendicular to the surface of theconductor. (1)

Right, because mutual force acting between twopoint charges is proportional to the product ofmagnitude of charges and inversely proportionalto the square of the distance between them i.e.independent of the other charges. (1)

Since, torque (t) on the dipole in electric field E is

t =p x E

12.

13.

It I=pEsin9

For minimum torque,Itl= 0

~ pEsin9 =0

~ sin9=0~ 9=00,1800 (1)

10

14. Given,Length 2a = 4 ern = 4 x 10-2 mAngle, 9 = 60°Torque, 't = 413NmCharge, Q= 8 X 1O-9CWe know that, 't = Q(la) E sin9Electric field,E = 't 413 NC1

Q(la)sin9 8 x 10-9 x4 X 10-2 x sin 60° (1)

E = 2.5X 1010NC1

:.Potential energy,U = - pE cos9 = - Q(la) E cos9

U = - 8 X 10-9 X 4 X 10-2

413 x cos60°x----~~----~-------8 x 10- 9 X 4 x 10 2 x sin 60°

-413=--1=-41.J315. Refer to Ans. 14 (1)

U = -pEcos9

= -4xl0-9 x 2x 10-2 X B.J3 x cos60°4xlO-9 x 2xlO 2 x sin 60°

= _ B.J3 1 = -81.J3 (1)

16. Refer to Ans. 14 (1)9 -2 6.J3 x cos 60°

U= -2xlO- x l x IO x--....,,---------o;-----2xlO-9 xi xro 2 x sin 60°

= _ 6.J3 1 = -61.J3 (1)

. a(i) Work done in rotating the dipole, W = r 2't d9Ja,

In turning the dipole from direction parallelto electric field to direction opposite toelectric field is angle will be 0° to 1t.

.. W = CpEsin9 d9= pE [-cos9]~ = 2pE (1)

(ii) We know that, 't = pEsin9

If 9 = 2:, then 't is maximum2

i.e. 't = pEsin2: ~ 't = pE (maximum)2 (1)

17.

18. According to question, the charge on innersurface = - Q (1)

According to question, the charge on outersurface = + QElectric field at point ~ is given by (1)

E=-Q-41tEo r/

o ehapterwise eBSE Solved Papers PHYSICS

19. Electric field lines are always perpendicular to thesurface of conducter

Plate ('12)

(1)

20. According to question, for unstable equilibrium,the angle between p and E is 91 = 180°Finally, for stable equilibrium, 92 = 0° (1/2)

Required work doneW = pE(cos91 - cos92) (112)

= 3xl0-8 xl03(cos180° - cos O")[":cos1800=-1 and cosoo=+l]

W = - 6 x l 0-5 J (1)

21. According to Coulomb's law, the magnitude offorce acting between two stationary point charges

is given by F = (ql q2) (...!..-)41tEo r 2

For given ql q2' F oc(:2 )1The slope of F versus 2' graph depends on ql q2'r

Magnitude of ql q2 is higher for second pair.1:. Slope of F versus 2 graph .r

For (1 uc, -3 )lC)pair of charges

(1)

For (1 uc, 2 )lC)pair of charges

(~)--Corresponding to second pair (1 )lC, - 3JlC) isgreater. Higher the magnitude of product ofcharges ql andQ2' higher the slope. (1)

22. When two identical conducting charged spheresare brought in contact, then redistribution ofcharge takes place, i.e. the charge is equallydivided on both the spheres.When C and A are placed in contact, charge of Aequally divides in two spheres. Therefore, chargeon each A and C = + 20. (1)

CHAPTER 1 Electric Charges and Fields

Now. C is placed in contact with B. then chargeon each A and C becomes 2f.2 + (-lOQ) = - 4Q

2When A and B are placed in contact. then chargeon each A and B becomes

2f.2 + (- 4Q) = _ Q

2 rn23. For stable equilibrium. the angle between p and E

61 = 0°.For unstable equilibrium. 62 = 180°. (1)

Work done in rotating the dipole from angle 61 to 62W = pE(cos61 - cos62) = pE(cosO° - cos1800)W = 'lpE (1)

24. Electric r _,d intensity. E = 106 NC-I

Work done. W = 2 X 10-23 JWork done in rotating the dipole from stableequilibrium position to unstable equilibrium position.

W = pE(cos61 - cos62)W = pE (cos 0° - cos 180°) = 2pE

Wp=-

2E

2 X 10-23 = 10-29 C- mz x 106

25. Let two point c.iarges (11 and q2 are situated atpoints A and B have position vectors Ii and r2.

E2

Aq

1_---\1---E1

--- /:/ I

/ 1r /' B

/,// q2_ r2

,/

AP= r -rland BP= r - r2Electric field intensity at point P due to ql'

E[=__I_ ....3.L AP41t€o IBPI3

Similarly. E2=_I_.~BP41t€o lAP 1

3 (1)

:. Net electric field intensity at point P,

E= EI+ E2_ 1 r ql q2 1---l-- (r-rl) + --(r-r2)J

41t€o Ir_r113 Ir-ri (1)

11

26. According to principle of superposition of electricfields. E (electric field) at a point due to system of three

1 r q. q2 • q3' 1charges. E = --l-2-f1P + -2-f2P + -2-f3PJ

41t€o f1P f2P f3P

(1)

Here. 'I = lrn, '2 = 3m and '3 = 9m

E~:~[~~~I~'+I~i+-l= .:,l~lq 9 r . a] 1 9q -I

= 41t€o x"8 Lusmg S_ = 1="; = 41t€o . gNC (2)

27. Let one of two charges is x ~c.Therefore. othercharge will be (7 - x) IlC.

By Coulomb's law. F = _1_. ql q241t€o r 2 (1)

1= 9 xro" X (x x 10- 6) (7 - xl xlO-6

(0.3 )2

9 xlO-2 = 9 X109-12 x (7 - x) ~1O= x(7 - x)

:. x 2_ 7x+ 10= O~ (x- 2) (x- 5)= 0

x = 2 IlC or 51lC (1)

Therefore. 'charges are 2~Cand 5~C28. (i) By electrostatic induction. charge induces on

the plates and opposite nature of chargeappears on the surface facing each other.Therefore. they start attracting towards eachother. (11

(1J

(il) P1 P2-0 +0 -0 +0

-+

-+ -+

29.

Field lines must be perpendicular tv the plates.Also. equispaced field lines exist between twoplates as electric field between them isuniform. (11

To find the electric field intensity at a point dueto two charges. first of all find the individualelectric fields due to both charges and then findthe resultant field using vector addition.

!---X1_!B• u __

-0(X1.0)

Auu_u. I

+0 0(-X2.0) _

I---X2~

12

Electric field intensity at 0 due to + Q charge,1 QEl =-- X--2 (towards B) ... (i)

41tEo (x2)

Electric field intensity at 0 due to - Q charge,

E2=_I_x~ (towardsB) ... (ii)41tEo (Xl )2 (1)

.: El and E2 act along the same direction.:. Net electric field intensity at 0,

E = El + E2 (towards B)=_I_x~+_I_x~

41tEo (X2)2 41tEo (XI)2

E = 4!J :i + :/ ]

30. (i) Electric field at a point on the equatorialline of an electric dipole.Consider an electric dipole consisting of twopoint charges + q and - q separated by a smalldistance 11B = 2/ with centre at 0 and dipolemoment, p = q(2l) as shown in the figure.

