- CHAPTER 02

Electrostatic PotentialAnd Capacitance

Chapter Analysis w.r.t. Lost 3 Year's Board 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 befocussed 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 Answer (1 mark) lQ lQ lQ lQ

Short Type I Answer (2 marks) lQ

Short Type II Answer (3 marks) lQ lQ 1 Q lQ 1 Q._ v_ ~Q ..J

Long Answer (5 marks) IQ 91

Value Based Questions (4 marks) , ...

• In 2015, in Delhi set, one numerical question of 2 marks and one numerical question of 3marks based on Potential Difference were asked.In All India set, only one numerical question based on Combination of Capacitor wasasked.

• In 2016, in Delhi set, one question of 1mark based on Potential Difference, onequestion of 3 marks based on Parallel Plate Capacitor were asked. In All India set, onequestion of 1marks based on Dipole Moment, one question of 3 marks based onCombination of Capacitor and one numerical question of 5 marks based on Dielectricand Capacitor were asked.

• In 2017, in Delhi set, one question of marks based on Electrostatic Shielding and onequestion of 3 marks based on Electric Dipole were asked. In All India set, one question of1 marks based on Polarity of the Capacitor and one question of 3 marks based onParallel Plate Capacitor were asked.

On the basis of above analysis, it can be said that from exam point of view Potential Difference,Dielectric and Capacitor, Electric Dipole and Parallel Plate Capacitor are most importantconcepts of the chapter.

[TOPIC 1] Electrostatic Potential andElectrostatic Potential Energy

1.1Electrostatic PotentialThe electrostatic potential at any point in anelectric field is equal to the amount of workdone in bringing the unit positive test chargewithout acceleration from infinity to thatpoint.

. . Work done (W)Electrostatic potential, V = -----'---'-Charge (qo)

Its SI unit is volt (V) and 1V = 1 rc' and itsdimensional formula is [ML2T-3 A-I].It is a scalar quantity.

NOTE Electrostatic potential is a state dependentfunction as electrostatic forces are conservativeforces.

Electrostatic PotentialDifferenceThe electrostatic potential difference betweent,"yopoints in an electric field is defined as theamount o[,work done in moving a unitpositive test charge from one point to theother point against electrostatic force withoutany acceleration (i.e. the difference ofelectrostatic potentials of the two points inthe electric field).

V-V - WABB A-

qo

where, WAB is work done in taking charge qofrom A to B against of electrostatic force.Also, the line integral of electric field frominitial position A to final position B along anypath is termed as potential differencebetween two points in an electric field, i.e.

VB - VA = - f: E·dl

NOTE As, work done on a test charge by theelectrostatic field due to any given chargeconfiguration is independent of the path, hencepotential difference is also same for any path.

For the diagram given below, potential differencebetween points A and B will be same for any path.

A~B

4

Electrostatic Potentialdue to a Point ChargeElectrostatic potential due to a point charge q at any point

P lying at a distance r from it is given by V = _1_.141tEo r

The potential at a point due to a positive charge ispositive while due to negative charge, is negative.• When a positive charge is placed in an electric field,

it experiences a force which drives it from points ofhigher potential to the points of lower potential. Onthe other hand, a negative charge experiences a forcedriving it from lower potential to higher.

Electrostatic Potentialdue to Electric Dipole• Electrostatic potential due to an electric dipole at any

point P inclined at an angle 3 whose position vectoris r w.r.t. mid-point of the dipole is given by

V = _1_.p cos e41tEo r2

p

9-q 0 P +q

11+-,--- 2a ----+II

Electrostatic potential at point P due to a short dipole

( j is gt b 1 p.i-a< < r IS given y V = -_. ~41tEo r:

where, e is the angle between rand p.

CHAPTER 2 : Electrostatic Potential And Capacitance 37

Important Results• The electrostatic potential on the perpendicular

bisector due to an electric dipole is zero.• Electrostatic potential at any point P due to a

system of n point charges ql' qz,.··, qn whoseposition vectors are r1, r2, ... , rnrespectively, isgiven by

V=_I_ i .sc:41t£o i;llr-r;!

where, r is the position vector at point P w.r. t.the origin.

• Electrostatic potential due to a thin chargedspherical shell carrying charge q and radius Rrespectively, at any point Plying

(i) inside the shell is V = _I_.fL, for r « R41tEo R

(ii) on the surface of shell is V = _1_. fL,41tEo R

for r= R

(iii) outside the shell is V= _1_.Il, for r » R41tEo r

where, r is the distance of point P from thecentre of the shell.

• Graphical representation of variation of electricpotential due to a charged shell at a distance rfrom centre of shell is given below:

V1 q

V=-·-4m:o r

O~--.~:R---r--.-----

Variation of potential due to charged shellwith distance r from its centre

1.2 Equipotential SurfaceA surface which has same electrostatic potentialat every point on it, is known as equipotentialsurface.

For a single charge q, the potential is given by

v=_I_.1l41tEo r

The shape of equipotential surface due to(i) line charge is cylindrical

(ii) Point charge is spherical

Fieldlines

, - kquiPotentialsurfaces

(spherical in shape)

Different properties of equipotential surface are givenbelow:

(i) Equipotential surfaces do not intersect eachother as it gives two directions of electricfield at intersecting point which is notpossible.

(ii) Equipotential surfaces are closely spaced inthe region of strong electric field and vice-versa.

(iii) Electric field is always normal toequipotential surface at every point of it anddirected from one equipotential surface 'ii.tfH,higher potential to the equipotential surfaceat lower potential.

(iv) Work done in moving a test charge from onepoint of equipotential surface to other is zero.

Relation Between Electric Fieldand Potential GradientRelation between electric field and potentialgradient is given by

dV . oV oV oVB=-- i.e. E =-- E =-- E =--

dr x ox' y oy' z oz

where, negative sign indicates that the direction ofelectric field is from higher potential to lowerpotential, i.e. in the direction of decreasing potential.NOTE (i) Electric field is in the direction of which the

potential decreases steepest.

(ii) Its magnitude is given by the change in themagnitude of potential per unit displacementnormal to the equipotential surface at the point.

38 o ehopterwise eSSE Solved Papers PHYSICS

Potential Energy inan External Field

(i) Potential Energy of a Single Charge Potentialenergy of a single charge q at a point withposition vector r, in an external field is qV(r),

where V(r) is the potential at that point due toexternal electric field E.

(ii) Potential Energy of a System of TwoCharges

U = q1V(rj) + q2V(r2) + qlq241tEOr12

where, ql ,q2 = two point charges at positionvectors r) and r2, respectivelyV (r)) = potential at r) due to the external fieldV(r2) = potential at r2 due to the external field

1.3 Electrostatic Shielding• The process which involves the making of a region

free from any electric field is known as electrostaticshielding. Conductor

It happens due to the fact that no electric field existinside a charged hollow conductor. Potential insidea shell is constant. In this way, we can alsoconclude that the field inside the shell (hollowconductor) will be zero.

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 1o 1 Mark Questions

1. Does the charge given to a metallic spheredepend on whether it is hollow or solid? Givereason for your answer. De!l!L2017

2. A point charge + Q is placed at point 0as shown in the figure. Is the potentialdifference (VA - VB) positive, negativeor zero?

+0·-------------- •..-----·o A BDeihl 2016, Foreign 2016, Deihl 2011

3. A charge q is moved from a point Aabove a dipole of dipole moment p to apoint Bbelow the dipole in equatorialplane without acceleration. Find thework done in this process.

A,,,,,,,- q .--------~--- ----4 +q,,,,,,,

B All Indio 2016

4. Why are electric field linesperpendicular at a point on anequipotential surface of a conductor?All India 2015C

5. Two point charges q and -2q are keptd distance apart. Find the location ofpoint relative to charge q at whichpotential due to this system of chargesis zero. All India 2014C

6. The figure shows the field lines of apositive charge. Is the work done bythe field is moving a small positive

. charge from Q to P positive ornegative?

Foreign 2014

7. For any charge configuration,equipotential surface through a pointis a normal to the electric field.Justify.Deihl 2014

8. What is the geometrical shape ofequipotential surfaces due to a singleisolated charge? Deihl 2013, All India 2010 C

CHAPTER 2 : Electrostatic Potential And Capacitance 39

9. What is the amount of work done inmoving a point charge around a circulararc of radius r at the centre of whichanother point charge is located?All India 2013 C

10. Two charges 21lC and -21lC are placed atpoints A and B, 5 em apart. Depict anequipotential surface of the system.Delhi 2013C

11. Why electrostatic potential is constantthroughout the volume of the conductorand has the same value as on its surface?Delhi 2012

12. Why the potential inside a hollowspherical charged conductor is constantand has the same value as on its surface?Foreign 2012

13. Why there is no work done in moving acharge from one point to another on anequipotential surface? Foreign 2012

14. A hollow metal sphere ofradius 5 em ischarged such that potential on its surfaceis lOV. What is the potential at the centreof the sphere? All India 2011

15. Can two equipotential surface intersecteach other? Justify your answer. Delhl2011C

16. Draw equipotential surfaces due to asingle point charge. All India 2011C

17. Name the physical quantity whose 81 unitis JC-1. Is it a scalar or a vector quantity?All India 2010

18. What is the work done in moving a testcharge q through a distance of 1 em alongthe equatorial axis of an electric dipole?All India 2009

19. Define the term potential energy forcharge q at a distance r in an externalfield.All Indio 2009

20. The potential due to a dipole at any pointon its axial line is zero. Correct or Wrong?All India 2009C

21. What is the electric potential due to anelectric dipole at an equatorial point?All India 2009

o 2 Marks Questions22. Two point charges ql and q2 are located at

rl and r2' respectively in an externalelectric field E. Obtain the expression forthe total work done in assembling thisconfiguration. Delhi 2014C

23. Two closely spaced equipotential surfacesA and Bwith potentials V and V +OV;(where OV is the change in V) are kept oldistance apart as shown in the figure.Deduce the relation between the electricfield and the potential gradient betweenthem. Write the two importantconclusions concerning the relationbetween the electric field and electricpotential. Delhi 2014C

B V+ sv-/--,V

24. Calculate the amount of work done todissociate a system of three charges, twoof lllC and one of -41lC placed on thevertices of an equilateral triangle of side10 cm. All India 2013C

25. A test charge q is moved withoutacceleration from A to C along the pathfrom A to B and then from B to C inelectric field E as shown in the figure.

(i) Calculate the potential differencebetween A and C.

(ii) At which point (of the two) is theelectric potential more and why?

A ·,:.-----~-~D]--r;B" ..,.••... :

5 crri> :3 em" I..... :

''''C E

All India 2012; Foreign 2009

26. Draw a plot showing the variation of(i) electric field (E) and (ii) electricpotential (V) with distance r due to apoint charge Q. Delhi 2012

40

27. Two uniformly large parallel thin plateshaving charge densities + cr and - cr arekept in the X Z- plane at a distance dapart. Sketch an equipotential surfacedue to electric field between the plates. Ifa particle of mass m and charge - qremains stationary between the plates.What is the magnitude and direction ofthis field? Delhi 2011

28. A dipole with its charge -q and +qlocated at the points (O,-b, 0) and(O,+b,O)is present in a uniform electricfield E. The equipotential surfaces of thisfield are planes parallel to the YZ-plane.

(i) What is the direction of the electricfield E?

(ii) How much torque would the dipoleexperience in this field? Delhi 2010C

29. Two point charges 31lC and -31lC areplaced at points A and B, 5 em apart.

(i) Draw the equipotential surfaces ofthe system. All India 2011C

(ii) Why do equipotential surfaces getclose to each other near the pointcharge?

30. Find out the expression for the potentialenergy of a system of three charges ql ,q2and q3 located at r1 ' r2 and r3 withrespect to the common origin O. Delhi 2010C

31. Draw three equipotential surfacescorresponding to a field that uniformlyincreases in magnitude but remainsconstant along Z-direction. How arethese surfaces different from that of aconstant electric field along Z-direction?All India 2009

32. (i) Can two equipotential surfacesintersect each other? Give reasons.

(ii) Two charges - q and + q are locatedat points A (0,0, - a) and B(O, 0, +a)respectively. How much work isdone in moving a test charge frompoint P (7,0,0) to Q (- 3,0, O)?Delhi 2009

o Chopterwise CBSE Solved Papers PHYSICS

33. (i) Write two characteristics ofequipotential surfaces.

(ii) Draw the equipotential surfaces dueto an electric dipole. All India 2009C

o 3 Marks Questions34. (i) Derive the expression for the electric

potential due to an electric dipole at apoint on its axial line.

(ii) Depict the equipotential surfaces dueto an electric dipole. Delhi 2017

35. Define an equipotential surface. Drawequipotential surfaces

(i) in case of a single point charge(ii) in a constant electric field in

Z-direction. Why the equipotentialsurfaces about a single charge are notequidistant?

(iii) Can electric field exist tangential toan equipotential surface? Give reason.All India 2016

36. (i) Depict the equipotential surfaces for asystem of two identical positive pointcharges placed a distance d apart.

