electrostatics pdf

8/13/2019 Electrostatics PDF

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Preface

We begin our study of electrostatics in this chapter by examining the nature ofelectric charge .We will find that electric charge is quantized and that it obeys aconservation p rinciple . We canturn to a discussion of the interactions of electric charges that are at rest in our frame of reference,called electrostatic interactions. Such interactions are exceedingly important.

They hold atoms, molecules and our bodies together and have numerous technological applications.Electrostatic interactions are governed by a simple relationship known as Coulomb's Law andare most conveniently described by using theconcept of electric field .

We will explore all these concepts in this chapter and expand on them in the following chapters.In later chapters we will expand our discussion to include electric charges in motion. While thekey ideas of electromagnetism are conceptually simple, applying them to practical problems willmake use of all of your mathematical skills, especially your knowledge of geometry and integralcalculus. For this reason you may find this chapter and those that follow to be more mathematicallydemanding than previous chapters. The reward for your extra effort will be a deeper understandingof principles that are at the heart ofmodern physics and technology.

This book consists of theoritical & practical explanations of all the concepts involved in the chapter.Each article followed by a ladder of illustration. At the end of the theory part, there are miscellaneoussolved examples which involve the application of multiple concepts of this chapter.

Students are advised to go through all these solved examples in order to develope bettter understanding of the chapter and to have better grasping level in the class.

ELECTROSTATICS

Total number of Questions in this chapter are :

(i) In chapter Examples ....................... 61

(ii) Solved Examples ....................... 22

Total no. of questions ....................... 83


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Electrostatics , deals with the study of chargesin rest. These stationary charges occurs due tofriction of two insulating bodies, therefore it isoften called frictional electricity

1. FUNDAMENTAL FORCE OF THE NATURE(i) Behind every process occurring in the nature,

there is one or the other force acting.(ii) Different forces are divided in FOUR parts

based on their nature(A) Gravitational force(B) Electro-magnetic force(C) Nuclear force(D) Weak force

(iii) Comparative analysis of forces

Force Nature Range Rel. StrengthGravitational Attraction very large 1Electro- Attraction very large 1036Magnetic or

RepulsionNuclear Attraction very less 1039Weak Unknown very less 1014

Important point s :

(i) Gravitational force is the weakest whilenuclear force is the strongest force of thenature

(ii) Nuclear force does not depend upon charge.It acts equally between proton-proton, protonneutron and neutron-neutron.

(iii) There are weak forces acting in -degradiation in radio-activity.

(iv) A stationary charge produces electric filedwhile a moving charge produce electric aswell as magnetic field.

(v) Moving charge produce electric field as wellas magnetic field but does not radiate energy

while uniform acceleration.(vi) Accelerated charge produce electrid field aswell as magnetic field and radiate energy.

2. CHARGE

Property of a substance by virtue of which it canrepel or attract another charged substance.Charges are of two types.(a) Posit ive charge : Lesser number of electrons

than number of protons.

(b) Negative charge : More number of electronsthan number of protons

Impor tan ts Poin ts : Only , e lec t ron i sresponsible for a substance to be chargedand not the proton.

2.1 Properties of Charge :(i) Like charges repel while unlike charges attract

each other.(ii) Charge is quantized in nature i.e. The

magnitude of charge possessed by differentobjects is always an integral multiple of charge of electron (or proton) i.e. q = newhere n = 1 , 2 , 3 ........

(iii) The minimum possible charge that can existin nature is the charge of electron which hasa magnitude of e = 1.60 207x10-19 coulomb.

This is also known as quantum of charge or fundamental charge.(iv) In an isolated system the algebraic sum of

total charge remains constant. This is thelaw of'Conservation of charge'.

Note : The fact that electric charge is an integralmultiple of electronic charge wasexperimentally proved by Milliken. Unit of charge 1 coulomb = 3 109 e.s.u.= 1/10 e.m.u.,in cgs e.s.u. (state coulomb)

Charge

Ex.1 Two spheres of the same metal (in allrespects) are taken. One is given a positivecharge of Q coulomb. and other is given thesame but negative charge. Which sphere willhave a higher mass.

Sol. Negatively charged sphere will have a higher mass. This is due to increase in number of electron to make it negatively charged.

Ex.2 Which of the following charge is not possible:(A) 1.6 10 18c (B) 1.6 10 19 c(C) 1.6 10-20 c (D) None of these

Sol. (C) 1.6 10 20 c, because this is 1/10 of electronic charge and hence not an integralmultiple.

Ex.3 How many electron are present in 1 coulombcharge.

Sol. q = ne q = 1c e = 1.6 10 19 c n = ? n = q/e = 6.25 1018 electrons.


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Ex.4 Identify X in the following nuclear reactions(i) H11 + Be94 X + n(ii) N157 + H11 He42 + X

Sol. (i) Charge on the nucleus of x = Q(H) + Q (Be) = 1 e + 4e = 5e[ Q (n) = 0 ]Therefore X will be Boron(The atomic no. of B = 5]

(ii) Q (X) + Q(He) = Q(N) + Q(H) Q (X) = 7 e + 1 e - 2e = 6e X is carbon

3. COULOMB'S LAW

The force of attraction or repulsion between twostationary point charges is directly proportionalto the product of charges and inverselyproportional to the square of distance betweenthem. This force acts along the line joining thetwo. If q1 & q2 are charges in consideration r,the distance between them and F, the force actingbetween them

Then , F q1 q2F 1/r 2

F 221

r qq

F k 221

r qq

, where k = constant.

k = K41

0=

K109 9 N1 m2 coulomb 2

where ,

0 = Electric permittivity of vacuum or air = 8.85 x 10 12 coul2 N 1 m 2 and

K = Relative permittivity.= Dielectric constant= Specific inductive capacity

[Newton's law for particles is analogous tocou l om b ' s l aw fo r r e s t cha rge s . Thedif ference is that Newto n's law giv esattraction force while coulomb's law givesattraction as well as repulsion force]

Note:(i) Coulomb's law is applicable to point charges

only. But it can be applied for distributedcharges also

(ii) This law is valid only for stationary chargesand cannot be applied for moving charges.

(iii) This law is valid only if the distance betweentwo charges is not less than 10 15 m.

(iv) K = 1 for air or vacuum,

for conductors> 1 for any other medium.

Medium KVacuum/air 1'Water 80Mica 8Glass 5-10Metal

Note : Be aware of k and KK is d ielectric constant and k electrostaticconstant

k = K41

0DirectionDirection of the force acting between two chargesdepends upon their nature and it is along the line joining two charges.

21F = force on q2 due to q1

21F = 2120

1221Kr 4r qq

...(A)

12F = Force on q1 due to q2

12F = 212120

2121 r Kr 4r qq

...(B)

Note:

1. | 21r | = | 12r | = 1 (unit vectors)

2. 21r = 12r ...(C)

3. 12r = r 214. Values of q1 & q2 are put with sign while

using this formula

5. From (A) , (B) and (C)F 12 = F 21


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RR

d

+++

+

--

---

-- -+

+ +

+ - -

-

4. PRINCIPLE OF SUPERPOSITION

The resultant force acting on a charge due to agroup of charges is equal to the vector sum of individual forces.