Ea

e eA ---- -~-- B-q 0 +q

I • 2/-------+i

Resultant electric field intensity at the point Q,

EQ = EA + E81 qHere, E A = -- . ---c;:-'-~

41tEo (x2 + /2)

and E8 =_I_. __ q_41tEo (x2 + /2) (1)

On resolving EAand E8 into two rectangularcomponents, me vectors E A sine and E8 sine areequal in magnitude and opposite to each otherand hence, cancel out.The vectors E A coss and E 8 coss are acting alongthe same direction and hence, add up.:.EQ = E A cosa + E8 cosS= 2E A coss

[':EA=E8

o Chopterwise CBSE Solved Papers PHYSICS

1 2iJ/= 41tEo . (x2 + /2)3/2

But, the dipole moment Ip 1=q x 21

.. EQ = _1_. Ipi41tEo (x2 + /2)3/2

The direction of E is along QE II BA. i.e .opposite toAB. In vector form, we can rewrite as

EQ= -p41tEo(X2 + /2)3/2 (1)

(ii) The orientation of the dipole(a) In stable equilibrium, , p is parallel to E. i.e.

e = 00 (1)

(1)

___ 2 I-+< E-q • • +q

p

(b) In unstable equilibrium, p is anti-parallelto E i.e. S = 1800

___ 2 I-+< E

+q. .-qp

31. Dipole in a uniform external field

-qE(1)

o-----:::--+q E,,,,,,

,,,

According to the figure, if we consider an electricdipole consisting of charges -q and +q and oflength 2a placed in a uniform electric field Emaking an angle Swith electric field, then Forceexerted on charge -q at A = -q E (opposite to E) (1)

Force exerted on charge +q at A = q E (along E)Hence, the net translating force on a dipole in auniform electric field is zero. But the two equal


and opposite forces act at different points andform couple which exerts a torque 'to't = Force x Perpendicular distance between thetwo forces

't = qE(AN) = qE(2asine)'t = q(2a)Esine~'t = pEsine~ 't = P xE

Pairs of perpendicular vectors(a) ('t,p) (b) (t; E)

32. According to question, suppose that the ring isplaced with its plane perpendicular to the X-axisas shown in figure. Consider small element dl ofthe ring.

dl

dlAs the total charge q is uniformly distributed so,the charge dq on element dl is dq = ~. dl

2na

dq = .s: dl cosa = dEcose (where,cose =~)2na r2 2

Since, only the axial component gives the net E atpoint P due to charge on ring.

E 2""So, f dE = f dE coss

o 0

2"" kq dl x= f -·-x-

o 2na r2 r

2"" 2""= kqx ~ f dl = ~. ~ f [l]~""

2na r30 2na r3

0

= kqx. 1 .2na [.:r2 = X2 + a2]2na (x2 + a2)3/2

5 E= kqx0,(x2 + a2)3/2 (1)

Now, for points at large distances from the ringx» a.

This is same as the field due to a point chargeindicating that for far-off axial point, the chargedring behaves as a point charge.

13

33. Refer to Q. 3l.

34. (i) The magnitude,

IEABI=~l_x!L=E41t£o a2

IEAcI=_l-x 2£[=2E41[£ a2

o EAB

(1/21

(1/2)

(1)

+qL- ~-2qB C

(1)

Enet = ~E AB 2+ EAc2 + 2EABEAC cose

= ~(2E)2 + E2 + 2 x 2E x E x ( - ~)

= ~4 E2 + E2 - 2E2 = E-J3

We know that, E = ---q-241tEoa

_ q-J3Enet--

42

1tEoa

(ii) Direction of resultant electric field at vertex.(1/2)

... (i) (1/21

So,

tana = E AB sin1200E AC + E AB cos 120°

Ex -J32

35.

2E + E x (-~)

1 -I ( 1 )tan a = -J3 ~ a = tan -J3

a = 30°For electric dipole momentRefer to Ans 6. (1)

Consider an electric dipole AB consists of twocharges +q and -q separated by a distance 2a. Wehave to find electric field at point P onequipotentialline separated by a distance r. (1)

Electric field at point P due to charge + q~ =_l_x q

41tEo [~(r2 + a2)]2

=_l_x q41t£o (r2 + a2)

(1)

(with side AC)

(1)

14

Along AP,E, E, sin a

E2 sina-<~)E2! \

(~~/ I \ ..J(r 2 + a2), I \

/ lr \I I \

/ : \, I \'a I ,).

B_~--a----,6-~a-_aJ~A

Electric field at point P due to charge - q

E2= _1_ x --q- along PB47tEo r2 + a2

On resolving ~ and E 2 into rectangularcomponents, we get resultant electricfield at point P.

E = EJ cos a + E2 cos a= _l_x q cos a+ _l_x q cos a

47tEo (r2 + a2) 47tEo (r2 + a2)

=2x_l_x q x a47tEo (r2 + a2) ~(r2 + a2)

= _1_ x --,._q.:...2a~."..,.,.47tEo (r2 + a2)3/2

E=_l_x P47tEo (r2 + a2) 3/ 2

[But qx2a = P]

Ifr> > a,then E = _I_x!...

47tEo r3 (1)

36. Refer to Ans 33. (1)(i) whens = 0; 1: = oand p and E are parallel and the

dipole is in a position of stable equilibrium. (1)

(ii) When a = 180°, 1: = 0 and p and E areanti-parallel and the dipole is in a position ofunstable equilibrium. (1)

37. (i) Electric field lines due to a conducting sphereare shown in figure.

*v.-- __ -'--ConductinQ

sphere havingnegative charge

o Chapterwise CSSE Solved Papers PHYSICS

(ii) Electric field lines due to an electric dipole areshown in figure.

(1'1.)

38. Equal charge of opposite nature induces in thesurface of conductor nearer to the source charge. (1)

~ L-~~ ~_q~~~~~~~~~~+q

(2)

(1 Va)

Electric lines of forces should fall/normally 90°away on/from the conducting plate. (1)

39. (i) 1: = pEsinaIn vector notation, 1: = P x E51unit of torque is newton-metre (Nom) and itsdimensional formula is [ML2 T-2]. Torque isalways directed in plane perpendicular to theplane of dipole movement and electric field.Case 1 Ifa = 0°, then 1: = 0The dipole is in stable equilibrium.Case 1 If a = 90°, then 1: = pE (maximumvalue)The torque acting on dipole will be maximum.Case 3 ire = 180°, then 1: = 0 (1)

The dipole is in unstable equilibrium.(ii) If the field is non-uniform there would be a

net force on the dipole in addition to thetorque and the resulting motion would be acombination of translation and rotation.

1: = P x E(r)


(iii)

Net torque acts on the dipole depending on thelocation, where r is the position vector of thecentre of the dipole. (1)

(a) E is increasing parallel to p, then 9 = 0°. So,torque becomes zero but the net force on thedipole will be in the direction fo increasingelectric field and hence it will have linearmotion along the dipole moment.

E

Force on-q• Force on +q

0·..··..-- .-q p +q

Direction of net force= ---~Direction of increasing field= ---~

(b) E is incrasing anti-parallel to p. So, thetorque still remains zero but the net force onthe dipole will be in the direction ofincreasing electric field which is opposite tothe dipole moment, hence it will have linearmotion opposite to the dipole moment. (1)

E

.......- .+q p -q

Forceon+q~--~

Direction of net force= ...••---Direction of increasing field= ---~

Forceon-q

40. (i) Electric field due to dipole at axial pointWe have to calculate the field intensity E at apoint P on the axial line of the dipole atdistance op= x from the centre 0 of the dipole.

A a B EA E8•.------4------- ...•.----... III

-q +q P1+--2/~

~r-/-+lI~ II

I~ r+t II

Resultant electric field intensity at the point Pis

Ep=EA +EBThe vectors E A and E B are collinear andopposite... Ep=EB-EA

Here, EA=_l_._q_;47tEo (r+ 1)2

15

1 qEB=--·--

47tEo (r_1)2

_ 1 [q q 1Ep

- 47tEo (r-1)2-(r+1)2J

1 4q x I= 41tEo' (r2 _/2)2

1 '2pEp=--'---

47tEo (r2 _/2)2

If the length of dipole is short i.e. 2i«r, then

Ep=-'2p--47tEo ·r3

The direction of E p is along BP produced.1Epoc-r3

(ii) E oc';. As r will increase, E will sharplyr

decreases. The shape of the graph will be asgiven in the figure. (1)

So,

o r-+

(ill) When the dipole were kept in a uniform electricfield Eo. The torque acting on dipole, t =P x E (1)

I+q

----+-+----l-E

(a) If9=0°, then t=O, pilEI

(1)

--------l--E

-q •..------- .• +q

The dipole is in stable equilibrium.(b) If9=180°, then t=O, p 11-£ (1)

IE

+q •.. ------- .• -q

The dipole is in unstable equilibrium.

