CHAPTE R

Elecric Charges
and Field

1.1 FRICTIONAL ELECTRICITY electrica for such substan ces. In fact, the Greek name for
amber is elektron which is the origin of all such words :
1. What is frictional electricity ? When is a body said electricity, electric force, electric charge and electron.
tobe electrified or charged ?
Frictional electricity. If aglass rod is rubbed with a For Your Knowledge
silk cloth, or a fountain-pen with a coat-sleeve, it is able
to attract small pieces of paper, straw, lint, light feathers,
Amber is a yellow resinous (gum like) substance
found on the shores of the Baltic sea.
etc. Similarly, a plastic comb passed through dry hair
can attract such light objects. In all these examples, we Both electric and magneticphenomena can be derived
from charged particles. Magnetism arises from
can say that the rubbed substance has becomeelectrified
charges in motion. The charged particles in motion
or electrically charged. It is because of friction that the
exert both electric and magnetic forces on each other.
substances get charged on rubbing.
Hence electricity and magnetism are studied together
The property of rubbed substances due to which they
as electromnagnetism.
attract light objects is called electricity. The electricity

developed by rubbing or friction is called frictional or static 1.2 ELECTRIC CHARGE

electricity. The rubbed substances which show this property
3. What is electric charge ? Is it a scalar or vector
are said to have become electrified or
of attraction
quantity ? Name its SI unit.

electrically charged.
Electric charge. Electric charge is an intrinsic property
2. Give a historical view of frictional electricity. From of the elementary particles like protons, etc.,
electrons,

where did the term electricity get its origin ? of which all the objects are made up of. It is because of
Historical view of frictional electricity. In 600 B.C., these electric charges that various objects exert strong
electric forces of attraction or repulsion on each other.
Thales ofMiletus, one of the founders of Greek science,

if a piece of amber is rubbed with a Electric charge is an intrinsic property of elementary

first noticed that
of matter which gives rise to electric force between
then acquires the property of attrac-
particles
woollen cloth, it
various objects.
ting light feathers, dust, lint, pieces of leaves, etc.
Electric charge is a scalar quantity. Its SI unit is

personal doctor to
In 1600 A.D., William Gillbert, the (+ e) and an
conlomb (C).A proton has a positive charge
-Iof England, made a systematic study (-è), where
Queen Elizabeth has a negative charge
amber. In his book electron
behave like
of the substances that
name e=1.6x 10 coulomb
De Magnete (on the magnet), he introduced the
(1.1)
 of egal uberod rt
Large-scale matter
that coests
tal M hene s rgedes od The t

electrons and
peotos is ectrially harge
bdy hasa
excsof clectr, theeslts
negative
an
ia potie darge
and an exoese of proons
fros apd upg Being eer ey
be
harped d r Thene l
13 ELECTROSTATICS plastic

4 Whatsdetrotatics 7 Metn e f its

important applicats
Easr 2. Ma asd. bed wi

Electrostatics Electrstatics s he stuy f ciecrie made o foxct teo sml pih bls peirtye
eadsen
for

we swfacare sosperded bysk

the fooes, feds and balls)
e t
charges at rest. Here study
12 Sme
ned w
with static charges
balls neped each obe, asseen in Fg
potentials associated
aplastic od wi
and pih balls touched
Applications of electrostatic The attraction
have many indus fur are fond to nepel ea oer (Fi 12
o

Bs
repulsion between charged bodies
Some of these are as follows: seen that a pih ball toched nd wi ss rats
trial applications
anoherpi bull toached wita plaic od JFi 123
1. In electrostatic loudspeaker.
and powder
2. In electrostatic spraying of paints
coating
3. In flyash collection in chineys
4. In aXerox copying machine.

of a cathode-ray tube used in

5. In the design
television and radar

14 TWO KINDS OF ELECTRIC CHARGES

that ()thereare
5. Ho ill youshowexperimentally

only tuo kinds ofelectric

charges and lke charges ()
other ?
charges attract each
repel and unlike
years ago,
Two kinds of electric chargesAbout 100 harges
showed that electric
Charles Du Fay of France
two kinds. The following
onvarious objects areofonly fact
simpleexperiments
prove this OActn
ExPERIMENT 1

12 sepe an snike canges

and suspend it from a Fig Like changes

() Rub a glass rod with silk

Bring
rigid supportby
means of asilk thread. we sone that e

charged rod near

it The wo From the aboe experiets fom
another similarly
on a glass rod is denee
rods repel each other [Fig
1.1(a)) charge produced
charge produced
o
a plastic od. Also

oached agas od as
e hurge

produced on a pith ball

an pitt hl
fron the dharge peoduced
different
Clas a plastic nod We can concde hut
touched with

1 There are alytu ndsofctrac chrg-p

Atractaom
nd negatine

unlie churgs tact ck

er
nd
Lledrges ngel
2 is known as hefndaemeetall
2.

The statement
of electrostatics
demcsrate hut te
also
PRastic The aboeexperimenits bas
traserned from the ods
o e pitth
harges
on contact
are

We say thut the puh bals hae be

by prperty
electrifed or changnd contact This

tekonds
dotinguihes te crges calod the polerity

le f
is
Fig 1.1 Lâe chasges epel and chages ofcharge
attract each othe.
 ELECTRIC CHARGES AND FIELD 1.3

6. What are vitreous and resinoUS charges What turns out to be negative in this convention. It would

?
was wrong with this nomenclature? haye been more convenient if electrons were assigned

Vitreous and resinous charges,Charles Du Fay used the positive charge. But inscience, sometimes we have to
live with the historical conventions.
terms vitreous and resinous for the two kinds of charges.

1. The charge developed on gass rod when rubbed with Different substances can be arranged in a series in

silk ons called vitreous Such away that if any two of them arerubbed together,
charge (Latin virlum =glass).
then the one occurring earlier in theseries acquires a
2. The charge developed on amber when rubbed with
positive charge while the other occurring later acquires
wool was called resinous charge (amberis a resin).
a negative charge
But later on, these
were found to be terms
1. Fur 2. Flannel 3. Sealing wax
misleading. For example, a ground glass rod develops
4. Glass 5. Cotton 6. Paper
resinous electricity while a highly polished ebonite rod
7. Silk 8. Human body 9 Wood
develops vitreous electricity.
10. Metals 11. Rubber 12 Resin
7. What are positive and negative charges ? What is
Amber 15. Ebonite
the nature 13. 14. Sulphur
of charge on an electron in this convention ?
16. Guta parcha
Positive and negative charges. Benjamin Franklin
(1706-1790), an American pioneer of electrostatics a positive charge when rubbed
Thus glass acquires

introduced the present-day convention by replacing with silk but it acquires negative charge when rubbed
the terms vitreous and resinous by positive and with

:
flarnnel.

negative, respectively.According to this convention

1. The charge developed on a glass rod when rubbed 1.5 ELECTRONICTHEORY OF FRICTIONAL
with silk is called positive ELECTRICITY
charge.
2. The charge developed on a plastic rod when rubbed 8. Describe the electronic theory of frictional
with wool is called negative charge.
electricity. Are the frictional forces electric inorigin ?
The above convention is consistent with the fact
Electronic theory of frictional electricity. All matter
thatwhen two opposite kinds charges are brought In made
of is of atoms. An atom consists of a small central
contact, they tend to cancel each others effect. According nucleus containing protons and neutrons, around
to this convention, the charge on an electron is negative.
which revolve a number of electrons. In any piece of
Table 1.1 gives a list of the pairs of objects which get matter, the positive proton charges and the negative
charged on rubbing against each other. On rubbing, an electron charges cancel each other and so the matter in
object of column I will acquire positive charge
while bulk is electrically neutral.
that of column II will acquire negative charge.
The electrons of the outer shell of an atom are
lable 1.1 Two kinds of charges developed on rubbing loosely bound to the nucleus. The energy required to
remove an electron from the surface of a material is
Column Column II
I
called its 'workfunction. When two different bodies
(Positiye charge) (Negative charge) are rubbed against each other, electrons are transferred
Glass rod Silk cloth from the material with lower work function to the
Flannel or cat skin Ebonite rod material with higher work function. For example,

Amber rod when a glass rod is rubbed with a silk cloth, some
Woollen cloth
electrons are transferred from glass rod to silk. The
Woollen coat Plastic seat
glass rod develops a positive charge due to deficiency
Rubber shoes
Woollen carpet of electrons while the silk cloth develops an equal

negaive charge due

excess of electrons.
to The
Obviously, any two charged objects belonging to
total charge of the glass rod and silk cloth
is
the same column will repel each other while those of Combined

two different columns will attract each other. stillzero, as it was before rubbing i.e., electric charge is

conserved during rubbing.

For Your Knowledge Electric origin of frictional forces. The only way by

and negative charges which an electron can be pulled away from an atom is
Benjamine'schoice of positive is

purely conventional one. However, it is unfortunate to exert a strong electric on it. As electrons are
force

that the charge on an electron (which is so important actually transferred from one body to another during
to physical and chemical properties of materials) rubbing, so frictional forces must have an electric origin.
 PHYSICS X

For Your Knowledge

The cause of charging is the actual transfer of elec
trons from one material to another during rubbirng.
Protons are not transferred
during rubbing.
The material with lower work function loses electrons
and becomes positively charged.
(a)
As an electronhas a finite mass,
(h)
therefore, there always
occurs some change in Fig. 1.3
Neutralisation of a (a) positively
mass during charging, The charged,
mass of a positively charged body (b) negatively charged, earthed bndy.
slightly decreases
due to loss of some electrons. The from the mains is
electricity
The mass ofa negatively
charged body slightly houses using a supplied to our
increases due to gain in some three-core wiring: live,
neutral arnd
electrons. earth wires. The live
wire red in colour brings in the
current. The black neutral
wire is the return wire. The
1.6 cONDUCTORS AND INSULATORS green earth wire is
connected to a thick metal plate
buried deep into the earth.
9. How do the conductors The metallic bodies of the
difer from the insulators ? electric
Why cannot we electrifya appliances such as electric iron, refrigerator, TV,
metal rod by rubbing it while etc. are connected to the
holding earth wire. When any fault
it in our hand ? How we charge occurs or live wire touches the
can it ? metallic body, the charge
Conductors. The substances through which clectric flows to the earth and the person who happens to touch
charges can flow easily are called conductors. They contain the body of the appliance does not receive any shock.
a large number of free electrons which make
them
good conductor of electricity. Metals, human and animal 1.7 ELECTROSTATIC INDUCTION
bodies, graphite, acids, alkalies, are conductors.
etc. 11. What is meant by electrostatic induction ?
Insulators. The substances through which electric
charges Electrostatic induction. As shown in Fig. 1.4, hold a
cannot flow easily are called insulators. In the atoms of conducting rod AB over an insulating stand. Bring a
such substances, electrons of the outer shell are tightly
positively charged glass rod near its end A. The free
bound to the nucleus. Due to the albsence of free charge electrons of the conducting rod get attracted
carriers, these substances offer high resistance to the towards
the end A while the end
Bbecomes electron deficient.
flow of electricity through them. Most of the non The closer end A acquires a negative charge while the
metals like glass, diamond, porcelain, plastic, nylon, remote end B acquires an equal positive charge. As
wood, mica, etc. are insulators.
soon as the glass rod is taken away,the charges at the
An importantdifference between conductors and ends A and Bdisappear.
insulators is that when some charge is transferred to a Conducting rod

conductor, it readily gets distributed over

surface. On the other hand,if somecharge is put on an
insulator, it stays at the same place. shall discuss We
its entire

EA B
this distinguishing feature in the next chapter. Excess of Deficiency of
charged
A metal rod held in hand and rubbed with wool rod
electrons electrons

Positively
does not develop any charge. This is because the glass

human body is a good conductor of electricity, so any Insulating

charge developed on the metal rod is transferred to the stand

earth through the human body. We can electrify the

rod by providing it a plastic or a rubber handle and induction.
Fiq. 1.4 Electrostatic
rubbing it without touching its metal part.

10. What is meant by earthing or grounding in Electrostatic induction is the phenomenon of

household circuits ? What is its importance ? temporary electrification of a conductor in which opposite

Earthing and safety. When a charged body is charges appear at its closer end and similar charges aypear at

brought in contact with the earth (through a connecting its farther end in the presence of a nearby charged body.
conductor), its entire charge passes to the ground in The positive and negative charges produced at the

the form of a momentary current. This process in which a ends of the conducting rod are called induced charges and
body shares its charges with the earth is called grounding or the charge on the glass rod which induces these
earthing. charges on conducting rod is called inducing charge.
 ELECTRIC CHARGES AND FIELD 15
12. Describe hoe tuometal splheres can beopposilely 13. How can you charge a metal sphere positiuely
carged by induction. wilthout touching it ?

Charging of two spheres by induction. Figure 1.5 Charging of a sphere by induction. Fig, 16 shows
shows the various steps involved in inducing,opposite the various steps involved in inducing a positive
charges on two metal spheres. charge on a metal sphere.
(a) Hold the metal sphere on an insulating stand.
Bring a negatively charged plastic rod near

it.
The free electrons of the sphere are repelled to
the farther end. The near end becomes posi
tively charged due to deficit of electrons.

(a) (b) When the far end of the sphere is connected to

the ground by a connecting wire, its free

electrons flow to the ground.

(c) When the sphere is disconnected from the

ground, its positive charge at the near end

(c) (d) remains held there due to the attractive force of
the external charge.
(d) When the plastic rod is removed, the positive

charge spreads uniformly on the sphere.

(e)

Plastic

Fig. 1.5 Two metal spheres get oppositely charged by induction. rod

(a) Hold the two metal spheres on insulating stands

(a (b)
and place them in contact, as shown in Fig. 1.5(a).

(b) Bring a positivelycharged glass rod near the left

sphere. The free electrons of the spheres get attracted

towards the glass rod. The left surface of the left sphere
develops an excess of negative charge while the right (C) (d)
side of the right sphere develops an excess of positive
However,all do
charge.
not collect at the left face.
of the electrons of the spheres
As the negative charge
Fig. 1.6 Charging y induction.

begins to build up at the left face, it starts repelling the Similarly, the metal sphere can be negatively charged
new incoming electrons. Soon an equilibrium is by bringing a positively charged glass rod near it.

established under the action of force of attraction of the

rod and the force of repulsion due to the accumulated For Your Knowledge
electrons. The equilibrium situation is shown in
Gold-leaf electroscope. It is a device used for detecting
Fig. 1.5(b).
an electric charge and identifying its polarity. It
(c) Holding the glass rod near the left sphere, sepa
consists of aconducting rod passing through
vertical
rate the two spheres by a small distance, as shown in a rubber stopper fitted in the mouth of a glass vessel.
Fig, 1.5(c). Thetwo spheres now have opposite charges. Two thin gold leaves are attached to lower end of the
(d) Remove the glass rod.The charges on the spheres rod. Whena charged object touches the metal knob at
get redistributed.Their positive and negative charges the outer end of the rod, the charge flows down to the
face each other, as shown in Fig, 1.5(d). The two leaves. The leaves

spheres attract each other. diverge due to

Metal
-Metal rod
(e) When the two spheres are separated quite apart, repulsion of the like knob
-Rubber stopper
the charges on them get uniformly distributed, as charges they have
shown in Fig, 1.5(e). received. The degree -Glass vessel

of divergence of the
Thus the two metal spheres get charged by a
leaves a Gold
gives leaf
process called charging by induction. In contrast to the
measure of the
process of charging by contact,here the glass rod does Tin foil

amount of charge.
not lose any of its charge.
 I:6 PHYSICS-XI|

18 BASIC PROPERTIES OF ELECTRIC CHARGE The cxperimental fact that electric charges occur in

discrete amounts instead of continuous amounts is called

It is observed from experiments that electric charge auantization of electric charge. The quantization of electric
has following three basic properties :
charge means that the total charge (q) of a body is always
1. Additivity 2. Quantization 3. Conservation. integral mulliple of a asic quantum of charge le), Le,

We shall discuss these properties indetail in the wheren =0, ± 1, ± 2, ±3,....

next feww sections.
Cause of quantization. The basiccause of quanti

zation of electric charge is that during rubbing only añi

1.9 ADDITIVITY OF ELECTRIC CHARGE
number
integral of electrons can be transferred from

What do you mean by additive nature of electric

14. onebody to another.
charge ? Ouantization of electric charge is an experi

Additive nature of electric charge. Like mass, mentally verified law :

electric charge is a scalar quantity.Just as the mass of 1. The experimental laws of electrolysis discoy
an extended body is the sum of the masses of its ered by Faraday first suggested the quanti

individual particles, the total charge of an extended zation of electric charge.