(ii) Deduce the expression for the potentialenergy of a system of two point chargesql and q 2 brought from infinity to thepoints with positions r1 and r2respectively, in presence of externalelectric field E. Delhi 2Dl0

o Explanations1. In case of metallic sphere, charge given to it is

mostly resides on its surface. Therefore, there is nodifference whether the sphere is hollow or solid. Asin both the cases, the charge that will reside will besame. (1)

2. According to question,+0. ._uu ..

o A B---(1-• (2----+

r2> r1 (given)

Potential at point A due to charge + Q,(VA) = kQr1

Potential at point B due to charge + Q, (VB) = kQr2

(1I2J

CHAPTER 2 Electrostatic Potential And Capacitance 41

1 1As VA OC - and VB OC - and '2>'] so, VA> VB'] '2

Thus, (VA- VB) is positive. (1/2)

3. According to questions,A, ,, ,, ,, ,, ,

r .: ',r" Y ", ,, ,, ,, ,

-<:;' x x '+q

BTotal potential at point A due to + q charge,

VA = kq = kqr !.

(x2 + i)2

Total potential at point A due to - q charge,V~ = - kq = - kq

, !.(x2 + i)2

So, net potential = VA + V~ = 0

Similarly at point B, potential will be O.So, net work done = l1V x q = O. (1)

4. Electric field is always normal to theequipotential surface at every point, because nowork is done, as W = qO(VA - VB) => VA - VB = 0hence, W = 0.

If the field were not normal to the equipotentialsurface, it would have a non-zero componentalong the surface. So, to move a test chargeagainst this component, a work would have to bedone. But there is no potential difference betweenany two points on an equipotential surface andconsequently no work is required to move a testcharge on the surface. Hence, the electric fieldmust be normal to the equipotential surface atevery point. (1)

5. According to the question, q A = q and qB = -2qA P B. . .

qA qBI~ d .1

I+-X~f+-a-x __

kqA kqBVPA =-- => VPB =--x (d-x)

As, the potential due to system of charges is zero,kq 2kqhence VpA + VPB = 0 => - = --x (d-x)

d-x=2x=> 3x=d=> x=~3

6. Work done by charge is given byW = q (potential at Q - potential at Pl.where, q = small positive charge (1/2)

The electric potential at a point distant, due tothe field created by a positive charge Q is given by

v=_l_q_41tEo r

'p <'Q => Vp > VQ (1/2)

So, work done will be negative.7. No work is done in moving the test charge from

one point of an equipotential surface to the other.(1/2)

.. WB-WA=o=-fE.dl => E·dl=O

Hence, e i dl (1/2)

8. The equipotential surfaces produced by a singlepoint charge are spherically symmetrical chargewhich are concentric. In concentric spheres whichare as shown below in the figure, the lines offorce point radially outwards, so they areperpendicular to the equipotential surfaces at allpoints.

Equipotentiol surfoces for 0 point charge

9. The amount of work done in carrying a charge onequipotential surface is always zero (potentialdifference is zero).

10. Given, qA=2IlC=2xlO-6CqB =-2IlC=-2xlO-6Cand ,=5cm

A +-x---;(5-X)-BI--- 5em-----.j

.. Potential.

V2XlO-6 -2XIO-6

= +-------:;-41tEoXxlO-2 41tEO(5-x) X 10-2

(1)

42

2xlO-6 2xlO-641t£oxX lO-z 411:EO(5-x) xlO-z

x=5-x ~x=2.5cm

[':V = 0]

Since, electric field intensity inside the conductoris zero. So, electrostatic potential is a constant.

B E = _ L\Vut,L\r

E = 0, L\V = 0 or Vz - VI = 0, V2 = VI

The potential at every point inside the conductorremains same. (1l

Electric field inside the hollow spherical chargedconductor is zero. So, no work is done in movinga charge inside the shell. Thus, potential isconstant and therefore, equal to its value at thesurface, i.e. V = _1_. !L

411:~ R ruAn equipotential surface is a surface at whichevery point electric potential is same.As, work done in moving a charged particle fromone point to another is defined as

L\W = q(L\V)

On an equipotential surface, the potentialremains constant. So, L\V = zero~ Work done, L\W = 0 (1l

14. Refer to Ans. 2 [Ans.lOV] rn15. (i) No, two equipotential surfaces cannot

intersect each other because two normals canbe drawn at intersecting point on two surfaceswhich give two directions of Eat the same pointwhich is impossible. (112l

(ii) Also, two values of potential at the same pointis not possible. (112l

16. Equipotential surfaces due to a single pointcharge are concentric sphere having charge at thecentre.

11.

12.

13.

,,> \7 Equipotential,\x surfaces

6--1.-+--+-'--_ Line ofEforces

- -_ ...

17. rc' is the 51 unit of electric potential. It is ascalar quantity. (1I2+1I2l

o ehapterwise eBSE Solved Papers PHYSICS

m

18. The electric potential at any point on equatorialline due to an electric dipole is equal to zero.No work is done in moving a test charge alongequatorial line, as given below:

W = qL\V= q(VB - VA)

VB = VA = 0 ~ W = 0 m19. The electric potential energy at any point lying at

a distance r from the source charge q is equal tothe amount of work done in moving unit positivetest charge from infinity to that point withoutany acceleration against electrostatic force. (1l

20. Wrong, as potential due to an electric dipole iszero on equatorial line in spite of axial line.The potential due to a dipole at any point onequatorial line is zero, not on axial line. (1l

21. Zero, as potential on equatorial point due tocharges of electric dipole, is equal in magnitudebut opposite in nature and hence, their resultanthuro. ru

22. Work done in bringing the charge ql from infinityto position rl

WI = ql V(rd ... (i) (1/2)

Work done in bringing charge q2 to the position rzWz = qzV(rz) +.3J.!J..L ... (ii) (1I2l

411:Eor12

Hence, total work done in assembling the twocharges.

(1)

W=WI + WzFrom Eqs. (i) and (ii),we get

W =ql V(rd+ qzV(rz)+~411:~r12 (1l

23. Work done in moving a unit positive charge alongdistance 8Z,

IEd8Z=VA-VB =V-(V+8V) =-OV E=-8V8t (1)

(i) Electric field is in the direction in which thepotential decreases steepest. (1/2)

(ii) Magnitude of electric field is given by thechange in the magnitude of potential per unitdisplacement normal to the equipotentialsurface at the point. (1/2)

24. Given, charges ql = lfJ.C qz = -1fJ.C and q3 = -4fJ.C

Each side of equilateral triangle, r = 10 emPotential energy, U = Total work done to assemblethe three charges U = WI + Wz + W3

U = _1_[1 x 10-6(-4 X 10-6) +

411:Eo 010

CHAPTER 2 : Electrostatic Potential And Capacitance 43

1X10-6(l X10-6)+ -4 X10-6(1X10-6)]010 010

1 ~C

10Cm/\10Cm

+1~C~-4~C10cm

U = _1_ X10-12[-4xlO+ 1O-4xlOj41t£o

U = -9 X109 X10-12 x 70 ~ U = -0. 630J

Work done to dissociate the system of chargesW = -v ':;.630J (1)

25. (I}',: blectric field intensity and potential differenceare related as

E = _L\VM

L\V = -EM ~ Vc- VA = -4E (1)

[.: By pythagoras theorem.ac? = AB2 + BC2]

~ AB2 = 52 - 32 ~ L\r=M ~L\r= 4

(ii) As Vc - VA = 4E is negative.. Vc< VAPotential is greater at point A than point C, aspotential derreases along the direction ofelectric field. (1)

26. Electric fit..J due to a point charge, .1 Q 1

E=---,Eo<,41tEo r2 r:

A graph showing variation of electric field (E)and electric potential (V) with distance (r) areshown below. (1)

VE5.0 r-r....,.---,-..,--.,.-.--.-,---,--.--,4.54.0

:::. 3.5GJ =i 3.0:2 ~ 2.5Q)-~ ci: 2.0n 1.5

&l 1.00.5 ------- _

O~~~~~~~~~~0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0Distance (r) _

Variation of electrostatic potential V and electricfield E with distance r

V=_1_ s.4~ r 1---- 1/r 1

E=_1_ q --1/r24~ (i

(1)

Potential due to a point charge,

V=_l_g ~Vo<~41tEo r r

The variation of electrostatic potential withdistance, i.e. V 0< ~ and also the variation of

r

electrostatic field with distance, i.e. E 0< ~ •

r

Due to a single charge, F 0< ~ and E 0< ~ butr r

V 0< ~,where, r is the distance from the charge.r m

27.

+0 / + + + + + + + II J-Y XEquipotentialsurface v-----;.v-~-O------/'d

{- I 1 Z

d/2r-- ---.-01/+ + + + + + + I

The equipotential surface is at a distance d/2 fromeither plate in XZ-plane. -q charge experiences aforce in a direction opposite to the direction ofelectric field.., - q charge balances when

qE=mg

E= mgq

The direction of electric field along verticallydownward direction. The XZ-plane is so chosenthat the direction of electric field due to two platesis along vertically downward direction, otherwiseweight (mg) of charge particle could not bebalanced. (1)

Y

z(1)

Equipotentialsurface --

----X

E (Verticallydownward)

28. The direction of electric field is perpendicular tothe equipotential surface.(i) The direction of electric field is along X-axis as

it should be perpendicular to equipotentialsurface lying in yz- plane.Length of the dipole = 'lb

44 o ehapterwise eSSE Solved Papers PHYSICS

As dipole's axis is along the Y-axis.:. Electric dipol~ moment,

~q(~j ru(ii) Electric field, E = E i

t=pxE=q(2b)JXElA A A

= + 2qbE(j x i) =2qbE(-k)

Torque, It I = 2qbE (1)

Alternative methodE is directed along X-axis.

Dipole moment p = q(2b) from (0, =b, 0) to (O,b,O),i.e. along Y-axis. (1)

.: Angle between p and E is 90°:. Torque on dipole, = t max =pE sin 90° = q(2b)E x l. . Torque, t = 2qbE (1)

29. (i)

(1)

(ii) Equipotential surfaces get closer to each othernear the point charges as strong electric field isproduced there.

E = - ~v and E cc _ J...-~r ~r

For a given equipotential surface, small Mrepresents strong electric field and vice-versa. (1)

30. Let three point charges ys.. q2 and q3 haveposition vectors r I' r2 and r23

r3, respectively.

Potential energy of thecharges qi and q2'U

l2= _1_. qi q2 X

41tEo IrI21=_I_ ...3J...iL

41tEo Ir2 - rl ISimilarly,

U23

=_I_.3..1..3L =} U3I

=_1_. qi q3

41tEo Ir3 - r21 41tEo Ir3 - ril(1)

:. Net potential energy of the system,U = UI2 + U23 + U3I

U = _1_ [3J..!!.L + .sas: + .s.ss: lJ41tEo Ir2 - rl I [r 3 - r21 Ir 3 - rl I (1)

31. Electric field is always directed from higher tolower potential and perpendicular to theequipotential surfaces.The figure is shown as below:

Z

(1)

32.

In case of constant electric field along Z-direction,the perpendicular distance between equipotentialsurfaces remains same, whereas for field ofincreasing magnitude, equipotential surfaces getcloser as we go away from the origin.In both cases, surfaces be planes parallel toXY - plane. (1)

(i) Refer to Ans. 15. (1)

(ii) .: Every point on X-axis is on equatorial line ofelectric dipole (system of two unlike charges).:. Potential on it is equal to zero.:.No work is done in moving a test charge.

W = q ~v = q x 0= 0. [.: ~v = 0]Z

B, +q (0, 0, a)r,

" Q (-3, 0, 0)(7,0,0) /0J------y

, I

/ II I

P -; :~'t-q (0, 0, -a)X

33.

.. Work done in moving a charge on anequipotential surface is zero. (1)

(i) (a) Two equipotential surfaces do not intersecteach other as normals at intersecting pointson two surfaces will give two directions ofelectric field which is impossible. (1/2)

(b) Closely spaced equipotential surfacesrepresent strong electric field andvice- versa. (1/2)

(ii) Refer to Ans. 29 (i). (1)

CHAPTER 2 : Electrostatic Potential And Capacitance 45

34. (i) Let electric potential to be obtained at point Plying on the axis of dipole at distance d fromthe centre of the dipole.>+- 21----+<

~--------------------4

--q +q P

d .1

Potential at P due to +q charge=_I_._q_41tEo (d - ~

Potential at P due to - q charge1 -q-----

41tEo (d + ~Total potential at P due to dipole=_q_[_I_-_I_] q x 21

41tEo (d - ~ (d + ~ 41tEo (d2 - /2)

P41tEo (d2 - /2)

where, p = q . 11if!« d,

1 Pthen, V = --2'41tEo d (1'12)

(ii) Equipotential surfaces of a dipole are asshown below. Potential of points lies onperpendicular bisector surface will be zero.

~ @Y;~:C~tential

~- - -- -~. .