321 FFFFq2

q 1q

3F

1F

2F

Principle of superposition

Ex.5 Five equal charges 'q'are placed at fivevertices of a regular hexagon. What will bethe resultant force ona charge 'Q' placed atthe centre of thehexagon given that the distance of a corner from centre is d.

Sol. Suppose , the same charge 'q' was placed atsixth corner also Then

0FFFFFFF 654321

(Note that resultant is zero due to symmetryof hexagon. This is applicable for anyREGULAR geometry)

654321 FFFFFF

= 20d4

qQ

and direction of force will be opposite to6F .Ex.6 A point charge q1 exerts a force F on q2. An

equal charge q3 is now kept near q2. Theresultant force on q2 due to q1 will be -

Sol. F. here superposition principle is to beapplied carefully. The force on q2 due to q1will remain same although resultant force on

q2 will change since 31 FFFEx.7 Find the ratio of electrostatic and gravitational

force acting between two electrons -

Sol. Fe = 20 r e.e.

41

; Fg = G. 2r m.m

g

eFF

= 20

2

0 m.G)4(e.

41

1043

Note : 1. ge

F

F

for proton - proton = 1036

2.g

eFF

for proton electron = 1039

Ex. 8 Force F is acting between two charges. If asheet of glass ( r = 6) is placed between thetwo charges, what will be the force.

Sol. F = 221

0 r qq.

41

F' = 2210 r

qq.K4 1

F' = KF = 6

F

Note : We can conclude that if there is a metallicmedium (conducting) between two charges, forcewill be zero since K =.

Ex.9 Two charged spheres of radius 'R' are kept ata distance 'd' (d >2R). One has a charge +qand the other - q. The force between them

will be

(1) 22

0 dq

41

(2) > 22

0 dq

41

(3) < 22

0 dq

41

(4) None of these

Sol. (2) Redistribution of charge will take placedue to mutual attraction and hence effectivedistance will be less than d.

Note : In the example above, if both had thecharge '+q', the answer would have been (3)because now mutual repulsion will result intoincrease in effective distance.

Ex.10 How should we divide a charge 'Q' to getmaximum repulsion between them

q

q16 2q

qq 5 4

3


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Sol. Let (q) & (Q q) be the two parts .

F = 20 r

)qQ(q4

1

For maximum F

dqdF

= 0 20 r

q2Q4

1 = 0

q = 2Q

hence Q should be divided in two equal parts.

Ex.11 3 10 19 C and 10 6 C are placed at(0 , 0, 0) and (1, 1,1) respectively. Find theforce on second in vector form

Sol.

21F = 1221221

0 r

r qq

41

k)01( j)01(i)01(r 12 = k ji

| 12r | = 222 111 = 3

12r = |r |

r

12

12 = 3

k ji

21F = 3)10(103109 6199 . 3

k ji

= 3 10 16 )k ji( Newton.Ex.12 Three charges (each q.) are placed at the

corners of an equilateral triangle. Find outthe resultant force on any one charge due toother two.

Sol. F = 60cosFF2FF 212221

ButF 1 = F2 = 22

0 aq

41

F = 22

0 aq3

41

Ex.13 Two charges 1 c and 5 c are kept at adistance 4cm. The ratio of magnitude of forceexperienced by first to the second will be -

Sol. 1 : 1

2112 FF

= | 12F | = | 21F |

5. ELECTRIC FIELD

A charge produces something called an electricfield in the space around it and this electric f ieldexerts a force on any charge placed in it.Note : The electric field doesnot exert force on

source charge.5.1 Electric field Intensity -

Force experienced by a unit positive chargeplaced in an electric field at a point is calledelectric field intensity at that point. It is alsoknown as electric field simply. Let q0 be thepositive test charge placed in an electric field.If F is the force experienced by this charge,then

E = Electric field intensity =00q q

Flim0

(i) Unit : Newton / coulomb or volt/metre(ii) This is a vector

quantity and itsdirection is thesame as force on thepositive test charge.

(iii) SinceE is the forceon unit charge, forceon charge q is -

F = qE .(iv) Dimension is [M1 L1 T 3 A 1](v) Electric field due to a point charge is

E = r .r kq2

(vi) Direction of electric field due to positivecharge is away from charge while direction of electric field due to negative charge istowards the charge.

F2 F1

Q Q

600

F

Q

a

a

a

600 300


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Special point

(a) If q1 and q2 are at a distance r and both havethe same type of charge, then the distance 'd'of the point from q1 where electric field is

zero is given by d = 211

qqr q

. This pointwill lie between line joining q1 & q2.

(b) If q1 and q2 have opposite charges thendistance 'd' of the point 'p' from q1 whereelectric field is zero is given by

d =21

1

qq

r q

, [|q1| > |q2|]

(c) Three charges +Q1, +Q2 and q are placed ona straight line. If this system of charges isin equillibrium, charge q should be as given

22121

QQQQq

5.2 Principle of superposition for electric fieldintensity -Resultant electric field intensity at a point p dueto a number of charges is vector sum of individual

electric field intensities 321p EEEEElectric field is represented by electric lines of forcesThe resultant two electric fields E1 + E2 is givenby E = cosEE2EE 212221 . If the resultant

field E, makes an angle with E1 thentan = cosEE

sinE21

1

5.3 Electric lines of forces :The electric field in a region can be representedby drawing certain curves known as electric linesof force.An electric line of force is that imaginary smoothcurve drawn in an electric field along which a freeisolated unit positive charge moves.

Properties -(i) Electric lines of force start from a positive

charge and end on a negative charge .(ii) No two lines of force can intersect each other.

If they does so then at the point of intersection twotangents could be drawn ,which gives two directions of electric at thesame point , which is impossible.

(iii) The tangent drawn at any point on line of force gives the direction of force acting on apositivecharge placed at that point.

(iv) These lines have a tendency to contract in lengthlike a stretched elastic string. This actuallyexplains attraction between opposite charges.

(v) These lines have a tendency to separate fromeach other in the direction perpendicular totheir length. This explains repulsion betweenlike charges.

(vi) Intensity of electric field is given by thenumber of electric lines of force in a unitarea at that point.

(vii) Lines of force of a uniform field are paralleland at equal distance.

(viii) Unit positive charge givesK4 lines in a

medium of dielectric constant K.(ix)Important : Electric lines of force can never

enter the conductor, because inside theconductor the intensity of electric filed is zero.

(x) Important : Lines of force leaves the surfaceof conductor normally.

Electric field

Ex.14 Charges of 3e and 9e are placed at adistance r. What is the distance of the pointform 9e where electric field is zero.

Sol. Putting the values in above formula

d =21 qq

r q = e3e9

r .e9

=3

3 1

r


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Systematically : E1 = 20 xe9.

41

E2 = 20 )xr (e3.

41

, E1 = E2

2xe9 = 2)xr (

e3

x = 13r 3

or 13r 3

x = 13r 3

is not possible since x < r

Ex.15 Which is true ?(A) EA < EB > EC (B) EA > EB > EC(C) EA > EB < EC (D) EA < EB < EC

Sol. (B) Number of electric lines of force in unitarea is maximum at A and least at C .

so EA > EB > EC .

Ex.16 A metal sphere is placed in an uniform

electric field which one is a correct electricline of force-

Sol. (4) Only 4 is normal to the conducting surface.

Ex.17 A charge particle is free to move in electricfield will it always move along the electriclines of force.