[TOPIC 2] Electric FluxArea VectorIt is the vector associated with every area elementof a closed surface and taken in the direction ofthe outward normal. Consider the diagram givenalonside

Here, dS is the area vector in the direction of theunit vector nnormal to the surface area llS.

Representation of orea vector

2.1 Electric FluxElectric flux linked with any surface isproportional to the total number of electric fieldlines that normally pass through that surface. It isa scalar quantity.SI unit of electric flux is N_m2C-1 or JmC-1 or Vm.

CGS unit of electric flux is dyne-em 2/ stat-c.

Different Conditions for theElectric Flux Linked with a Surface

(i) When surface is held normal to the directionof uniform electric field E, then iltPE = EllS

e = 0° A

-+--+----+ n (6S)E

Electric flux through normal orea

(ii) When area vector of surface makes an anglee with the direction of uniform electric fieldE , then iltPE = EllS cos e.

In vector form, iltPE = E· ilSPositive A

normal n E

65

Eelectric Flux through on angle e(iii) Closed surface S lying inside the

non-uniform electric field EThe total electric flux linked with the closedsurface S is tP ::;1/.dS

Closed surface S

dS... :.:.1 .•.•••••. 1.•. 1 •. ,.1.1.>.1.:.

Electric flux through a closed surface sThe surface integral of electric field over theclosed surface represents total electric fluxlinked with the surface.

2.2 Gauss' TheoremThe total electric flux linked with closed surface

Sis tPE =1 E-dS= s.eo

where, q is the total charge enclosed by the closedGaussian (imaginary) surface.

Applications of Gauss' Theorem(i) Electric field due to infinitely long

uniformly charged wire with linearcharged density A..We have consideredcylindrical Gaussian surface.

+---- - ----

--

----- -+ ------

E+

+ p

+ +-Gaussiansurface

(ii)

From Gauss' law,

fE -dS = ...i = AlEo Eo

Al~ E x 21trl =-

Eo

or E=_A_21tEor

Here, r is the perpendicular distance fromthe charged wire.

(a) Electric field due to a thin infinite planesheet of charge with uniform surfacecharge-density o at any nearby point

+ + +

E+/ \+

I I E+1 :+

'- I

++++~------ Gaussian

surface

E=~2Eo

[for infinite plane sheet of charge]o

E=-Eo

[for near charged conducting surface](b) Electric field intensity due to two

equally and oppositely charged parallelplane sheet of charge at any point

and

17

E =~ [between the two plates]Eo

and E = 0 [outside the plates](iii) Electric field due to a thin charged

spherical shell of radius R at a distancer from its centre.To find the field at a distance r from the centreof the spherical shell, we consider a sphericalGaussian surface of radius r centered at theshell and then Gauss' law is applied.(a) For point lying outside the shell

(r> R)Since E and dS are in the same direction,

fE-dS =...iEo

E 4 2 c X (41tR2)x ttr = ---'----"-Eo

crR2E=--

Eo~

E=_l_.!!....41tEo r2

(b) Field at a point inside the shell (r < R)

Here, the charge inside the Gaussiansurface shellAs, q =0 ~ E =0

(c) Field at a point on the surface (r = R)

On putting r = R,

E = q c41tEo R2 Eo

where, surface charge density,

o =-q-41tR2

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 2o 1 Mark Questions

1. How does the electric flux due to a pointcharge enclosed by a spherical Gaussiansurface get affected when its radius isincreased. Delhi 2D16

2. What is the electric flux through a cube ofside 1 em which encloses an electricdipole? All India 2015

3. Consider two hollowconcentric spheres S1and S2 enclosingcharges 2Q and 4Qrespectively, asshown in the figure.(i) Find out the ratioof the electric flux through them. (ii) Howwill the electric flux through the spheresS1change if a medium of dielectricconstant e, is introduced in the spaceinside S1 in place of air? Deduce thenecessary expression. All India 2014

4. Two charges of magnitudes -2Q and + Qare located at points (a, 0) and (4a, 0),respectively. What is the electric flux dueto these charges through a sphere of radius3a with its centre at the origin? All India 2013

5. A charge q is placed at the centre of acube of side l.What is the electric fluxpassing through each face of the cube?All India 2010; Fareign 2010

6. Figure showsthree pointcharges, +2q,-qand +3q. Twocharges +2q and-q are enclosedwithin a surface S. What is the electricflux due to this configuration through thesurface S? Delhi 2010 ~'''c

4Q

oS1

S +2q•+3q

7. If the radius ofthe Gaussian surfaceenclosing a charge is halved, how does theelectric flux through the Gaussian surfacechange? All India 2009.2008

8. (i) Define the term 'electric flux'. Writeits 81 unit.

(ii) What is the flux due to electric fieldE = 3x 10' i NC-1 through a square ofside 10cm, when it is held normal toE ? All India 2015C

o 2 Marks Questions9. Given a uniform electric field E = 5 x 10" i

NC -l, find the flux of this field through asquare of 10 em on a side whose plane isparallel to the YZ-plane. What would bethe flux through the same square if theplane makes an angle of 30° with theX-axis? Deihl 2014

10. Given a unifor ,1 electric field E = 2 x 10" iNC -1, find the flux of this field through asquare of side 20 ern, whose plane isparallel to the YZ-plane. What would bethe flux through the same square, if theplane makes an angle of 30° with theX-axis? Farelgn 2014

11. Given a uniform electric field E = 4 x 10"iNC-1. Find the flux of this field through asquare of 5 em on a side whose plane isparallel to the YZ-plane. What would bethe flux through the same square if theplane makes an angle of 30° with theX-axis? Delhi 2014C

12. A sphere S1 of radius r1enclosed a net charge Q.If there is anotherconcentric sphere S2 ofradius r2(r2 > r1)

enclosing charge 2Q,find the ratio of theelectric flux through S1 and S2. How will theelectric flux through sphere S1 change if amedium of dielectric constant K isintroduced in the space inside S2 in place ofair? All India 2014


13. Two concentric metallic spherical shells ofradii R and 2R are given charge Q1 and Q2respectively. The surface charge densitieson the outer surfaces of the shells are equal.Determine the ratio Q1: Q2'Foreign2013

14. A thin straight infinitely longconduction wire having charge density Ais enclosed by a cylindrical surface ofradius r and length I, its axis coincidingwith the length of the wire. Find theexpression for the electric flux through thesurface of the cylinder. AllIndio2011

15. Two charged conducting spheres of radii r1and r2 connected to each other by a wire.Find the ratio of electric fields at thesurfaces of the two spheres. Delhi2011C

16. Show that the electric field at thesurface of a charged conductor is givenby E = ~ n, where a is the surface

EOcharge density and n is a unit vectornormal to the surface in the outwarddirection.AllIndio2010

17. A spherical conducting shell of innerradius ~ and outer radius R2 has acharge Q. A charge q is placed at thecentre of the shell.

(i) What is the surface charge densityon the (a) inner surface, (b) outersurface of the shell?

(ii) Write the expression for the electricfield at a point to x> R2 from thecentre of the shell.Alllndio 2010C

18. Define electric flux. Write its SI unit.A charge q is enclosed by a sphericalsurface of radius R. Find the electricflux. AllIndio2009

19. A sphere S1 of radius r1 encloses a chargeQ, if there is another concentric sphereS2 of radius r2(r2 > r1) and there are noadditional charges between S1 and S2'Find the ratio of electric flux through S1and S2- AllIndio2009

19

20. Draw the shapes of the suitable Gaussiansurfaces while applying Gauss' law tocalculate the electric field due to

(i) a uniformly charged long straight wire.(ii) a uniformly charged infinite plane

sheet. Deihl2009C

21. A uniformly charged conducting sphere of2.4 m diameter has a surface charge densityof 80.0 J.l.Cm-2.

(i) Find the charge on the sphere.(ii) What is the total electric flux leaving

the surface of the sphere? Delhl2009C22. A point charge causes and electric flux

-3 x 10-14 N m2 c' to pass through aspherical Gaussian surface.