body is the algebraic sum (i.e., the sum taking into 2. Millikan's oil drop experiment in 1912 on the
account the positive and negative signs) of all the measurement of electric charge further estab
charges located at different points inside it. Thus, the lished the quantization of electric charge.
electric charge is additive in nature. 17. Can we ignore the quantization of electric

Additivity of electriccharge means that the otal charge ? Ifyes, under what conditions ?
charge ofa system is the algebraic sum ofall the individual When can we ignore the quantization of electric
the system.
charges located at different pointsinside
charge.While dealing with macroscopic charges (9 = ne),
Ifa system contains charges 4,,4,-.,qpr then its
can ignore the quantization of electric
charge. This
we
very small and n is very
total charge is is because the basiccharge e is
so q behaves as if it
large in most practical situations,

were continuous i.e., as if a large amount

of charge

a system containing four For example, when we switch on a 60 W

The total charge of were flowing.
through its filament
2 uC, -3 uC 4 uC and-5uCis bulb, nearly 2x 10 electrons pass
charges or structure of charge
per second. Here the graininess
q=2 uC-3uC+ 4 uC- 5 uC=-2
uC.
the bulb does not flicker with the
does not show up i.e.,

OF ELECTRIC CHARGE Quantization of charge becomes

1.10 QUANTIZATION entry of each electron.
level, where the
charges
a physical important at the microscopic or hundreds of e
What meant by quantization of order of a few tens
15. is involved are of the

quantily ?
quanti For Your Knowledge
Quantization of a physical quantity. The
vary conti charge or basic quantum of
quantity imeans that it cannot The smallest amount of

zation of aphysical Its

on an electron or a proton.
value but it can change disconti
anyarbitrary charge is the charge
nuously to have
of values. For e= 1.602192 x 10-19 C
one of only a discrete set exact magnitude is
nuously to take any cannot be explained on
(ground, of electric charge
example, a building can have different floors Quantization
or even modern
ground floor upwards but it
the basis of classical electrodynamics
second, etc.) from the
first, properties
the value in-between. Thus the
and chemical
have a floor of physics. However, the physical
cannot cannot be explained
the electric charge of a molecules and bulk matter
energy of an electron in atom or of atoms,
of electric charge.
system is quantized. The minimum
amount by which a withoutconsidering
the quantization

energy physics have indi

quantum. in high
physical quantity can change is called its Recent discoveries
and
particles like protons
that the elementary
meant by quantization of electric charge ? cated
16. What is built out of
more elementary
charge ? neutrons are themselves (2/3)e and
What is the cause of quantization of electric
which have charges
called quarks,
units,
ound in future,
Quantization of electric charge. It is
quark-model is established
e Even
(-1/3) if
the
hold. Only
experimentally that the electric charge of any body, of charge will still
the quantization e
large or small, is always an integral multiple of a will reduce from to e/3.
quantum of charge
certain minimumamount of charge. This basic charge is a universal law of nature. Like charge,
>Quantization
is the charge on an electron, which is denoted by eand of an electron are
also
energy and angular momnentum
has magnitude 1.6x 101 coulomb. Thus the
charge on of mass is yet to be
quantized. However, quantization
an electron is -e on a proton is +e and that on
established.
a-particle is +2e.
 AND 1.7
1GAELECTRIC CHARGES FIELD

Examples based on Example 4. How much positive and negative charge is

Quantisationof Electric Charge therein a cup of water ? [NCERT)

Solution. Suppose the mass of water contained in a

Formulae Used
cup is 250 g. The molecular mass of water is 18 g.
1. q=ne Number of molecules present in 18 g of water
2.

Units Used
Mass

qand
transferred

e are in coulomb, n
during charging

is pure
= m, x

integer.
n

.
of water
Number
= Avogadro's
of molecules present
number = 6.02
in a cup (or 250 g)
1023

Constants Used
n= 6.02
x 10x 250 =8.36 x 1024
e= 1.6 x 10C, m, =9.1 x 10 kg
18

Each molecule of water (H,O)contains 2 +8=10

Example 1. Which is bigger - acoulomb or a charge on an electrons as well as 10 protons.
?How many electronic charges formone coulomb of
electron Total number of electrons or protons present in a
charge ? [Haryana01] cup of water,
Solution. One coulomb of charge is bigger than the n=nx 10 =8.36 >x 1025
charge on an electron.
Total negative charge carried by electrons or total
Charge on one electron, e=1.6 x 10-19C
positive charge carried by protons in a cup of water,
Number of electronic charges in 1 coulomb, q=ne
.:.
1C
11= =6.25 x 1018 =8.36 x 102 x 1.6 x 10-19 C=1.33 x 10? C
e 1.6x 10-19 C

Example 2.A comb drawn through person's hair on a dry Problems For Practice
day causes 10 electrons to leave the person's hair and stick
to the comb. Calculate the charge carried by the comb. 1. Calculate the charge carried by 12.5 x 10 electrons.

Solution. Here n =102, e=1.6 x 10-19 C [CBSED 92]

=10x x 10-19 =1.6x 10° C (Ans. 2 x 1010 C)

q= ne 1.6

2. How many electrons would have to be removed

As the comb has excess of electrons,
from a copper penny to leave it with a positive
.. Charge comb =-1.6 x C. charge of 10 C?
on 10
(Ans. 6.25 x 10 electrons)
Example 3.Ifa body gives out 10 electrons every second, 3. Calculate the charge on an alpha particle. Given
how much time is required to get a total charge of 1 C from charge on a proton =1.6x 10-19 C.
it ? [NCERT]
(Ans. + 3.2 x 10-1 O
Solution. Number of electrons given out by the 4. Calculate the charge on Fe nucleus. Given charge on
body in one second =10 a proton = 1.6 x 10o-19 C (Ans. + 4.16 x 10s c)

Charge given out by the body in one second 5. Determine the total charge on 75.0 kg of electrons.

=ne =10x 1.6x 10-19 C (Ans. - 1.33 x 10

=1.6x 10-10 C 6. How many mega coulombs of positive (or

x 10-10 C negative) charge are present in 2.0 mole of neutral

Time required to get a charge of 1.6
hydrogen (H,) gas ?
=1s
7. Estimate the total number of electrons present in

Time required to get a charge of 1 C

1
100 g of water. How much is the total negative
x 10 s charge carried by these electrons ? Avogadro's
x 10-10
S=6.25
1.6 number= 6.02 x 10 and molecular mass of water

6.25x 109
= 18. (Ans. 5.35 x 10° C)
years =198.18 years.
365 x 24x 3600 HINTS
Thus from a body emitting 10 electrons per
3. An alpha particle contains 2 protons and
second, it will take nearly 200 years to get a charge of 2 neutrons.
1C from that body. This shows how large is one
q=2e.
coulomb as the unit of charge.
 PHYSICS- XIl

Fe nucleus contains 26 protons and 30 neutrons. 4. Electric charge is conserved during the

9426e phenomenon of pair produclion in which a y ray

Total mass 75.0 25 photon materialises into an electron positron pair.

5. H

Mass of an electron 9x 10-1 3 7-ray elcctron positron

zero charge (-e) (+o)

25
q-e =
3
x 10 × l6x 10 19 --1.33 x 10l C 5, In annilhilation of matter, an electron and a post

tron on coming in contact destroy each other.

6. Number of molecules in 2.0 mole of H, gas y-ray photons, each of
producing two energy
2.0 x 6.02 x 1023
0.51 MeV.
Aseach H, molecule contains 2electrons/protons, so
electron + positron 21- rays
n=2 2.0 x x 6.02 x 10 -24.08 x 1023
(-e) (+e) zero charge

q= ne =24.08 x 10 x 1.6x 1019

=0.3853 x 10° C=0.3853 MC. For Your Knowledge

[1 MC= 10°CI Conservation of charge implies that electric charges
7. Proceed as in Example can be created or destroyed always in the form of equal
4.
opposite but never in isolation. For example,
and pairs

in the beta decay of a neutron (zero charge), a proton

1.11 coNSERVATION OF CHARGE
(charge + )and an electron (charge - e) are produced.
18. State the lawof conservation of charge.Give some Total charge remains zero before and after the decay.

examples to illustrate this law. law of

The law of conservation of charge is an exact

Law of conservation of charge. If some amount of nature. It is valid in all domains of nature. Even in the

matter is isolated in a certain region of space and no domains of physics, where mass changes
high energy
enters or leaves this region by moving
of
matter either and the law of conservation
into energy vice-versa,

across its boundary, then whatever other changes may charge strictly holds good.
total charge will not
occur in the matter inside, its

change with time. This is the law of conservation of 112 ELECTRIC CHARGE VS MASS
charge which states Compare the properties of electric charge with
:
19.

1. The total charge of anisolated system remains constant.

those of mass of a body.
can neither be created nor destroyed,
2. The electric charges
of
lable 1.2 Comporison of the properties
they canonly be transferred fromone body to another.
electric charge and mass
The law of conservation of charge is obeyed both in

microscopic processes. In fact, charge Mass

large scale and Electric charge
charge of
conservation is a global phenomenon
i.e., total
Electric charge may be Mass of a body is
1.

the entire universe remains constant. positive, negative or zero. always positive.

Examples : 2. Electric charge is always Quantization of mass is

not yet established.

1. When a glass rod is rubbed with a silk cloth, it quantized :g=ne
positive charge. But at the same Charge on a body does Mass of abody
develops a 3.
increases with its speed.
negative not depend on its speed.
tíme, the silk cloth develops an equal
not conserved by
of the glass rod and Mass is
charge. Thus the net charge 4. Charge is strictly
itself as some of the mass
the silk cloth is zero, as it was before rubbing. Conserved.
may get changed into
2. The rocksalt ionises in aqueous solution as
energy or vice versa.
follows : Gravitational forces
5. Electrostatic forces
NaCl Na + CI between two masses are
between two charges
zero before and the may be attractive or always attractive.
As the total charge is after

ionisation, so charge is conserved. repulsive.

Gravitational forces
3. Charge is conserved during the fission of a 6. Electrostatic forces
bodies
between different between different
nucleus by a neutron. never cancel out.
charges may cancel out.

Ba Kr +3 n+ Energy 7. A charged body always A body possessing

may not have any
(0 92) posseSses some mass. mass
Total charge before fission +
net charge.
= Total charge after fission (56 + 36 +3 x 0)
 ELECTRIC CHARGES AND HELO
19
20. 1How does the speed of an clectrically harqed where E, is called permitlivity of free space. So we can
particle affect its ()mass and (ii) charge ? expressCoulomb's law in SI units as

Effect of speed on mass and clectric charge. Aording

F=

1
to the special theory of relativity, the mass of a body
increases with its speed in accodance with the relation:

Units of charge.(i) The SI unit of charge

is coulomnb. In

the aboveequation, if 4, =4, =1Candr=lm, then

1
F= -9x10" N

where, mh = rest mass of the body, c= speed of light,

So one coulomb is that amount of charge that repels an
and m= mass of the body when moving with speed v. aual and similar charge with a force of 9 x 10 when N
As v <c, therefore, m> My placed in vacuum at a distance of one metre from it.

In contrast to mass, the charge on a body remains (ii) In electrostatic cgs system, the unit of charge is

constant and does not change as the speed of the body known as electrostatic unit ofcharge (e.s.u. of charge) or
changes. statcoulomb (stat C).
or one statcoulomb charge
One e.s.l. of charge is that

U3 coULOMB'S LAW OF ELECTRIC FORCE zwhich repels an identical charge in vacuum at

with a force of 1 dyne.
a distance of

21. State Coulomb's law in electrostatics. Express the one centimetre from it

same in SI units. Name and define the units of electric l coulomb =3x 10 statcoulomb

charge. =3x 10 e.s.u. of charge

Coulomb's law. In 1785, the French physicist (ii) In electromagnetic cgs system, the unit of
Charles Augustin Coulomb (1736-1806) experimentally charge is abcoulomb or electromagnetic unit of
measured the forces between small charged
electric charge (e.m.u.of charge).
spheres by using a torsion balance. He formulated his 1 coulomb =10 abcoulomb = 10 e.m.u. of charge
observations in the form of Coulomb's law which is
electricalanalogue of Newton's law of Universal
Gravitation in mechanics. For Your Knowledge

Coulomb's law states that the force of attraction or A torsion balance is a sensitive device to measure force.

repulsion between two stationary point charges is (i) directly

When the linear sizes of charged bodies are much smaller

proportional to the product of the magnitudes of the two than the distance between them, their sizes may be ignored
charges and (ii) inversely proportional to the square of the
and the charged bodies are called point charges.
distance between them. This force acts along the line joining
Coulomb's law is valid only for point charges.
the two charges.
In SI units, the exact value of the combination 4n s,is
10
CN'm²
Fig. 1.7 Coulomb's law.

where cis the speed of light in vacuum having the exact

If two point charges 4, and 4, are separated by x 10 msl
value 2.99792458
distance r, then the force Fof attraction or repulsion
Electrostatic force constant,
between them is such that
k= 8,98755 x10° Nm'c=9x 10 Nm²c2
and
Fa442 Permittivity freespace,
of
E = 8.8551485 x101? c²Nm-2
2 F=k
SI unit
-9x102c'N'm,
of permittivity

where constant of proportionality,called electro

k is a coulombx coulomb
static force constant. The value of
k depends on the CN'm?
newtonx metre
nature of themedium between the two charges and the
measure F, 4, 4, and r. The unit C'N'm is usually expressed as farad per
system of units chosen to
metre (Fm),
For thetwo charges located in free space and
in SI

More strictly, the Sl unit of charge coulomb is equal to

we have
1
units,
in terms of
Iamperesecond, where 1 ampere is defined
Nm' C2
1
k -9x 10"
themagnetic force between two current carrying wires.
 PHYSICS-X1

114 COULOMBS LAW IN VECTOR FORM Rangeof coulornbian forces.Coulombian forces act
The ratio (E/E,)of tM
over an enormous rarnge of separationis (r, fronm
22. Write Coulomnb's law vector form. What 15
permittiity (E,) of fre
in is the nucear dimensions (r= 10 m)to macroscopic dis
importance expressing it in ector form ? (6,jor delectric costa
tances as large as 10 m. Inverse square over

of
is valid

Coulomb's law in vector form.As shown in Fig. L8, this range of separation to a high degree of accuraey
considertwo positive point charges q, and q, placed in Limitations of Coulomb's law. Coulomb's law is
vacuum at distance r from each other. They repel cach not applicable in all situations. can defin
other.
It is valid only under So one
the following conditions : forces between charg
1. The electric charges must be at rest The dielectric con

2. Theelectric charges must be point charges ie, medium my be define

the extension of charges must be much smaller charges placed some di

than the separation between the charges. between the same eo c
distance apart in fhe &
Fig. 1.8 Repulsive coulombian forces for q, 4, >0. 3. The separation between the charges must be
Clearly. when a
greater than the nuclear size (10m), because
In vector form,Coulomb'slaw may be expressed as
constant k is placed
for distances <10m,
the strong nuclear force
between them becom
Ey =Force on charge q, due to 4,
dominates over the electrostatic force.

vacuum.That is,
1
1.15 DIELECTRIC CONSTANT:
RELATIVE PERMITTIVITY
ofa medium? Hence the Coulon
where =2, is a unit vector in the direction from 4, 24. What do you mean by permittivity

between tuo mav be written as

r,
Define dielectric constant in terms forces

of
to G2
charges.

Similarly, Fp =Force on charge 4, due to 4, Permittivity :An introduction. When two charges
the force
are placed in any medium other than air, KvacuL
1

2 between
of
them
the
is greatly
medium which
affected.

determines
Permittivity is a

the electric force

K
property
(wat
between two charges situated in that medium. For example,
K

where f, =21,is a unit vector in the direction from 4, the force between two charges located some distance Examples based
in water is about 1/80th of the force between
apart
Coulomb's Law
to them when they are separated by same distance
41 in air.

The coulombian forces between unlike charges This is because the absolute permittivity of water is Formulae Used
of
(9,4, <0) are attractive, as shown in Fig. 1.9.
about 80 times greater than the absolute
permittivity 1

air or free space.