Sol. No. If the particle has its initial velocity = 0,then it will move along the lines of force butif the initial velocity makes some angle withlines of force, the resultant path will not bealong the lines of force.

Ex.18 Find E at point P.

Sol. E1 (due to 5 c) = 9 109 2

6

)1.0(

105

= 4.5 106 N/C

E2 (due to 3.6 c) = 9 109 26

)06.0(106.3

= 9 106 N/C Ex = E1x + E2x = E1cos + 0

= 3.6 105 N/C Ey = E1y + E2y = E1 sin + E2

= 6.3 106 N/C

E = 2y2x EE = 7.3 106 N/C

Note: To avoid mistakes, always take the givenquantities in SI units and final answer will alsobe in the SI units.

Ex.19 The given charge Q is positive or negative ?

Sol. Q is a positive charge because lines arestarting from it.These lines are supposed to terminate atinfinity (and not at negative charge). If Qwas negative.

8cm

10cm6cm

+5 c 3.6 c

E2

PE1

1 1

2 2

3 3

4 4

Q


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E

A B

6. ELECTRIC POTENTIAL

Work done in bringing a unit positive charge frominfinity to any point is termed as potential at thatpoint i.e. if W = work done in bringing a positive charge q0

from infinity to that point, then , V =0q

W

(i) Electric potential at infinity is taken to bezero.

(ii) It is not path dependent quantity if simplydepends upon the starting and end points.

(iii) It is a scalar quantity.(iv) Unit : Volt or Joule/Coulomb

(v) Dimension : [M1

L2

T 3

A 1

](vi) Potential due to a positive charge is positiveand potential due to a negative charge isnegative, hrere potential being positive andnegative implies whether work is done on thecharge or done by the charge respectively.

(vii) Potential due to a point charge Q at a

distance r is V = r q

41

0

V r

1

(viii) Total potential at a point due to a group of charges is scalar sum of individual potentials Vp = V1 + V2 +....Vn

(ix) Electric field is gradient of electric potential

at that point. E = dr dv

Note : The negative sign implies that directiono f e l e c t r i c f i e l d i s i n t he d i r ec t i on o f decreasing p otantial.

(x) Work done in bringing a charge Q from infinityto that point isW = QV where V is potential at that point.

(xi) Potential of earth is taken to be zero.[Allthough the earth's negatively chargedsphere, yet its potential is zero. It is because

of its large capacitance (C = 4 0R] C = Vq

V = q/C, V 0 , as C is too large]

Electric Potential

Ex.20 Can metal sphere of 1cm radius held a chargeof 1 coulomb ? [Air gets ionized at theElectric field of E max = 3 x 10

6 volt/m]

Sol. No. The potential of a metal sphere of radius1cm is given by

V = r q

41

0 = 9 109 2101

1= 9 1011

The potential is much greater then needed toionise the air and hence the charge leakesto surrounding air.[Air gets ionized at thepotential of 3 x 10 6 volt]

Ex.21 In the given diagram VA < VB since directionof E is from B to A.

Ex.22 Infinite number of same charge q are placedat x = 1 , 2 , 4 , 8 ...... What is the potentialat x = 0 ?

Sol. V =04

1

.....

8q

4q

2q

1q

=04 q

211

1=

04 q2 = 02 q

[ a + ar + ...... =ar 1

a r < 1

Ex.23 If the alternative charges are unlike, thenwhat will be the potential ?

Sol. Then , V =

......

8q

4q

2q

1q

41

0

=

....

8q

2q.....

2q

1q

41

0

=

411

121

411

14

10 = 3

q24

10

[ a + ar + ar 2 ..... r 1a r ]


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Ex.24 A charge +q is fixed at each of the pointsx = x0 , x = 3x0 , x = 5x0 ... ad inf. on thex-axis, and a charge -q is fixed at each of the points x = 2x0 , x = 4x0 , x = 6x0 ......ad inf. Here x0 is a positve constant. Takethe electric potential at a point due to acharge Q at a distance r from it be Q/4 0r.Then , the potential at the origin due to theabove system of charges is

(A) 0 (B) 2logx8q

00

(C) (D)00 x4

2logq

Sol. Total potential caused by +q , at the origin

V1 =......

x5

kq

x3

kq

x

kq000

=0x

kq [1 + 5

131

+......... ]

Total potential caused by q , at the origin

V2 = .....x6kq

x4kq

x2kq

000

= .....61

41

21

xkq

0

Net potential at the origin V = V1 + V2

= .....51

41

31

211

xkq

0

=0x

kqlog (1 + 1)

[ log(1 + x) = 1 12

x2 +13

x3 14

x4 + ......... ]

= 0xkq

log2 = 00 xq

41

log2. Hence answer is (D)

6.1 Potential difference :

The work done in taking a charge from one pointto the other in an electric field is called thepotential difference between two points.Thus , if w be work done in moving a charge q0form B to A then the potential difference is givenby-

VA VB =0

qW

(i) Unit of potential difference is volt.(ii) This is a scalar quantity(iii) Potential difference does not depend upon

Co-ordinate system(iv) Potential difference does not depend upon

the path followed. This is , because electricfield is a conservative force field and workdone is conservative force field does notdepend upon path followed.

Ex.25 In the following fig. Along which path the workdone will be maximum in carrying a chargefrom A to B in the presence of any another charge

Sol. Same for all the path

[Because the work done doesn't depend uponthe path]

Ex.26 A charge 20 C is situated at the origin of X-Y plane. What will be potential differencebetween points (5a, 0) and 3a , 4a)

Sol. Distance between (0 , 0) & (5a ,a),

r 1 = 0a25 2 = 5a

V1 = a5kq

Distance between

(0, 0) & (3a , 4a) r 2 = a16a9 2 = 5a

V2 = a5kq

V1 V2 = 0


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6.2 Relationship between electric potential andintensity of electric field

(i) VA = A

dr .E , VA = electric potential at

point A .(ii) Potential difference between two points in an

electric field is given by negative value of lineintegral of electric field i.e.

VB VA = B

A

dr .E

(iii)E = V = grad

= (gradient) =

kx

jy

ix

Ex = xV

, Ey = yV

, Ez = zV

(iv) If v is a function of r only , then E = dr dV

(v) For a uniform electric field , E = r V and

it's direction is along the decrease in thevalue of V.

Ex.27 Electric potential for a point (x , y , z) isgiven by V = 4x2 volt . Electric field at point(1 , 0 , 2) is-

Sol. E = dxdV

= 8x

E at (1, 0 ,2)= 8 V/mMagnitude of E= 8V/m direction along x axis.

Ex.28 Electric field is given by E =2x

100potential

difference between x = 10 and x = 20 m.[PET '89 ,94]

Sol. E = dxdV

dV = EdxB

A

dV = B

A

dx.E

VB VA = 20

102x

100= 5 volts

Potential difference = 5 volt.

Ex.29 The potential at a point (x , 0 ,0 ) is given as

V =

32 x500

x1500

x1000

. What will beelectric field intensity at x = 1m ?