(i) Calculate the value of the point charge.(ii) If the radius of the Gaussian surface is

double, how much flux would passthrough the surface? Foreign2D09

o 3 Marks Questions23. Two infinitely

large plane thinparallel sheetshaving surfacecharge densities a1and a 2(a1 > a 2) areshown in thefigure. Write themagnitudes and directions of the net fieldsin the regions marked II and III. Foreign2014

24. A hollow ycylindrical box oflength 1m andarea ofcross-section25 em 2 is placed Zin a three-dimensional coordinate system asshown in the figure. The electric field in theregion is given by E = 50 xi, where E is inNC-1 and x is in metre.Find

(i) net flux through the cylinder.(ii) charge enclosed by the cylinder.

Delhi2013

+

++++ a1

+ +++ B

II

III+

O~~r------+-r+X

20

25. (i) State Gauss's law.(ii) A thin straight infintely long

conducting wire of linear chargedensity A. is enclosed by a cylindricalsurface of radius r and length l. Itsaxis coinciding with the length of thewire. Obtain the expression for theelectric field, indicating its direction,at a point on the surface of thecylinder. Deihl 2012

26. State Gauss' law in electrostatics. A cubewhich each side a is kept in an electric fieldgiven by E = cxi as shown in the figure,where C is a positive dimensional constant.Find out

y

z(i) the electric flux through the cube.

(ii) the net charge inside the cube.Foreign 2012

27. Using Gauss' law, obtain the expression forthe electric field due to uniformly chargedspherical shell of radius R at a pointoutside the shell. Draw a graph showingthe variation of electric field with r, forr » Rand r < R. All Indio 2011

28. Use Gauss' law to derive the expression forthe electric field between two uniformlycharge parallel sheets with surface chargedensities o and -cr, respectively. All Indio 2009

29. A positive point charge (+q) is kept in thevicinity of an uncharged conducting plate.Sketch electric field lines originating fromthe point on to the surface of the plate.Derive the expression for the electric fieldat the surface of a charged conductor.All Indio 2009

30. State Gauss' law in electrostatics. Usingthis law, derive an expression for theelectric field due to a uniformly chargedinfinite plane sheet. Delhr2009

o chnpterwlse eBSE Solved Papers: PHYSICS

31. State Gauss' law in electrostatics. Usingthis law derive an expression for theelectric field due to a long straight wireof linear charge density A. elm. Delhi 2009

o 5 Marks Questions32. (i) Use Gauss's theorem to find the

electric field due to a uniformlycharged infinitely large plane thinsheet with surface charge density cr.

(ii) An infinitely large thin plane sheethas a uniform surface charge density+O', Obtain the expression for theamount of work done in bringing apoint charge q from infinity to apoint, distant r, in front of thecharged plane sheet. All India 2017

33. (i) An electric dipole of dipole momentp consists of point charges +q and-q separated by a distance 2a apart.Deduce the expression for theelectric field E due to the dipole at adistance x from the centre of thedipole on its axial line in terms ofthe dipole moment p . Hence, show _that in the limitx» a,E ~ 2p (41tEo.~?).

(ii) Given the electric field in the regionE = 2xi, find the net electric fluxthrough the cube and the chargeenclosed by it. All India 2015; Delhi 2015

y

/ /

1/ 1/a

x

z34. (i) Define electric flux. Write its SI

unit. Gauss' law in electrostatics istrue for any closed surface, nomatter what its shape or size is.Justify this statement with the helpof a suitable example.

(ii) Use Gauss' law to prove that theelectric field inside a uniformlycharged spherical shell is zero.Delhi 2015

35. (i) Deduce the expression for the torqueacting on a dipole of dipole moment pin the presence of uniform electricfield E.

(ii) Consider two hollow concentricspheres, 81 and 82 enclosing charges2Q and 4Q respectively as shown inthe figure. All Indio 2014

36. Using Gauss' law, deduce the expression forthe electric field due to a uniformly chargedspherical conducting shell of radius R at apoint

(i) outside the shell(ii) inside the shell

Plot a graph showing variation of electricfield as a function of r > Rand r < R.(r being the distance from the centre ofthe shell) All Indio 2013

37. (i) Define electric flux. Write its SI unit.(ii) A small metal sphere carrying charge

+ Q is located at the centre of aspherical cavity inside a largeuncharged metallic spherical shell asshown in the figure.Use Gauss' lawto find the expressions for the electricfield at points PI and P2•

Delhl2012C

38. (i) Define electric flux. Write its SI unit.(ii) Using Gauss' law, prove that the

electric field at a point due to auniformly charged infinite plane sheetis independent of distance from it.How is the field directed if(a) the sheet is positively charged?(b) negatively charged? Deihl 2012

21

39. (i) State Gauss' law. Use it to deducethe expression for the electric fielddue to a uniformly charged thinspherical shell at points(a) inside the shell and(b) outside the shell.

(ii) Two identical metallic spheres A andBhaving charges + 4Q and -lOQ arekept a certain distance apart. A thirdidentical uncharged sphere C is firstplaced in contact with sphere A andthen with sphere B. Then, spheres Aand B are brought in contact andthen separated. Find the charges onthe spheres A and B.All Indio 2011C

o Explanations1. According to question, electric flux ($) due to a

point charge enclosed by a spherical Gaussiansurface is given by

$= E·A

k (k')$ = 3... 41tr2 = kq- 41t .: E = 3.. and A = 41tr2

r G"(112)

So, there is no effect of change in radius on theelectric flux. (112)

2. Since, according to the Gauss' law ofelectrostatics, electric flux through any closedsurface is given by

$E=fE'dS.

=!L ...(i)Eo

where, E = electrostatic fieldq = total charge enclosed by the surface

Eo = absolute electric permittivity of free spaceSo, in the given case, cube encloses an electricdipole. Therefore, the total charge enclosed by thecube is zero. i.e. q = O.Therefore, from Eq. (i). we have

$E =!L = 0Eo

i.e. electric flux is zero. (1)

22

3. (i) According to Gauss' theorem,</>= r.q oc r.q

EoE

=> ~=~=!.</>S2 2Q+ 4Q 3

(ii) If the medium is filled in S!, then</>SI = r.q = 2Q

EOEr EoEr

4. Gauss' theorem states that the total electric fluxlinked with closed surface S is </>E = </>E·dS = .!L

Eowhere, q is the total charge enclosed by the closedGaussian (imaginary) surface.

-20 +0(a,O) (3a, 0) (4a, 0)

Charge enclosed by the sphere = -2Q

Therefore, </>= 2Q (inwards)Eo

5. By Gauss' theorem, total electric flux linked witha closed surface is given by </>= .!L

Eowhere, q is the total charge enclosed by the closedsurface.:. Total electric flux linked with cube, </>= .!L

EoAs charge is at centre, therefore, electric flux issymmetrically distributed through all 6 faces.Flux linked with each face = .!.</>= .!. x .!L = -q-

6 6 Eo 6Eo (1)

6. Electric flux through the closed surface S is</>s= r.q = + 2Lj - q =.!L => </>s=.!L

Eo Eo Eo Eo (1)

Charge + 3 q is outside the closed surface S,therefore, it would not be taken into considerationin applying Gauss' theorem.

7. Refer to Ans 1.As the radius is half

• = EA ~ • = (~i'4'(~)'= kq 4.So, there will be no change in flux due to changein radius. (1)

o Chopterwise C8SE Solved Papers PHYSICS

(1/2)

8. (i) The electric flux through a given area heldinside an electric field is the measure of thetotal number of electric lines of force passingnormally through that area. Its SI unit isNm2Cl or V-rn or JmCI.