Dielectric constant or relative permittivity. Accor

+41 F2
-42 ding to Coulomb's law, the force between two point 2. ed
charges 9, and g,, placed in vacuumn at distance r trom
Units Used
each other, is given by
Fig. 1.9 Attractive coulombian forces for g, 4, < 0. 1
..1) 3are in cou

Importance of vector formn. The vector form of cou Constant Used

1
information placed same
lomb's law gives the following additional same two charges
:
are k=
When the
in any medium other than vacuum, the
distance apart
1. As h =-p therefore F =-F: force between them becomes

Thismeans that the two charges exert equal and .2) Example 5. The electr
opposite forces on each other. So Coulombian Fmed 4TE charged ions C
positiely

forces obey Newton's third law of motion.

permittivity or just
ehen they are separat
The quantity e is called absolute
2. As theCoulombianforces act along , F, or nermitivity of the intervening medium. Divids
electrons are misstng f

Solution. Here F
Le, along the line joining the centres of two equation (1)by equation (2), we get
charges, so they are r=5A
1
central forces.

1
23. What is the range over which Coulombian forces Fyac
As F=
can act ? State the limitations of Coulomb's law in
med
1 412 4T
electrostatics.
41cE
 ELECTRIC CHARGES AND FIELD 111
forces act
(r), from The ratio (s/ c)ofthe permittivity (s)of the medium to the
9x 10 x qx4
.. 3.7x 10-9
opic dis permittivity (E,) free space is called relative permitlivity (5x 1010

of
walid over (E,)or dielectric constant (K)of the mediun. Thus
accuracy.
given

or
3.7 x 10x25x 10-20 =10.28x 1038
Fvat
b's law is E, Or K= 9x 10

nly under
Fmed
or q=3.2 x 10-19 c
So one can define dielectric constant in terms of
Number of electrons missing from each ion is
forces betweencharges as follows :
3.2 x 10-19
The dielectric constant or relative permittivity ofa 2.
arges i.e., 10-19
medium may be defined as the ratio of the force between two L.6x
h smaller
ges. charges placedsome distance apart in free space to the force Example 6. A free pith-bal A of 8g carries a positive

must be betoeen the same two charges cohen theyare placed the same charge of 5x 108 c What must be the nature and

, because
lear force
distance apart in the given medium.

Clearly, when a material medium of dielectric

magnitude of charge that
pith-ball B fixed cm below 5
should

the
be given

former ball
to a second

so that the upper

constant K is placed between the charges, the force ball is stationary ? (Haryana 01]
betweenthem becomes 1 / K times the original force in Solution. The pith-ball Bmust be of positivecharge
vacuum. That is, i.e., of same nature as that of A, so that the upward

Fvac
force of repulsion balances the weight of pith-ball A.
Fmed = K
When the pith-ball Aremains
anedium?
Hence the Coulomb's law for any material medium stationary,
ween two
may be written as
F= mm8 91

o charges
the force
= Fmed
1

2 1 912
Mh8 = 5cm
ivity is a K (vacuum)= 1

ectric

example,
force
K (water)
K (air) =1.00054

= 80.
But m =8g=8
9, =5x 1o-
x 10
c
kg
B) 42

e distance
Examples based on r=5 cmn =0.05 nm Fig. 1.10
between
nce in air. Coulomb's Law 9x 10 x 5x 10 x 42 _8x10- x9.8
water is
Formulae Used (0.05)²
ittivity of
1
8x9.8x (0.05) x 10-4
1. Fac
2 or
9x 5
ky. Accor
1

wo point 2. Fmed = 4.36 x 10 /C (positive).

nce r from

Units Used
Example 7. Aparticle of mass m and carrying charge - q

is moving around a charge + q, along a circular path of

..(1) 44 are in coulomb, F in newton and r in metre. radius r. Prove that the period of revolution of the charge -4,

Constant Used about + 4, is given b

ced same 1
k= 9x10 Nmo-2 16
uum, the T=
9192

betweentwo
.(2) Example 5. The electrostaticforce of repulsion

is 3.7 x 10 N,
Solution. Suppose charge - 4, moves around the
charged ions carrying equal charges
+4
positively
with speed v along the circular path of
when they are separated by a distance of 5 A. How many charge
ity or just radius r. Then
electrons are missing ?
from each ion
Dividing
N, Force of attraction between the two charges
Solution. Here F =3.7 x 10
= Centripetal force
r=5Å=5x 10 m, =4, =9(say)

As F= 1
9192
4,

1 912 mu
Or
V4nE, mr
9142

4E r
 112 PHYSICS-XI

The period of revolution of charge - 4, around +q, (b) ()When charge on each sphere is doubled, and
will be the distance between them is halved, the force of
Solution. Let the charge
T= 2r -2r 16n',mr3 repulsion becomes
Or T and B be q. Ifthe separation
24,24;
9192
=k." 16 k 12 electrostatic force betweern sp
(r/2'
Example8. Tuo F =k. 20
mass
identical
10g and charge 2.0 x 10Care placed
charged particles
on a
each having a
horizontal
-16x 1.5x 10 =0.24 N. 2
table with a separation L betwcen them (i) The force between two charges placed in a When sphere C is touch
such that they stay medium of dielectric constant k
in limited equilibrium. If the coefficient
is given by charge q/2 each, because b
friction between

of
1
each particle and the F= 1 9142
Force on C due to
table is 0.25, find the value of
2

A
L.

IUseg = 10 ms] JEE Main June 22]

4TEg K
(q/2)5
Solution. For equilibrium For water, K =80 =k
of the charged particles, (r/2y
Mutual 1.5 x 10-2
electrostatic force= Force of Force on C due B
water =air
friction to
K 80
kq'
Fmg N1.9x 10 -k9-9/
L= k x4
Example 10. Suppose the
identical sizes. A
=1.875 x 10

third
spheres

sphere of the
A and B in

same size but uncharged

N.
Example 9have
Since these
therefore net force on C
(r/2
forces

is brought in contact with the first, then brought in contact

9x10 with the second, and finally removedfrom both. What is F 2q

Vo.25x0.01 ×10 x2
x 10-7, m new force of repulsion between A and B ? [NCERT)
the =k.

Solution. Charge on each of the spheres A and B is Example 12.Tuo identi

3
x10° x2 x 10- mn =12x 10m=12 cm. q=6.5x 10 -7 c distance r from each other
When a similar but uncharged sphere C is placed in line joining the above tu

Example 9. (a) Two insulated charged copper spheres A Contact with sphere A, each sphere shares a charge charges are in equilibriun

and B have their centres separated by a distance 50cm. /2, equally. posifion of the charge q
of
What is the mutual force of electrostatic repulsion if the Charge=0 4/2 9/2 Suppose
Solution.
charge on each is 6.5 x 10C? The radii of A and B are A the manner,as shown
negligible compared to the distance of separation. Also
4/2 34/434/4
compare this force with their mutual gravitational attraction
if each weighs 0.5 kg.
B

9/2 34/4
(0) What is the force of repulsion if (0) each sphere is

charged double the above amount, and the distance between

them is halved ; (ii) the two spheres are placed in water ?
The charge q wil
(Dielectricconstant of water = 80). [NCERT] Fig. 1.11
exerted on it by the
Solution. (a) Here 4, =4, =6.5 x 107 C, Now when
the sphere C (with charge q/2) is placed
opposite.
in with sphere B (with charge q), the charge is
contact
r=50 cm =0.50 m redistributed equally, so that
Using Coulomb's law,
Chargeon sphere B or
X=r
New force of repulsion between A and B is
.:.
Since the charg
=9x 109 6.5 x 10 x6.5x 1n-7 N 3q q
1 charge at C, so it w
(0.50) F= 4 2
by the charge atE
= 1.5 x 10 N.
opposite to that of
q

The mutual gravitational attraction,

x 1.5 x 10 N =0.5625 x 102 N .:. Force of rep
8
R2 - 5.7x 10 N.

6.67 x 10- x 0.5 x 0.5

Example 11. Two similarly equally charged identical metal
Or Q.4
k
=6.67 x 10 N spheres A and B repel each other with a force of 2.0 x 10 N.
(0.5)
(r/2)
A third identical uncharged sphere C is touched to A, then

placed at the midpoint between A and B.Calculate the net Example 13. Tw
Clearly, F << Fuir
electrostatic force on C. (CBSE OD 03] distance 'a' apart.

placed on the line j

 ELECTRIC CHARGES AND FIELD J13
Solution. Let the charge on each of the spheres A
cauilibriun ? In which case the equilibrium
and B be q. If the separation
between A and Bisr, then
will be stable
and in which unstable ?
electrostatic force between spheres A
and B will be
Solution. Suppose the three charges are placed as
=2.0 x 105 N shown in Let the charge q be
2 Fig. 1.13. positive.

When + 4e
sphere C is touched to A, the spheres share
charge q/2 each, because both are identical.
.. Force on C due to A a-X

=k.(q/2)² AC
(r/2) 2' along Fig. 1.13

Force on C due to B For the equilibrium of charge + q, we must have

Force of repulsion F between + 4e and + q
-k,2:9/2 along BC
(r/2)² 2 = Force of repulsion E, between + e and + q
Since these forces act in opposite directions,
or
4ex q 1 exq
therefore net force on Cis
4ne, (a -x

F=k. =k -2.0 x 10S N, along BC. Or

4 (a -x)²=x?
Or 2 (a- x) =#x
Example 12.Two identical charges, Q each, are kept at a 2a
distancerfrom each other. A or 2a
third charge q is placed on the 3
line joining the above two charges such that all the three

charges are in equilibrium. What is the magnitude, sign

As the charge q is placed between + 4e and + e, so
and
position the charge q? [CBSE OD,98]
only x =2a/3 is possible. Hence for equilibrium, the
of
charge q must be placed at a distance 2a/3 from the
Solution. Suppose the three charges be placed in charge + 4e.
the manner, as shown in Fig. 1.12.
We have considered the charge q to be positive. If
K
we displace it slightly towards charge e, from the

A -C equilibrium
increase and a net force (E,
position, then F, will

-E)
decrease and F,
will act on g towards
will

left i.e., towards the equilibrium position. Hence the

Fig. 1.12
equilibrium of positive q is stable.

The charge q will be in equilibrium if the forces Now if we take charge q to be negative, the forces E
exerted on it by the charges at A and C are equal and and E, will be attractive, as shown in Fig. 1.14.

opposite.
+4e +e
k Or x=(r-x'
(r-x? F F,

X=r-X Or -a-x

Fig. 1.14
Since the charge at A is repelled by the similar

charge at C, so it will be in equilibrium if it is attracted

the sign charge q should be The charge -q will still be in equilibrium at

by the charge q at B, i.e., of

Q. x=2a/3. However, if we displace charge - q slightly

opposite to that of charge
towards right, then F will decrease and E, will
.:. Force of repulsion between charges at A andC
increase. A net force (E, - E) will act on -q towards
- Force of attraction between charges right i.e., away from the equilibrium position. So the
at A andB equilibrium of the negative q will be unstable.

Q.4 =k Q.Q
Example 14 Two free' point charges + 4e and + e are
(r/2' placed a distance 'a' apart. Where should a third point charge
system mnay be
Example 13.Two point charges + 4e and + e are fixed' a q be placed between them such that
the entire

ofq
a'apart. Where should a third point charge g be in equilibrium ? What should
be the magnitude and sign ?
distance
What an will it be ?
the two charges so that it may be in type equilibrium
placed on the line joining
of
I14 PHYSICS

XI
Solution. Supposethe charges are placed as shown Example 16. A charge Qie to be divnded on tuo ohjects
What shoud be the vales of the charges on the tuo objects
Now (4, +4,=(4, -42)
in Fig. 1.15.
- (2 x 10
44e so that the force between the objects can be maximum

?
- 16x 10
Solution. Let q and Q-qbe the charges on the two

objects. Then force between the two objects is 4, +4, = 4x 10

q(Q-) On solving equations (1)

1
F= and
4 3 10 C

Fig. 1.15
which are the initial charge
where r is the distance between the two objects.
As the charge + e exerts repulsion F on charge + 4c,
For F to be maximum, Example 18. Tuo small sp
so for the equililbrium of charge + 4e,the charge - 4 and charge q coulamb are
must exert attraction F on +4e. This requires the each l metr
insulating threads
charge q be dq
to negative. 9 is the angle, euch thread

For equilibrium of charge +4c, - 1 d

Or (qQ-)- 0 equilibrium has been attained

F= F' -(4mgl sin

1 4exe d
1 4exq or (9Q-g)=0 Solution. The given sit
4TtE0 dq
x Each of the spheres A a
ex
Or Q-2q =0 following forces
Or
or () its weight mg,
2
For equilibrium of charge
() the force of rep
-4 i.e, the charge should be divided equally on the two 1
Attraction F, between+4eand -q objects.
F=
=Attraction E, between + e and -q Example 17. Two identical spheres, having charges of
1 4ex exq other with a force of 0. 108 Nwhen
1
opposite sign attract each
x separated by 0.5 m. The spheres are connected by a conduc

or x'=4 (a -x ting wire, which then removed,and thereafter they repel each

other with a force ofO.036 N. What were the initial charges

x=2a/3
on the spheres ?
ex? e 4a 4e
Hence Solution. Let + 4, and -4, be the initial charges on
9
the two spheres.
The equilibrium of the negative charge q will be
unstable.
(a) When the two spheres attract each other,

Example 15.Two point charges of charge values Q and q

F=k T2 ie., 0.108 =9 x 10. n2
(0.5)
are placed at distances x and x/2 respectively from athird

in the same 0.108 x (0.5)

charge of charge value 44, all charges being -3x 10 -12
straight line. Calculate the magnitude and nature
of charge 9x 10
such that the net force experienced by the charge q is zero. As the torces are in

(b)When two
Q,
[CBSE D 98] the spheres are connected by the
sphere A can be rept
wire, they share the charges equally. A AOCtaken in the sa
Suppose the three charges are placed as

shown
Solution.
in Fig. 1.16. : Charge on each sphere = 91+(-42)_1-92
2 2
F mg
4 AC OC
them
A B Force of repulsion between is
Or F = g

k
Fig. 1.16
F= From ()and (in), v

the charge Qmust

For the equilibrium of charge
q,

that of qor 4q, so that the forces

have the same sign as 109
0.036 = 9x 4nE A

F and Fp are equal and

opposite. i.e.,

(0.5)° But AC=l sin 9,

As 0.036 x (0.5) x4 1

1 4qx9 1
=4x 1012
9x 10 4nE 4I sin
(x/2'
4nE, (x/2)' 47
Or
Q= 44. 41 -9 =2 x 10-6
 ELECTRIC CHARGES AND FIELD 1.15

Now

On
(4,+t 9,)'

solving equations
t95
-(4,-4,) + 494:
-(2 x 10+4x3x
=

=
16x

4x 10 6
10

(i)
12

and (ii), we
10

get
12

...i)
1.

2.
Problems For
Obtain the dinensional formula

Calculate coulomb
Practice

force
of

(Ans.
between two
of 3.2 x 101 m
: M 'L TA?)
u-particles
in air.
separated by a distance
4, =3x 10 C and =106C
9, [CBSE OD 92)
which are the charges on the two spheres.
initial (Ans. 90 N)
Example 18. Tw0 having nass m kg
small spheres each 3. Calculate the distance between two protons such
and charge q coulomb are suspended from a the repulsive between them is
point by that electrical force
insulating threads each metre long but of
negligible mass. If equal to the weight of either. [CBSE D 94]
l
0 is the angle, each thread makes with the vertical when (Ans. 11.8 cm)
equilibrium has been attained, show that
4. How far apart should the two electrons be, if the
q =(4 mgl sin 0 tan 0) 4n E (Punjab 95] force each exerts on the other is equal to the weight
Solution. The given of the electron ? Given that e =1.6 x 101C and
situation is shown in Fig. 1.17.
Each of the spheres A and B is upon by the
acted m, = 9.1 x 103 kg. (Haryana 02]

following forces: (Ans.5.08 m)

() its weight mg, (ii) tension T in the string 5. A pith-ball Aof mass 9 x 10 kg carries a charge of
5uC. What must be the magnitude and sign of the
(iii) the force of repulsion F given by
charge on a pith-ball B held 2 cm directly above the
F= 1

..i)
pith-ball A, such that the pith-ball A remains
4nEn AB? stationary ?

(Ans. 7.84 pC, sign opposite to that of A)

6. Two identical metal spheres having equal and
similar charges repel each other with a force of
103 N when they are placed 10 cm apart in a medium
of dielectric constant 5. Determine the charge on
each sphere. (Ans. 23.9 x 10 C)

7. The distance between the electron and proton in

hydrogen atom is 5.3 x 101 m.Determine the magni

tude of the ratio of electrostatic and gravitational
F force between them.