Sol. E = V = zV

kyV

jxV

i

or iEx + jEy + kEz = zVk

yV j

xVi

= xV

zV0

yV

Comparing both sides

Ex = xV

= 32 x5000

x1500

x1000

x

= 43 x50003

x15002

x1000

For x = 1 , (Ex ) = 5500 V/m

Ex.30 In the following fig , what will be the electricfield intensity at r = 3

Sol. For 2 < r < 4, V = 5 volts

E = dr dV

= 0

Note : In the above problem, what will value of E atr = 6 ?

at r 2 = 7m V2 = 2 voltat r 1 = 5m V1 = 4 volt

E =

12

12r r VV

=

5742

=1 volt/metre

Ex.31 An oil drop 'B' has charge 1.6 10 19C andmass 1.6 10 14kg. If the drop is inequilibrium position, then what will be thepotential diff. between the plates.[Thedistance between the plates is 10mm]

54321

0 1 2 4 5 6 7 8 9 10

V ( v o

l t s )

3r (meter)


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11

Sol. For equilibrium, electric force = weight of drop qE = mg

q. dV

= mg

V = qmgd

= 19314

106.110108.9106.1

V = 104 volt[When a charged particle is in equilibriumin electric field, the following formula is oftenused qE = mg]

6.2 Equipotential Surface -(i) These are the imaginary surface (drawn in an

electric field) where the potential at any pointon the surface has the same value.

(ii) No two equipotential surfaces ever intersects(iii) Equipotential surfaces are perpendicular to

the electric field lines(iv) Work done in moving a charge from a one

point to the other on an equipotential surfaceis zero irrespective of the path followed andhence there is no change in kinetic energy of the charge.

(v) Component of electric field parallel toequipotential surface is zero.

(iv) Nearer the equipotential surfaces , stronger the electric field intensity

Ex.32 Some equipotential surfaces are shown infig.What is the correct order of electric fieldintensity ?

A

B

C

Sol. EB > EC > EA , because potential gradientat B is maximum.

7. POTENTIAL ENERGY OF CHARGED PARTICLE IN

ELECTRIC FIELD

(i) Work done in bringing a charge from infinityto a point against the electric field is equalto the potential energy of that charge.

(ii) Potential energy of a charge of a point isequal to the product of magnitude of chargeand electric potential at that point i.e.P.E. = qV

(iii) Work done in moving a charge from one pointto other in an electric field is equal to changein it's potential energy i.e. work done inmoving Q from A to B = qVB qVA

= UB UAV A VB

A B(iv) Work done in moving a unit charge from one

point to other is equal to potential differencebetween two points.

Note : Circumference of the circle in aboveexample can be considered as equipotentialsurface and hence work done will be zero.

7.1 Potential Energy of System :

(i) The electric potential energy of a system of charges is the work that has been done in

bringing those charges from infinity to near each other to form the system.(ii) If a system is given negative of it's potential

energy , then all charges will move to infinity.This negative value of total energy is calledthe binding energy.

(iii) Energy of a system of two charges

PE = dqq

41 21

0

(iv) Energy of a system of three charges

PE = 31

1323

3212

210 r

qqr

qqr qq

41

(v) Energy of a system of n charges.

PE =

n

ji1 j ij

jn

1ii

0 r q

q4

1.21

+ ++ + + + + +

- - - - - - -

+

10cmB

A


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12

Note : Method to find energy of a system of ncharges.

(a) Find the PE of each charge relative to allother charges.

(b) Add these all(c) Divide the addition by 2 and resultant will be

the potential energy of the system.

7.2 Work Done in An Electric Field -(i) If electric potential at a point is V then

potential energy (PE) of a charge placed atthat point will be qv.

(ii) Work done in moving a charge from A to Bis equal to change is PE of that charge WAB= work done from A to B= PEB PEA = q (VB VA)

(iii) Work done in moving a charge along a closedsurface in an electric field is zero.

(iv) Total energy remains constant in an electricfield i.e. KEA + PEA = KEB + PEBKE = Kinetic energyPE = Potential energy

(v) A free charge moves from higher PE to lower PE state in an electric field. Hence(a) a + ve charge will move form higher

potential to lower potential while,(b) a ve charge will move form lower

potential to higher potential

(c) Work done for displacement throughr

for a charge experiencing a forceF = W = F . r

Electric Potential Energy

Ex.33 A charge Q is placed at the centre of a circleof a radius 'r'. Work done in taking a chargeq from A to diametrically opposite point B.

Sol. Potential energy of q at A

= UA =r

Qq

4

1

0

PE of q at B

= UB = r Qq

41

0

Work done = UB UA = 0Ex.34 What will be change in potential energy of

q3, in moving it along CD for the following fig.

Sol. Potential energy of q3 at C[where q1 = 2 10-8C ,q2 = 0.4 10 8 C , q3 = 0.2 10 8 C]

Uc = k 1qq

8.0qq 2331

[ BC = 22 6080 cm = 410 cm= 102 cm = 1m]

Potential energy of q3 at D,

UD = k 2.0qq

8.0qq 2331

UD UC = kq2q3 11

2.01

= 9 109 0.4 10 8 0.2 10 8 4= 2.88 10 7 Joule

Ex.35 In the following fig, where the charge 'q' mustbe kept , so that the potentail energy of thesystem will be minimum.?

q1 q3r 13

q2

r 32r 12

A BQr

80cm

B D

A80cm

60cmq2q1

q3

9cmq

8q(9x)

2qx


13/28

13

Sol. Suppose the charge q is placed at distancex from 2q.Potential energy of the system

U = k 222 109q8q2

10)X9(qq8

10Xqq2

For U to be minimum XU

= 0 which gives

X = 3cm.Ex.36 The charges of 10 c each are kept at three

corners of an equilateral triangle of 10cmside. What is the potential energy of thesystem ?

Sol. PE of 1 = U1 = r q

41

r q

41 2

0

2

0

PE of 2 = U2 = r q

41

r q

41 2

0

2

0

PE of 3 = U3 = r q

41

r q

41 2

0

2

0

PE of system= 12

(U1 + U2 + U3) = r q.

43 2

0

= 2

269

1010)1010(1093

= 27 Joule.Ex.37 A charge +q is placed at

the centered of a circles.What will be the amountof work done in carryinga charge q' from B to Cin the fig.

Sol. Zero. Because circular path is a equipotentialsurfaceHence VB VC = 0

W = q' (VB VD ) = 0Ex.38 An electron (mass m , charge e) is

accelerated through a potential difference of V volt. Find the final velocity of electron.

Sol. KEi = 0PEi = eV1KEf = 1/2 mv2

PEf = eV2 finalf initiali

KEi + PE

i = KE

f + PE

f

0 + eV1 = 21 mv2 + eV2

2

1 mv2 = e (V2 V1 ) = eV

v =meV2

[When an electron is acclerated throughpotential difference 'V' the following f ormula

is generally used21

mv 2 = eV]

Ex.39 A charge q moves along the path PQRS inan electric field E which is directed towards

positive X-axis. P, Q , R , S, have thecoordinates (a , b , 0) , (2a , 0 ,0 )(a , -b , 0) , ( 0 , 0 ,0 ) respectively. Whatis the work done by electric field in thisprocess ?

Sol. E = Ei , F = qE = q.Ei

displacement =PS

= (0 a)i + (0 b) j+ (0 0)k = a i b j

W = PS.F = iqE.) jbia( = qEa

8. MOTION OF A CHARGED PARTICLE IN AN

ELECTRIC FIELD

(i) Charged particle experience force in anelectric field.