(ii) According to the question, E = 3x 103 iNCl.Area of the square, ~ S = 100 em? = 10-2 m 2Hence, electric flux through the square,

</>= E.~S = (3x 103).10-2i = 30Nm2c-1 (1)

9. Given, electric field intensityE= 5x103iNC-1

Magnitude of electric field intensityIE 1= 5x103NCI

Side of square, S = 10 ern = 0.1 mArea of square, A= (0.1)2 = 0.01 m2

The plane of the square is parallel to the YZ-plane.(1)

Hence, the angle between the unit vector normalto the plane and electric field is zero.i.e., a = 0°:. Flux through the plane,</>= IE Ix A cos a=></> = 5 X 103 x 0.01 cos 0°

</>=50Nm2CI (1)

If the plane makes an angle of 30° with the X-axis,then a = 60°:. Flux through the plane,

</>= IEIx A x cos 60°

= 5x 103 x 0.01 x cos 60°= 25 Nm2C1

10. Refer to Ans. 9. (Ans. 40 Nm2 Cl) (2)

11. RefertoAns. 9. (Ans. 5 Nm2CI) (2)

12. According to Gauss' law,Flux through Sl'

</>1 =!{ ... (i)Eo (112)

Flux through S2'

</>2= Q+ 2Q= 3Q ... (ii)Eo Eo

Ratio of fluxes = ~</>2

From Eqs. (i) and (ii), we get

=!{x~=!.Eo 3Q 3 (1)

There is no change in the flux through SI withdielectric medium inside the sphere S2. (1/2)

(112)

(1)

13. According to the question, surface charge density,o '" constant = _Q-

41tR2

[Let Q1 and <22 are two charges]Hence,Charge Q1 = 41tR1cr

Charge Q2 = 41t(2R)2cr .,. (ii)

On dividing Eq. (i) withEq. (ii),

Q1 _ 41tR2cr _ IQ2 - 41t(2R)2cr -"4

14. A thin straight conductingwire will be a uniform linearcharge distribution.Let q charge be enclosed bythe cylindrical surface.:. Linear charge density,

A=IJ..l

q = AI ... (i) (1)

... (i)

T ++

I +

1 +---- +---,

By Gauss' theorem,:. Total electric flux through Charge enclosed bythe surface of cylinder the cylindrical surface

 = 3-. [Gauss'theorem]Eo

 = "AIEo

15.

[From Eq. «n(1)

When two charged conducting spheres areconnected, then charge flows between the two,till their potentials become same.Electric potential on the surface of connectedcharged conducting spheres would be equal.i.e. \'I = V2

__ I_~= __ q241tEo rl 41tEo r2

[Assuming ql and q2 are charges on the spheresconnected to each other and rl, r2 are their radii.]

~ = q2 ~ ~ =!i ...(i)rl r2 q2 r2 (1)

Now, ratio of electric field intensitiesI ql

EI_~'?E2 - _1__.1J..2

41tEo r22

5J.. = (~) x Ii =!i x r~E2 q2 rl

2 r2 rl-

5J.. =!iE2 rl

[From Eq. (i)]

(2)

(1)

23

16. Let q charge be uniformly distributed over thespherical shell of radius r.""--+-- .

, "', "'/ ,/+ +\dS "

f \ nfII\\\-f,

'........ ..-/"./---.:!-_ ....:. Surface charge dcnvity on spherical shell

qo=--~41tr

... (i)(1)

.: Electric field intensuy on the surface of sphericalshell E=__1_.3.."

41tE(I r2(1/2)

[.: E acts along radially outward and along I1J

E=(4:,i)"Eo

... (ii)(112)

17. Here, two points are importantl. Charge resides on the outer surface of

spherical conductor (skin effect).2. Equal charge of opposite nature induces in the

surface of conductor nearer to source charge.

(i) (a) Charge produced on inner surface dueto induction = - q:. Surface charge density of inner surface

- -q- 41tR2

I+ + + +q

+0

+ .147""'-'>-E:;;;;;lr x --4j--------E+ p

When charge -q is induced on inner walls,then equal charge + q is produced at outersurface.

(b) Charge on outer surface = q + Q:. Surface charge density of outer surface

- q + Q- 41tR; (1)

24 (21 Chapterwise cast Solved Papers PHYSICS

(ii) Electric field intensity at point P separated by adistance x (x > R2)

E = _1_ x (q + OJ41tEo X2

[along CP and away from spherical shell] (1)

Whole charge is assumed to be concentrated at thecentre. (1)

18. The total electric flux linked with a surface isequal to the total number of electric lines of forcepassing through the surface when surface is heldnormal to the directionof electric field, <Il= EA.If surface is placed innon-uniform electric Efield, then electric field Ais shown by the figuregiven along side. Sincea charge q is enclosedby closed surface, therefore total electric fluxlinked with the closed surface

<Il=,c E·dS=!Lt, EoThe SI unit of electric flux is Nm2 c' . (2)

19.1 According to Gauss's theorem, total flux passing. through a closed surface, <Il=.!L = fE.dS

EoAs, SI and S2 are concentric spheres of radii rl andr2respectively, then

<Ill= il. ... (i)Eo

<Il2= q2 ... (ii)Eo

.. On dividing Eq. (i) with Eq. (ii).~=il.xEo =il.=g=l<Il2 Eo q2 q2 Q

Hence, <Ill: <Il2= 1 : 120. The surface that we choose for application of

Gauss' theorem is called Gaussian surface. Weusually choose a spherical Gaussian surface.(i) Electric field due to a long straight wire of sheet

L--... Uniformly chargedstraight wire

Cylindrical __

Gaussiansurface ----

(1)

(ii) Electric field due to a plane sheet of chargeUniformly chargedinfinite plane sheet

EEAnA' ......•..------l A'-~----~

r - tJ.SCylindrical Gaussian

surface

\,1II

tJ.S-r-

(1)

21. (i) Radius of sphere, R = ~ = 2.4 = 1.2 m2 2

Surface charge density,o = 80 x 10 -6Cm-2

22.

o = -q- or q= 41tR2C;41tR2

Charge on the sphereq = 4x 3.14X(1.2)2 X 80 X 10-6 C

q = 1.447 X 10-3 C(ii) According to Gauss' theorem,

Electric flux, <Il= .!LEo

<Il= 1.447 x 10-3 Nm2 C-I8.85 X 10-12

<Il= 1.63 X 108 Nm2 C-I

(i) By Gauss' theorem,total electric fluxthrough closedGaussian surface isgiven by

(1)

(1)

<Il=.!LEo

.. q = <IlEoBut electric flux passing through the surface

<Il= - 3 x 10- 14Nm2 C-I

.. q = (- 3 x 10-14)x 8.85 x 10- 12

=- 26.55 x 1O-26Cq = - 2.655 x 10- 25C (1)

(ii) Electric flux passing through the surfaceremains unchanged because it depends onlyon charge enclosed by the surface and isindependent ofits size. (1)

CHAPTER 1 Electric Charges ond Fields

23. According to the figures, A and Bare tWQthinplane parallel sheets of charge having uniformdensities oland a 2 with 01 > O2 (1)

<p = E x area of the end faces of the cylinder

Ex2A= oA ~ E=~EO 2£0

+

+ + EB+ + --01 + --++ + EA III

+ A II + BI

In region IIThe electric field due to the sheet of charge A willbe from left to right (along the positive direction)and that due to the sheet of charge B will be fromright to left (along the negative direction). (1)

Thm:O:'f ,:g[:~rhav1

~ E = -(01 - (2) (along positive direction)21"0

In region IIIThe electric fields due to both the charged sheetswill be from left to right, i.e. along the positivedirection. Therefore, in region III, we have

E=~+~21"0 21"0

1E=-(ol +(2)Eo

(along positive direction)

24. (i) Given, E = 50 x iand t!.S = 25 cm2 = 25 x 10-4 m2

y

O'~-+-+---------1~-+XA

nBz +--+

As the electric field is only along the X-axis, soflux will pass only through the cross-section ofthe cylinder.Magnitude of electric field at cross-section A,

EA=50Xl=50NC-1

Magnitude of electric field at cross-section B,EB=50x2=100NC-1 (1/2)

The corresponding electric fluxes are<PA = EA . ~S = 50 x 25 X 10-4 x cos1800

=_ 0.125 Nm2C-1

25.

25.