Perpendicularbisector of dipole (1'12)

Equipotential surfacehaving potential equalto zero at all points

35. Any surface that has same electric potential atevery point on it is called equipotential surface.(i) Equipotential surface in case of single point

charge (1)

E--t-++-t Equipotentialsurfaces

(ii) Equipotential surfaces when the electric fieldis in Z-direction.

yEpuipotential

SUrfr.,s

J--+-+--t--lr--t-+Z

ExThe equipotential surfaces due to a single pointcharge is represented by concentric sphericalshells of increasing radius, so they are notequidistant. (1)

(ill) No, the electric field does not exist tangentiallyto an equipotential surface because no workdone in moving a charge from one point to otheron equipotential surface. This indicates that thecomponent of electric field along theequipotential surface is zero. Hence, theequipotential surface is perpendicular to fieldlines. (1)

36. (i) The figure is shown as below:Equipotentialsurfaces

Equipotentiol surfoces of two identical positive charges(1)

(ii) By definition, electric potential energy of anycharge q placed in the region of electric field isequal to the work done in bringing charge q frominfinity to that point and given by

U = qV (112)

where, V is the electric potential (as potential atinfinity is assumed to be zero) where, the chargeq is placed. Now, considering the electricpotentials at positions rl and r2 as ~ and V2,

respectively. Therefore, total potential energy ofthe system of two charges ql and q2 placed atpoints with position vectors rl and r2 in theregion of E is given by U = work done in bringingcharge q from infinity to that position in E isequal to work done for charge q2 from infinity tothat position in E + work done to that of charge q2

at these positions in presence of ql. (1)

i.e. U = ql ~ + q2V2 ~U = _1_ . .3J.iL41tEo Ir2 - rl I (1/2)

[TOPIC 2] Capacitance2.1 Conductors and

InsulatorsConductor contains a large number of free chargecarriers to conduct electricity while insulator doesnot contain any free charge carriers to conductelectrici ty.Examples of conductors are metals and graphite.Examples of insulators are plastic rod and nylon.NOTE Insidea conductor,the electrostaticfieldiszero.

Free Charges and Bound ChargesInside a Conductor

(i) In a metal, the outer (valence) electrons arefree to move. These electrons are free formoving within the metal but not free to leavethe metal. These free electrons are freecharges inside a conductor and are the causeof conducting the electricity by conductors.

(ii) The bound charges are those positive ionswhich are made up of nuclei and the boundelectrons remain in their fixed positions.

NOTE (I) Insidea conductor, the electric field is zero.'(il) Theinteriorof a conductor can have no excess

charge in static situation.(iil) Electricfieldjust outside a charged conductor

is perpendicular to the surface of theconductor at every point.

2.2 Dielectrics andPolarisation

(i) When a conductor is placed in an externalelectric field, the free charge carriers adjustitself in such a way that the electric fielddue to induced charges and external fieldcancel each other and the net field insidethe conductor is zero.

In case of dielectric however, the opposingfield so induced does not exactly cancel theexternal field.

- -.:- ~ ++ - - ---- ~ ++-~ + Eo + Eo

_ l:in + _ E,n +--Eo+E,n=O++ - - --Eo+ Ein .=0++

Conductor DielectricBehovioursof a conductorand a dielectric

in on externalelectricfield

(ii) A net dipole moment is developed by anexternal field in either case, whether polaror non-polar dielectric. The dipole momentper unit volume is called polarisation andit is denoted by p.

p=X,E

where, X, is called electric susceptibilityof the dielectric medium.

2.3 CapacitorA capacitoris a device which is used to storeelectrostatic potential energy or charge. Itcomprises of two conductors separated by aninsulating medium.

Capacitance of a condurtorIf charge q is given to an insulated conductor, itleads to increase its electric potential by V suchthat q ex V ~ q = CV

where, C is known as capacitance of a conductor.The capacitance depends on the shape, size andgeometry of conductor, nature of surroundingmedium and presence of other conductor in theneighbourhood of it.

. . 1 coulombIts SI urnt IS farad (F). Here, 1 farad = ----I volt

Farad is a very large unit of capacitance. So, I!F isusually taken.

CHAPTER 2 : Electrostatic Potential And Capacitance 47

Parallel Plate CapacitorThe most common among all capacitors is parallelplate capacitor. It comprises of two metal plates ofarea A and separated by distance d filled with airor some other dielectric medium. The capacitanceof air filled parallel plate capacitor is given by

C - £oA0- d

Surface x / Area A

Charge~ 1 . Idensity .~~ ++++++++++

E

j j j j j j j d

Surface Jc-----;----Icharge y Idensity -= Earthed

(- 0)

When a dielectric of dielectric constant K is filledfully between the plates, then

C = 1(.4£0 = KCod

2.4 DielectricWhen a dielectric slab is introduced between theplates of charged capacitor or in the region ofelectric field, an electric field E p induces inside thedielectric due to induced charge on dielectric in adirection opposite to the direction of appliedexternal electric field. Hence, net electric fieldinside the dielectric gets reduced to (E 0 - E p)'where, Eo is external electric field. The ratio ofapplied external electric field and reduced electricfield is known as dielectric constant K of

dielectric medium, i.e. K = EoEo - Ep

Dielectric ConstantIf C vacuum be capacity of a condenser with vacuumor air between its plates and Cdielectric be thecapacity with dielectric between the plates, the

dielectric constant K is defined as K Cdielectric.Cvacuum

Dielectric constant is also known as specificinductive capacity of the dielectric.

Dielectric StrengthThe dielectric strength is equal to that maximumvalue of electric field that can exist in a dielectricwithout causing the breakdown of its insulatingproperty.

Dielectric and CapacitorWith the advent of dielectric in capacitor, netelectric field inside the dielectric gets reduced,consequently potential difference across(charges disconnected) capacitor decreases.Hence. capacitance of capacitor increases as

C oc ~ and new capacitance becomes KC o-V

Some Important Points• The capacitance of a parallel plate capacitor

partially filled with a dielectric medium ofdielectric constant K is given by

I--t-lI--d--l

C = £oA(d - t + t/K)

where, A = area of each plated = separation between two platest = thickness of dielectric medium

K = dielectric constant of dielectric medium• Capacitance of spherical capacitor

C=41t£Ok(~) b-a• Capacitance of an isolated spherical conductor

of radius r is given byC=41t£or

48

(i) Capacitance of cylindrical capacitorL

C

-A21tEoKLlog 0 (bI a)

where,a = radius of inner coaxial cylinderb = radius of outer coaxial cylinderL = length of coaxial cylinder

9iiO Introduction of dielectric medium into the(charges disconnected) parallel platecapacitor leads to change in physicalquantities as listed below.

Before introduction After introductionof dielectric slab of dielectric slab

Charge (q) No change (charge - q )

Electric field (E) Decreases ( E' = ~)

Potential difference (V) Decreases ( V' = f)Capacitance (G) Increases (C' = KG)

Electrostatic energy (U) Decreases ( U' = ¥)(ill) Introduction of dielectric medium in a

charged capacitor connected with abattery.

Beforeintroduction ofdielectric slab

Increases, Charge q' = qK

No change, Electric fieldE'=E

Increases, Electrostaticenergy U' = KU

After introduction ofdielectric slab

Potential difference(V)

No change, Potentialdifference (V)

Increases, Capacitance(C'=KG)

Capacitance (C)

Charge (q)

Electric field (E)

Electrostaticpotential energy (U)

o ehopterwise eBSE Solved Papers PHYSICS

Conductor and CapacitorWhen a metallic conducting slab is partially filled ina capacitor, then capacitance of conductor becomesC = EoA

d - twhere, t = thickness of metallic plate andd = separation between two plates.

NOTE The dielectric constant of a metallic conductor is infinity.

2.5 Combination of CapacitorsTwo or more capacitors can be connected by twodifferent ways:

Series Combination of CapacitorsThe charge on each capacitor is same for any valueof capacitance and equal to the charge across thecombination.

'------flll-----....JV

The potential difference across the combination isequal to the algebraic sum of potential differenceacross each capacitor, i.e. V = VI +V2 +V3

The potential is divided across capacitors in inverseratio of their capacitances, i.e.

I I 1VJ :V2 :V3 =-:-:-CI C2 C3

The equivalent capacitance is given by1 1 1 I-=-+-+-C CI C2 C3

The equivalent capacitance of n identical capacitorsconnected in series each of capacitance C is given by

CC, =-

n

Parallel Combination of CapacitorsThe potential difference across each capacitor issame for any value of capacitance and equal to thepotential difference across the combination.

q, +_ C,+-v

C

+-v

vThe total charge on combination is equal to thealgebraic sum of charges on each capacitor, i.e.q = q, + q2 + q3The equivalent capacitance (C) is given byC =C, +C2 +C3The total charge on the capacitors is divided in theratio of their capacitances,i.e. q ex C ~ q,: q2 :q3 = C, : C 2 : C 3

The equivalent capacitance of n identical capacitorsconnected in parallel combination is Cp = nC.

2.6 Energy Stored in a CapacitorElectrostatic energy stored in a (parallel plate)capacitor is given by

1 2 l IU=-CV =-=-qV2 2C 2

where, q = charge on capacitor,C = capacitance,V = potential difference across capacitor.

The energy stored per unit volume in an electricfield E is known as energy density. It is given by

I 2UE = -£oE

2

Common PotentialWhen two capacitors of different potentials areconnected by a conducting wire, then charge flowsfrom capacitor at high potential to the capacitor atlow potential. This flow of charge continues tilltheir potentials become equal, this equal potentialis called common potential.

. C v: +C VCommon potential, V = 'I 2 2. C, +C2

PREVIOUS YEARS'EXAMINATION QUESTIONSTOPIC 2o 1 Mark Questions

1. Predict the polarity of the capacitor inthe situation described below:

I_s _N-.Jlm - m -:------j_S __ N 1

c

All India 2017

2. Why does current in steady state notflow in a capacitor connected across abattery? However, momentary currentdoes flow during charging or dischargingof the capacitor. Explain. All India 2017

3. The given graph shows the variation ofcharge q versus potential difference V fortwo capacitors ~and C2. Both thecapacitors have same plate separationbut plate area of C2 is greater than that~. Which line (A or B) corresponds to C1and why? All India 2014C

qB

A

v4. Distinguish between a dielectric and a

conductor. Delhi 2012

5. Define the dielectric constant of amedium. What is its unit? Deihl 2011C

6. A metal plate is introduced between theplates of a charged parallel platecapacitor. What is its effect on thecapacitance of the capacitor? Farelgn 2009

50

o 2 Marks Questions7. Calculate the potential difference and the

energy stored in the capacitor C2 in thecircuit shown in the figure. Givenpotential at A is 90 V, C; = 20IlF,C2 = 30IlF and c; = 151lF.

A --I r------1 ~ ~ B

C1 C2 C3 -+All Indio 2015

8. A parallel plate capacitor of capacitance Cis charged to a potential V. It is thenconnected to another uncharged capacitorhaving the same capacitance. Find out theratio of the energy stored in the combinedsystem to that stored initially in the singlecapacitor. All Indio 2014

9. Two parallel plate capacitors ofcapacitances C; and C2 such that C1 = 2C2are connected across a battery of V volt asshown in the figure. Initially, the key (k) iskept closed to fully charge the capacitors.The key is now thrown open and adielectric slab of dielectric constant K isinserted in the two capacitors tocompletely fill the gap between the plates.Find the ratio of

(i) the net capacitance and(ii) the energies stored in the combination

before and after the introduction of thedielectric slab. Delhi 2014C

k

~.) 1: 1V T IC

1 TC2

10. Two parallel plate capacitors ofcapacitances C1and C2 such that C1 = C2/ 2are connected across a battery of V volts asshown in the figure. Initially, the key (k) iskept closed to fully charge the capacitors.The key is now thrown open and adielectric slab of dielectric constant K isinserted in the two capacitors to completely

o Chapterwise CBSE Solved Papers PHYSICS

fill the gap between the plates. Find theratio of

(i) the net capacitance and(ii) the energies stored in the combination

before and after the introduction of thedielectric slab. Delhi 2014C

k

~.) 1: 1V T IC

1 TC2

11. Find the charge on the capacitor asshown in the circuit. Foreign 2014

6IJ.FIIII

10n 20n

II

2V

12. A slab of material of dielectric constant Khas the same area as that of the plates ofa parallel plate capacitor, but has thethickness d/2, where d is the separationbetween the plates. Find out theexpression for its capacitance when theslab is inserted between the plates of thecapacitor. Delhi 2013

13. Determine thepotential differenceacross the plates ofthe capacitor C1ofthe network shownin the figure.(assume, E2 > ~)All Indio 2013

14. A parallel plate capacitor, each of platearea A and separation d between the twoplates, is charged with charges +Q and-Q on the two plates. Deduce theexpression for the energy stored incapacitor. Foreign 2013

CHAPTER 2 : Electrostatic Potential And Capacitance

15. A network of fourcapacitors, each ofcapacitance 15~F, isconnected across abattery of 100 V, asshown in the figure. Findthe (i) net capacitanceand (ii) the charge on thecapacitor C4• All India 2012C

16. Deduce the expression for the electrostaticenergy stored in a capacitor of capacitanceC and having charge Q.How will the

(i) energy stored and(ii) the electric field inside the capacitor

be affected when it is completelyfilled with a dielectric material ofdielectric constant K? All India 2012

17. 1 ~F capacitance connected to a battery of6 V. Initially switch S is closed. Aftersometime S is left open and dielectricslabs of dielectric constant K = 3 areinserted to fill completely the spacebetween the plates of the two capacitors.How will the (i) charge and (ii) potentialdifference between the plates of thecapacitors be affected after the slabs areinserted? Delhi 2011

~~6 V T C1 T 1 IlF T 1 IlF

18. Net capacitance of three identicalcapacitors in series is 11lF. What will betheir net capacitance, if connected inparallel?Find the ratio of energy stored in thesetwo configurations, if they are bothconnected to the same source. All India 2011

19. Figure shows two identical capacitors C1

and C2, each of 2 ~F capacitance,connected to a battery of 5 V. Initially,switch S is closed. After sometime, S isleft open and dielectric slabs of dielectricconstant K = 5 are inserted to fill

51

completely the space between the plates ofthe two capacitors. How will the (i) chargeand (ii) potential difference between theplates of the capacitors be affected afterthe slabs are inserted?