Given m, =9.1 x 10 kg, m, =1.67 x 10 kg,

e=1.6x10-1Cand G=6.67 x 101 Nm² kg.
Fig. 1.17 (Ans. F / F = 2.27 x 10)
8 Two identical metallic spheres, having unequal,
As the forces are in equilibrium, the three forces on
opposite charges are placed at a distance 0.90 m
sphere A can be represented by the three sides of
apart in air. After bringing them in contact with
A AOC taken in the same order. Hence
each other, they are again placed at the same
F mg distance apart. Now the force of repulsion between
AC OC AO them is 0.025 N. Calculate the final charge on each
AC of them. [CBSE D 02C|
F=mg X
OC (Ans. 1.5x 10°C)

From (i) and (ii), we have 9. A small brass sphere having a positive charge of
AC 1.7x 10 Cis made to touch anothersphere of the
1
=mg x OC same radius having a negative charge of 3.0x 10C.
AB? Find the force between them when they are
But AC=I sin 0, OC=lcos 0, AB=2 AC=2lsin 0 separated by a distance of 20 cm. What will be the
sin 0 I
force between them when they are immersed in an
=mg x
3?
412 sin? 0 cos 0 l
oil of dielectric constant
4n E
N)
(Ans. 1.1x 10 N; 0.367 x 10
=(4 mg I sin 0 tan 0) 4n Eg'
 I.l6 PHYSICS- X1

(10. The sum of two point charges is 7 C. They repel

each other with a force of 1 N when kept
30 cm
threads 05
long On m
charging the balls
they are found
apart in free space. Calculate the value to repel equaly
of each 0.2 m.
each ofher to a
charge. Calculate the charge distance of
ICBSE091 on each ball

11. Two
(Ans. 5C 2iC) (FHaryana 20021
point charges q, 5x 10 °C 3x10 C HINIS
are located at
and q,
positions (1 m, 3 m, 2m)
and (3 m,
(Ans. 2.357
10G
5 m, 1 m) respectively. Find the forces 1. F= 1
h and
using vector fornm of Coulomb's law.
AT AT
IAns.
+2 -k)N,
hp=-5x 10 (2i

E=5x102i+2-k)N 2. Here 4, =4, -2e3.2x 10"Cr= x 1015

12. Three equally
3.2 m
charged smallobjects are placed F=
shown in Fig. 1.18. The object A exerts an
as
electric
force on object B equal
to 3.0 x 10N.
9x10 x 3.2 x 1019 x3,2 x 1019
B -90 N.
(3.2 10 5

2cm 1 cm
3. For a proton, m = 1.67 x 10 kg,
9=+e= 1.6 x 1019 c.
Fig. 1.18 Weight of proton = Electrical repulsive force
(i) What electric force does Or 9x9
C exert on B? mg = k.
(ii) What is the net electric force on B?
[Ans.(i) 12.0x 10°N,along BA 2kq'9x10 x(1.6 1019 x

(ii) 9.0 x 10° N,

mg 109.8 1.67 x
along BA]
13. Two 23.04
identical

ing a charge
metallic spheres A and B, each carry
16.36
x 10 = 0.014
q, repel each other with a force F. A
third or r= 0.118 m
mnetallic sphere C of the same size, but = 11.8 cm.
un
charged, is successively made to touch
the spheres 4.
exe
m, g k.
A and B, and then removed away. What is the force
=
2
of repulsion between and B?
A
(Ans. 3F/8) 9x 10 x(1.6 x 10-19y
14. Two point charges
+9eand + e are kept at a distance -25.84
9.1× 10Sx9.8
a from each-other. Where should we place a third
r= 5.08 m.
charge q on the line joining the two charges so
that
it may be in equilibrium ? 5. The pith-ball B must have charge opposite to
that of
A so that the upward force of attraction balances
3

Ans 4 from +9echarge the weight of pith-ball A.

When the pith-ball A remains

15. Two point charges of values q and 2q are
electric
stationary,
kept at a distance d apart from each other in air. A
2cm
third charge Q
is to be kept along the same
line in
F= m8
such a way that the net force acting on q and 2q is
zero. Calculate the position of charge Qin terms of q
and d.
[CBSE D 981 But
(Ans.At a
m =9x10 kg,
distance of (V2 - 1) d from charge g)

16. A charge q is placed at the centre of the line joining

4-5uC=5 x 10C, Fig 1.19

two equalcharges Q Show that the system of three r 2 cm = 0.02 nm

charges will be in equilibriumn if q = -O/4. 9x 10" x5 x 10 xq2

ICBSEOD 051 -9x10 x98
(0.02)
17. Two pith-balls each weighing 10 'kg are or

suspended from the same point by means of silk 4,-7.84 x 10 C=7.84 pC.
 ELECTRIC CHARGES AND FIELD 1.17

6. F=
1

.103 9x 10
5 x(0.10)
xq' 1 4142
2
q= 23.9 x10 C
9x10 x5x 10 x3x10 (2í +2j-k)
7. Proceed as in illustrative problem on page 1.18.
3 3

8. Thetwo spheres will share the final charge equally. - 5x10 (2í +2,-kyN
Let q be the charge on cach sphere.

F=
1
492 =0.025 Also, F - - =-5x10(2í+ 2í -kyN.
N
12. Here AB =2cm =0.02 m, BC - 1Cm =0.01 m
or
9x 10 xqxq
=0.025 B
A
(0.90) FgC FRA

2cm cm

1
0.025 x (0.90)2
=225 x 10-14
9x 10 Fig. 1.20

or 10 6 C.
q=1.5 x
Let q be the charge on each object.

9. Charge shared by each sphere 9x9

FBA
4nE, (AB)
(17-3)x 109
2
=7x 10c 9x10° xq
3.0x10-6
9x 10° x(7x10 (0.02
=1.1x10N
(0.20)
gf-10c
9x10° x(7x10
3 x(0.20)
0.367 x105 N
(i) FRC = 19x9-9x10 x
4x 1019
3x(0.01)²
AnE (BC)?
10. Here F= N, 1 r=30 m = 12.0 x10 N,along BA.
As F=k 2 (ii) Net force on charge at B,

10 F= Fgc- FRA =(12.0- 3.0) x 10°

1= 9x x492
(0.30)
=9.0 ×10°N, along BA.
13. Proceed as in Example 10 on page 1.12.
or
495 = 1o-11
14. Force between +9eand q =Force between + e and q
But 4,+ q,=7uC=7x10c 9ex4k. ex
k.
Now (4 -42 =(4 + 44) -Ag,42 (a-x?
=49x1012_4x10-11 or
3 1
or x= 3a/4
a-x
=9x1012
Q
or 4-4 =3x10° =3uC ...(i)
15. For equilibrium of charges q and
must have sign opposite to thatof q or 24.Suppose
24, the charge

On solving ()and (i), we get it is placed at distance x from charge q.

4 =5uC and 4, =2uC 2q

)m d-x
11. Here -í+3f+2k )m, -(3 +5,+ Ê

Fig. 1.21
h-i-(3i+ 5,+k)-(i+ 3j +2ê ) For equilibrium of charge q.

-(2í + 2j-k)m

I-2'+2+(-1 =3m
For equilibrium of charge 24.

2í+2-k
3
k9X24Qx 24 .1)
(d-x)?
 PHYSICS-XI1

From (i)and (i). we get 116 coOMPARING ELECTROSTATIC AND

GRAVITATIONAL FORCES
25. Give acomparison of the electrostatic and orai
2x-(d-) tational forces.

Or V2r=d-N Electrostatic force vs gravitational

or
1

V2+1 d-(V2-
1)d static force is the
charges at rest while the
force

gravitational
of attraction or repulsion
force
force.

betuween

isthe force
o
Electva

of
ie, the charge Qmust be placed at a distance of 1troction between two bodies by virtue their mioee

of
(V2-)d from the charge q. Similarities :
16. Suppose the three charges are placed as shown in
Fig. 1.22.
1. Both forces obey inverse square law ie.,
1
Fa
A
2. Both forces are proportional to product of
masses or charges.
Fig. 1.22
3. Both are central forces i.e., they act along the line

Clearly, the net force on charge g is zero. So

joining the centres of the two bodies.
it is in

equilibrium, the net force on other two charges 4. Both are conservative forces ie., the work done
should also be zero. against these forces does not depend upon the

Total force on charge Q at path followed.

point B is
1 QQ 1 5. Both

4n E (2r)²
q20 forces can operate in vacuum.

Dissimilarities:

or 1 qQ 1 1. Gravitational force is attractive while electro

4n E (2x)' static force may be attractive or repulsive.

2. Gravitational force does not depend on the
or nature
9=-Q/4.
17. In A OCA of forces, we have
of themedium while electrostatic force depends on
the nature of the medium between the two charges.
F mg T 3. Electrostatic forces are much stronger than
AC OC OA gravitational forces.

Illustrative Problem. Coulomb's law for electrical force

between tuocharges and Neuwton's law for gravitational force
between two masses, both have inverse-square dependence on
the distance between charges/masses.

(a) Compare the strength of these forces by determining

the ratio of their magnitude (i)for an electron and a
proton and (i)for two protons.

(b) Estinmate the accelerations and

0.1 m0.1 moB due
for electron proton

C to the electrical force of their mutual attraction

when they are 1 (-1010m) apart.

Á

mg How much is the electrostatic force stronger than

the gravitational force ?
Fig. 1.23

AC (a) (1) From Coulomb's law, the electrostatic force

F= mg X
OC between an electron and a proton separated by

1 AC distance r is

OC ke

9 10" g 10 9.8 x0.1

Negative sign indicates that the force is attractive.
(0.2) 0.5-(0.1?
From Newton's law of the corresponding
9=2.357 10C gravitational attraction
gravitation,
is
 ELECTRIC CHARGES AND FIELO 119
-G 26. Give fuo eramples which illustrate that the

)
electrical forces are enormously stronger than the gravi
where m, and n, are the masses of the proton and tational forces.
electron.
Examples A plastic comb passed through hair
Hence
can easily lift a piece of paper upwards The electro
ke? static attraction between the comb and the piece of
paper overcornes the force of gravity exerted by the
Gm, m, entire earth on the paper.
But k =9x 10 Nmc?, c=1.6x 10 C, (i) When we hold a book in our hand, the electric
m, =1.67 x 10 kg, m, =9.1 x 10 kg,
(frictional) forces between the palm of our hand and
the book easily overcomethe gravitational force on the
G=6.67 x 10 Nm² kg2 book due to the entire earth.

the words of Feynman,

9x 10 (1.6x 10y
x
In if you stand at arm's

6.67 x 10x1.67 x 1027 9.1x 1031 length from your friend and instead of being electri
x

cally neutral each of you had an excess of electrons

= 2.27 x 1039
over protons by just one per cent, then the force of
(a) (i) Similar to that in part (),the ratio of the repulsion between you would be enough to lift the
magnitudes of electric force to the gravitational force entire earth.

between two protons at a distance ris given by

1.17 FORCES BETWEEN MULTIPLE CHARGES:
ke? 9x 10 x (1.6 x 10-19y
THE SUPERPOSITION PRINCIPLE
Gm,m, 6.67 x 10- x (1.67 x 10273
27. State the principle of superpositionof electrostatic
=1.24 x 1036 forces. Hence write an expression for he force on a point
Thus the large value of the (dimensionless) ratio of charge due to a distribution of N -l point charges in
the two forces indicates that the electrostatic forces are terms of their position vectors.

enormously stronger than the gravitational forces. Principle of superposition of electrostatic forces
Coulomb's law gives force between two point
(b)Themagnitudeof the charges.
electric force exerted by a The principle of superposition
proton on an electron is equal to the magnitude of the enables us to find the
force exerted by an electron on a proton. The magni- orce ona point charge due to point charges. a groupof
tude of this force is This principle is based on the property that the forces
with which two charges attract or repel each
other are
F= ke' 9x 1o x (1.6 x1019,?
not affected by the presence of other
charges.
(1010,2
The principle of superposition states that when a
[:: r=1 =10 mn] number charges are interacting, the total force ona given
of
-2.3 x 10N charge is the vector sum of the forces exerted on it due to all

Acceleration of the electron due to the mutual other charges. The force between tuo charges is not afected

attraction with the proton, by the presence of other charges.

2.3 x 10N -2.5 x 102 ms 2

As shown in Fig. 1.24, consider N point charges

m 9.1x 10 kg
413 3e N placed in vacuum at points whose

Acceleration of the proton due to the mutual

position vectors W.r.t. origin O are h 5 5 N

attraction with the electron, respectively.

2.3 x 10N -1.3 x 10 ms2

According to the principle of superposition, the
total force on charge 4, is given by
m 1.67 x 10 kg

Clearly, theacceleration of an electron or a proton

due tothe electric force is much larger than the accele
ration due to gravity. So, we can neglect the effect of
gravitationalfieldon the motion of the electron or the
where Fp.
charge 4, by the
.
N are the

individual charges
forces exerted

443
on

4N
proton. respectively.
 PHYSICS-XII
I.20
force on ath charge
4Y F=Total

4TEig b-1
ba ba
12
N.
where a =1,2,3, ..,
for each choice of a. tho
It may be noticed that
because
the value a. This is
summation on b omits
over other charges. The
summation must be taken only
in a simpler Way as
aboveexpression can be written
follows :
93
=Total
É force on charge q due tomany point

charges
Fig. 1.24 Superposition principle :Force on
charge q, exerted by 4, and q,: 7-7
According to Coulomb's law, the force exerted on 4Te0 all pont |7-7
charges
charge 4, due to 4, is
1
q92
F2 Ae 22T2 Examples based on
Principle of Superposition of
1 9192 Electric Forces

Formulae Used

-4142

F=R+ +2 F E, cos 0
where fo =12=a unit vector pointing from 4, to Units Used
Forces are in newton, charges in coulomb and

q, and p l;-zl=distance of 4, from 4,

distances in metre.

Hence the total force on charge 4, is

Example 19.An infinite number of charges each equal to
1 uCare placed along x-axis
4 at x =lm,x =2 m, x =4 m,
9192
13 t.... + N x=8m and so on. Find the total force on a charge of 1 C
placed at the origin. [IT95]
or Solution. Here q =4 uC=4x 10° C, q% =1C
4Te i=2 1i Bythe principle of superposition, the total force
In terms of position vectors, acting on a charge of 1 C
placed at the origin is

1 1
1 F= q40
9142 + 4,93 +.....

1
+ 49N -9x 10 x 4x 1o x1
Sum of the infinite geometric progression
or

,
4Tio
a 1 4
i- 2

In general,force F, on ath charge q, located at due 4

to all other (N-1)charges may be written as :. F=9x 10 x 4x 10x= 4.8 x 10 N.
3
 ELECTRIC CHARGES AND FIELD 121
Example 20. Consider three charges q,, 9,cach equal to

4,
qat the vertices of an equilateral triangle of side What is the

l.
force on a charge Q (with the same sign as q) placed at the
centroid of the triangle ? [NCERT)
Solution. Suppose the given charges are placed as
shown in Fig. 1.25(a).

Fig. 1.26
ÁO
Let Q be the charge required to be kept at the
F
BO
B centroid G. Then,
42
(a) (6) F =Force at A due to the charge at B

Fig. 1.25 1
along BA
Let AO= BO =CO =r
Force on charge Q due to
1 4
q,y E,= Force at A due to charge at C = along CA
1 Q4;
AO
F+=2 cos 30°,along GA = W3.-1 GA
2 along
Force on charge Q due to 4,

1
Q42 BO Force at A due to charge at G
Ane, BO2 1 1 1 3Q4
Force on charge Q due to q 4TEn AG 4nE, (l/N3) 4E
This must be equal and opposite to (E+ E,).

or
By the principle of superposition, the total force on
nd
charge Q is
Example 22. Consider the charges q, q and -g placed at
the vertices of an equilateral triangle, as shown in Fig. 1.27.
al to What is the force on each charge ? [NCERT)
m, Q9
1[AO +BO+CO] [: 4 =42=43 =9l
=4 Solution. The forces of attraction or repulsion
of 1 C between different pairs of charges are shown in
[TT 95) As shown in Fig. 1.25(b), the angle between each Fig. 1.27. Each such force has magnitude,

pair of the unit vectors AO, BO and CO is 120 so they

1
d force
F=
form a triangle of cyclic vectors. Consequently,

AO+ BO + CO =0 439
Hence =0i.e, the total force on charge Q is zero.
Example 2Y.Three point charges +q each are kept at the

vertices of an equilateral triangle of side ".Determine the

so
magnitude and sign of thecharge tobe kept at its centroid
that the charges at the vertices remain in equilibrium.
(CBSE F 2015)
B
Solution. At any vertex, the charge will be in

due to the 424

equilibrium if the net electric force

remaining three charges is zero.