(11

dS

+ rEtdS

-+-r-+

<P = fs E· dS = f EdS cos 90°+ f EdS cos 0°+ f EdS cos 90°Upper plane face Curved surface Lower plane face

<P = 0 + EA + 0 or <p = E . 21trl

But by Gauss's theorem,q 'AI

<p=-=-Eo Eo

where, q is the charge on length I of wire enclosedby cylindrical surface S and 'A is uniform linearcharge density of wire.

'AI 'AE x 21trl = - ~ E = --. Eo 21tEor

(1)

2 o Chopterwise CBSE Solved Papers PHYSICS

Thus, electric field of a line cbarge is inverselyproportional to distance directed normal to thesurface of charged wire. (1)

26. Gauss' law states that the total electric fluxthrough a closed surface is equal to ~ times, the

Eomagnitude of the charge enclosed by it is.

$=~Eo

Here, Eo is the absolute permittivity of the freespace and q is the total charge enclosed.

Also, $=J E· dS=~Js Eo

where, E is the electric field at the area element dSNow, the electric field E= Cxi is in X-directiononly. So, faces with surface normal vectorperpendicular to this field would give zero electricflux, i.e. $ = E dS cos 90° = 0 through it.

y

(Sanivurl J_Ir..rt-·

,; ,

z

~ So, flux would be across only two surfaces.Magnitude of E at left face,

EL = Cx = Ca [x = a at.left face]Magnitude of E at right face •

ER = Cx = Cm = mC [x = m at right face]Thus, corresponding fluxes are

$L = E L . dS = E L dS cos 9=-aCxa2, [As,9 = 180°]

[':9 = 0°]$R = E R . dS = mC dS cos 9= mCa2 = m3C

(i) Now, net flux through the cube is= $L + $R = - alC + m3C = a3C Nm2C-' (1/2)

(ii) Net charge inside the cubeAgain, we can use Gauss' law to find totalcharge q inside the cube.Wehave $ = ~

Eoq = $Eo q = alCEOcoulomb

(1/2)

or (1)

27. Let us consider charge + q be uniformlydistributed over a spherical shell of radius R. Let Eis to be obtained at P lying outside of sphericalshell. (1/2)

/-----' ~ Concentric// + + +", spherical

,/ + + '\ Gaussian" \ surface1+\, I

: + III

'. +\\,,

"

/PChargedspherical

shell.: Eat any point is radially outward (if charge q ispositive) and has same magnitude at all pointswhich lie at the same distance r from centre ofspherical shell such that r> R.Therefore, Gaussian surface is concentric sphere ofradius r such that r> R. (112)

Ir:~~~~~-:--~f·E

I: S1 [d' 'E

1i : r -J. dS1

I I

! ------ ± -----...... ::" - + , I

~ S3 ~E'........ + ... ' 0----.. + ..--- es,I+-- r ---+I

Cylindrical Gaussian surface far line charge

Since, Gaussian surface enclosed charge q inside it.By Gauss' theorem,

fE.dS=~Eo

=> fE dS cos 0° = s.Eo

[.: Eand dSare along the same direction]

fdS = ~ [.: Magnitude of E is same at everyEo point on Gaussian surface]

Ex 41tr2 = ~Eo

E =_1_.3..41tEo r2

(1)


Now, graphE 1 q

E=-'-41t€o R2

Variation of E with r for a spherical shellof charge q (1)

28. Let us consider two uniformly charge, largeparallel sheets carrying charge densities + o and- o respectively, are separated by a small distancefrom each other.By Gauss' law, it can be proved that electric fieldintensity due to a uniformly charged infinite planesheet at any nearby is given by

E=~2£0

... (i)(1)

The electric field is directed normally outward fromthe plane sheet if nature of charge on sheet ispositive and normally inward if charge is of negativenature. +cr -(1

+++--r _

+ P E1 ---+ E2

+

+ ~r

Let r represents unit vector directed from positiveplate to negative plate.Now, Electric Field Intensity (EFI) at any point Pbetween the two plates is given by(i) ~ = + ~ r [due to positive plate]

2Eo

(ii) E2 = + ~ r [due to negative plate]2£0

(1)

:. Electric field intensity at point P is given byE=~+E2

cr· cr·=-r+-r2Eo 2Eo

2i1

~ E=~rEo

Thus, a uniform electric field is produced betweenthe two infinite parallel plane sheet of chargewhich is directed from positive plate to negative~~. (1)

29. Starting from the charge +q, the lines of force willterminate at the metal plate, including negativecharge on it. At all positions, the lines of force willbe perpendicular to the metal surface.

uncharged conducting plate

Refer to Ans 14. (3)

30. Gauss' law refer to Ans 25. (1/2)

Let us consider a large plane sheet of charge havingsurface charge density cr.For figure see Ans (ii). (1/2)

Let electric field is to be obtained at a point P at adistance r from it. It is obvious that Gaussiansurface will be a cylinder of cross-sectional area Aand length 2r with its axis perpendicular to planesheet of charge. Now, applying Gauss' law over theclosed Gaussian surface.

(1)

f -E·dS+ f E·dS=~CSA CSA Eo

rEdS cos 0° + rEdS cos 90° = !LJCSA JcSA Eo

[as Eand dSare along the same direction by at CSAEperpendicular to dS]

r dS=!L ~Ex2A=!LJCSA Eo Eo

E=-q-=~2AEo 2Eo

Electric field intensity = ~2£0

The direction of E is normal to the plane sheet anddirected away from sheet when charge on plate ispositive and vice-versa.Closed cylinder comprises of two caps and CurvedSurface Area (CSA). (1)

... (i)

[From Eq. (i)]

28

31. (i) Refer to Ans 25 (i). (112)(ii) Let us consider a long straight wire carrying + q

charge on its length I and linear charge densityACm

A=iI

=> q=AI ... (i)Let electric field intensity is to be obtained at adistance r from it. Since, magnitude of E due to longcharged wire is same at every point which lie at thesame distance from the wire. So, Gaussian surfacewill be a cylinder of radius r and length I such thatwire lies along the axis as shown in figure..: Angle between E and dS is 90° at caps, whereas 0°at any point on curved surface of a cylinder.Now, applying Gauss' theorem

fE.dS=~Eo (1)

i E·dS+ J E·dS= Al [From Eq. (i)]CSA CSA Eo

[CSA=Close Surface Area]

i E dScos 90°+ J E dS cos 00= Al(SA . <;SA Eo

(SI and S3 are caps and S2 represents CSA)

0+ f EdS= Al [':cos900=0]CSA Eo

Ei dS=AICSA Eo

[.: E is a constant at every point at CSA]u r r 1

E x Znrl = Eo l': JCSA dS - = 21trlJE=~

21tEo rl

E=_A_21tEo r (112)

32. (i) According to the question, o is the surfacecharge density of the sheet. From symmetry, Eon either side of the sheet must beperpendicular to the plane of the sheet, havincsame magnitude at all points equidistant fromthe sheet. We take a cylinder of cross-sectionalarea A and length 2r as the Gaussian surface.On the curved surface of the cylinder, f and nare perpendicular to each other.

o Chopterwise CBSE SolvedPopers PHYSICS

Therefore, the flux through the curved surfaceof the cylinder = O. (1'12)

1-

-'.t't- + A

+ ++ nA

n

(ii)

Flux through the flat surfaces = EA + EA = 2EAThe total electric flux over the entire surface ofcylinder

E=2EATotal charge enclosed by the cylinder, q = crAAccording to Gauss's law,

fE.dA=(h=~Eo

2AE crAEo

E =~2Eo

E is independent of r, the distance of thepoint from the plane charged sheet. E at anypoint is directed away from the sheet forpositive charge and directed towards the sheetin case of negative charge. (1'12)

Surface cbarge density of the uniform planesheet which is infinitely large =+ cr. Theelectric potential (V) due to infinite sheet ofuniform charge density + o

V= -crr2£0

The amount of work done in bringing a pointcharge q from infinite to point, at distance r infront of the charged plane sheet.

, 1 -CJr o r-q'W=q x V= q '-- =---Joule2Eo 2Eo

or

33.