1 I -{~rV C'I_2

_

IlF C

_'r·F

All India 2011; Delhi 2011

20. What is the area of the plates of 2Fparallel plate capacitor having separationbetween the plates is 0.5 cm?Alllndia 2011

21. Two identicalparallel plate (air) Tcapacitors CI andC2 havecapacitance Ceach. The space 1between theirplates is now filled I-- d ----lwith dielectrics asshown in the figure. If the two capacitorsstill have equal capacitance, then obtainthe relation between dielectric constantsK, KI and K2. Foreign 2011

22. You are given anair filled parallelplate capacitor CI.

The spacebetween its platesis now filled withslabs of dielectricconstants KI andK 2 as shown in I-- d ----lfigure. Find the capacitance of thecapacitor C2 if area of the plates is A anddistance between the plates is d. Foreign 2011

23. Figure shows a sheet ofaluminium foil of ~negligible thickness placedbetween the plates of acapacitor. How will itscapacitance be affected, if

(i) the foil is electrically insulated?

T1/2

1

52

(ii) the foil is connected to the upperplate with a conducting wire?Foreign 2011

24. Distinguish between polar and non-polardielectric. All Indio 2010 C

o 3 M'crks Questions25. Two identical

parallel platecapacitors AandB areconnected to abattery of Vvolts with the switch S is closed. The switchis now opened and the free space betweenthe plates of the capacitors is filled with adielectric of dielectric constant K. Find theratio of the total electrostatic energy storedin both capacitors before and after theintroduction of the dielectric. All Indio 2017

26. A 12 pF capacitor is connected to a 50 Vbattery. How much electrostatic energy isstored in the capacitor? If anothercapacitor of 6 pF is connected in serieswith it with the same battery connectedacross the combination, find the chargestored and potential difference acrosseach capacitor. Oelhi 2017

27. Two parallel plate X y

capacitors X and Y C Jhave the same area ofplates and sameseparation between + -them, X has air 15 Vbetween the plateswhile Y contains a dielectric medium ofe, = 4.

(i) Calculate the capacitance of eachcapacitor, if equivalent capacitanceof the combination is 411F.

(ii) Calculate the potential differencebetween the plates of X and Y.

(iii) Estimate the ratio of electrostaticenergy stored in X and Y. Oelhi 2016

28. In the following arrangement ofcapacitors, the energy stored in the 611Fcapacitor is E.

o Chapterwise cast Solved Papers PHYSICS

(iii)

-IIJ+6~"F

311F capacitor 12JlFtotal energy drawn from the batteryForeign 2016

Find the value of thefollowing

(i) energy stored in1211F capacitor

(ii) energy stored in

29. Find the ratio of the potential differencesthat must be applied across the paralleland series combination of two capacitors C1and C2 with their capacitances in the ratio1 : 2, so that the energy stored in these twocases becomes the same. All Indio 2016

30. Two capacitors of unknown capacitances C1

and C2 are connected first in series andthen in parallel across a battery of 100 V. Ifthe energy stored in the two combinationsis 0.045 J and 0.25 J respectively, thendetermine the value of C1and C2. Also,calculate the charge on each capacitor inparallel combination. All Indio 2015

31. Calculate the potential difference and theenergy stored in the capacitor C2 in thecircuit shown in the figure. Givenpotential at A is 90 V, c;. = 2011F,C2 = 30 I1Fand Ca = 1511F.Delhi 2015

A--1~~~C1 C2 C3 -=-

32. (i) Obtain the expression for the energyst6red per unit volume in a chargedparallel plate capacitor.

(ii) The electric field inside a parallel platecapacitor is E. Find the amount of workdone in moving a charge q over a closedrectangular loop abeda. Delhi 2014

33. (i) Derive the expression for thecapacitance of a parallel platecapacitor having plate area A andplate separation d.

(ii) Two charged spherical conductors ofradii ~ and R2 when connected by aconducting plate respectively. Findthe ratio of their surface chargedensities in terms of their radii.Delhi 2014

CHAPTER 2 : Electrostatic Potential And Capacitance 53

34. In a parallel plate capacitor with airbetween the plates, each plate has anarea of 6 x 10-3 m2 and the separationbetween the plate is 3 mm.

(i) Calculate the capacitance of thecapacitor.

(ii) If this capacitor is connected to 100 Vsupply, what would be the change oneach plate?

(iii) How would charge on the plates beaffected if a 3 mm thick mica sheet ofK = 6 is inserted between the plateswhile the voltage supply remainsconnected? Foreign 2014

35. A sphere 81 of radius r1 encloses a netcharge Q. If there is another concentricsphere 82 of radius r2(r2 > r1) enclosingcharge 2Q.

(i) Find the ratio ofthe electric fluxthrough sphere 81 S2

and8z·(ii) How will the

electric fluxthrough sphere 81change, if a medium of dielectricconstant 5 is introduced in the spaceinside 81 in place of air? All India 2014

36. A capacitor of unknown capacitance isconnected across a battery of V volt. Thecharge stored in it is 360llC . Whenpotential across the capacitor is reducedby 120 V, the charge stored in it becomes120IlC.Calculate

(i) the potential Vand the unknowncapacitance C.

(ii) what will be the charge stored in thecapacitor, if the voltage applied hadincreased by 120 V? Delhi 2013

37. A capacitor of 200 pF is charged by a300 V battery. The battery is thendisconnected and the charged capacitor isconnected to another uncharged capacitorof 100 pF. Calculate the differencebetween the final energy stored in thecombined system and the initial energystored in the single capacitor. Foreign 2012

38. A network of four capacitors each of 121lFcapacitance, if connected to a 500V supplyas shown in the figure.

C2

Determine(i) the equivalent capacitance of the

network and(ii) the charge on each capacitor.

HOTS;All India 2010

39. A parallel plate capacitor is charged by abattery. After sometime, the battery isdisconnected and a dielectric slab with itsthickness equal to the plate separation isinserted between the plates. How will

(i) the capacitances of the capacitor,(ii) potential difference between the

plates and(iii) the energy stored in the capacitors be

affected? Justify your answer in eachcase. Delhi 2010

40. A parallel plate capacitor, each with platearea A and separation d is charged to apotential difference V. The battery used tocharge it remains connected. A dielectricslab of thickness d and dielectric constantK is now placed between the plates.What change if any will take place in

(i) charge on plates?(ii) electric field intensity between the

plates?(iii) capacitance of the capacitor?

Justify your answer in each case. Delhi 2010

41. A parallel plate capacitor is charged to apotential difference V by a DC source. Thecapacitor is then disconnected from thesource. If the distance between the platesis doubled, state with reason, how thefollowing will change? Delhi 2010

(i) Electric field between the plates(ii) Capacitance

(iii) Energy stored in the capacitor.

54

42. Show that the capacitance of a sphericalconductor is 41teo times the radius of thespherical conductor. Delhi 2010

43. Find the ratio of the potential differencesthat must be applied across the paralleland the series combination of two identicalcapacitors, so that the energy stored in thetwo cases becomes the same. Foreign 2010

44. (i) How is the electric field due to acharged parallel plate capacitoraffected when a dielectric slab isinserted between the plates fullyoccupying the intervening region?

(ii) A slab of material of dielectricconstant K has the same area as theplates of a parallel plate capacitorbut has thickness.!. d, where d is the

2separation between the plates. Findthe expression for the capacitancewhen the slab is inserted betweenthe plates. Foreign 2010

45. (i) Plot a graph comparing the variationof potential V and electric field E dueto a point charge Q as a function ofdistance R from the point charge.

(ii) Find the ratio of the potentialdifferences that must be appliedacross the parallel and the seriescombination of two capacitors, C1 andC2 with their capacitances in theratio 1 : 2, so that the energy storedin the two cases becomes the same.Foreign 2010

46. Four capacitors of values 6 ~F, 6 ~F, 6 ~Fand 2 ~F are connected to a 6 V battery asshown in the figure.

6 ~F...---410::---; It----;

o ehopterwise eBSE Solved Papers PHYSICS

Determine the(i) equivalent capacitance of the network.

(ii) charge on each capacitor. Delhi 2010C

47. A parallel plate capacitor is charged by abattery. After sometime, the battery isdisconnected and a dielectric slab ofdielectric constant K is inserted betweenthe plates. How would

(i) the electric field between the plates?(ii) the energy stored in the capacitor be

affected? Justify your answer.Afllndia 2009

48. Three identical capacitors C1 , C2 and c, ofcapacitances 6 ~F each are connected to a12 V battery as shown below:

C:r---Ic~C'

Find(i) the charge on each capacitor

(ii) the equivalent capacitances of thenetwork

(iii) the energy stored in the network ofcapacitors. Delhi 2009C

49. The equivalent capacitance of thecombination between points A and Binthe given figure is 4 ~F.

A - II II -820~F C

(i) Calculate the capacitance of thecapacitor C.

(ii) Calculate the charge on eachcapacitor if a 12 V battery isconnected across terminals A and B.

(iii) What will be the potential dropacross each capacitor? Delhi 2009

50. Two parallel plate X y

capacitors X and Y have [.]the same area of theplates and sameseparation between them. +_X has air between the 12 Vplates while Y contains adielectric medium of e, = 4.

CHAPTER 2 : Electrostatic Potential And Capacitance 55

(i) Calculate the capacitance of eachcapacitor if equivalent of thecombination is 4JlF.

(ii) Calculate the potential differencebetween the plates of X and Y.

(iii) What is the ratio of electrostaticenergy stored in X and Y? Deihl 2009

51. A system ofcapacitorsconnected asshown in thefigure has atotal energy of160 mJ storedin it. Obtain the +value of the ---- ..•200 Vequivalentcapacitance of this system and the valueof Z. All India 2009C

[ZI 5 Marks Questions52.

53.

(i) If two similar large plates, each of areaA having surface charge densities +oand - o are separated by a distance din air, find the expression for

(a) field at points between the twoplates and on outer side of theplates. Specify the direction of thefield in each case.

(b) the potential difference between theplates.

(c) the capacitance of the capacitor soformed.

(ii) Two metallic spheres of radii Rand2R are charged, so that both of thesehave same surface charge density cr.If they are connected to each otherwith a conducting wire, in whichdirection will the charge flow and why?All India 2016

(i) Explain using suitable diagrams, thedifference in the behaviour of a(a) conductor and(b) dielectric in the presence of

external electric field. Define theterms polarisation of a dielectricand write its relation withsusceptibility.

(ii)A thin metallic ra'2'6spherical shell ofradius R carries a Q' Ccharge Q on its -surface. A point 2

charge Q is placed at its centre C and2

an other charge + 2 Q is placedoutside the shell at a distance x fromthe centre as shown in the figure.Find(a) the force on the charge at the centre

of shell and at the point A,(b) the electric flux through the shell.

All India 2015

54. (i) Derive the expression for the energystored in parallel plate capacitor .Hence, obtain the expression for theenergy density of the electric field.

(ii) A fully charged parallel plate capacitoris connected across an unchargedidentical capacitor. Show that theenergy stored in the combination isless than stored initially in the singlecapacitor. Delhi 2014

(i) Obtain the expression for thepotential due to an electric dipole ofdipole moment P at a point x on theaxial line.

(ii) Two identical capacitors of platedimensions l x b and plate separationd have dielectric slabs filled inbetween the space of the plates asshown in the figure.

~I~ 1 ~I~-.;f :r~~>~~~~i".~:,)

I+-- 1/2 ~I• 1/2 --+l

Obtain the relation betweendielectric constants K, s, and K2•

All India 2013

55.

56

56. (i) A parallel plate capacitor is charged bya battery to a potential. The battery isdisconnected and a dielectric slab isinserted to completely fill the spacebetween the plates.How will(a) its capacitance(b) electric field between the plates and(c) energy stored in the capacitor be

affected? Justify your answergiving necessary mathematicalexpressions for each case.

(ii) (a) Draw the electric field lines due to aconducting sphere.

(b) Draw the electric field lines due to adipole. All India 2011

(i) Deduce the expression for the energystored in a charged capacitor.

(ii) Show that the effective capacitancesC of a series combination of threecapacitors ~ , C2 and Ca is given byC= ~ C2 Ca .