Fig. 1.27
I20 PHYSICS XI

1-Total force on ath charge

where a 1,2,3, ,N
It may be noticed that for each choice ofa he

summation on b omits the value This is becaue

a.
summation must be taken only over other charges The
above expression can be written in a simpler way as

follows :

Total force on charge qdue to many point

Fig. 1.24 Superposition principle :Force on charges q

charge q, exerted by q, and q,.

According to Coulomb's law, the force exerted on 4no all point|

charges
charge q, due to 4, is
1 492 T2
Examples based on
Principle of Superpositionof
Electric Forces

Formulae Used

1
4192
4T
F-JR++2h H cos 0

where p-5 -a unit vector pointing from 4, to Units Used

Forces are in newton, charges in coulomb and

distances in metre.
4, and rp l-z-distance of 4,from 4,

Hence the total force on charge 4, is Example 19. An infinite number of charges each equal to
4uC are placed alongxaxis at x=1m, x =2 m, x4m,
1
4;N fN X =8m and so on. Find the total force on a charge of 1 C
placed at the origin.

Solution. Here q =4 C= 4x 10C,4, -1C

4Tb0 i-2 By the principle of superposition, the total force

acting on a charge of 1Cplaced at the origin is

In terms of position vectors,

1
F=
4,42 + 4,4%

+ 4,4N -9x10'x4x10 "t

Sum of the infinite geometric progression
or 1

4nlig 2 Ir 3
4
Ingeneral, force F, on ah charge 4, located at rdue
Fe9x 10 x 4 x 10, 3 4.8 x 10 N.
to all other (N - 1) charges may be written as
 ELECTRIC CHARGES AND FIELD I21
Example 20.Consider three charges q,,4:4, cach equal to
qat the vertices of an equilateral triangle side What is the

of
l.
forcc ona chargeQ(with the same sign as g) placed at the
centroid of the triangle ? [NCERT|
Solution. Supposethe given charges
are placed as
shown in Fig. 1.25(a).

CO

Fig. 1.26
AO
Let Q be the charge required to be kept at the
B BO centroid G.Then,
43
(a) (b)
=Force at A due to the charge at B
Fig. 1.25 1 q BA
Let AO= BO =CO=r 4E 2along
Force on charge Q due to 4 = Force at A due to charge at C = along CA
1 091
AO 2
AO?
GA =V3.
4nEn 1
i+E, =2 cos 30°, along along GA
Force on charge Q due to q

1
BO Force at A due to charge atG
4re BO2 1 Q4 1 304
on charge
4ne (/N3
Force due
Q to 43
47E, AG 4rE
1 Q93 -Co
This must be equal and opposite to (E+ E).
By the principle of superposition, the total force on 3Q9 =-V3q or
Q
charge Qis
Example 22.Consider the charges q, q and -q placed at

As shown in
[A0+ BO +CO)
Fig. 1.25(6),
:
the angle between
4 =44=45 =9l

each
the vertices
What

between
Fig.
is

1.27.
of an

different
equilateral

the force on each

Solution. The forces

triangle,

charge

of
pairs of charges are
Each such force has magnitude,
?
as shown in Fig.

attraction or repulsion
shown
1.27.

[NCERT)

in

pair of the unit vectors ACO, BO and CO is 120°, so they

1
F=:
form a triangle of cyclic vectors. Consequently,

A0+ BO + CO -0 439
Hence =0ie.,the total force on charge Qis zero.

Exanple 21.Three point charges +q each are kept at the

vertices of an equilateral triangle of side ".Determine the

so
magnitude and sign of the charge to be kept at its centroid

that the charges at the vertices remain in equilibrium.

|CBSE F 2015]

B
Solution. At any vertex, the charge will be in

equilibrium if the net electric force due to the 429

remaining three charges is zero.
Fig. 1.27
 1.22 PHYSICS-XIl

By theparallelogram law, the net force on charge

JU80y (180 2 180 180 con 120 N

4, is 180,/1+ 1+2x(-1/2) N-180 N

E-+r+2FxF cos 120 BC Let the resultant

force F,. Ther

force F make an angle p with the

V2F212) BC -FBC
tan f
L, sin 120
180+
180 sin
180 cos 120
120

120
h+Lcos
where BC is a unitvector along BC.
180 3/2 V3
Similarly.total force on charge q, is

180+ 180(-)
-FAC
B= 60
where AC is a unit vector along AC. F is parallel to BC.
ie., the resultant force
Total force on charge q, is
Example 24. Four equal point charges each 16 uC are
a square of side 0.2 m.
H=F+ F+2FxF cos 60° n =V3 Fn placed on the four corners
Calculate

of
the force onanyone of the charges.
where n is a unit vector along the direction bisecting
Solution. As shown in Fig, 1.29, suppose the four

corners of the square ABCD.

LACR charges are placed at the
calculate the total force on 4
Example 23. Charges of + 5uC, + 10 uC and -10uCare Let us

placed in air at the corners A, B and C of an equilateral

triangle ABC, having each side cqual to 5 cm. Determine the

resultant force on the charge at A. 0.2 m
charge at B repels the charge at A
Solution. The
with a force,

10 x (5x 109) x (10 x 109), m

(0.05)
U.2

= 180 N, along BA

Bg 0.2 m
A120°
+5 uC
60 Fig. 1.29

Here AB= BC= CD = AD =0.2 m

C
41 =92 93 =94 = l6 uC= 16 x 10

Force exerted on 4, by q, is

9x 10 x 16 x 10 x 16 x 10
B
+10 pC 5cm 10 C (0.2)

Fig. 1.28
= 57.6 N,along AD produced

Force exerted on 4, by 4, IS
A with a force
The charge at Cattracts the charge at
9x 10 x l6 x 10x l6 x 10

9x10 x(5x 10) x(10x10")N (0.2)° + (0.2y'

(0.05)
- 28.8 N, along BD produced
= 180 N, along AC, Force exerted on q, by 4, is
addition, the
By the parallelogram law of vector 9x 10° x 16 x 10x l6 x 10
of resultant force F on charge at A is
magnitude (0.2)

F-5'+h +2h,cos 0 - 57.6 N, along CD produced

 ELECTRIC CHARGES AND FIELD 1.23
20° N
As F and E; are perpendicular to each other, so Net force on charge at A will be zero if

their resultant force is

9x10 x qx 2 x 10 6 =2,34
with the
F-R+R-57.6 + 57.6 9.1x v3'
-57.6/2 -81.5 N, in the direction of E: 2.34 x 0.01 x 3
Hence total force on q, is or =3.9 x 106C= 3.9 uC.

18 x 10
F= + F= 28.8 + 81.5

=110.3 N, along BD produced. Problems for Practice

Example 25. Three point charges of +2 C, -3 uC and
-3 uC are kept at the vertices A, B and C respectively of an 1. Ten positively charged particles are kept fixed on

equilateral triangle of side 20 cm as shown in Fig. 1.30. the x-axis at points x =10 cm, 20 cm, 30 cm, ...,

What should has a charge 1.0 x 10 C,

be the sign and magnitude of the charge to be 100 cm. The first particle

placed at the midpoint (M) of side BC so that the charge at A the second 8x 10C,third 27 x 10 C, and so on.
6 uC are
10* C. Find
remains in equilibrium ? [CBSE D 05] The tenth particle has a charge 1000 x
Calculate
the magrnitude of the electric force acting on a 1C
+2uC (Ans. 4.95 x 10° N)
A A+2uC charge placed at the origin.
the four
Charges q = mC, q, = 0.2 mCand q, =- 0.5 mC

,
2. 1.5
re ABCD. F

B and C respectively, as
are placed at the points A,
shown in Fig. 1.32. If r =12 m and = 0.6 m,
calculate the magnitude of resultant force on 4,

-3 uCA60° +q (Ans. 3.125 x 10 N)

60?,3uC
B 20cm C B
-3 uC -3 uC

Fig. 1.30 Fig. 1.31

Solution. As shown in Fig. 1.31, the force exerted on

F,
charge +2uC by charge at B,

1 4192
E= AA
B

91 92

9x 10 x2 x 106x3 x 10-6
Fig. 1.32
(0.20)

=1.35 N, along AB
3. Two equal positive charges, each of uC interact2
Force exerted on charge + 2uC by charge at C, with a third positive charge of 3
uC situated as
10C 9x 10 x 2 x 10 x3x 10-6 shown in Fig. 1.33. Find the magnitude and
direction of the force experienced by the charge of
(0.20) 3uC. (Ans. 3.456 x 10 N, along OC produced)
106 =1.35 N, along AC
Resultant force of E and E 2 C

ced
F=F+ E,+2 F,E cos 60°
3m
=1.35+ 1.35 +2x 1.35 x 1.35 x 0.5
3 uC
10-6

=1.35 x N3 =2.34 N,along AM o 4m

For the charge at A to be equilibrium, the charge q to
ced
be placed at point Mmust be a positive charge so that it 3m
exerts a force on +2uCchargealong MA.

10 Now, AM=20? - 102

=V300 = 10/3 cm
ed =0.1x 3m Fig. 1.33
 24 PHYSICS-XI|

HINTS
Similarly, force exerted by charge q, on qc.
1. By the principle of superposition,the total on
force F, 2.16x 10 N, along BC There is a d
the Cchargeplaced at the origin is
1 produced
Clearly, F,F (in magnitude) the above equa
The components of F and Ep along charge distrib
Y-axis will
cancel out and t added along X-axis. change the elec

:.Total force on 3 uCcharge, The test charg

4
%
F 2h cos 0 =2 x 2.16 x 10 53
not change the
1.0 x 10 8x 10-8 field as follows
-Ix9x10
(0.10) (0.20) -3.456 x 10 N, along CX. The electric
per unit
27 x 10% 1000 × 10-8 1.18 ELECTRIC FIELD force
+ +
positive cha
...
test
(0.30) (1.00)² 28. Briefly develop the concept of electric field.

=9x 10 x 10°[1+2+3+...+ 10]

Concept of electric field. The electrostatic force acts
between two charged bodies even without any direct
=9x55 x 10° =4.95 x 10° N. contact between them. The nature of this action
The electri
2. = 1 495 9x10 x 0.2 x 10 x1.5 x 103 at-distanCe force can be understood by introducing the
concept of electric field. direction is sar
(1.2)²
Source charge positive
=1.875 x 10° N, along AB produced
Test charge

Units
test ch

and
1 9293_9x10 x0.2 x 10 x0.5 x 103 electric field is

(0.6) newton per cou

= 2.5 x 10° N, along BC L AB metre (Vm

As F LE, So the resultant force on q, is The dimer
Fig. 1.35 A charged body produces an electric field around

it.
follows :
F=JE + -3.125 x10° N. Consider a charged body carrying a positive charge q
placed at point 0. It is assumed that the charge g (E] =
3. Here 94=98 =2uC=2 x 10° C, produces an electrical environment in the surroun
9c=3uC=3x106c ding space, called electric field.

To test the existence of electric field at any point P,

AC= BC=/3 + 4² =5 m
we simply place a small positive charge qo, called the
30. Gire the
A test charge at the point P. If a force F is exerted on the
Physical s

test charge, then we say that an electric field E exists at

5m the point P. The charge q is called the source charge as it
experienced b

ditferent point
3m produces the field E.
9c In general, E
4m
-X 29. Define electric field at a point. Give its units and
vectors. Each
dimensions.

Electric field. An electric field is said to exist at a point vector E (rn.So

3m
5m ifa force of electrical origin is exerted on a stationary charged By knowir
body placed at that point. Quantitatively, the electric fieldor determine the

B or the electric field strength E at a

The Coulomb
the electric intensity

point is defined as the force experienced by a unit positive charge g may

Fig. 1.34 test charge placed at that point, without disturbing the (0) The so

position of source charge. E(n)at

Force exerted by charge 44 on 4c
1 44 9c As shown in Fig. 1.35, suppose a test charge 0 (ii) The val
(AC? experiences a force F at the point P. Then the electric
4Teg force or
at that point will be
9x10 x 2x 10 x3x 100 field

52
= 2.16 x 10 N, along AC produced
Electrosta
 ELECTRIC CHARGESAND FIELD
1.25

There isa difficulty in defining

4c the electric field by Thus an electric field plays an intermediary role in
the above
n
equation. The test charge go may disturb the
Dduced theforces between two charges
charge distribution of the source
charge and hence
field Charge.
Charge Electric
change the electric field E which we want to measure.
Y-axis will concept of electric field is
It is in this sense that the
The test charge qo must be small enough so that it does eofu Electric ficld is a characteristic of the systern of

not change the value of E. lt is better to define glectric charges and independent of the test charge that
is we
field as follows : field.
place ata point to determine the

The electric jicld at a point is defined as the electrostatic Examples based on

force per unit test charge acting on a vanishingly Field
small Relation between Electric
positive test charge placed at that point. Hence
Strength and Force
field. Formulae Used
cforce acts
E=lim
any direct

his action The electric field E is a vector quantity whose

ducing the direction is same as that of the force F exerted on a
Units Used

newton, charge in coulomb and

positive test charge. When force is in

metre, electric field strength is in

distance in
Units and dimensions of electric field. As the
newton per coulomb (NC)or equivalently in
force per unit charge, so unit
electric field is its SI is
volt per metre (Vm).
newton per coulomb (NC).It is equivalent to volt per

metre (Vm). Example 26.Calculate the voltage needed to balance an oil

E can be determined as drop carrying 10electrons when located between the plates
The dimensions for
d around it.
follows : of a capacitor which are 5 nm apart. The mas of oil drop is

3x 10-16 kg. Take g=10 ms s-2 [CBSE OD 95C)

ive chargeq Force MLT-2
[E]= Solution. Here g=10 e=10 x 1.6 x 10-1 C
e charge q Charge
e surroun d=5 mm =5x 10 m, m=3 10 x kg, g =10 ms
MLT-2
A.T 1s
ny point P
E
,called the
30. Give the physical significance of electric field.
erted on the mg
Physical significance of electric field. The force
at
experienced by the test charge
Éexists at qo is different

charge as it Fig. 1.36

different points. So E also varies from point to point.
When the drop is held stationary,
In general, E is not a single vector but a set of infinite

ts units and Upward force on oil drop due to electric field

vectors. Each point r is associated with a unique

=Weight of oil drop
Cist at a point vector E (r), So electric field is an example of vectorfield.
qE =mg
nary charged
By knowing electric field at any point, we can
V
ectric field or
determine the force on a charge placed at that point. or q. = mg
d
gth E at a The Coulomb force on a charge gn due to a source
3x 10-x 10x 5x1o-3 =9.375 V.
nit positive charge q may be treated as two stage process : .. V= mg d

definite field 10 x 1.6 x 10-19

turbing the (i) The source charge g produces a

É() at every point r Example 27. A positively charged oil drop is prevented
charge under by applying a vertical electric
he electriC
9o

(i) Thevalue of É(r) at any point r determines the

from
field
falling

of 100 V. If
gravity

he mass of the drop is 1.6x 108find

force on charge 4, at that point. This force is the number of electrons carried by the drop.