(ii) Magnitude of force on a charge q in anelectric field E is F = qE

(iii) Direction of force on a positive charge issame as direction of electric field while it isopposite to direction of electric field in caseof negative charge.

+q

B

C

B

Y

Q

PE

S

E


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14

Uniform Electric Field Case : 1

Initial velocity is zero or in the direction of electric field -

F = qE

acceleration a = mqE

v = u + at

Distance travelled in time 't' = S = ut +21 at2

Case : 2Initial velocity is perpendicular to electricfield -Distance travelled in X direction = ut

Distance travelled in Y direction =21 at2

where a = mqE

Locus of the path followed -

Y = 22

uax

21

(a parabola)

(iv) Accelerating a charge q through a potentialdifference V results in

(a) decrease in PE = qV(b) increase in KE = qV

(v) In a non uniform electric field electronaccelerates and translates also.

Motion of a charged particl e in anelectric field

Ex.40 An electron is accelerated through 10eV,what will be the velocity acquired by electron.

Sol. We know accelerating charge q through vpotential difference increase in K.E. = qV

12

mv2 = 10eV

v = meV20

=2 x 1.6 x 10 x 10

9.1 x 10

16

-31 m/sec

Ex.41 A particle having a charge of 1.6 10 19Centers midway between the plates of aparallel plate capacitor. The initial velocity of particle is parallel to the plates. A potentialdifference of 300 volts is applied to thecapacitor plates. If the length of the capacitor

plate is 10cm and they and separated by2cm. Calculate the greatest initial velocityfor which the particle will not be able to comeout of the plates. The mass of particle is12 10 24kg.

Sol. The situation is shown in fig .

Here E = ddifferencePotential

= 100/2300

= 15000 mV

As the particle does not come , its maximumdeflection y = 1 cm = 10 2 m

We know that y =2

ux

mqE.

21

or

u2 =2x.

myqE.

21

2

224

19

101

)10)(1012()15000)(106.1(

21

= 108

u = 10 4 m/s

x

y

+ + + + + + + + + + + +

- - - - - - - - - - -

10cm

mq

y =

1 c mu

E


15/28

15

Note : When an electron enters normally to

electric field, it path becomes parabolic,

whi le in magnet ic f ie ld pa th becomes

circular

9. ELECTRIC FLUX

(i) It is denoted by ''.(ii) It is a scalar quantity.(iii) It is defined as the total number of lines of

force passing normally through a curvedsurface placed in the filed.

(iv) It is given by the dot product ofE and normal

infinitesimal areads integrated over a closedsurface-

d = E . ds

=

ds.E = Eds cos

where = angle between electric field andnormal to the area

(v) (a) if = 0, = Eds (maximum)

(b) if = 900, = zero(vi) Unit : (a) Newton - metre2 / coulomb.

(b) Volt - meter (vii) Dimension : [M L3 T 3 A 1 ](vii i)Flux due to a positive change goes out of

the surface while that due to negative changecomes into the surface.

(ix) Flux entering is taken as positive while fluxleaving is taken as negative

(x) Value of electric flux is independent of shapeand size of the surface.

(xi) Flux is associated with all vectors.(xii) If only a dipole is present in the surface then

net flux is zero.(xiii) Net flux of a surface kept in a uniform electric

field is zero.(xiv) Net flux from a surface is zero does not

imply that intensity of electric field is alsozero.

10. GAUSS'S LAW

This law states that electric fluxE through anyclosed surface is equal to 1/0 times the netcharge 'q' enclosed by the surface i.e

E =

ds.E = 0q

Note :The closed surface can be hypothetical and thenit is called a Gaussian surface.If the closed surface enclosed a number of charges q1 , q2 ........... qn etc. then

=

ds.E =0

q=

0n21 )q....qq(

Flux is -(i) Independent of distances between charges

inside the surface and their distribution.(ii) Independent of shape , size and nature of

surface.

ds

E

S

)

Q

q2q1

q3 q4


16/28

16

(iii) Dependent on charges enclosed by surface,their nature and on the medium.

(iv) Net flux due to a charge outside the surfacewill be zero.

(v) If Q = 0 , then = 0 but it is not necessarythat E = 0

(vi) Gauss law is valid only for the vector fieldswhich obey inverse square law

(vii) Gauss's and coulomb's law are comparable.

Note -(i) A charge q is placed at the centre of a cube,

then .......................

(a) Total flux through cube =0

q

(b) Flux through each surface =06

q

(ii) A charge q is placed at the centre of a faceof a cube ,then total flux through cube

=02

q

How ? A second cube can be assumed adjacent tothe first cube total flux through both cubes

=0

q, So flux through each cube =

02q

(iii) Now , q is placed at a corner then the flux

will be08

q

Gauss's Law

Ex.42 A hemispherical surface of radius R is keptin a uniform electric field E such that E is

parallel to the axis of hemi-sphere , Net fluxfrom the surface will be -

Sol. =

ds.E = E . R2. = (E) (Area of surface perpendicular to E) = E. R2.

Ex.43 A rectangular surface of length 4m and breadth2m is kept in an electric field of 20 N/c. Anglebetween the surface and electric field is 300.What is flux thought this surface ?

Note : Angle between surface andE

is given to be30. This is not the '' used in our formula ''is the angle between normal to surface and

E . So here = 90 30 = 60

Sol. = EA cos = 20 8cos60 = 80 V-m

Ex.44 In the following, find out the emerging electricflux through S1 and S2 where[q1 = 1 c , q2 = 2 c , q3 = 3 c]

Sol. 1 = 01q = 12

6

1085.810

= 1.13 105 V.m

2 = 031 qq = 12

6

1085.810)32(

= 11.3 105 V.m

Ex.45 A charge 'q' is placed at the centre of a cubeof side 'a'. If the total flux passing throughcube and its each surface be and 2resepctivley then 1 : 2 will be -(A) 1 : 6 (B) 6 : 1

(C) 1 : 6a2 (D) 6a2 : 1Sol. (B) When q is placed at the centre of cube

then total flux passing through cube is

1 = 0q

and flux through each surface is2 = 06q

1 : 2 = 6 : 1

q

E

q 1S 1

S 4q 4

S 2q 3

q 2


17/28

17

Ex.46 If charges q/2 and 2q are placed at the centreof face and at the corner, of a cube. Thentotal flux through cube will be -

(A)q

20

(B)q

0

(C)q

6 0(D)

q 8 0

Sol. (A) Flux through cube , when q/2 is placedat the centre face , is

1 = 022/q

=04

q

Flux through cube , which 2q is plaed at thecorner of cube , is

2 = 08q2 =

04q ,

Total flux = 1 + 2

=04

q+

04q

=0

q21

Ex.47 Flux entering a closed surface is 2000V-m.Flux leaving that surface is 8000 V-m. Findthe charge insidesurface.

Sol. Net flux = out in = (8000-2000) = 6000V-m

=0

q

q = (6000) (8.85 10 12

) = 0.53 c

11. APPLICATION OF GAUSS'S LAW

11.1 Electric field due to a charged conductingsphere/ Hollow conducting or insulatingsphere.

(i) In all the three type of spheres, chargeresides only on the outer surface of thesphere in order to remain in minimumpotential energy state.