~ C2 + C2 c, + Ca ~ All India 2010C

58. (i) Show that in a parallel platecapacitor, if the medium between theplates of a capacitor is filled with aninsulating substance of dielectricconstant K, its capacitance increases.

(ii) Deduce the expression for the energystored in a capacitor of capacitance Cwith charge Q. Delhi 20D9C ' "

59. A small sphere of radius a carrying a .positive charge q is placed concentricallyinside a large hollow conducting shell ofradius b (b > a). This outer shell has chargeQ on it. Show that if these spheres areconnected by a conducting wire, charge will

r, always flow from the inner sphere to the, outer sphere irrespective of the magnitude

of the two charges. All India 2009

[II( .

57.

o Explanations1. The polarity of the capacitor shown below.

~-----o-~----~-+ -B __

,

o Chnpterwlse eBSE Solved Papers PHYSICS

From Lenz's law, induced current produces samepolarity as that of approaching pole. So, plate Awill have +ve polarity and plate B will have -vepolarity. (1)

2. In steady state, electric flux between plates of acapacitor is constant.So, displacement current is,

id =Eo dellE and dcj>E =0 =? id =0dt dt

So, there is no current between plates when steadystate is reached.During charging, flux is increasing.•• dcj>E * 0

dtHence, a displacement current exists in the

. hi h i dcj>Ecapaotor w c IS ld = Eo -.dt (1)

3. Line B corresponds 'to C1 because slope (q versus V)of B is less than slope of A. (1)

4. Dielectrics are non-conductors and do not havefree electrons at all. While conductor has freeelectrons which makes it able to pass theelectricity through it. (1)

5. Dielectric When a dielectric slab is introducedbetween the plates of charged capacitor or in theregion of electric field, an electric field Ep inducesinside the dielectric due to induced charge ondielectric in a direction opposite to the directionof applied external electric field. Hence, netelectric field inside the dielectric gets reduced toEo - Ep, where Eo is external electric field. Theratio of applied external electric field and reducedelectric field is known as dielectric constant K

of dielectric medium, i.e. K = EoEo - Ep

and it is a dimensionless quantity. (1)

6. If a metal plate is introduced between the plates ofa charged parallel plate capacitor, then capacitanceof parallel plate capacitor will become infinite. (1)

7. For a series combination of three capacitors Cl' C2

and C3' the equivalent capacitance Ceq will be1 11111 II-=-+-+- =? -=-+-+-

Ceq C1 C2 C3 Ceq 20 30 15

C1=20 JlF C2=30 JlF

C~90V

CHAPTER 2 : Electrostatic Potential And Capacitance 57

~_1_ = 3+ 2+ 4 ~C = 601lF= 20JlFCeq 60 eq 9 3

(1)

Charge on equivalent capacitor C20 ~

Q=C V=-x10-6x90 ~:Jeq 3 :j: :::~ Q= 600llC

Charge on each capacitor issame as they are in series. 90 VNow, potential drop across C2

V2

= ~ = 600 X 10-6 = 20VC2 30 X 10-6

Hence, work done stored as electric potentialenergy U of capacitor is U = ~c2vf

2

U = ~ x 30 X 10-6 x (20)2 = 6 X 10-3 J2 ru

8. Let q be the charge on the charged capacitor.2

:. Energy stored in it is given by U = !L2C

When another uncharged similar capacitor isconnected, then the net capacitance of the systemis given by C' = 2C (1)

The charge on the system remains constant. So,the energy stored in the system is given by

, q2 q2U =-=- [':C'=2C]

2C' 4C

Thus, the required ratio is given byU' _ q2 14C _ 1U - q2 12C - '2

9. (i) Given, C1 = 2C2

Net capacitance before filling the gap withdielectric slab is given by

Ciniti.1=C1 + C2

Ciniti.1=2C2 + C2 = 3C2

Net capacitance after filling the gap withdielectric slab of electric constant K

... (i)

[from Eq. (i)]... (ii)

Ciniti.1=KC1 + KC2 =K(C1 + C2) [from Eq. (ii)]

Clin•1=3KC2 ••• (iii)

Ratio of net capacitance is given by

Ciniti.1= 3C2 = ~ [from Eqs. (ii) and (iii)]Clin•1 3KC2 K (1)

(ii) Energy stored in the combination beforeintroduction of dielectric slab

dUiniti.1= 3C

2

... (iv)

Energy stored in the combination afterintroduction of dielectric slab.

dUfin•1 = 3KC

2

Ratio of energies stored

... (v)

Uiniti.1.:Ulin•1

[from Eqs. (iv) and (v)](1)

10. Given, C1 = C2 •.. (i)2

Hence, Cimtial= C1 + C2 = C2 + C2 = 3C2 •.. (ii)2 2

Net capacitance after filling the gap withdi 1 . KC - 3KC2Ie ectnc , initial- --

2 (2)

11. Total current through the circuit is given by1= VIRHere, V=2V ~ R=(10+20)n=30n

.. 1=2 =~A30 15

Voltage across IOn resistor =1(10) = 10/15 = '!:V3

Charge on the capacitor is given byQ = CV =(6 x10-, x 2/3 = 41lC (2)

(1)

12. Initially, when there is a vacuum between two

plates, then capacitance of the plate is Co= Eo A,d

where, A is the area of parallel plates.Suppose that the capacitor is connected to abattery.jm electric field Eo is produced. Now, ifwe insert the dielectric slab of thickness t=dI2,the electric field reduces to E .Now, the gap between plates is divided in twopans, for distance t, there is electric field E andfur the remaining distance (d - t) the electric fieldis Eo. (1)

If V be the potential difference between the platesof the capacitor, then V =Et + Eo (d - t)

V=E: + E~d=~(E+Eo) [-: t=~J

~ V=~(;+Eo)

=~(K+1) [AS, ~=KlNow, Eo=!!....=-q- ~V=~.-q-(K+1)

Eo EoA 2K EoA

Ne know that, C=!L= 2KEoAV d(K +1) (1)

58

13. Potential difference,-q +EI-3....-E

2=0 or 3....+3....=E

1-E2

C1 C2 C1 C2 (1)r~El~~_rq

Vi = -q and V2=+qC1 C2

Now,

14. The work done incharging thecapacitor is storedas its electricalpotential energy.Hence consideringa capacitor ofcapacitance C, initially whose two plates areuncharged, let Q and -Q are charges on the twoplates and produces a uniform electric field,

E = ~ between the plates and a potentialEo

difference

+0-j~l-(1)

... (i)v=iC

If a charge dq is transported in steps from negativecharged plate to positive charged plate, till chargesrises to +Q and -Q, thenwork done, dW = dq. VFrom Eqs. (i) and (ii), we get

dW = dq( ~) ~ W = f dW ',T ,"

This work done is stored as electrical potential U ofthe capacitor,

Qq Q2

U = W = Ic·dq = 2C

Q2 1 2 1 .U = - = -' CV = - QV [.: Q = CV]

2C 2 2 (1)

,)... (ii)

15. (i) According to the diagram given in thequestion

1 II 1C1~t-JC3L1~V-l

o ehapterwise eSSE Solved Papers PHYSICS

(1)

Here C1, C2 and C3 are in series, hence theirequivalent capacitance C' is given by

1 1 1 1-=-+-+-C' C1 C2 C3

Redrawing the circuit as shown below1 1 1 1-=-+-+-C' 15 15 15

C' = ~IlF3

~ C'=5IlFSince, C' and C4 are inparallel,:.C = C' + C4 = 51l + 151l = 20ll

(ii) Since, C' and C4are in parallel potentialdifference across both of them is 100 V.:. Charge across C4is ~ = C4 X V = C4 X 100C=15 X 10-6 x 100

= 15xlO-4C~ =1.5mC

C'r;g(1)

(1)

16. Expression for the energy stored in a capacitor:If we consider a capacitor of capacitance C,initially whose two plates are uncharged thepositive charge will be transferred from plate 2 toplate 1 bit by bit. During this process, thepote~tiabPifference between the two plates willbeY =-.

If a smafi' additional charge det be transferredfrom plate 2 to plate 1, the work done will be,

dW = V'·det = Q' . dQ'C

The total work done in transferring a charge Qfrom plate 2 to plate 1 will be

QQ' ,W = f dW = f -' dQ

o C

=[~;I=~'~This work done is stored as electric potentialenergy U of the capacitor,

1 Q2 1 2U=-·-=-·CV

2 C 2(i) Energy stored will be decreased or energy

stored will become ~tirnes the initial energy.K

(ii) Electric field would decrease or

E'=~K

CHAPTER 2 : Electrostatic Potential And Capacitance 59

17. According to the diagram are when the switch S isclosed, the two capacitors CI and C2 in parallel willbe charged by the same potential difference V.

S

-=- 6VT_So, charge on capacitor CI

ql = CIV= 1 x 6 = 6~and charge on capacitor C2

q2 = C2V=1 x 6= 6~Hence total q = ql + q2 = 6 + 6 = 12 ~When switch S is opened and dielectric isintroduced. Then,

... (i)

... (ii)(1)

s

rzz: 6V A Be',

Capacity of both the capacitors becomes K timesi. e. C{ = C~= KC = 3x 1 = 3~ (as CI '= C2 ')

Capacitor A remains connected to battery.. V'I = 'V = 6V

q{ = KtJI = 3 x 6~ = 18~Capacitor B becomes isolated:. q~ = q2 or C~V; = C2V2 or (KG)V; = CV

or V2=(f)=~=2V

18. If n identical capacitors, each of capacitance Careconnected in series combination give equivalentcapacitance, C, = ~ and when connected in

nparallel combination, then equivalentcapacitance, Cp': nC

Also, for same voltage, energy stored in thecapacitor is given by

U = ~CV2 [for constant]2

UocC

c, =1 ~F

I . binati C Cn senes com matlon, s =-n

In parallel combination, Cp = nC

According to the problem,

C = n C, = 3x 1~F = 3~FFor each capacitor,

[·:n = 3]

19.

In parallel combination,

Cp = nC = 3 x 3 = 9 ~F

c, = 9~F

For same voltage, U cc C~ Us = Cs ~

Up c,Us _ 1 _ 1 ~Up - (3)2 -"9

or Us : Up = 1 : 9

When a dielectric medium of dielectric constantK is introduced,(i) in an isolated (not connected with battery)

capacitor, then total charge on capacitorremains same.

(ii) in a capacitor connected with battery, thenpotential difference across the capacitorremains same as that of potential differenceacross battery.

Two identical capacitors CI and C2 get fullycharged with 5 V battery initially.So, the charge and potential difference on bothcapacitors becomes

q = CV = 2 X 10-6 X 5 V = 10 ~C

and V = 5VOn introduction of dielectric medium of K = 5.(1/2)

For C1 (continue to be connected with battery)potential difference of CI' (Vi = 5 VCapacitance of C; = KC = 5 x 2~F = 10 ~Charge, q' = C'V' = (10~F) (5 V) = 50 ~CFor C2 (disconnected with battery)

Charge, q' = q = 10 ~CPotential difference,

V'=~=~=l VK 5

(1)

(1)

(1)

(1Y2)

20. According to the question, separation betweenthe plates = d = 0.5em = 0.5x 1O-2m andcapacitance C = 2 FDielectric constant

d = O.5cm = 0.5 x 10-2 mEo = 8.854 X 10-12 2 -lm-2

C= EoAd

A = Cd = 2 x 0.5 X 10-2

Eo 8.854 x 10 12

= 1.13 X 109 m2 (2)

.er

21. The capacity of condenser is proportional to thearea and inversely proportional to the distance

60

between its plates. If a medium of dielectricconstant K is filled in the space between theplates, its capacity becomes K times the capacitywhen there is air between the plates.After inserting the dielectric medium, let theircapacitances become C; and C;.

For Cz

C; = KC ... (i) (112)

C' - x, Eo (A/2) _K.=...2E-,,-o.:....(A...:../2....:..)2- +d d

C2 acts as if two capacitors each of area A/2andseparation d are connected in parallelcombination.

C; = EoA (Kl + K2)d 2 2

C; = C (Kl : K2)

[-:... (ii)

C - Eo A]d (1/2)

According to the problem,

C; = C;

22. After introduction of dielectric medium ofdielectric constants K, and K2, capacitor acts as ifit consists of two capacitors, each having plates ofarea A and separation ~ connected in series

2combination for

EoAC1 =--dI I I

C;=(KIEOA)+ (K2EOA)d/2 d/2

I I (I I)-=--- --+-C2 (E~A) 2Kl 2K2

... (i)

I _

The capacitors will be in series.

23. (i) The system will be equivalent to two identicalcapacitors connected in series combination in

IZl Chopterwise CBSE Solved Papers PHYSICS

which two plates of each capacitor have. separation half of the original separation.Thus, new capacitance of each capacitor

C'=2C [-: coc~]C and C' are in series

C =2CX2C=Cnet 2C + 2C

(1)

Cnet = C (Original capacitor) (1)

(ii) System reduces to a capacitor whoseseparation reduces to half of original one.:. New capacitance, C' = 2C (1)

Polar dielectricsA polar molecule which has permanent electricdipole moment (p) in absence of electric fieldalso polar molecules are randomly oriented.e.g. Water, alcohol. Hel. NH}Non-polar dielectricsA non-polar molecule having zero dipole momentin its normal state.Non-polar molecules have symmetrical shapese.g. any non-conducting material.