Solution. Here m= 1.6x 10g=1.6 x 10°kg,

E=100 Vm-l
Electrostatic force = Charge x Electric field.
 I26 PHYSICS-XI1

As the oil drop is positively Let T be the tension

Solution. (a)The upwa
charged, the field E must act in the thread and 0 be force ef on the electron

vertically upwards. When the drop the angle itmakes with

remains stationary, vertical, as shown in Acceleration of the e

Upward force of electric field mg Fig. 1.39. When the bob

is in equilibrium,
As -, s=uf +-at
-Weight of oil drop Fig. 1.37 2
gE or ncE = mg T sin 0 qEi Time of fall of the ele

-
Sin

Number of electrons carried by the oil drop, T cos 0 mg 25 2sm

1.6x 10x 10 T sin9 q
- 1012 tan 0
16x 1019x100 T cos 0 mg
m -2.9 10s.
Example 28.A sinple pendulum consists asmall splhere 2x 10x2 10 -0.51 x
of
(b)Thedownward field
of mass m suspended by a thread length The sphere 80× 10"9,8 Fig. 1.39
of
1.
on the proton.
carriesa positioe charge q. The pendulum is placed in a
0= 27
uniform electric field of strength E directed vertically or
doenoards. Find the period of oscillation of the pendulum =
qE 2x 10x2 x 10
=8.81 10 N.
Also, T

due to the electrostatic force acting on the sphere, neglecting sin 0 sin 27 Time of fall of the prot

the effect of the gravitational force. [CBSE OD 19]

Example 30.An electron moves a distance of 6 cm when 24 25m
Solution. Restoring force, F=-qEsin0 accelerated
Jrom rest by an electric field of strength

For small 0, sin 0-0= 2x 10 NC-. Calculate the time of travel. The mass and
charge of electron are 9x 10 kg and 1.6x 10 C -1.25 10 s.
respectively. [CBSE D 91] Thus the heavier partie
Solution. Force exerted on the electron by the through the same distan
A electric field,
situation of free fall unde
F= eE fall is independent of the

acceleration due to gravit

.. Acceleration,
has been ignored.
F eE 1.6x x2 x 104
a=-= 10-19
=0.35 x 10l6 ms2 Example 32.An electron
9x 10-31 fwo large parallel metal

pl
Now u =0,s =6.0 cm =0.06 mn, a =0.35 x 10°ms 20 mm. The upper plate has

sin to the lower plate. Hou lonm

qE As s=ut +at?
qE cos 0
2 the upper plate ? Take

Fig. 1.38 qE
0.06 =0 +x 0.35 x 10l6 x
Solution. Here V=
0.06 x 2 e
ma = Or t= =0.585 x 10s. =L8x 10 Ckg
V0.35 x 106
Lpward force on the
or d'xqEy=-ox
ml
Example 31. An electron falls through a distance of 1.5 cm field is

in a uniform electric field of magnitude 2.0 x 10 NC

F=eE =
This equation represents SHM of angular frequency, (Fig. 1.40(a)]. The direction of the field is reversed keeping

its magnitude unchanged and a proton falls through the Acceleration,

gE
same distance [Fig. 1.40(b)]. Compute the time of fall in each
F eV 18 x 10l
Vml
case. Contrast the situation (a)with that of 'free fall under 0.02
2 ml [NCERT ;CBSE
T= = 21 gravity' 18C]
we
Using, s =at',

Example 29.A pendulum ofmass 80milligram carrying a t= 25 24

charge of 2 x 10* Cis at rest in a horizontal uniform

electric fieldof2 x 10* Vm.

Find the tension in the thread
Example 33.A stream of

pendulum and the angle it makes with the vertical.

of the
of 3 x 10 msis deflectea

Solution. Here m= 80mg =80 x 10 kg,

of 0.1 m in uniform a ele
(a) (b)

q=2 x 10C, E=2 x 10 Vm! Fig. 1.40

Determine elm of electron:
 ELECTRIC CHARGES AND FIELD 1.27
Solution. (a) The upward
force eE on the electron.
field exerts a downward Solution. Here U, =3x 10 ms,
y=2 mm =2 10 m, x x =0.1 m,
Acceleration of the electron, a, =
1m,
E= 18 V cm 1800 V m

1 eE
As u =0, s=ut+-at= 1ma = eE or a= and t=
2
.. Time of fall of the electron is
y=
2 m v
|2s 2sm, 2x 1.5x 10 x9.1 x 10-31

1.6x 10x 2.0 x 10 Or e 2yv, 2x2x10- x9x 1014

Ex? 1800 x (0.1
= 2.9 x 10 s.

(b)The downward field exerts a downward force eE - 2x 10"Ckg1.

on the proton. Example 34. An electric field E is set up betveen the two
eE
parallel plates of a capacitor, as shoun in Fig. 1.41. An
electron enters the field symmetricaly between the plates

Time of fall of the proton is

with a speed vo. The length eachplate is l. Find the angle of

of
deviation of the path of the electron as it comes out of the field.
2s |2x 1.5x 10-×1.67 10-27
(2sy x

1.6x 10-19. x 2.0x 10

=1.25x 107 s.
Thus the heavier particle takes a greatertime to fall
through the same distance. This is in contrast to the
situation of 'free fall under gravity' where the time of
fall is independent of the mass of the body. Here the
Fig. 1.41
acceleration due to gravity 'g', being negligibly small,
has been ignored. Solution. Acceleration of the electron in the

Example 32.An electron is liberated from the lower of the upward direction,
two large parallel metal plates separated by a distance of a= eE
20 mm. The upper plate has apotential of +2400 V relative
to the lower plate. How long does the electron take to reach
Time taken to cross the field, t=
upper plate ? Take of electrons 1.8 x 1o" Ckg-1.
the
Upward component of electron velocity on
Solution. Here V = 2400 V, d=20 mm =0.02 m, emerging from field region,
eEl
-1.8 x 10" C kg

Upward force on the electron exerted by electric

Horizontal componentremains same, V, =Vo
field is
eV If 0 is the angle of deviation of the path of the
F= eE = d electron, then
.. Acceleration, 0= Py eEl
tan Or 0 = tan
eV 1.8x 10 x 2400 ms 2 mug
ms=2.16 x 10l6

md 0.02
Example 35. A charged particle, of charge 2uC and mass
Using, s =at,
2 we get 10 milligram, moving with a velocity of 1000 m/s entres a
NC-
10 directed
2s 2d 2 x 0.02 10s.
uniform electric field of strength

S=1.4 x
perpendicular to its direction of motion. Find the velocity

V2.16 x 1o16 after 10 s. [CBSE SP 11]

and displacenent, of the particle

a normal to the
Example 33. A stream of electrons moving with velocity Solution. The velocity of the particle,

of 3 x 10 ms!is deflected by 2mm in traversing a distance direction of field.

of 0.1 m in a uniform electric field of strength 18 V cn. U. = 1000 ms, is constant

Determine elm of electrons.
 1.28 PHYSICS- XI1

19 10
cE L6x 10 x3.34
F
The velocity of the particle, along the direction of
3. ()a 27
m 1.67x 1

field, after 10 s, is given by

q, 2x 10x10'x 10 -3.2 10!ms the

-2000ms
4, ta,t=0+

1
10× 106 the
s0}at'
()
The net velocity after 10 the
s,
|2002 -3.54 107
p-vI000)+(2000) -1000 V5 ms ! V3.2 x 10T sym
Sam
Displacement, along the -axis, after 10 s, (i)The field must act vertically downwards so that

X=1000x 10 m 10000 the positively charged protorn falls downward. 12

m
thrown against the field, so
Displacement alongy-axis (in the direction of field) 4. As the particle is
after 10 s, F qE 40x10x10 4x10 ms 2
100x 106 po
yet4y? =
1

(0) +1y212x10x10
2 m ,10x10 x(10)
7=+
11

2as
m

Con
-10000 m
0 -(200)' +2(-4×10')xs
Net displacement, 200 ×200 hav
m 0.5

m.
8x104 orig
r=yr'y? - (10000° +(10000) =10000V/2 m.
3.2 x 1019 x16x10 =8x102ms2
5. a= F_qE
11 6.4x1027
Problems For Practice
-i=2as
1. Calculate the electric field strength required to just u'-0=2 x8 x102 x2x102
supporta water drop of mass 10° kg and having a
v=4/2 x10 ms1.
charge 1.6 x 10-19 c. [CBSE OD 99]

106 NC-l)
(Ans.6.125 x
1.19 ELECTRIC FIELD DUE TO A POINT CHARGE
2. In Millikan's experiment, an oil drop of radius
31. Obtain an expression for the electric field
10 cm remains suspended between the plates What
of 5e at a point at a distance r from a charge q.
which are 1 cm apart. If the drop has charge
intensity

between is the nature of this field ?

over it, calculate the potential difference
the plates. The density of oil may be taken as Electric field due
a point charge. to A single point
As shown
gem -3 charge has the simplest electric field.
(Ans.770 V) in
1.5

consider a point charge q placed the 0.

3. A proton falls down through a distance of 2cm in a Fig. 1.42, at origin

x 10°NC-,
uniform electric field of magnitude 3.34
the (ii) the poir
Determine (i) the acceleration of electron
Source
Test
time taken by the proton to fall through the
charge
Cou
of the electric charge
distance of 2 cm, and (iii)the direction chai

field. Mass of a proton is 1.67 x 10kg. Fig. 1.42 Electric field of a point charge.

(Ans. 3,2 x 10 ms-2 ,3.54x 10 -s, at a point P at a

We wish to determine its electric field
vertically downwards)
distance r from it. For this, imagine a test charge 4
4. A positive of 100 mg is thrown
charge particle in placed at point P. According to Coulomb's law, the
whe
opposite direction to a uniform electric field of
and
force on charge 4, is

strength 1x10°NC.If the charge on the particle is

elec

40 uCand the initial velocity is 200 ms,how much

distance it will travel before coming to the rest
? JEE Main June 22] (Ans. 0.5 m)
momentarily where is a unit vector in the direction from q to o

5. A a-particle of mass 6.4 x 10 kg and charge

Electric field at point P is
3.2 x10-1C is situated in an electric field of

16x 10VmIf the particle starts from rest, find its 1

velocity at the end of 2x 10 m path.

(Ans. 4/2 x10° ms-l)
The magnitude of the field E is
fiel
HINTS
1 is e
V E=
2 User
3 pg=ne each

are
 ELECTRIC CHARGESAND FIELD .29
Clearly, Exl/ This means that at
the spherical surface
all points on Hence, the electric field at point Pdue to the systern
drawn around the point charge, of N charges is
the magnitudeof E is
same and does not depend on
the direction of r. Such a
field is called
spherica lly
symmetricor radial field, ie., a field which looks 1
the
same in all directions when seen
from the point charge.

120 ELECTRIC FIELD DUE TO A SYSTEM

OF POINT CHARGES or

32. Deduce an expression for the electric

field at a
point due to a system of point charges. N
Electric field due to a system of point charges. L
Consider a system of N point charges 41 92 N
having position vectors r, z, ..y with respect to the
origin 0. We wish to determine the electric field at

E;
E,+
91

Fig. 1.44 Electric field at a point due to a system of

Notations used in the determination of electric
Fig. 1.43 charges is the vector sum of the electric
feld at a point due to two point charges. fields at the point due to individual charges.

point P whose position vector is r. According to In terms of position vectors, we can write

Coulomb's law, the force on charge test 4o due to 1

charge 4, is
1 4neo
i=1|7-77-l
1
4Te0

is a unit vector in the direction from q, to P

where r,p

and ro is the distance between q, and P. Hence the

Examples based on
electric fieldat point P due to charge 4, is
Electric Fields of Point Charges
Formulae Used
1
1, E=
Similarly, electric field at P due to charge 4, is

1
2. By the principle of superposition, electric field

47k0 due to a number of point charges,

According to principle of superposition of electric

fields, the electric field at any point due to a group of charges
Units Used
is equal to the vector sum of the electric fields produced by
coulomb and r in metre; E is in NC-!
each charge indvidully at that point, when all other charges When q in
is
are assumed to be absent. or Vm
 I30
PHIYSICS
COmpie 36. Assuming that the charge on an alom is Exarmple 39. Tuv point charges 16 andpCare jC

of
istribuied uniformly in a sphere of rodius 10 hat & cm
m placed apart in air. Determine the posilion of lhe pornt
twll be the electric field at the
surface of the gold at ehich the resultant field is zero.
atom
For gold.Z-79
Solution. Let P be the point at distarice x cm from
Solution. The charge may be
assumed to becon where the net field is zero.

A,
Centrated at the centre of the
sphereof radius 10 m
m. q-Ze-79 x 16x 10C 4,16uC -9p

E
9x10 x 79 x 1.6x 1019
(10 0

-1138 x 10 NC1 Fig. 146

Example 37. Tuo point charges A
+8x 10C and -8x 10C
and Bof magnitude

distance d apart.
respectively
The
are placed ata
clectric field at the middle point
At point P, E + E =0
O
betueen the charges is 6.4x
NC 10 Find the distance 'd
kx 16x 10 kr(-9) 10
=0
betueen the point charges A and
B. (JEE Main June 22] (xx 102} [(8-x) 1o
Solution.
16 9
-810 C
-8x 10 c 6-x
-B 3
d/2 or

32
kq 8k
d
or
I=m,
7 32 cm
32
=
7 cm, both
At x E, and E, will be in the same
d2škg_8x9x 10 x8x10-6
=9 direction,therefore, net electric field cannot be zero.

Eo 6.4x 10 Hence x =32 cm

d=3m. i.e., electric field is zero at a point 24 cm to the right of
Example 38. Tuo point -9HC charge.
charges of +luC and +4uC are
kept 30 cm apart. How far from the +luC charge on the line Example 40. Tuo point charges q, =+ 0.2 C and
joining the two charges, will the net electric field be zero ?
42 =+0.4 C are placed 0.1 m apart. Calculate the electric
[CBSE OD 20] field at
Solution.
(a) the midpoint between the charges.
92 =+4 uC
(b)a point on the line joining 4, and q, such that it is
A E 0.05 m away from 4, and 0.15 m away from 4,
0.30- x
Solution. (a) Let O be the midpoint between the

Fig. 1.45 two charges.

The electric field at point P will be zero if

91 =+0.2 C -+04C

1 1x1o6 1 4x 10-6
A
EE 0.1 m
4n (0.30-x?

(0.30- x =4y? or 0.30 - x=+2x

Fig. 1.47
or X=0.10 m or X=-0.30 m
Atx=-0.30 n i.e., at 30cm to the left of q,, both E, Electric field at O due to q,

and E, will be in the same direction, therefore net kq, 9x10 x0.2
electric field cannot be zero. Net electric field will be E -72 x 10l NC!,
zero at x=+0.10m ie,at 10cm to the right of +1uC (0.05)

charge. acting along A0

 ELECTRIC CHARGES AND FIELD 1.31

Electric field atO due to q The clectric field vector E, at A due to the negative
ky, 9x10 x0.4
charge 4, points towards the right and it has a
(0.05) magnitude,

-144 x 1o" NC, 9x 10× 10 8

acting along BO E, NC-3.6x 10' NC1
Net field at O =E, -E (0.05

-7.2 x 10" NC,acting along BO. Magnitude of the total electric field at A L

(b) Electric E, = E + E,
field at P due to q, =7.2 10 NC1
3.6x 10 +3.6x 1o
9x10 x 0.2
kq,

(0.15)
, acting along AP
E, is directed towards theright.
Electric field at P due to b The electric field vector E,at B due to the positive

kq, 9x 10 x 0.4 charge q, points towards the left and it has a

E, = acting along BP
(0.05)° magnitude,
9x10 x 10-8
NC-=3.6 10 NC1
9 =+0.2C 2=+0.4 C E E, =
x

-p (0.05)?
A E,
m The electric field vector E, at B due to the negative
-0.1 0.05 m
charge q5 points towards the right and it has a

Fig. 1.48 magnitude,

9x10 x 10-8
Net electric field at point P is E, = NC- =4 x 10° NC-l
0.2 0.4 (0.15)?

E= E, + E, =9x 10 Magnitude of the total electric field at B

(0.15) (0.05)

E, =E -E, = 3.2 x 10 NC1

-1,52% 1o" NC,acting along AP.
E, is directed towards the left.