Case: 1 OP = r R

E = r r q

4q

20 = r

r R12

2

0

( = surface charge density)

Case: 2 r = R E = r 0

Case: 3 r < R E = 0i.e. At point interior to a conducting or ahollow sphere, electric field intensity is zero.

(iii) For points outside the sphere , it behaveslike all the charge is present at the centre.

(iv) Intensity of electric field is maximum at thesurface

Imp.

(v) Electric field at the surface is alwaysperpendicular to the surface.

(vi) For points, near the surface of the conductor,

E =0

perpendicular to the surface

(vii) Graphically ,

Electric potential

Case: 1 r < R

Vin = RQ

41

0

in out

+++

+++++

++ + + ++ +

++

++++

++

++++

qp

OR

r

Er

E = 0r > R r

E 12r

r < R r = R


18/28

18

Case: 2 r = R

Vsurface = RQ

41

0

Case : 3r > R

Vout = r Q

41

0

(i) For points interior to a conducting or a hollowsphere, potential is same everywhere andequal to the potential at the surface.

(ii)

(iii) at r = , V = 0

Note : Here , we see that E inside the sphere is

zero but V 0. So E = 0 does not implyV = 0. This presents a good example for it.Similarly V = 0 doesnot imply E = 0

Application of Gauss's Law

Ex.48 When a charged conductor Q is placed insidea hollow conductor P , in such a way that ittouches P , then-(A) charge will flow from Q to P(B) Opposite charge will induced on the outer

surface of P(C) whole of the charge of Q will transfer on

the internal surface of P(D) whole of the charge of Q will transfer onthe outer surface of P

Sol. (D) To keep minimum potential energy wholeof the charge of Q will transfer on the outer surface of P.

(B) For a point p at a distance r from centre o.11.2 Electric field due to solid insul ating sphere

A charge given to a solid insulating sphere isdistributed equally throughout its volume

Electric Field

Case: 1 r > R (point is outside the sphere)

E = r r Q

41

20

Case: 2 r = R (point is at the surface)

E = r RQ

41

20 = Emax = Esurface

Case: 3 r < R (point is inside the sphere)

E = r RQ

41

30r

=03r

Ein r at r = 0 , E = 0

(i) Graphically

(ii) Again , for points outside the sphere , itbehaves as all the charge is present at the

centre(iii) For points outside , it obeys inverse square law(iv) Intensity of electric field at infinity is zero.(v) Intensity at the surface is maximum and is

equal to 20 R

Q4

1

(vi) Again , it is perpendicular to the surface atthe surface .

(vii) Intensity is zero at the centre and for pointsinside the sphere, it is directly proportionalto distance of the point from the centre

r > R r

E 1

2r

r < R r = R

E

rEr

V = RkQ

V = r kQV

r < R r = R r> R

pQ

o


19/28

19

Electric Potential

Case: 1 r > R

Vout

=04

1

r

Q

Case : 2 r = R

Vsurface =04

1RQ

Case : 3 r < R

Vin =04

13

22

R2)r R3(Q

Vcentre =3

2

1

4 0 R

Q (Imp)

Vcentre = 3/2 Vsurface

(i) Graphically

(ii) Again , Ecentre = 0 , but Vcentre 0.(iii) Electric potential at infinity is zero.(iv) Electric potential is maximum at the centre

Ex.49 A solid insulating sphere of radius R is givena charge . If inside the sphere at a point thepotentail is 1.5 times that of the potential atthe surface, this point will be -(A) At the centre

(B) At distance 3/2R from the centre(C) potential will be same inside and on the

surface of sphere , so given information isinadequate.

(D) Insulating bodies can not be given chargeSol. (A) Potential at the centre of insulating

sphere is given by

Vin = 322

0 R2)r R3(Q

41

.......(1)

and on the surface ,

Vsurface = RQ

41

0.......(2)

given that, Vin = 23 Vsurface

322

R2)r R3(Q = R

Q23 r = 0

Hence the point will be at the centre.Ex. 50 Two concentric spheres

of radii r & R (r < R)are given the chargesq and Q respectively.Find the potentialdifference between twospheres.

Sol. Potential at the inner sphere = potential dueto inner + potential due to outer sphere

V1 = RQ

41

r q

41

00

(potential at points inside is same everywhereand is equal to potential at the surface).Potential at outer sphereV2 = potential due to inner + potential dueto outer sphere

= RQ

41

Rq

41

00

potential difference = V1 V2

=

Rq

r q

41

0

V =

R1

r 1

4q

0

Note : Here , we see that ' V' depends onlyon the charge of inner sphere .

Ex.51 In the following fig, of charged spheres A , B& C whose charge densities are , - &and radii a, b & c respectively what will bethe value of VA & VB.

P a r a b o l a

h y p a r a b o l a

V0

V = 322

R3)r R3(kQ

V = RkQ

V

Or < R r = R r > R

0V23

C B

B b

A+

+ c

a


20/28


21/28

21

(iii) Intensity of electric field does not depend uponthe distance of points from the sheet for thepoints in front of sheet i.e.. There is anequipotential region near the charged sheet.

(iv) Potential difference between two points A& B at distances r 1 & r 2 respectively is

VA VB = )r r (2 120 11.5 Electric field d ue to in fini te charged metal

sheet

(i) Intensity at points near the sheet

= E = r 0

where

= surface charge density

(ii) E is independent of distanceof the point from the sheet andalso of the area of sheet i.e. There is anequipotential region near the sheet.

(iii) Direction of electric field is perpendicular tothe sheet.

(iv) Potential difference between two pointA (r 1) and B (r 2) (r 1 < r 2) near the sheet is

V = VA VB =

0 (r 2 r 1)

Note : The difference in the cases (D) & (E) is thatin (D) was only on one side of the sheetwhile here is there on both sides, becauseit is a metal sheet .

11.6 Electric f ield due to two infinite paral lelplates of charge

(i) Both plates have same type of charge

EO = 1E + 2E = 2 0

+ 2 0

=0

EP = 1E + 2E = 2 0

2 0

= 0

ER = 1E + 2E = 2 0

+ 2 0

=0

(ii) Two plates have opposite type of charge

EO = 1E + 2E = 2 0

2 0

=0

EP = 1E + 2E = 2 0

+ 2 0

=0

ER = 1E + 2E = 2 0

2 0

= 0

Note : In this case , we will have an uniform electricfield between the two plates directed frompositive to negative charged plate. Electricfield intensity is zero elsewhere.

11.7Electric field d ue to charged ring : Q chargeis distribut ed over a ring of radius R.(i) Intensity of electric field at a distance x from

the centre of ring along it's axis -

E = 2/3220 )xR(Qx

41

= )xR(Qx

41

220

E = 20 r cosQ

41

and it's direction will be

along the axis of the ring

++++++++

++++++++

E

r

++++

+++

++++

+++

1 2

1 2

P

. . . RO

2E

1E

2E

2E

1E

1E

2

+++++++

1 2

1

O.

P

. .

R

1E2E 1E1E

2E2E

O

Rr

xdE

dE cos axis of the loopp

dl


22/28

22

(ii) Intensity will be zero at the centre of thering.