25. The given figure is shown below.

24.

When switch S is closed, the potential differenceacross capacitors A and B are same

i.e. V = QA = QB

C C(1) Initial charges on capacitors

QA = QB = CV •When the dielectric is introduced, the newcapacitance of either capacitor

C'=KCAs switch S is opened, the potential differenceacross capacitor A remains same (V volts).Let potential difference across capacitor B be V'.When dielectric is introduced with switch S open(i.e. battery disconnected), the charges oncapacitor B remains unchanged, so

QB = CV= C'V'

V'=!:.. V =!voltc' K

(1)

(1)

(1)

CHAPTER 2 : Electrostatic Potential And Capocitonce 61

26.

Initial energy of both capacitors

U = ~CV2 + ~CV2 = cv2122

Final energy of both capacitors

UI = ~C'V2 + ~C'V'2 = ~(KC) V2 + ~(KC) (~)22 2 2 2 K

1 2 [ 1 1="2CV K+"KJ

= ~ cv2( K~+ 1)

U, _ CV2 2K

UI ~CV2 (K~+I) - K2+1

Energy stored in capacitor =~ C1V2

2

= ~ x 12 X 10-12 x (50)2 J2

= 6x 25 X 10-10 J = 15 x 1O-9J

With other capacitor 6 pF in series.. C xC 6 x 12Total capacitance (C) = _I __ 2 = -- pF

C1 + C2 6 + 12

_ 12 x 6 _ 4 F---- P18 n)

in each capacitor is same and isCharge storedgiven by

Q = CV = 4 X 10-12 X 50 C = 2 X 1O-IOC

Each of the capacitors will have charge equal to Q= 2 X 1O-IOC

Potential on capacitors with capacitance 12 pF is=!l = 2 X 10-

10V = 16.67V

C1 12 x 10 12

Potential on capacitor with capacitance 6 pF is

2 x 10-10

V = 33.33 V6 x 10-12

27. According to question, let the capacitance of X beC, so capacitance of Y = e, C = 4C [.: e, = 4]

(I) E . I . Cx4C1 qmva ent capacitance = --C+4C

(X and Yare in series)

4C2

4C d it i h 4C 4= - => - an It IS gIven t at - = ~5C 5 5

So, 4C = 20~ = capacitance of Y. 20

Capacitance of X = C = - = 5~4 (1)

(1)

(ii) Charge flowing through the capacitor is given by4C 4x 5q = CV = - x 15 = -- x 15 = 60 ~5 5

Now, let the potential difference betweenplates of capacitors X and Yare Vx and,Vy,

respectively. .

So, Vx = !L = 60 = 12 V and Vy = !L = 60 = 3Vc, 5 c, 2,~ (1)

(iii) Electrostatic energy stored in capacitance

X(Ex) = ~Cv; I ••• (i)

Similarly for Y,Ey =~4CV; ... (ii)2

From Eqs. (i) and (ii), we get1 2

. E -CVx V2 12x12RatIO = -L = _2__ = _x_ = = 4: 1Ey ~4CV2 4V; 4x3x3

2 y (1)

28. (i) As given in the question, energy of the 6 ~capacitor is E. Let V be the potential differencealong the capacitor of capacitance 6 ~F. Fromthe mathematical formula,

Since, ~ CV2 = E2

~X6xlO-6xV2=E=>V2=~xI06 ... (i)2 3

Since, potential is same for parallelconnection, the potential through 12 ~capacitor is also V. Hence, energy of lzucapacitor is

EJ2 = ~ x 12x 10-6 x V2 [From Eq. (i)]2

= ~ x 12 X 10-6 x ~ X 106 = 2EY 3 m

(ii) Since, charge remains constant in series, thecharge on 6~ and 12 ~ capacitors combinedwill be equal to the charge on 3 ~ capacitor.Using the formula, Q = CV, we can write=> (6 + 12) x 10-6 x V = 3 X 10-6 X V'

V'=6V

Using Eq. (i) and squaring both sides, we getV'2= 12V2 => V'2 = 12E X 106

:. E3 = ~ x 3 X 10-6 x 12E X 106 = 18E2 m

(iii) Total energy drawn from battery isEtotal = E + EJ2 + E3

= E + 2E + 18E = 21E

(1) .

n)

62

29. Total energy stored in series or parallelcombination of capacitors is equal to the sum ofenergies stored in individual capacitors. In parallelcombination energy stored in the capacitor

1 2 1 2= -CIl'J. + -C2VJ ..• {i)2 2

In series combination energy stored in thecapacitor

... (ii)(1)

According to the question, energy in both the casesis same so,

(~C+~C)V,2= CIC2 V22 I 2 2 I 2(CI + C2) 2

VI2 _ CI C2 X 2 l'J. _ ~CI C2--~-"----~vi - 2(CI + C2){CI+ C2) V2 - CI + C2

S. = ~ ~ C2 = 2CIC2 2

l'J. = .J~CI-X-2"""C::-I= .fi CI = .J2V2 CI + 2CI 3 CI 3

But,

So,

30. When the capacitors are connected in parallel.. equivalent capacitance, Cp = CI + C2•

The energy stored in the combination of thecapacitors, Ep = ~CpV2

21 2Ep = - (CI + C2) (100) = 0.25J2

~ CI+C2=5XIO-5

When the capacitors are connected in series,equivalent capacitance,

Cs=~CI + C2

The energy stored in the combination of thecapacitors,

1 2s, =-CsV2

s, = ~ ~ (100)2= 0.45J2 CI + C2

~ X ~ (100)2 = 0.45J2 5xl0-5

CIC2 = 0.045 x 10- 4 X 5 x 10- 5 X 2= 4.5 X 10-10

{CI - C2)2 = {CI + C2)2 - 4CI C2~ (CI - C2)2 = 25x 10-10 - 4 x 4.5x 10-10

= 7xlO-10

... (i)

(1/2)

o Chopterwise CBSE Solved Papers PHYSICS

(2)

~ CI - C2 = 2.64 x 10- 5 ... {ii)

On solving Eqs. (i) and (ii). we getCI = 35 JlF and C2 = 15 JlF ('I. + 'I.)

QI =CIV=35XlO-6x100=35xIO-4C (112)

Q2=C2V=15XlO-6x100=15xIO-4C (112)

31. Consider the given figure20 JlF 30JlF 15 JlFA---ll---ll---l ~

C1 C2 C3 -=-Given, CI = 20JlF, C2 = 30JlF, C3 = 15JlFPotential at A = 90 VAs, we can see that capacitor C3 is earthed,therefore, potential across C3 will be zero.Since, capacitor CI' C2and C3 are connected inseries, therefore_1_= J.... + J....+ J....~ _1_=J....+ J....+ ~Ceq CI C2 C3 Ceq 20 30 15

~ 1 3+ 2+ 4 ~ _1_ = ~Ceq 60 Ceq 60

60 20~ Ceq = - ~ Ceq = - JlF

9 3~)Since, charge remains same in seriescombination,

(112) 20Q = CeqV ~ Q = - x 90

3~ Q= 600 JlC~ Q = 600 X 10-6 C~ Q = 6 X 10-4 C

.:.Potential difference across C2 = ~V2

~ V - Q ~ V _ 6 X 10-42 - C2 2 - 30xl0-6

~ V2 = 0.2XI02 ~ V2 = 20VAlso, energy stored in capacitor C2 is given by

1 2 1 2-6E=-C2V2 ~E=-x30x(20) x io2 2

~ E = ~ x 30 x 400 X 10-6 ~ E = 6000 X10-62

~ E = 6 X 10-3 J

So,

(1)

(1)

32. (i) The energy of a charged capacitor is measuredby the total work done in charging thecapacitor to a given potential.Let us assume that initially both the plates areuncharged. Now, we have to repeatedlyremove small positive charges from one plateand transfer them to other plate.

CHAPTER 2 : Electrostatic Potential And Capacitance 63

Now, when an additional small charge (dq) istransferred from one plate to another, thesmall work done is given by

dW = V'dq = f..dqC

[let charge on plate, when dq charge istransferred is q']The total work done in transferring charge Q isgiven by

iQ q' I iQW= -dq =- q'dqo ceo

= ~ [(q')2]Q = Q2

C 2 0 2C (1/2)

This work is stored as electrostatic potentialenergy U in the capacitor.

_ Q2 _ (CV)2U-----

2C 2C

U = ~ CV2

2 n)The energy stored per unit volume of space in acapacitor is called energy density.

~CV2u=_2 __

Ad

(1/2)

[':Q = CV]

(ii)

1 eoAv2~ U=---

2 d2A1 2Energy density, U = - eo E2

Total energy stored in series combination orparallel combination of capacitors is equal tothe sum of energies stored in individualcapacitor.i.e. U=UI +U2+U3+ ...Due to conservative nature of electric force,the work done in moving a charge in a closepath in a uniform electric field is zero. (1)

Parallel plate capacitor consists of two thinconducting plates each of area A held parallel toeach other at a suitable distance d. One of theplates is insulated and other is earthed. Thereis a vacuum between the plates.

Surface X,( Area A

Charge~ 1 Idensity .

++++++++++

lllEIII

33. (i)

Surface ~ - - - - 2

charge ~ L ydensity

d

II~ Earthed

Suppose, the plate X is given a charge of+q coulomb. By induction, -q coulomb ofcharge is produced on the inner surface of theplate Y and +q coulomb on the outer surface.Since, the plate Y is connected to the earth, the+q charge on the outer surface flows to theearth. Thus, the plates X and Y have equal andopposite charges.Suppose, the surface density of charge on eachplate is a. We know that the intensity ofelectric field at a point between two planeparallel sheets of equal and opposite charges isa/Eo, where Eo is the permittivity of free space.The intensity of electric field between theplates will be given by, E = ~

EoThe charge on each plate is q and the area ofeach plate is A. Thus,

a = ~ and E = _L ...(i)A EoA

Now, let the potential difference between thetwo plates be V volt. Then, the electric fieldbetween the plates is given by

E = ~ or V = Edd

Substituting the value of E from Eq. (i). we get

V=~EoA

:.Capacitance of the capacitor is

C = ~ = -q-- or C = EoAV qd/EoA d

where, Eo = R,85xIO-12 C2_Nm-2It is clear from this formula that in order toobtain high capacitance

(a) A should be large, i.e. the plates of largearea should be taken.

(b) d should be small, i.e. the plates should bekept closer to each other. (2)

(ii) Surface charge density is given by

a = -q-41tR2

After connecting both the conductors, theirpotentials will become equal.

Vi = V2

~ Kql = Kq2 [':for spherical conductors~ ~ 1 Kqv= __ i or V=-]

41tEo R R

~~=Rl ~~=ql/41tR~ ql (R2)2_R2q2 R2 a2 q2 /41tR~ q2 RI RI (1)

64

34. Given, area of each plate, A = 6 x 10-3 m2

Distance between plates d, = 3mm = 3 x 10-3 m

(i) Capacitance of paraliel plate capacitor is givenby

{ I

C=EoAd

Eo = 8.85XI0-12C2 N-lm-2

. 'llA[') I 885xlO-12X6xl0-3C =-.----~--

3xl0-3

I J I '1 C =;ol,;ZZ x 1O-11F (1)

(ii) Charge on parallel plate capacitor is given byQ=CV

Given, V = 100VNow, Q=17.7xl0-12 x 100

Q=1.77XI0-lOCGiven, K = 6Now, C'=KC.. O'=KQ

Q'=6 x 1.77 X10-10

t-

... (i) (1)

(ill)

[From Eq. (i)]

35. According to Gauss's theorem, total flux ~ = .!L.Eo

Applying the formula sphere of radius rl haselectric flux ~1' and sphere of radius r2 has electricflux ~2' then(i)~1 =~'~2=3Q ~=~

Eo Eo ~2 3(ii) if a medium of dielectric constant 5 is filled in

the space inside SI< the flux inside SI

~'1 = ~ = ~ [Replacing Q by ~1Eo]5Eo 5

36. (i) We have initial voltage, VI= V volt and chargestored, <4 = 360JlC

Ql=CVJCharged potential, V2= V-120

Q2=120JlCQ2=CV2

By dividing Eq. (ii) from Eq. (I), we get. Ql _ CVJ 360 _ V--- ~----

Q2 CV2 120 V -120

V=180V

... (i)

(1/2)

... (ii)

(1/2)-6

C = Ql = 360x1O = 2xlO-6 F = 2 JlFVI 180

Hence, the potential, V=180V and unknowncapacitance is 2 JlF. (Wo)

o ehapterwise eSSE Salved Paper> PHYSICS

(ii) Let Qbe the charge stored in the capacitorQ = CV= 2x 10-6 x 120

Q = 24 X 1O-5C (112)

37. Given, C = 200 pF = 200 x 10-12 F and V = 300 V

The energy (initially) stored by the capacitor isU· = .!.CV2 = .!.x 200 X 10-12 X 300 x 300

I 2 2

= 9 X 10-6)

The charge on the capacitor when charge through300 V battery is

Q=CV= 200 X 10-12 X 300

=6xlO-8C

= 60 X 1O-9C = 60nC (1)

(1)

When two capacitors are connected, they havetheir positive plates at the same potential andnegative plates also at the same potential. Let Vbe the common potential difference. By chargeconservation, charge would distribute but totalcharge would remain constant.Thus, Q = q+ q'

!!.. = q'C c

...L=~200 100

q = 2q'Thus, Q = 2q' + q' = 3q'

So, q,=!l= 60nC=20nC3 3

and q = 2q' = 40 nCThus, final energy

q2 q,2Uf = 2C+ 2C'

1 (40 X10-9)2 1 (20 X 10-9)2=-x +-x-'----~2 200xl0 12 2 100xl0 12

= 4 X10-6 + 2 X 10-6 = 6 X 10-6)

Difference in energy= final energy - initial energy=Uf - U,= 6 X 10-6 - 9 X 10-6

= - 3x 10-6

Thus, difference in energy is - 3 X 10-6). (1)

[1)

CHAPTER 2 : Electrostatic Potential And Capacitance 65

38. In a series combination, where there is nodivision of charge,

1 1 1 1-=-+-+-+ ...C, C1 C2 C3

In a parallel combination, where potentialdifference is same.