Example 41. Two point charges q, and q, of 10 Cand

Magnitude of each electric field vector, at point C
10Crespectively are placed 0.1 m apart. Calculate the

electric fields at points A, B and C shown in Fig. 1.49. of charges 4, and q, is

[NCERT] 9x 10 x 10-8 =9x 10° NC-I
E =
E, E, = (0.1°
The directions in which these two vectors point are
C 120° E shown in Fig. 1.49. The resultant of these vectors is
given by
60°

E, =NE+ E +2 E,cos 0
E,
=(9x 10 )'+ (9 x 10)+ 2x9x10° x9x 10' cos 120°
=9x 10°1+1+2(-1/2) NC-=9x10° NC-1
+10°c
B --10c Since E, and E, are equal in magnitude, so their

0.05 m -H- 0.05 m - 0,05 m - resultant E, acts along the bisector of the angle
k
between E, and E,, i.e, towards right.
Fig. 1.49

Example 42. ABCD is asquare of side 5

m. Charges of

Solution. The electric field vector E, atA due to the

D
+ 50 C, -50 C and + 50C are placed at A, and C

positive charge 4, points towards the right and it has a respectively. Find the resultant electric field at B.

magnitude, Solution.Electric field at B due to + 50 C charge at

9x10x108 NC
E =kg, (0.05)²
A is
50

NC
E =k=k=2k,along AB
=3.6x 10
I32 PHYSICS 1
Exarple 43. For
charges44-4-4 are placed
A Sm pectielyaf the four corners A, B, C and D of a square of
+ 50 C side 'a' Calculate the electric field at the centre of the square

Panjab

5m Solution. Let E,, Ep and E, be the electric felds

5m
at the centre O ofthe square due to the chargrs at A, B
Cand D respectively. Their directions are as shown in
Fig. 1.51(a).
+50 C 50 C

5m Tig 152

A
Fig. 1.50 Their magnitudes are

1 6a
Electric ield at B due to -50 C charge at Cis
50
-2 k, along BC
De

D
Electric field at B due to + 50 C charge at D is (b
(a)
50

Component of
E, =k.
s
E,along x-axis
+5
=k,

=2k
along DB Fig. 1.51

Since all the charges are of equal magnitude and at

The magniude of the

the same distance r from the centre O, so

(as it acts along x-axis)

Component of E, along r-axis =0 E = Ey = e - Ep =k 4 2-2

k

Component of E, along x-axis

(as it acts along y-axis)

E
2 :+-) =E52 =

(fhe
. same
1 k Because E, and act in the direction, so 5eld
=E cos 45° = k.
their resul tant is
resultant

Total electric field at B along x-axis

E =E + E 92kg_ 4k tanß =
Ee-E cos
E, =2k +0 + Similarly, resultant of E, and Ey is
Now, E = Eg + Ep = 4kq B= tan
343
5
Component of E, along y-axis =0
Component of E, along y-axis =2k
Now, the resultant of E,and E will be
Problems For Prac
Component of E along y-axis
1
Ey =E sin 45° =k. 1 An electron is separa
V2 2 distance of 0.53 A C
=4V2 k9 of the electru
But the components of E, and E, act in opposite direc location

tions, therefore, total electric field at Balong y-axis 2. Determine the electr
directed parallel to AD or BC, as shown in Fig. 1.51{b)
=2k - muceus at a distance
cos B = ..B=45°
E V2
Resultant electric field at B will be 3. Two point charges
ie, the resultant field is inclined at an angle of45° with AC. of 6a. Find
distance

E-E+ Example 44.Two point charges +6q and -8qare placed at bwo chargeswhere
the vertices 'B and 'C' of anequilateral triangle ABCof sade (Ans. At

a' as shown in Fig. 1.52(a). Obtain the expression for ()the 4 Two point charges

magnitude and (i) the direction of the resultant electric field -2x10Crespec
=3k =3x9x 10 NC=2.7 x 1010 NC 1 at the vertex A due to these tuo charges. [CBSE OD 14C) Calculate the electr

joning the two char

If the resultant field Emakes angle Bwith x-axis, then Solution. ()As shown in Fig. 1.52(b), the fields at
(Ans 9000
tan ß=
Ey 2-1/N2) k =0.4776 or ß= 25.5°, point A due to the charges at Band C are E and EC
E, (2+1/ N2 ) k
respectively.
 ELECTRIC CHARGES AND FIELD 133
EpA 5. Two pointcharges +q and -2q are placed at the

vertices 'B'and 'C' of an equilateral triangle ABC

of side 'a'as shown in Fig. 1.53. Obtain the

20
expression for () the magnitude ard (ii) the

of the resultant electric field at the vertex

direction

A due to these two charges. (CBSE OD 14C]

Fac 1 qv3 30° with AC]

(Ans. () (ii)

A
+6qe 8q + 6qe

Fig. 1.52 (a) (b)

Their magnitudes are

+9 B C -24

EgA = 4TE 1 6q
=6E, where E =
1
9
a
Fig. 1.53

1 8q of electric field at
-8E 6. Find the magnitude and direction

point Pin Fig. 1.54.

The magnitude of the resultant field is +4

Epet =ERA+EÁC +2Ega EAc cos120° Eg

= /(6E' +(8E'+2 x6Ex8Ex

1
=EV52 qN52
4E a
the resultant field makes an angle ß with AC, then B C
(i) If

ERA Sin 120° 6Ex (V3/2) _343 Fig. 1.54

tanß =
EAc t EgA COS120°
SE+6E
5 Ans. E=
4E,
1
-, a
along BP produced

3/3 7. Three charges,each equal to q are placed at the three

B= tan- 3V3
.:
5 corners of a square of side a. Find the electric field
at the fourth corner.
Ans. (2V2 1) +
Problems For Practice
HINTS
1. An electron is separated from the proton through a
1. Electric field at the location of the electron,
field at the
distance of 0.53 Ä. Calculate the electric

of the 5.1x 10 NC-) 1 9x 10 x 1.6 x 10l9

location electron. (Ans. E= =5.1 x10" NC-1
field produced by a helium (0.53 x 10i02
2. Determine the electríc

from it.
and r=1=10 10 m.
nucleus at a distance of q=+
1
2e
2. Here
(Ans. 2.88 x 10! NC-)
3. Suppose the electric field is zero at distance x from
charges + q and + 4q are separated
by a
3. Two point the charge + 4. Then
on the line joining the
distance of 6a. Find the point
two charges where the electric field is
zero.

from charge + q)
4n E (6a - )
(Ans. Ata distance 2a
2x 10°C and (6a-x) =4r or 6a -x=2r
Two charges 4, and 4, of
point
Or
are placed 0.4 m apart.
4.
-2 x 10C respectively
at the centre of the line Electric field is zero at distance 2a from the
field
Calculate the electríc
[CBSE F 94C]
joining the two charges. charge +4
(Ans.9000 NC, towards the -vecharge) 4. Proceed as in Exercise 18 on page 1.80.
 I34 PHYSICS-XI|

5. Proceed as in the solution of Example 44 on

page 1.32.

6. Here E, and Ec are equal and opposite and hence

cancel out.
where r= is a unit

BP =a sin 45° =n/N2

vector pointing from the
Hence E= Ep= smallcharge dq towards
ArE, (a/N2) the point charge 4o- By dg
1 the principle of super
along BP produced. position, the total force
Ane, a
on a point charge
on charge g, will be the Fig. 1.56 Force
7. Refer to Fig. 1.55. 4o due
vector sum of the forces to a continuous charge

distribution.
exerted by all such
small charges and is
D given by

Or 1

34. Name the different types of continuOus charge

C
distributions. Define their respective charge densities.
Fig. 1.55 Write expression for the electric field produced by each
type of charge distribution.
Hence write expression for
the electric field of a general source charge
distribution.

Differenttypes of continuous charge

distributions.
There are three types of continuous
+ charge distributions:
(a) Volume charge distribution. If is a charge distri
bution spread over a three dimnensional volume or region V of
1
-(2/2 + 1)
space, as shown in Fig, 1.57. We define the volumecharge
density at any point in this volume as the charge contained
per unit volume at that point, i.e.,

dq
1.21 CONTINUOUS CHARGE DISTRIBUTION p=
dy
33. What is
a continuous charge
distribution ? How The SI unit for p
can we is coulombper cubic metre (Cm).
the force on a point
calculate
charge q due to a For example,
continuous charge distribution ? if a
charge q is distributed
Continuous charge distribution. In practice, over the entire volume
we
deal with charges much greaterin
magnitude than the of a sphere of radius R,
charge on an electron, so we
can ignore the quantum then its volume charge
nature of charges and imagine
that the charge is spread
in a region in a density is
continuous manner. Such a
charge
distribution is known as a continuous
charge distribution.
p= 4 9Cm3 dV +
Calculation of the force ona
charge due to a conti TR3
nuous charge distribution. As 3
shown in Fig. 1.56,
consider a point charge q, lying
near a region of contin The
uous charge distribution. This charge Con
continuous charge distri-
bution can be imagined to tained in small volume
consist of a large number of dV is
dq =p dV
small charges dq.
According to Coulomb's law,
force on point charge 4, the
due to small charge dq is dq =p dV ig.1.57 Volume charge
distribution
 ELECTRIC CHARGES AND FIELD 1.35
Total electrostatic force exerted on charge g, dueto density at any point on this line as the charge per unit
the entie volume V isgiven by length of the line af that point, ie,

dq
dl

Electric field due tothe volumecharge distribution

The SIunit for .is Cm
at the location of
charge q, is

0
(b) Surface charge dL
distribution. It is a charge dq = dL
distribution spread over a two-dimensional surface S in
space, as shown in Fig. 1.58. We define the surface
charge density at any point on this surface as the charge per
unit areaat that point, i.e.,

dq
Fig. 1.59 Line charge distribution.
ds
The SI unit for is Cm For exanmple, if a charge q is uniformly distributed
over a ring of radius R, then its linear charge density is

2= Cm-1
2TR
The charge contained in small length dLis
+
dq =dL
Total electrostatic force exerted on charge q, due to
the entire length Lis given by

dL
dq =o dS
Electric field due to the line charge distribution at
Fig. 1.58 Surface charge distribution.
the location of charge 4, is

For example, if a charge qis uniformly distributed

over the surface of a spherical conductor of radius R,

then its surface charge density is

The total electric field due to a continuous charge

Cm2
4rR2
distribution is given by

The charge contained in small area dS is

dq =odS 1
or Econt
Total electrostatic force exerted on charge q% due to
V L
the entire surface S is given by
General charge distribution. A general charge distri
bution consists of continuous as well as discrete charges.
Hence total electric field due to a general charge
Electric field due to the surface charge distribution distribution at the location of charge q,is given by
the location of charge 4, is
=E
at
E total discrete cont

1 1
N
dSi. total

(c) Line charge distribution. If is a charge

distribution alongT oHe-dimensionaleurve or line L in
space, as shown in Fig. 1.59. We define the line charge
 PHYSICS-XI

four drops of radius 0.02 m and eacl

In all the above caseslr is a variable unit xample d6 Sixty

carrving a charge of S uCare combined to form a bigger drom

vector direted from each point of the volume, surtace Find hoe the surface density of electrification will changeif
or line charge distribution towards the location of the Ho charge is lost. EE Main June 221

point chage Soution. Volume of each smalldrop

Examples based on n(002y m'
Continuous ChargeDistributions
Formulee Used Volume of 64 small drops

1 Volune charge density, p Let Rbe the radius of the bigger drop formed. Then
dV

2. Suriace charge density, dq (002)64

o

,
dS
Or
R'(002yx4'
3 Linear change density, A
dL R0.02x4 =0.08 m
4 ForeeNerted on a charge due to a continuous Charge on small drop 5uC-5x 10C
charge distribution
Surface charge density of small drop,
5x10
Cm2
4
4n(0.02)
5. Electric field due to a continuouscharge distribution,
Surface charge density of bigger drop,
5x 10x 64
Cm 2
Units Used 4n(0.08)²
-6
pis in Cmin Cm,à in Cm and Ein NC1, 5x 10 4n
(0.08)-1:4
4n(0.02) 5x 10x64 4

Example 45. A charged spherical conductor has a surface Example Obtain

47. the formula for the electric field due to
density of 0.7 Cm.When its charge is increased by 0.44 C, a long thin wire of uniform linear charge density àwithout
the charge density changesby 0.14 Cm". Find the radius of using Gauss's law.
the sphere and initial charge on it.

,
Solution. Electric field of a line charge from
Solution. o Coulomb's law. Consider an infinite line of charge with
uniform line charge density as shown in Fig. 1.60.
n first case =9
0.7
4n *)
We wish to calculate its electric field at
any point Pat a
distance y from it. The charge on small element dx of
the line charge will be
In second case
= hdx
0.7 +0.14 -9+0.44 d

4n

9t0.44 dEy
0.84 ..(0)
4nr
Dividing (i) by (), we get,
d E,
0.84 +0.44 or
07 514044
Initial charge, q-2.2 C.

From (), r-22 Vox4n Vo7x4x

22 x7
0.5 m. dx

y07x4x22
Fig. 1.60 A section of an infinite line of charge.
 ELECTRIC CHARGES AND FIELD 137
The electric field at the point P due to the charge
element dq will be

dE - 1
dg 1
hdx 11 dE.sin 9

dE cos 9
The field dE has two dE cos
components:
dE, =-dE sin 9 and dE, = dE cos 0
dE sin 9
The negative sign in x-component
indicates that
d E, acts in the negative x-direction. Every charge ele
ment on the right has a

out.
corresponding charge element
on the left. The x-components of
elements will be equal and
two such charge
opposite and hence
The resultant field E gets contribuions only
cancel

froms
.
element
The magnitude
dl at the field
Fig. 1.61

of the
point
field

P is
d E produced by the

y-componentsand is given by
dq k dl
x=+o
2Tua2
E= E, =[aE,- fcos9dE
As shown in Fig. 1.61, the field dE has two
X=00 components :
1 dx
=2 cos 0. 1. the axial component dE cos 0, and
x=0 2. the perpendicular component dE sin 0.

Since the perperndicular components

cos 0 dx of any two
diametrically opposite elements are equal and
2 e0x=0 y'+x?
opposite, they all cancel out in pairs. Only the axial
Now *=y tan 0 components will add up to produce the resultant field

É at point P, which is given by

dx =y sec0 de
2rra

0 =/2
0 de E= dEcos 0
cos 0 y sec
0
[Only the axial components
2TE0
=0 y' (1+ tan'e) contribute towards E]
2ra
0=n/2 dl x 1
2 kqx

cos 0 do = (sin9jg/2 2Ta r 2Ta

0
2Teg Y
6=0 2eo y
T
-sin 0
2Tteo y kqx kqx 1

2ra 2rna (x+a'y32g2ra

or E
2Te, y
kqx 1
Or E=
Example 48.A charge is distributed uniformly over a ring

radius 'a'. Obtain an expression for the electric intensity E

of
Special Case
at a point on the axis of the ring. Hence show that for points
at large distances from the ring, it behaves like a point For points at large distances from the ring, x >>a
charge. [CBSE D 16, 20] k 1
Solution. Supposethat the ring is placed with its

plane perpendicular to the x-axis, as shown in Fig 1.61. This is the same as the field due to a point charge,
Consider a small element dl of the ring. indicating that for far off axial points, the charged ring

As the total charge q is uniformly distributed, the as a point charge.

behaves
charge dq on the element dl is Example 49. A thin semicircular ring of radius a is

dq =2T 9 dl charged uniformly and the charge per unit length is à Find
a the electric field at its centre. [CBSE PMT 2000, AIEEE 2010]
 PHYSICS-X1
138
cach HINTS
two symmetric clements
Solution. Consider 4
of the two 1 Use o
of length dlat A and
B The electric fields
4nr
PO get cancelled while
to
elements perpendicular q4nr a
Use
PO added. 2.
those along
Electric
get

at field O
due to an element of length dl
is
3. Surface area of cube 6r16 0.01-0%, m2

4 4
dE= 1 d4cos 0 |Along Poj 4
3
23 2 22/2:2
1 2dl
COs 0
4n2 24 22 22
1 (ad0)cos 0 =ad9]
[dl
1.22 ELECTRIC DIPOLE
4Ten a ? Define dipole moment
dipole
35.What isan electric

Give some examples of

electric
and give its SI unit.
or point dipoles ?
dipoles. What are ideal
and opposite charges
dE Electricdipole. A pair of equal
dl dipole.
is called an electric
separated by a small distance
the strength of an
Dipole mornent. It measures
an electric dipole is a
P A The dipole moment

of
electric dipole.
is either charge times the separation
vector whose magnitude
along
charges and the direction is
dE between the two opposite
the positive charge.
the dipole axis from the negative to

As shown in consider an electric dipole

Fig. 1.63,

separated by dis
consisting of charges + q and -q and
is called dipole axis.
Fig. 1.62 tance2 a. The line joining the charges
2a +4
Total electric field at the centre O is

p
n/2 z|2 1 hcos 9d0
E= (dE=2 Fig. 1.63

-n/2

1 Dipole moment =Either charge x a vector drawn

1 1= from negative to positive charge

or p =qx2 a

Droblems for Practice Thus the dipole moment p is avector quantity. Its

1 Auniformly charged sphere carries a total charge direction is along the dipole axis from -q to + q and its
of 2r x 1012C. Its radius is 5 cm and is placed in magnitude is

vacuum. Determine its surface charge density.

p=qx2a
(Ans.2 x 1010 Cm2)
to electrify a The SI unit of dipole moment is coulomb metre (Cm).
2. Whatcharge would be required
charge When both the charge q and separation 2a are finite,
sphere of radius 15 cm so as to get a surface
7 the dipole has a finite size (equal to 2a), a location
density Cm? 2 (Ans. 1.8 10 C) (midpoint between +q and -9), a direction and a
of
x

3. A metal cube of length 0.1 m is charged by 12 C. strength.

Calculate its surface charge density.