(iii) Intensity will be maximum at a distance2/R from the centre and

Emax= 332

. 041

. 2RQ

(iv) Electric potential at a distance x from centre,

V =)Rx(

Q4

1220

(v) Electric potential will be maximum at the

centre and Vmax = RQ

41

0

Electric Field

Ex.56 Two symmetrical rings of radius R each areplaced coaxially at a distance R meter. Theserings are given the charges Q1 & Q2respectively , uniformly. What will be the workdone in moving a charge q from center of onering to centre of the other.

Sol. Work done = q (V2 V1) potential at thecentre of first ring

V1 = 222

01

0 RRQ

41

RQ

41

= 2QQR4 121

0

potential at the centre of second ring

V2 = 221

02

0 RRQ

41

RQ

R41

=

2QQ

R41 1

20

work done = q (V2 V1 )

=

2QQ

2QQ

R4q 1

22

10

W1 2 =

121)QQ(

R4q

210

Note : Work done in moving the same charge fromsecond to the first ring will be negative of the work done calculated above i.e.

W 1 =

211)QQ(

R4q

210

.

11.8 Uniformly charged semi - circular arc

Ecentre = R2 0

where l = linear charge density = RQ

Vcentre =1

4 0 RQ

Note : In all of the above discussion , is taken tobe a positive charge and accordingly thedirection of electric field is decided. If wasnegative , all the directions would have beenopposite to what they are.

12. ELECTRIC DIPOLE

(i) A system consisting of two equal andopposite charges separated by a smalldistance is termed an electric dipole.

Example : Na+Cl - , H+ Cl - etc.

(ii) An isolated atom is not a dipole becausecentre of positive charge coincides withcentre of negative centres. But if atom isplaced in an electric field, then the positiveand negative centres are displaced relative toeach other and atom become a dipole.

(iii) DIPOLE MOMENT : The product of themagnitude of charges and distance betweenthem is called the dipole moment.(a) This is a vector quantity which is directed

from negative to positive charge.(b) Unit : Coulomb - metre (C-M)

Q1 Q2

q A

2R

qB

R

1

R


23/28

23

(c) Dimension : [M0 L1 T1A1]

(d) It is denoted byp that is p = dq 12.1 Electric field due to a dipole

(i) There are two components of electric field atany point

(a) Er in the direction ofr

(b) E in the direction perpendicular tor

Er = 30 r cosP2.

41

E =

30 r sinP.

41

r

p

E r E

q qO

E

(ii) Resultant

E = 22r EE = 30r 4

P 2cos31

(iii) Angle between the resultantE and r is

given by = tan-1

r EE

= tan-1

tan21

(iv) If = 0 , i.e point is on the axis -

Eaxis = 30 r P.

41

= 0 , i .e . along the axis.(v) If = 90 , i.e. point is on the line bisecting

the dipole perpendicularly

Eequator = 30 r P.

41

(vi) So, Eaxis = 2Eequator (for same r)

(vii) Eaxis = 2220 )r (Pr 2.

41

Eequator = 2/3220 )r (P2.

41

where P = q . (2l)

(viii) V = 20 r cos)2(q.

41

= 20 r cosP.

41 = 20 r

r .P.4

1

= 30 r r .P.

41

where is the angle between P and r . V can also be written as

V = r 1.P

41

0 because

r 1 = 2r

r

(ix) If = 0, Vaxis = 20 r .4qd

(x) If = 90, Vequator = 0(xi) Here we see that V = 0 but E 0 for points

at equator (xii) Again, if r >> d is not true and d = 2l ,

Vaxis = )r (P.

41

220

Vequator = 0

Note :(i) This is not essential that at a point, where

E = 0 , V will also be zero there eg.insid e a unif ormly charged sphere, E = 0but V 0

(ii) Also if V = 0 , it is not essential for E tobe zero eg. in equatorial position of dipole V = 0, but E 0

12.2 Electri c Dipol e In an Electric Field - Uniform Electric Field(i) When an electric dipole is placed in an

uniform electric dipole , A torque acts on itwhich subjects the dipole to rotatory motion.This is given by = PE sin or

= EP

2l

Q

Ol90 +q p

ll

q


24/28

24

Electric Dipole

Ex.57 An electron and a proton are placed atdistance of 1. What will be dipole momentof so formed dipole

Sol. p = qd = 1.6 10 19 1 10 10 = 1.6 10 29 coulomb metre

Ex.58 E is the intensity of electric field at distancex (axial condition) from the centre of anelectric dipole. If the same intensity is at apoint distance x' on perpendicular bisector of dipole from its centre , then relation betweenx & x' will be -(A) x' = x (B) x' = x/2(C) x' = x/22/3 (D) x' = x/21/3

Sol. (D) Given, Eaxis = Eequvatorial

k 3xp2 = k 3'x

p

x' = x/21/3Note : All these are valid only if d < < r .otherwise

Ex.59 An electric dipole is placed in a uniformelectric fieldE . What must be the anglebetween E and dipole, so that dipole hasminimum potential energy ?

Sol. Zero. U = PE sin

U to be minimum = 0 Umin = 0

Ex.60 A dipole with dipole moment p is placed in anelectric field E. The dipole is displaced fromits equilibrium position AB to A'B' as shown infig. Now what will be the work required , sothat the point A' coincids with B.

(A)2

)32(pE2 (B)2

)32(pE

(C)2

)32(pE2 (D)2

)32(pE

(ii) Potential energy of the dipole

U = PE cos = E.P

Cases :

(a) If = 0 , i.e. P || E = 0 andU = PE , dipole is in the minimum potentialenergy state and no torque acting on it andhence it is in the stable equilibrium state.

(b) For = 180, i.e. P and E are in oppositedirection , then = 0 but U = PE which ismaximum potential energy state. Although itis in equilibrium but it is not a stable stateand a slight perturbation can disturb it.

(c) = 900 , i.e. P E , then = PE (maximum) and U = 0

Note :

(a) There is no net force acting on the dipole ina uniform electric field.

(b) Dipole can only perform rotatory motion.(c) If dipol e is placed in a nonun iform electric

field , it performs rotatory as well astranslator motion because now a netforce also acts on the dipole along withthe torque. (important)

12.3 Work do ne in rot ating on electric d ipole in

an electric field(i) To rotate the dipole by an angle from the

state of stable equilibrium W = PE (1cos).(ii) Work done in rotating the dipole from

1 to 2 in an uniform electric field

W = PE (cos 1 cos 2)

(iii) Work done in rotating the dipole through 1800from stable equilibrium state

W = 2PE = 2 (potential energy)

E

B'

AB

q

+q

+q

q

O

30

A


25/28

25

Sol.(B) When the dipole is rotated such that itacquires a new position A'B' form positionAB then 1 = 300 Now if dipole is rotatedthrough 1800 from its position AB then

2 = 1800Now from figure work done in rotating thedipole form position A'B' so that the point A'coincide with B, isW = pE [cos1 cos 2]

= PE [cos300 cos1800]

= PE )1(23

= PE

2

23

13. FORCE ON THE SURFACE OF A CHARGED

CONDUCTOR(i) If surface charge density on a surface is,

then electric field intensity at a point near

this surface is0

.

(ii) When a conductor is charged then it's entiresurface experiences an outward forceperpendicular to the surface.

(iii) The force per unit area of the charged surfaceis called as the electrical pressure ,

Pelectrical. = 0

2

2 N/m2.(iv) The direction of this force is perpendicular to

the surface.