Cp = C1 + C2 + C3··•

(i) Here, C1 ,C2 and C3 are in series, therefore,their equivalent capacitance

1 1 1 1-=-+-+-C' C1 C2 C3

C'=~=~=4~F3 3

(ii)

Now, C' and C are in parallel combination.

Cnot= C' + C= 4 ~F+ 12~F= 16~F

Cnot= 16~F (1)

Being C' and C are in parallel, 500 V potentialdifference is applied across them.

:. Charge on C'ql = C'V

= (4~F)x 500 = 2000 ~C

39.

:. C1 ,C2 and C3 capacitors each will have2000 ~Ccharge.

Charge on C4' q2 = C X V

=12x500= 6000~C (1)

On introduction of dielectric slab in an isolatedcharged capacitor.(i) The capacitance (C') becomes K times of

original capacitor asC = Eo A and C' = K eo A

d d n)(ii) Charge remains conserved in this

phenomenon.cv= C'V'

V' = cv = CV [refer part (i)]C' KC

V'=~K

Potential difference decreases and become ~K(1)times of original value.

(ill) Energy stored initially,2

U=!L2C

(1)

Energy stored later,2

U,=-q-2/.KC)

[·:C' = KC]

40.

where, K = dielectric constant of medium

~ u,=~(ct) ~U'=~(U)~U'=~XUK 2C K K

The energy stored in the capacitor decreasesand becomes ~ times of original energy.

K (1)

On introduction of dielectric slab to fill the gapbetween plates of capacitor completely whencapacitor is connected with battery. ,j

(i) The potential difference V between capacitorsis same due to connectivity with battery andhence, charge q' becomes K times of originalcharge as

q' = C'V' = (KC) (V) = K(CV) = Kqq' = Kq n)

(ii) Electric field intensity continue to be the sameas potential difference and separation betweentwo plates remain unaffected as

E=~.d (1)

(ill) The capacitance of capacitor becomes K timesof original capacitor.

. . C' = KC = K eo Ad (1)

41. After disconnection from battery and doublingthe separation between two plates(i) Charge on capacitor remains same.

i.e. CV = C'V'

~ CV = ( %) V' ~ V' = 2V

Electric field between the platesE'= V' = 2V

d' 2dE'=..!::.=E

d~ Electric field· between the two platesremains same. (1)

(ii) Capacitance reduces to half of original value asCoo.!. ~ C'=~

d 2 (1)

(ill) Energy stored in the capacitor beforedisconnection from battery

q2U1 =-

2C

66

Now, energy stored in the capacitor afterdisconnection from battery

2 2 2U-q - q -q

2- 2(C') --(C)-C2x -2

=> U2= 2(~)= 2UI U2= 2UI

Energy stored in capacitor gets doubled to its(tJ initial value. (1)

42. As we know, to determine the electric field at anypoint at distance r from centre if we applyGa~ss' tth~orem

E = q41tEo . r2

Hence, V = f E· dr = --q-41tEo . r

where.e, = 8.854 x 10-12 C2 N-I m "

The capacitance of the spherical conductorsituated in vacuum is given by

C=!!...=--q-V _1_.:i

41tEo r

=> C = 41tEor.Hence, the capacitance of an isolated sphericalconductor situated in vacuum is 41tEo times of itsradius.

43. Let VI and V2 are the potential differences acrossthe series and parallel combination of twoidentical capacitors each of capacitance C.

Equivalent capacitance in series combinationC =!:.., 2 n)

Equivalent capacitance in parallel combinationc, = 2C

According to the question,

U, = Up

~ C V2 = ~ C V22 s 2 p p

V/ _ Cp _ 2C

Vp2-C,-(%)=>

,") )

2V,=4=>V,=2Vp2 Vp

V,:Vp=2:1

o Chapterwise CBSE Solved Papers PHYSICS

44. (i) The total charge on the capacitor remainsconserved on introduction of dielectric slab. Also,the capacitance of capacitor increases to K timesof original values.

.. CV = C'V'CV = (KG) V' => V' = ..!:::.

K:. New electric field,

E'= :' =(V~K)=(~)~=f

:. On introduction of dielectric medium, newelectric field E' becomes ~ times of its original

Kvalue. (1)

(ii) .: Capacitance of a parallel plate capacitorpartially filled with dielectric medium is givenby

C= EoA(d - t + t/K) (1)

where, t is the thickness of dielectric medium.

Here, t = ~2

C= EoA

d-~+~2 2K

C= 2EoAK(K + 1) d (1)

45. (i) The graph comparing the variation ofpotential V and electric field. Refer to Ans. 26,Topic -1. (1)

(ii)

(1)

Let CI = C and C2 = 2C

:. Equivalent capacitance

In series, C = 2C x C = 2C2

= 2C, 2C+C 3C 3

In parallel, Cp = 2C + C = 3C

. . Vp and V, are potential difference across thefinal capacitor in parallel and series combinationrespectively, to have same potential energy.

Up = U,

~ C V2 = ~ C V2

2PP 2"

Vp _ Kv, V Cp

11= /2C/3) = ~(3C) V"9

Vp:V,=.fi:3 (1)

(1)

(1)

CHAPTER 2 : Electrostatic Potential And Capacitance 67

46. In series combination, charge on each capacitor isthe same.(i) All capacitors of 61lF are in series combination,

then equivalent capacitance is given by1 1 1 1-=-+-+-C' C1 C2 C3

or C' = ~ = 6 IlF = 21lFn 3

C' and 21lFcapacitors are in parallel combination.:. Equivalent capacitance

Ceq= C' + 21lF= 21lF + 21lF

Ceq= 41lF (1)

(ii) Since, C' and 2 IlFare in parallel combination,therefore, same potential difference 6 V isapplied on them.

:. Charge on C'q' = C'V = (21lF) x 6 V = 121lC

The charge across the each capacitor of 61lFcapacitor is same and equal to charge across thecombination i.e. 121lC

Charge on 21lF capacitor q = CV = (21lF) (6 V)= 121lC (112 x 4 = 2)

(2)

(1)

47. (i) Refer to Ans. 44 (i).(ii) Refer to Ans. 39 (iii)

48. (i) The equivalent capacitance of C1 and C2connected in series

1 1 1-=-+-C' C1 C2

C'=~=3IlF2

:. Charge, q' = C'V = (3IlF) 12 = 361lC:. Charge on each capadtor of C1 and C2is 361lCCharge on C3'

q3 = C3V = (6IlF) x 12= nllc

q3 = 72IlC

(ii) Equivalent capacitance of network,

Ceq=~+ C3C1 + C2

=6X6+66+ 6

=3+6=9IlFCeq= 91lF

(1)

(1)

(iii) Energy stored in the network of capacitorsq,2 q,2 q2

U = U1 + U2 + U3 = - + - + -2C1 2C2 2C3

': C1 = C2 = C3 = 61lF.. U = _1_ [q,2 + q" + q2]

(l21lF)

= _1 _ [(361lq2 + (361lq2 + (nllC)2](l21lF) I, t.J

U = 6481l) .... ilIJ· (1)

49. According to the question,Capacitors of 20 IlF and C are connected in series.(i) The equivalent capacitance J ,J

(4 IIF) = (20 IlF) x Cr- 20+ C

(20 + C) = 5C 4C = 20C = 51lF

(ii) Charge on capacitor (equivalent)q = (4IlF) x12= 481lC

same charge 481lClies on both the capacitors. (1)

(ill) Potential drop across 20 IlF capacitor

Vi =..i. = 481lC = 2.4 VC1 20llF

Potential drop across 5 IlF capacitor

V2

= ..i. = 481lC = 9.6 VC2 5j.1F

(1)

50. A dielectric medium is inserted between theplatesof a condenser in place of air, its capacitybecomesx times of original one.The capacitance of two capacitors are

C = Eo A = C [say]x d

C = 4EoA = 4 Cy d

(i) According to the problem,Ceq= 41lF

c, xCy.. Ceq=---c, + Cy

4IlF=CX4C=4CC+ 4C 5

[.: For series combination, C = ~ lJC1 + C2

C = 51lF=> Cx=C=5IlF

Cy = 4C = 4 x 51lF= 20 IlF (1/2 x 2)

68

(ii) Charge across the combinationq = Ceq V

= (4~F) x 12

=48~C.. Same charge, i.e. 48 ~C lies on each capacitorbeing in series combination.

.. Potential difference across Cx'Vi =.!!....= 48 ~C = 9.6 V

c, 5~FPotential difference across Cy'

V2

=.!!....= 48 ~C = 2.4 VCy 20 ~F (1/2 x 2)

(ill) Energy stored in capacitance X-2

U =-q-x 2Cx

'~y stored in capacitor Y,2

U =!Ly 2Cy

[ Since, in series, the charge will remain same]

. Ux = ~ x 2Cy = Cy = 20 ~F = 4• • 2

Uy 2Cx q c, 5~F

Ux:Uy=4:1

51. Given, total energy =160mJ

Let equivalent capacitance of the combination ofcapacitors is C.According to the question,

~ CV2 =160X10-3 J2

~ x C X (200)2 = 160 X 10-3

2=> Equivalent capacitance, C = 8 ~F

=>

.: Equivalent capacitor of 7 ~F and 3~F connectedin parallel = 7 + 3 = 10 ~F

:. 1O~F,1O~Fcombinationof7~F, 3~F and l sufare in series combination. Therefore, theirequivalent capacitance

1 1 1 1 3+ 3+ 2 8-=-+-+'-C' 10 10 15 30 30

C' _15 F-4~But, C' and Z are in parallel combination,therefore equivalent capacitance

~ + Z = C [from Eq.(i)]4

(2] Chapterwise CBSE Solved Papers PHYSICS

52.

~+Z=8 =>Z=8-~ _32-154 4 4

17Z = - ~F => Z = 4.25 ~F4 m

(i) According to question

+++ E+ ~- •+ P Q

+A + - A

I+-- d--+l

(a) Electric field due to a plate of positivecharge at point P = ~

2Eo

Electric field due to other plate = ~2Eo

Since, they have same direction, so

E =~+~=~2Eo 2Eo Eo

(1)

Outside the plate, electric field be zerobecause of opposite direction. (1)

(b) Potential difference between the plates isgiven by

V _. Ed= odEo

(c) Capacitance of the capacitor is given by(':Q = CV)

C = g = aA Eo = EoAV ed d

(ii) According to question,

(1)

(1)

(1)

(1)

Potential at the surface of radius R,

= kq [':q = 0 x 4R2]R

ko4rcR2=> --- = ok4rcR = 4kCJ1tR

RPotential at the surface of radius 2R,

= kq [':q = 0 x 4rc(2R)2 = 16CJ1tR2]2R

So, ko16rcR2

= 8kCJ1tR2R

Since, the potential of bigger sphere is more. So,charge will flow from sphere of radius 2R to sphereof radius R. (2)

CHAPTER 2 Electrostatic Potential And Capacitance 69

53. (i) When a capacitor is placed in an externalelectric field, the free charges present insidethe conductor redistribute themselves in sucha manner that the electric field due to inducedcharges opposes the external field within theconductor. This happens until a staticsituation is achieved, i.e. when the two fieldscancels each other and the net electrostaticfield in the conductor becomes zero.In contrast to conductors, dielectrics arenon-conducting substances. i.e. they have nocharge carriers. Thus, in a dielectric, freemovement of charges is not possible. It turnsout that the external field induces dipolemoment by stretching molecules of thedielectric. The collective effect of all themolecular dipole moments is the net chargeon the surface of the dielectric whichproduces a field that opposes the externalfield. However, the opposing field is soinduced that does not exactly cancel theexternal field. It only reduces it. The extent ofthe effect depends on the nature of dielectric.Both polar and non-polar dielectric developnet dipole moment in the presence of anexternal field. The dipole moment per unitvolume is called polarisation and is denotedby P for linear isotropic dielectrics. (2'12)

P=XE(ii) (a) At point C, inside the shell.