(2)

(i) Electric field on an axial line of an electric dipole2d

E+q Eq ~.......:.-•.r p_--o._:q - - -r -_-q

! !

Let P be at distance r from the centre of the dipoleon the side of charge - q.Then, the electric field at point P due to charge- q of the dipole is given by

E = - q p-q 41tEo (r + a)2

where, p is the unit vector along the dipole axis(from -qtoq). (1)

Also, the electric field at point P due to charge +qof the dipole is given by

E = q p+ q 41tEo(r - a)2

The total field at point Pis

E =E+q + s.,

= 4~ [(r ~ a)2 - (r: a)2] pE _q_. 4ar p

41tEo (r2 _ a2)2

For x» a,

r = xE

q 4ax A-- P41tEo (x2 _ a2)2

E=~P41tEoX3

E=~41tEox3

y

(given)

(1%)

(ii)

/ /

/ /a

x

ZSince, the electric field has only x component, forfaces normal to X -direction, the angle between Eand tJ.S is ± 2:. Therefore, the flux is separately

2zero for each of the cube except the shaded ones.The magnitude of the electric field at the left face is

E 1= 0 (as, x = Oat the left face).The magnitude of the electric field at the rightface is E R = 3a (as, x = a at the right face). (1)

The corresponding fluxes are<l>L = EL· ~S= 0<l>R = ER . ~S = E RtJ.S cosa = E RtJ.S (": a = 0°)

~ <l>R =ERa2

Net flux (<1» through the cube= <l>L + <l>R

= 0 + ERa2 •

= ERa2

q = 2a (a)2 = 2a3

29

We can use Gauss' law to find the total charge qinside the cube.

=.i..Eo

= <l>E 0 = 2a3Eo (1%)

34. (i) Electric flux over an area in an electric fieldrepresents the total number of electric field linescrossing the area. The SI unit of electricflux is Nm2Cl

•

According to Gauss' law in electrostatics, thesurface integral of electrostatic field Eproduced by any sources over any closedsurface S enclosing a volume V in vacuum, i.e.total electric flux over the closed surface S invacuum, is 1 / Eo times the total charge (q)contained inside S, i.e.

<l>E = fE-dS = .i..s Eo

Gauss' law in electrostatics is true for anclosed surface, no matter what its shape orsize is.So, in order to justify the above stateme1nt,suppose in isolated positive charge q issituated at the centre 0 of a sphere of radius r.According to Coulomb's law, electric fieldintensity at any poipt P on the surface of the

sphere is E = -q-~41tEo r2

r E"n

where, r is unit vector directed from 0 to P.

Consider a small area element dS of the spherearound P. Let it be represented by the vectordS + n·dS.where, n is unit vector along out drawn normalto the area element.:. Electric flux over the area element,

d<l>E=E·dS

= (-q .i}(n.dS)41tEo r2

30

q d5"E·dS = --·l·r·n

41tEo rAs normal to a surface of every point is alongthe radius vector at that point, therefore,r n= 1

q dSE·dS = --'----,

41tEo r:Integrating over the closed surface area of thesphere, we get total normal electric flux overthe entire sphere,

G>E= YE. dS= -q-.,f 15=_'_7 ? X iotat areas. 41tfor- s 41tfor-

of surface of sphere.

= _q_(41tr2)41tEo 1'2

=!LEo

Hence, p RdS = .!L, which proves Gauss'S Eo

theorem. (2%)(ii) Electric field inside a uniformly charged

. spherical shell.According to Gauss' theorem

! E.dS=p E n·d5=J.-js s Eo

Epd5 =!LS Eo

E· 21tr2 =!LEo

E=-q-41tEor2

In the given figure, the point P where we have tofind the electric field intensity is inside the shell.The Gaussian surface is the surface of a sphere 52passing through P and with the centre at O. Theradius of the sphere 52 is r < R . The electric fluxthrough the Gaussian surface, as calculated inEq. (i), i.e. E x 41tr2. As, charge inside aspherical shell is zero, the Gaussian surfaceencloses no charge. The Gauss' theorem gives

E, p

or

=> ... (i)

o Chopterwise CSSE Solved Papers PHYSICS

E x 41tr2 = !L = 0Eo

. . E = 0 for r < R .Hence, the field due to a uniformly chargedspherical shell is zero at all points inside thes~l. ~~

35. (i) Dipole in a Uniform External field:Consider an electric dipole consisting of charge-qand +q and of length 2£lplaced in a uniformelectric field E making an angle 6 with electricfield.Force on charge -q at A = -q E(opposite to E)Electric dipole is under the action of two equaland unlike parallel force, which give rise to atorque on the dipole.

+qN : - - H- - - - - O:a"";-..o-------+ dE,,=,

-dE ---q-:::O

t = Force x Perpendicular distancebetween the forces

t = qE(AN)= qE(2£lsin6)t = q(2£l)Esin6t = pEsinat=PxE (2%

(ii) (a) Charge enclosed by sphere 51 = 2QBy Gauss law, electric flux through sphere

51 is = 26EoCharge enclosed by shere.

52 = 2Q+ 4Q= 6Q=> <1>1= 66/ EoThe ratio of the electric flux is

<1>1/ <1>1= 2QEo/ 6QEo=2/6=1/3

(b) For sphere 51 the electric flux is <1>'= 2Q/ e,'/G>I= 2 a / tr +6G>/Eo

= 6$/ Eo

= Eo.~er 3

.. e, > EO,< G>ITherefore, the electric flux through thesphere 51 decreases with the introductionof the dielectric inside it. (2%)

36. Electric field due to a uniformly charged thinspherical shell.

Gaussian surface

.> ------1 E//// -- "~dS

,,,,,,,,,,,,,, Chargedsphericalshell

--- -~' (1)

(i) When point P lies outside the sphericalshell Suppose that we have to calculateelectric field at the point P at a distancer (r > R) from its centre. Draw the Gaussiansurface through point P. so as to enclose thecharged spherical shell. The Gaussian surfaceis a spherical shell of radius r and centre 0. LetE be the electric field at point P. Then. theelectric flux through area element dS is givenby

d<\>= E·dSSince, dSalso along normal to the surface.

d<\>= E dS:. Total electric flux through the Gaussiansurface is given by

<\> = fsE dS = E fpNow, f dS = 41tr2

.. <\>=Ex41tr2 ... (i)

Since, the charge enclosed by the Gaussiansurface is q. According to Gauss' theorem

cjl=!LEo

From Eqs. (i) and (ii), we obtain

Ex 41tr2 =!LEo

E=_I_.!L41tEo r2

... (ii)

[forr> R 1(1)

31

(ii) When point P lies inside the sphericalshell In such a case, the Gaussian surfaceenclosed no charge.

According to Gauss law,

Ex 47tr2 = °i.e. E=O [forr<RJGraph showing the variation of electricfield as a function of r

E(NC-1)~.-~

E, ----LCR rIm) (2)

37. (i) Electric flux Refer to sol. 28 (2'10)

(ii) Using Gauss' theorem,2 Q I QE x 41tr, = - ~ E = -- -

Eo 41tEo r,2

Field at point P2 = 0, because the electric fieldinside the conductor is zero. (2'10)

(1)38. (i) Refer to Ans. 8 (i).

(ii) Refer to Ans. 28.The field directed

(2)

(a) Normally away from the sheet when sheetis positively charged.

(b) Normally inward towards the sheet whenplane sheet is negatively charged. (3)

39. (i) Refer to Ans. 26 and 27. (3)(ii) Refer to Ans. 22 of Topic 1. (2)

Value Based Questions (From Complete Chapter)04 Marks Questions

1. While travelling back to his residence inthe car, Dr. Pathak was caught up in athunderstorm. It became very dark. Hestopped driving the car and waited forthunderstorm to stop. Suddenly, henoticed a child walking alone on the road.He asked the boy to come inside the cartill the thunderstorm stopped. Dr. Pathakdropped the boy at his residence. The boyinsisted that Dr. Pathak should meet hisparents. The parents expressed theirgratitude to Dr. Pathak for his concern forsafety of the child.Answer the following questions basedon the above information

(i) Why is it safer to sit inside a carduring a thunderstorm?