13.1 Energy associated with the electric field(i) The energy stored per unit volume around a

point in an electric field E is given by

U =12 0 E2

This is also called energy density

(ii) If in place of vacuum some medium is presentthen U =

12 0 r E2.

(iii) For the electric field around a charged

conducting sphere U = Rq.

81 2

0 Where q = charge on sphere

R = radius of sphere(iv) The force of attraction per unit area between

plates of parallel plate capacitor is F =0

2

2q

(v) Energy associated with the electric fieldbetween plates of parallel plate capacitor is

U =

2

0E21

(Ad) where E =0

[These topics can be best studied in thechapter "Capacitance"](vi) Work done in charging a parallel plate

capacitor is stored as the electric fieldbetween plates.

13.2 Drop of a charged liqui d -If n identical drops each having a charge qand radius r coalesce to form a single largedrop of radius R and charge Q, then

(a) Charge will be conserved i.e. nq = Q

(b) Volume will be conserved i.e.

n .43 r 3 =

43 R3 or R = n1/3 r

(c) Potential of each small drops = V = r q.

41

0(d) Potential of large drop = V'

V' = RQ

41

0 = V' = n2/3 V

(e) Electric field at surface of small drop = E

E = 20 r q.

41

(d) Electric field at surface of large drop = E'

E' = 20 RQ.

41

E' = n1/3 E.

Forece on drop of a charged liquid

Ex.61 1000 equal drops of radius 1cm, and charge1 10-6 C are fused to form one bigger drop.The ratio of potential of bigger drop to onesmaller drop, and the electric field intensityon the surface of bigger drop will berespectively-(A) 100 : 1, 9 108 V/m(B) (10)1/3 : 1, 9 108 V/m(C) (100)1/3, 8 108 V/m(D) (1000)2/3 : 1, 9 106 V/m


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(5) If equal charge q is placed at points r, 2r, 4r,8r, ......... from a point 'P', then potential at'P' will be V= 2kq/r

(6) The work done in moving a charge in circular

orbit, in a electric field is zero.(7) If a charged particle of charge q and kinetic

energy E, moving about a nucleus of atomicnumber z, then the least distance betweennucleus and charged particle will be-For least distance of reach,potential energy = Kinetic energy

qr

ze4

10

= E

(where r is least distance of reach)[Note : E is kinetic energy, not the electricfield]

r = Eq)ze(

41

0 m

(8) If + q and q charges are placed at the endsof a diagonal of a rectangle, of side a & b,then potential difference between the ends of another diagonal will be

V = ab)ba(kq2

(9) A sphere of 1 cm radius, can not be givencharge of 1 coulomb, because the electricfield intensity at the surface of sphere will be9 1011. In air the electric field intensitygreater than 3 106 V/m, ionizes the air,and the charge of sphere starts leaking.

(10) The electric potential of a conductor is aelectric state, which ensures the direction of flow charge.

(11) If a positively charged conductor is connectedto earth, then the positive charge of conductor will flow to earth and there by the potential of conductor will be zero.

(12) If a negatively charged conductor isconnected to earth, then the negative charge(electron) will flow to earth and there by itspotential will be zero.

(13) If a charged conductor is placed inside ahollow spherical conductor and theconducters are connected by wire with each

Sol. (A) Let the potential of one smaller drop be Vthen potential of bigger drop, isV' = n2/3 V

V'V = n2/3

= (1000)2/3 = 100

V' : V = 100 : 1Also let the electric field on the surface of smaller drop be E then electric field on bigger drop is

E' = n1/3 E = n1/3 2r kq

= (1000)1/3 2269

)101(101109

= 9 108 V/m

[Note : also charge density bigger drop

' = 1/3 =3/1

2R4q

]

POINTS TO REMEMBER

(1) The charge density and intensity of electricfield is greater at the sharper end, but theelectric potential remains same at all thepoints.

(2) The workdone in carrying a point charge inelectric field, does not depend upon the path,because electric f ield remains conserved.

(3) Potential due to a monopole charge

V

x1

Potential due to dipole charge

V 2x1

Potential due to tetrapole charge

V 3x1

(4) If n equal drops of radius r and charge density form one big drop, then the charge density

of big drop ' =1/3


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other, the entire charge of charged conductor will come at the outer surface of outer conductor to have minimum potential energy.

(14) If an electron and proton are moving in a

uniform electric field, then the electric forceacting on them, will be same, but theacceleration of proton will 1/1836 times thatof electron, (Because the mass of proton is1836 times that of electron)

(15) The electric field inside a charged conductor is zero

(16) The electric potential of a charged conductor is same at inner and outer surfaces

(17) The dipole placed in a uniform electric fieldexperience torque, and the net force actingon it is zero.Therefore in uniform electric fieldthe dipole has rotatory motion only nottranslatory motion.

(18) When electric dipole is placed in non-uniformfield it experience torque as well as net force,then by it has rotatory as well as translatorymotion.

(19) The electric field due to electric dipole in endside on position on = 2 (electric field in broadside on position)

(20) The potential of earth is zero.(21) The work done in moving a charged particle

does not depend upon the path.(22) The best conductor of electricity is silver (Ag)(23) The bubble of soap always inflates, when it

is charged (negatively or positively)(24) The volume of air inside the soap bubble

remains constant, in the process of charging.(25) If two bodies having charges q1 and q2 are

brought in contanct and again separated, thennet charge on each of them will be

q = 2qq 21

(26) The electric field vanishes in a cavity madein a conductor. This is called electrostaticshielding. It implies that the electricinstrument can be protected from outsideelectric fields by placing it in a box made of a good conducting material.

(27) If a charged particle having a charge q andmass m is moving in an electric field betweentwo points having a potential difference of Vvolts, then the increase in kinetic energy of the body is

21 mv2 = Vq or v =

mqV

(28) Electrophorus is used to charge a body byelectrostatic induction.

(29) If E = 0 at any point then it is not necessarythat the electrostatic potential at that pointwill also be zero. It may be finite, as in caseof the interior point of a uniformly chargedconducting sphere, E = 0 but V 0.

(30) If V = 0 at any point, then it is not necessarythat the intensity of electric field at that pointwill also be zero, as in case of broad side onposition of a dipole, V = 0 but E 0

(31) If a small charged conductor is placed insideanother big and hollow charged conductor andboth are joined by a wire then the chargeflows from smaller conductor to bigger conductor because the potential of smaller conductor is more than that of bigger conductor.

(32) If two like charges are placed at some distancefrom each other, then the intensity of field willbe zero at any point on the line joining the twocharges, somewhere between the charges.

(33) If two unlike charges are placed at somedistance from each other, then the intensityof field will be zero at any point lying on theline joining the charges but outside thecharges. The neutral point is situated on theside of charge of smaller magnitude.

(34) Polar dielectrics are those dielectrics in whichthe centre of positive charge of a moleculedoes not coincides with the centre of negativecharge and hence they do not show a dipolemoment in the absence of electric field.However, they show a dipole moment whenthey are placed in external field.

(35) When two charged pith balls having chargesq1 and q2 are suspended from same pointwith then help of silk threads then consideringthe equilibrium of any one ball -


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Moment of Fe about 0 = Moment mg about O Fe OC = mg AC

mgFe = tan

mgx4

qq2

0

21

= tan

(36) In the above problem (x <

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