The electric field inside a spherical shell iszero. Thus, the force experienced bycharge at the centre C will also be zero..: Fe = qE (Einside the shell = 0):. Fe = 0

At point A, IFAI = 2Q[_1_ 3;]41tEo X2

3Q2F = ---2' away from shell

47tEox(b) Electric flux through the shell,

~= ~ x magnitude of the charge enclosedEo

by the shell.~ ~=~xg=~

Eo 2 2 Eo (2'12)

54. (i) In order to calculate the energy stored in thecharge configuration, suppose the conductors1 and 2 are initially uncharge. Let positivecharge be transferred from conductor 2 toconductor 1 in very small installments of each

till conductor 1 get charge +Q. By chargeconservation, conductor 2 would get charge-Q.

+Qt==J-Q+-+-+-+-+-

1 2

At every stage of charging, conductor 1 is athigher potential than conductor 2. Therefore,work is done externally in transferring eachinstallment of charge.:. Potential difference between conductors 1 and2 isi

C

.. Potential of condenser = i .C

Small amount of work done in giving anadditional charge dqto the condenser,dW=ixdq

C:. Total work done in giving a charge Q to the

condenser,

jq=Qq 1[q21q=QW= q=OC=C 2'1-1,=0

1 Q2~ W=--

C 2As, electrostatic force is conservative, this work isstored in the form of potential energy (U) of thecondenser.

IdU=W=--2C

Hence,

Q=CV

U = ~ (CV)2 = ~CV22 C 2

1CV=Q ~ U=-QV2

1 d 1 2 1U=-- =-CV =-QV2 C 2 2

Energy density (U) is defined as the total energyper unit volume of the condenser.

~CV2=_2_

Adi.e. U = Total energy (V)

Volume (V)

C = EoA and V = Edd

Using

70

(ii)

1 (EoA) (E 2d2) 1 2We get, U="2 d Ad ="2EoE

Here, E is the strength of electric field in thespace between the plates of the capacitor. (21/2)

Initial conditionIf we consider a charge capacitor, then its chargewould be given, q = CV

«c»A"'--H--B

and energy stored in it is given by1 2UI = -cv ...{i)2

When this charged capacitor is connected touncharged capacitor,

C1·V1e~Let the common potential be Vi, the charge flowfrom first capacitor to the other capacitor unlessboth the capacitor attain the common potential.~ QI = CVi and Q2 = CV2

Applying conservation of charge,Q=QI + Q2 ~ CV=CVI + CV2

V~ V = Vi + v, ~ VI =-- 2Total energy stored on both the capacitor

I 2 1 2 1 ( V )2 1 ( V )2U2 = -CVI + -CVI ~ U2 = -C - + -C -2 2 2 2 2 2

2CV2 I 2U2 = -- = -CV ... {ii)8 4

From Eqs. (i) and (ii). we getU2< UI

It means that energy stored in the combination isless than that stored initially in the singlecapacitor. (2112)

55. (i)-qo I o---------- .• p

o +q~------ r ------~

Let P be an axial point at distance r from the centreof the dipole. Electric potential at point P will be

o ehapterwise eSSE Solved Papers PHYSICS

V = Vi + V2= _1_. (-q) + _1_._q_

41tEo r + a 41tEo r - a

= _q_[_I 1_] = .s.i.:»:41tEo r - a r + a 41tEo r2 - a2

1 P= 41tEo . r2 _ a2

For a far away point, r » > a.. V = _1_.!... or V oc ~

41tEo r2 r2

Thus, due to a dipole potential at a point is V oc -; .r

(2112)

(ii) Let A -7 area of each plate and CI and C2 arecapacitance of each slab.

L ... 11 C C EoA ,et iruna y, I = = - = (,2d

[':p = q (2a)]

After inserting respective dielectric slabs.C; =KC

d C' - K Eo{A/2) K Eo{A/2) - EoA (K K ).an 2- 1---+ 2----- 1+ 2'd d 2d

C2' = ~ (KI + K,) ... (ii)2 -

From Eqs. (i) and (ii). we get"C 1CI =C2~KC=-{KI +K2)~K=-{KI +K2)

2 2 (2112)

56. (i) For (a) and (c), refer to Ans. 39{i) and (ill)For (b), refer to Ans. 44 (i). (3)

(ii) (a) Electric field lines due to a conductingsphere are shown in the figure.

---"-{V)----- - - ~ Conducting

sphere havingnegative charge (1)

(b) Electric field lines due to an electric dipoleare shown in the figure

p -E

(1)

CHAPTER 2 : Electrostatic Potential And Capacitance 71

57. As, charge on capacitor increases, we have towork more against electrostatic repulsion and thisamount of work done will be stored as potentialenergy in the capacitor.(i) We know that, q =CV ~ V =qt C

dW = Vdq = !idqC

where, q = instantaneous charge,C = instantaneous capacitance andV = instantaneous voltage

:. Total work done in storing charge from 0 to qq 2

is given by W = f !idq =!Lo C 2C (2)

(ii) In series combination of capacitors, samecharge lie on each capacitor for any value ofcapacitances.

+0 -0 +0 -Q +O-Q!=H!=H!=+- +- +-+- +- +-C1 C2 C3

V1 ------1- V2--+- V3

v(+) (-)

Capacitors in series combination (1)

Also, potential difference across the combinationis equal to the algebraic sum of potentialdifferences across each capacitor.i.e. V=Vi+V2+V} ... (i)where, Vi ' V2, V} and V are the potentialdifferences across C, ' C2, C} and equivalentcapacitor, respectively.

Vi =~C,

Similarly, V2 = ~ ~ V} = ~C2 C}

:. Total potential difference [from Eq. (i)]

V=~+~+~C, C2 C}

VIII 1111-=-+-+-~-=-+-+-q C, C2 C} C C, C2 C}

[.: ~ = 2.., where C is equivalent capacitance ofq C

combination]or 2.. = C2 C} + C} C, + C, C2

C C, C2 C}

C = C, C2 C}C, C2 + C2 C} + C} C,

58. (i) Refer to Ans. 39 (i). (3)

(ii) Refer to Ans. 32 (i). (2)

59. Let small sphere has charge q and radius a isplaced inside a outer shell of charge + Q andradius b.

Insulated___~_suspension

+0

(1)

Electric potential on the small sphere due to itsown charge q.

1 qVi =--'- ...(i)41tEo a

where, q = charge on small spherea = radius of small sphere

Similarly, electric potential on outer sphere due toits own charge

V2=_I_.Q ... (ii)41tEo b

where, Q = charge of outer shellb = radius of outer shell. (1 x 2 = 2)

Also, same potential V2 exists at every point insideouter shell due to its own charge, + Q.Now, net electric potential at inner sphere ofradius a.1'; = Electric potential due to its own charge

1 q 1 Q1'; = -- - + --- ... (iii)

41tEo a 41tEo bNet electric potential at outer sphere due to chargeon the both spheres

1 q 1 QVo =--'- + --'- ...(iv)41tEo b 41tEo b

.. 1'; - Vo = -q- (2.. - 2..) ... (v)41tEo a b

From Eqs. (iii) and (iv), we get1 1': a < b and - > - ~ 1'; - Vo > 0a b (2)

Thus, inner sphere has net potential higher thanpotential of outer sphere for every value of q and Q.Therefore, when they are connected by a wire,positive charge always flow from inner sphere (athigher potential) to outer sphere (at lower potential)irrespective of the magnitude of charge. (1)(1)

Value Based Questions (From Complete Chapter)rlI 4 Marks Questions

1. Immediatly after school hour, as Bimlawith her friends came out, they noticedthat there was a sudden thunderstormaccompanied by the lightning. They couldnot find any suitable place for shelter. Dr.Kapoor who was passing thereby in hiscar noticed these children and offeredthem to come in their car. He even tookcare to drop them to the locality wherethey were staying. Bimla's parents, whowere waiting, saw this and expressedtheir gratitude to Dr. Kapoor. All Indio 2015 C

(i) What values did Dr. Kapoor andBimla's parent displayed?

(ii) Why is it considered safe to be inside acar especially during lightening andthunderstorm?

(iii) Define the term 'dielectric strength'.What does this term signify?

••••• (i) Helpingattitude, kindness, concern for thechildren, awareness, application of knowledge.

(1)

(ii) Ascar is a goodconductor, its made up ofmetallic body so it will behave as a lightningconductor. (1)

(iii) The maximum electicfield that can exist in adielectricwithout causing the breakdown ofits insulating property is called dielectricstrength. It is usually expressed as volts/unitthickness. (2)

2. In Pradeep's classroom, the fan wasrunning very slowly. Due to which, histeacher was sweating and was restlessand tired. All his classmates wanted torectify this. They called an electrician whocame and changed the capacitor only,after which the fan started running fast.Answer the following questions based onthe above information.

(i) What energy is stored in the capacitorand where?

(ii) A thin metal sheet is placed in themiddle of a parallel plate capacitor. Whatwill be the effect on the capacitance?

(iii) What values did the classmates have?

Ans. (i) Electricalenergy is stored in the capacitor. It isstored in the dielectric. [1]

(ii) Noeffect. When the metal sheet is placed inthe middle the new arrangement is equivalentto a old combination of two capacitorseach ofplate separation t!.- and hence capacitance 2C.

2

Cs

= 2Cx2C =C2C+ 2C

(iii) Teamwork, concern, respect to teacher andresponsibility.

3. An old woman who had suffered from a'heat stroke was taken to the hospital byher grandson who is in class XII. Thegrandson has studied in Physics that, howto save person who is suffering from aheat stroke, regular beating of the heat isto be restored by delivering a jolt to theheart using a defibrillator, whose capacityis 70 IlF and charged to a potential of5000 V and energy stored is 875 J, 200 Jof energy is passed through a person'sbody in a pulse lasting 2 ms. The oldwoman gets paniced and refuses to betreated by defibrillator. Her grandsonthen explained to her the process thatwould be adopted by medical staff andhow the result of that would bring herback to normalcy. The woman was thentreated and was back to normal.Answer the following questions based onthe above information.

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

(ii) What will be the net charge of thecapacitor in defibrillator?

Ans. (i)Presence of mind, knowledge of subject,concern for his grandmother, empathy helpingand caring, [2]

(ii) The net charge of the capacitor in defibrillatoris givenby

(1)

[2]

[where, C= capacitance]

V = voltageQ = 70 X 1O-{i X 5000 = 35 X 10-2C [2]

Q=CV

CHAPTER 2 : Electrostatic Potential And Capacitance 73

4. Shikhaj was working on a project forscience exhibition. He considered acapacitance of 2 ~F having a capacity ofoperating under 1 kV potential. Whenhe reached to shop, he found that theshopkeeper is having a capacitors of1 ~F of 400 V rating. Shikhaj calculatedminimum number of capacitances of1 ~F each, so he could arrange to form acapacitor of 2 ~F.Answer the following questions basedon the above information.

(i) What are the calculations done byShikhaj?

(ii) What do you think of Shikhaj?Ans. (i) Total potential difference across each row

=1000 V

Potential difference across each capacitor=400V

:. Number of capacitors in series,

n = 1000 = 2.5 = 3400

Capacitance of capacitors in series, C, =!. IlF3

Let m be rows of capacitors, for anequivalent capacitance of 21lF, we have

mx!'=23

~ m=6Hence, total number of capacitance required

= mx n=6x3=18

Total 6 rows of capacitor in parallel withthree capacitors in each row. (1)

(ii) We think that Shikhaj knows the value ofmoney and have the ability to calculatescientifically in daily life. (2)

(1)

5. A man travelling in a car during heavy rainand thunderstorm, sees a boy standing .,under a tree. He immediately stops his carand asks the boy to get inside the car and,'ilsaves him from a possible natural calamity,Answer the following questions based on theabove information.

(i) What danger did the boy had whilestanding under the tree during theth understorm?

(ii) How is it safer inside the car duringsuch weather?

(iii) What according to you are the valuesdisplayed by the man to help the boy?

(iv) Give another example from everydaylife situations which represent displayof similar values.

Ans. (i) Trees often being much taller than surroundingstructures, will attract lightning and the tree canget on fire. The boy might get electrocuted. Also,if the lightning strikes the tree, then the treemight fall on the boy. (1)

(ii) During thunderstorms, it is safer to -sit inside acar because car has a metallic body that does notallow the lightning to hit the person sitting· ",,",inside the car. We know that charges alwaysreside on the surface of metal. Inside theconductor, charge is zero. If lightning strikes thecar during a thunderstorm, the charges will bedistributed on the surface of the metal body ofcar. No charges will exist inside the car. Hence,the person sitting inside the car isconsidered safe are the values displayed. (1)

(ill) Helpfulness; Compassion, General awareness,service to others are the values displayed. (1)

(iv) Advising a person who is doing repair work onair condition system without switching OFF themains.Stopping a person entering nuclear medicineward in a hospital without proper protection orprecautions. [1/2 + 112)