(ii) Which two values are displayed byDr. Pathak in his action?

(iii) Which values are reflected in parentsresponse to Dr. Pathak?

(iv) Give an example of similar action onyour part in the past from everydaylife. Delhi 2013

Sol. (i) It is safer to be set inside a car duringthunderstorm because the car acts like aFaraday cage. The metal in the car will shieldyou from any external electric fields and thusprevent the lightning from travelling withinthe car. (1)

(ii) Awareness and humanity (1)(iii) Gratitude and kindness felt obliged. (1)(iv) I once came across to a situation where a

puppy was stuck in the middle of a busy roadduring rain and was not able to go cross due toheavy flow, so I quickly rushed and helpedhim. (1)

2. During an endoscopic surgery, a surgeonsees the interior of the patient's body onthe viewing screen of a video monitor. Thesurgeon continues to do the surgery withthe help of other medical staff and one ofthe medical staff on noticing the surgeon'sgloved fingers coming within a fewcentimeters of a screen, pointing to a

particular part of the image, say inexplaining a surgical concern to othermedical staff, the gloves is contaminated.When one of the medical staff asks thesurgeon that whether his gloves wouldhave been contaminated, the surgeonanswers him later, after the completion ofthe operation.Answer the following questions based onthe above information

(i) What is learnt from the above?(ii) Can you find the bacterial source? If

yes, name it.(iii) Name the force which plays a role in

bacterial contamination.Sol. (i) Concentration and involvement in the work by

the doctor, reply by the doctor answering thequery without forgetting, observation of themedical attendant. [2]

(ii) Yes, name of the bacterial source is charge. [1]

(iii) Electrostatic force is the force involved.Bacteria from screen may contaminate theglove. 11]

3. A and B are two students in a class whohave been assigned to organise a Republicday function. They have also beeninstructed to invite personally more than 60members from all the nearby culturalorganisations and VIPs in their area. Whilestudent A arranged invitations using aphotocopier/fax, student B arrangesinvitations by writing to them individual.

Answer the following questions based onthe above information

(i) Which student's method would youadopt and why?

(ii) State the principle behind the sourceused by student A.

(iii) How this principle works?Sol. (i) Student A, because he is aware of the latest

technology and its applications. [1]

(ii) Electrostatic force The force of interactionbetween any two point charges is directlyproportional to the product of the charges andinversely proportional to the square of thedistance between them. [1]

(iii) A photocopier/fax is one of the manyindustrial applications of the forces ofattraction and repulsion between chargedparticles/bodies. Particles of black powder,called toner stick to a tiny carrier bead of themachine due to electrostatic forces betweenthem. The negatively charged toner particlesget attracted from carrier bead to a rotatingdrum, where an image [positively charged] ofdocument being copied has formed. A chargedsheet of paper then attracts the toner particlesfrom the drum to itself. Finally, they producephotocopy. [2]

4. Two persons are standing under a treeand another person near them is inside acar. They were arguing about going out·for a movie or to the beach, when alightning struck the tree with some force.The person inside the car notices hisfriends standing under the tree areaffected by lightning, he comes out andtakes them to the nearby hospital.

(i) Why the person in the car was notaffected by lightning?

(ii) What quality do you find the personinside the car?

(iii) Explain the process that takes placeduring lightning.

(iv) If the total charge enclosed by asurface is zero, what will be theelectric flux over the surface?

Sol. (i) Due to electrostatic shielding, the person inthe car was not affected (electric field insidethe metallic body is zero). [1]

(ii) Helping others; taking a quick decision as towhat is to be done, in case, he is unaware orunable to do on his own presence of mind. [1]

(iii) Lightning is a massive electrostatic dischargebetween electrically charged regions withinthe clouds or between a cloud and the earth'ssurface. [1]

(iv) As $= !L, therefore, electric flux over theEo

surface will be zero because net chargeenclosed by a surface is zero. [1]

5. The surface integral of electrostatic field Eprodi ced by any source over any clc - :-dsurface S enclosing a volume V invacuum, i.e. total electric flux over the

33

closed surface S in vacuum is 1/ Eo timesthe total charge Q contained inside S, i.e.

<l>E = fE. dS = QEo

The charges inside S may be pointcharges or even continuous chargedistributions. There is no contribution tototal electric flux from the charges outsideS. Further, the location of Q inside S doesnot affect the value of surface integral.Read the above passage and answer thefollowing questions

(i) What are the 8I units anddimensions of electric flux?

(ii) A closed surface in vacuum enclosescharges - q, + 3 q and +5 q. Anothercharge + 4q lies outside the surface.What is total electric flux over thesurface?

(iii) A point charge q lies inside aspherical surface of radius r. Howwill the electric flux be affected, ifradius of the sphere is doubled?

(iv) What values of life do you learn fromthis theorem?

Sol. (i) SIunits of electric flux ls'N _m2C-1.

Dimensional formula of electric flux is[M1t3r3A-1]. [1]

(ii) $ = Iqinside = - q + 3q + 5q = 7q [1]

Eo Eo Eo

(iii) As $ = !L, electric flux is not affected byEo

area/shape of the surface. So, the electric fluxremains unaffected. [1]

(iv) This theorem emphasis that total normal fluxfrom the surface depends only on algebraicsum of charges enclosed by the surface,irrespective of their location. The chargesoutside the surface do not affect the electricflux. In day-to-day life, the theorem impliesthat the knowledge you can import dependsonly on what you have stored within you. Youcannot emanate what you have not absorbedor what lies outside your reach?In the examination, you can write what youhave learnt by heart? Extraneous help fromoutside by unfair means will never serve yourpurpose. tn

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6. Geeta has dry hair. A combran through herdry hair attracts small bits ofpaper. Sheobserves that Neeta with oilyhair combsher hair, the combcouldnot attract smallbits ofpaper. She consults her teacher forthis and gets the answer. She then goes tothe junior classes and shows thisphenomenon as a Physics experiment tothem.Read the above passage and answer thefollowing questions.

(i) What according to you are the valuesdisplayed by Geeta?

(ii) Explain the phenomenon involved.Sol. (i) The values displayed by Geeta are curiosity,

leadership and compassion. (2)

(ii) Frictional electricity Frictional electricity isthe electricity produced by rubbing twosuitable bodies and transfer of electrons fromone body to other.Further, if the hair are oily or wet, then thefriction between the hair and comb reducesand the comb will not attract small bits ofpaper. (2)

7. As it is known that all matter is made upof atoms/ molecules. Every atom consistsof a central core, called the atomicnucleus, around which negatively chargedelectrons revolve in circular orbits. Everyatom is electrically neutral containing asmany electrons as the number of protonsin the nucleus. All the materials are

o ehapterwise eSSE Solved Papers: PHYSICS

electrically neutral, they contain charges,but their charges are exactly balanced.Read the above passage and answer thefollowing questions.

(i) Every body whether a conductor oran insulator is electrically neutral. Isit true?

(ii) Charging lies in charge imbalance,i.e. excess charge or deficit charge,comment.

(iii) How do you visualise this principlebeing applied in our daily life?

Sol. (i) Yes, it is true. Every conductor/insulator iselectrically neutral, as it contains equalamounts of positive charge and negativecharge. ~)

(ii) This statement is true. Charging lies in chargeimbalance. When a body loses some electrons,it becomes positively charged because it hasexcess of protons over electrons. The reverse isalso true. (1)

(ill) Nature/God has created the universe. Inoriginal, all bodies are neutral with no forces ofattraction/repulsion. When interests of any twopersons dash (i.e. two bodies are rubbedagainst each other), they become charged.From the charging, the forces ofattraction/repulsion arises, i.e. pulls andpressures of life.Nature/God wants us to live in peace withoutstress and tensions in life. We get charged overpetty things in life and invite all sorts of pulls,pressures and tensions. (2)

electric charges and fields - wordpress.com...[topic1] coulomb's law, electrostatic field and...

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