Examples of electric dipoles. Dipoles are common
(Ans. 2 x 10 Cm) in nature, In molecules like H,0, HCI, C,HOH
4. Two equal spheres of water having equal and CH,COOH, etc., the centre of positive charges does not
similar charges coalesce to form a large sphere. If fallexactly over the centre of negative charges. Such
no charge is lost, how will the surface densities of molecules are electric dipoles. They have a permanent
electrification change ? (Ans. a, :G, = 222) dipole moment.
 ELECTRIC CHARGES AND FIELD 139
Ideal or point dipole. We can think of a dipole in Electric field due to charge + qat point Pis
which size 2 a >0 and charge g - o
in such a way that
the dipole moment, p=qx 2a has a finite value. Such a E (towards right)

dipole of negligibly small size is called an 4ne, (r -a

ideal or point
dipole. Hence the resultant electric field at point P is
Dipoles associated with individual atoms or molecules
axial -4
may be treated as ideal dipoles. An ideal dipole is
specified only by its location and a dipole moment,as 1

it has no finite size.

4n6,(r-a' (r+a?|
1.23 DIPOLE FIELD 4ar

36. What is a dipole field ? Why does the dipole field at

large distance falls off faster than V2 ?
Or
1 2pr
Dipole field. The electric field produced by an electric
dipole is called a dipole field. This can be determined by
using (a)the formula for the field of a point charge Here p= qx 2a= dipole moment.
and
(6) the principle of superposition.
For r >> a, a' can be neglected compared to rr.
Variation of dipole field with distance. The total
1

charge of an electric dipole is zero. But the electric field (towards right)

of an electric dipole is not zero. This is because the

-
charges +q and q are separated by some distance,so
field at any axial point of the dipole
Clearly, electric
the electric fields due to them when added do not acts along the dipole axis from negative to positive
exactly cancel out. However,at distances much larger
than the dipole size (r >> 2a), the fields of a and + - charge 1.e,, in the direction of dipole moment p.
nearly cancel out. Hence we expect a
dipole hela to ral 1 25 FLECTRIC FIELD AT AN EQUATORIAL
off, at larger distance, faster than 1/ r, typical of the
POINT OF A DIPOLE
field due to a single charge. In fact a dipole field at
larger distances falls off as 1/ r. 38.Derive an
expressionfor the electric field at any
point on the equatorial line of an electric dipole.
1.24 ELECTRIC FIELD AT AN AXIAL POINT
at an equatorial point of a dipole. As
OF A DIPOLE Electric field

shown in Fig. 1.65, consider an electric dipole consis

Derivean expression for the
37. electric field at any ting of charges -q and + q separated by distance 2a
point on the axial line of an electric dipole. and placed in vacuum. Let P be a point on the equa
Electric field at an axial point of an electric dipole. torial line of the dipole at distance r from it.

As shown in Fig. 1.64, consider an electric dipole 1.e., OP =r

consisting of charges + qand - 4, separated by distance
2a and placed in vacuum.Let P be a point on the axial
line at distance r from the centre O of the dipole on the
side of the charge +4. equa

+4 P Ex

20

Fig. 1.64 Electric field at an axial point of dipole.

Electric field due to charge-g at point P is

(towards left)

o (r+
A B

where p is a unit vector along the dipole axis from -q

to +q. Fig. 1.65 Electric field at an equatorial point of a dipole.
 PHYSICS-XI1
I40
Electric field at point P due to +qcharge is
Clearly, Evial 2 Fna
dipole at a distance r
1 Hencethe electric field of a short
directedalong BP is fwice the electric feld at the same distance
alongits axis

equatorial line.
at point P due to-charge is along the
Electric field

126 TORQUE ON A DIPOLE IN A UNIFORM

along PA
ELECTRIC FIELD
directed
4

an electric
Derive an expression for the torque on
40.
Thus the magnitudes of E,and E,, are equal ie, field. Hence define
dipole placed in a uniform
electric
1

dipole moment.
E E
field. As
a uniform electric
Torque on a dipole in

Clearly, the components of E,and E,, normal to shown in Fig, 1.66(a), consider an electric dipole

will cancel out. The components consisting of charges +qand -4 and of length 2a

the dipole axis

parallel to the dipole axis add up. The total electric placed ina uniform electric field E making an angle 9

field Epa is opposite to p. with it. It has a dipole moment of magnitude,

equa
P=qx 2a
Fequa =-(E, cos 0+ E Cos 0) p
É =gE
Force exerted on charge + g by field

=-2 E_ cos p |E,=E,] (along E)

1

=-2.
Force exerted on charge -g by field E =-gE
E)
(opposite to
:cos0=
J+ Eotal =tqE -gE =0.
1
or qE

where p=2 qa, is the electric dipole moment. 2

If the point P is located far away from the dipole,

r>>a, then
1

Fequa

(a)

Clearly, the direction of electric field at any point

on the equatorial line of the dipole will be antiparallel

to the dipole moment p.

39.Give a comparison of the magnitudes of electric

fields ofa short dipole at axial and equatorial points.

Comparison of electric fields of a short dipole at

axial and equatorial points. The magnitude of the

electric field of a short dipole at an axial point at
distance r from its centre is (b)

1 2p Fig. 1.66 (a) Torque on a dipole in a uniform electric field.

Eaxial =
(b) Direction of torque as given by right hand serew rule.

Electric field at an equatorial point at the same on a dipole a

Hence the net translating force in
distance r is
zero. But the two equal and

Fequa
1P uniform electric field

opposite forces act at different

is

points of the dipole.

They form a couple which exerts a torque.

 141
ELECTRIC CHARGES AND FIELD

to E. In a
Torque = Either force x Perpendicular distance is parallel or arntiparallel
When the dipole
between the two forces to
field, ifp is parallel to E or antiparallel
distancer non-uniform

e distance
t=qEx 2a sin 0 =(qx 2a) E sin 0
E,the net torque on the dipole is zero (because
the

However, there
t= pEsin 0 (p=qx 2 a) forces on charges q become t linear).
when
net force on thedipole. shown in Fig. As 1.67,
is a
As the direction of torque is perpendicular to on the dipole in the
NIFORM p is parallel to E, a net force acts

both p and E ,so we can write

direction of increasing É. When
p is antiparallel to E,

E.
an electric
a net force acts in the direction of decreasing
ence define E
The direction of vector t is that in which a right
Force on -q
c field. As handed screw would advance when rotated from p to

tric dipole Force on +4

length 2 E. As shown in Fig. 1.66(b), the directionof vector t is
O---- p
0 perpendicular to, and points into the plane of paper. -4
an angle =
de,
When the dipole is released, the torque r tends to
Direction of net force

Direction of increasing field=

align the dipole with the field E i.e., tends to reduce

(a)

angle 0 to 0. When the dipole gets aligned with E, the

torque t becomes zero.

(along E)
Clearly, the torque on the dipole will be maximum
+q
E. Thus
when the dipole is held perpendicular to Force Force

on -q
Pposite
to E)
Tmax pE sin 90° =pE. on+q

Direction of net force =

Dipole moment. We know that the torque,
Direction of increasing field=

T= pE sin (b)
+gE
If E=1 unit, 0 =90°, thent=p
Fig. 1.67 Forces on a dipole (a) when p is parallel

Hence dipole momnent may be defined as the torque

acting on an electric dipole, placed perpendicular to a uniform to E and (b) when p is antiparallel to E.

electric field of unit strength.

A comb run through dry hair attracts small pieces

1.27 DIPOLE INA NON-UNIFORM of paper. As the comb runs through hair, it acquires

ELECTRIC FIELD charge due to friction. When the charged comb is

piece of paper, it
What happens when an electric dipole is held in
41.
a ougt eloser to an uncharged
paper induces a net dipole
POlarises the Plece of i.e.,
? What will be the force and the
non-uniform electric field moment in the direction of the field. But the electric
or anti-parallel to
torque whenthe dipole is held parallel
field due to the comb on the piece of paper is not uni
? Hence explain why does a comb run
the electric field
form.It exerts a force in the direction of increasing field
through dryhair attract pieces ofpaper.
i.e., the piece of paper gets attracted towards the comb.
Dipole in a non-uniformn electric field. In a non

uniform electric field, the + qand -qcharges of a dipole 42. Give the physical significance of electric dipoles.

experience different forces (not equal and opposite) at Physical significance of electric dipoles. Electric

slightly different positions in the field and hence a net dipoles have a common occurrence in nature. A
electric field. on the dipole in non-uniform field. Also, molecule consisting of positive and negative ions is an
force acts a
n
F

Moreover, a complicated array of

nd screw rule.
a net torque acts on the dipole which depends on the electric dipole.
charges can be described and analysed in terms of
location of the dipole in the non-uniform field.
a dipole in a electric dipoles. The concept of electric dipole is used
wo equal and (i) in the study of the effect of an
electric field on
of the dipole.
where is the position vector of the centre of the dipoleInsulator, and () in the study of radiation of energv
ue. r
from an antenna.
 T42 PHYSICS-XI1

ForYour Knowiedge Units Used 2pr

Charge g is in coulomb, distarice Ze in etre
In a unform electrie feld, an electric dipole expe
dipole momert p in coulomb metre (Cmy fiefd E
riences no net foroe buta non zero torqure
in NCor Vm 9 102x
As the on a
netforoe dipole in a uniform clectric feld [(0.04
is Pero therefore, no linear acoeleration is produced

Torgue on a dipole becomes zero when it aligns

Example 50.An electric dipole is placed at an ongle of 64y 144
itself

parallel to the field of mnagnitude 4x 10° NC 144 108

with an electric field

Torque on
perpendicular
a dipole is

to the field E
maximum when it is held experiences

4on, deternine
a torque

the
of8V3 Nn.If the
magnitude of either
length of the dipole

charge of the dipole

is
Example 54. Tun churges 1
apart Dtermine the electric field
In a nan-unitorm electric field, a dipole experiences a Solution. Here -60, E=4x 10°NC -8/3 Nm of the dipole 15 cm oy from its

non zero force and non zero torque. In the special case
2a-4cm = 0.04 m positioe th)a point
charge, 15 O
when the dipole moment is parallel or antiparallel to
passng through O and rormal
the field, the dipole experiences a zero torque and a As tpE sin 9 = qx2ax Esin 9

non zero force

Solution. Here q 10 uC
Anon-uniform or

an (2a)E sin 0
speifically Za-5mm-5x
E-field
increasing 8/3 r 15cm-15
may be represen
ted by field lines Direction of
0.04 x 4x 10 x sin60° (a)Field at the acial point

as shown. increasing 2p 2
E-field V3x 10x2 -10'C=1
Clearly, E<Eg < Ec 2x v3
mC.
The direction of the electric field at an axial point of an
electric dipole is same as that of its dipole moment Example 51. An electric dipole consists of tuo opposite
9x10x2
and at an equatorial point it is opposite to that of charges of magnitude 1/3 x 10 C, separated by2 cm. The (15
dipole moment. dipole is placed in an external field of 3x 10 NC. What
The strength of electric field at an axial point of a short maximum torque does the electric field exert on the dipole ? -2.66 10
dipole is twice the strength at the same distance on
the equatorial line. Solution. Here q =, 107C x 2a=2 cm =0.02 m, This field is directed alo
3 moment vector,Le. rom- to
At larger distances, the dipole field (E « 1/r')
decreases more rapidly than the electric field of a E=3x 10 NC1 B
point charge (E l/r).
Tmav =pE sin 90° = qx 2ax Ex 1 -10 uC
Examples based on 1
Fig.
Dipole Moment, Dipole Field and x 10 x 0.02 x 3 x 10 x1 = 0.02 Nm.
Torque ona Dipole (b) Field at the equatoria

Example 52. Caleulate the electric field due to an electric P

Formulae Used
dipole of 10 cm having charges ofl uCat an
length equatorial
1. Dipole moment, p= qx 2a; where 2a is the
point 12 cm from the centre of the dipole.
distancebetween the two harges. 9x10 x
Solution. Here uC- 10° Cr =12 cm =0.12 m,
q =1
2. Dipole field at an axial point at distancer from the (15
centre of the dipole is 2a=10 cm, =5cm =0.05 m a

1
2pr 1
2qa = 1.33 x 10
Exial
(+a'2
When r >> a, 1 2p
4nE

9x10 x2x 10x0.05 9x 100 This field is directed

3.Dipole field at an equatorial point at distance r

(0.12 +0.0s y2 (0.13)
opposite
of the
to

dipole
the direction

moment
from the centre of the dipole is = 4.096 x 10 NC vector, Le., tromm +q to-4
1
P as shown in Fig. L69.
Example 53.Tuo point charges, each of 5uC but opposite
insign, are placed 4 cm apart. Caleulate the electric field
When r >>4, intensity at a point distant 4 cm from the midpoint on the
axial line of the dipole. [Punjab02]
4. Torque,t =pE sin @ where 0is the angle between
Solution. Here q -5 10°C,2a
x =0.04 m,
P and E
a =0.02 m, r=0.04 m
 ELECTRIC CHARGES AND FIELD 143
2 pr 1 2(qx 2a)r Example 55. Tuo identical electric dipoles are placed

along the diagonals ofa square ABCDof side 42 mas shown

9x 10 x2 x 5x 10x0.04 x 0.04
in Fig 1.70(a). Obtain the magnitude and direction of the
net electric field at the centre O of the square. (CBSE OD 231
[(0.04)-(0.02' +4
B
144
&10° NC-1
144 x 10

Example 54. Tuo charges t 10 uC are placed 5.00 mm

apart. Determine the electric field at(a) a point P on the axis
of the dipole 15cmn avay from its centre O on the side of the

positive charge, (b) a point Q 15 cm away from O on a line

D
+4
passing through O and normal to the axis of the dipole. (a) (b)

[NCERT] Fig. 1.70

Solution. Here q =10uC = 10G

Solution. Refer to Fig.1.70(b).
2a=5mm =5x m 10 Distance of each of the dipoles ABor CD from the

r=15cm =15 x 10 m equatorial point O is

r= BC V2 1
(a) Field at the axial point P of the dipole is
2 2
2p 2xqx2a
Ep = Electric field at O due to dipole AB,
kqx 2a kqx v2
9x 10 x2 x 105x 5x 10-3
NC -1 E =(+a'/2
(15 x 10³
= 2.66 x 10 NC-, along AB.
of dipole Electric field
-
at
2O
kq, parallel
due to
to
dipole CD,
AB
This field is directed along the direction

moment vector, i.e., from -g to +q, as shown in Fig, 1.68. kqx v2

E, =
B Ep

-10 uC +10 uC
Fig. 1.68
=V2 kq,parallel to DC.
(b) Field at the equatorial point Q of the dipole is Eyne = E t E, =2V2kq
9x2a3 2424 parallel to ABor DC.
41

9x 10 105x 5x 10-3 NC-1

x
Example 56. Two identical dipoles are aranged in -y
(15x 102 plane as shown in Fig. 1.171(). Find the magnitude and the
electric field at the origin O. [CBSE F 23)
× 1o° NC,along
direction of net
- 1.33
BÁ.

This field is directed

opposite to the direction

of the dipole moment
vector, ie., fromn +q to 4

as shown in Fig, 1.69,

Fig. 1.71 (a)

B Solution.Dipole moment due to dipole BA is P,

A

10C + 10uC
and due to DC is p,,as shown in Fig. 1.171(b).
Fig. 1.69
 PHYSICS X1

m.
Hereg-2x 10C, 2a 3em-3 10

2.
E-2 x 10 NC

Tmax PE sin 90 qk2a x Ex1

-2x 10 x3 10x2x 10 -1.2 x
10 Nm
3. Here r>0
1 2p 1 2(qx 2a)

Fig. 1.71 (6) 9x10 ×2x0.2 102x10 x

(0.1)
Electric field at O due to p is

-3.6 x 10 NC
1

E =2along OB q= 100 C=10 C 2a 10cm 10 m

0.
4. Here
Cm
Electric field at O due to p, is p=qx 2a = 10×0.10 =10

E, -2, along OD
1 4
Clearly, E -E,-E-2
20 cm
Magnitude of the net field at the origin O is, 20 cm

E =VE+ E² =N2E=
1 2V2q

Directiopof E is 180°+ 45° or 225° to I-axis.

+ 100 C 100 uC
B
Problems For Practice 10cm
1. An electric dipole of dipole moment 4 x 10°Cm is Clearly,
field of 10- NC-1
placed in a uniform electric

making an angle of 30° with the direction of the (+a?j2 = 20cm = 0.20 m
field. Determine the torque exerted by the electric P
field on the dipole. (Ans. 2 x 10 Nm) Exqua

An electric dipole consists two opposite charges 9x 10 x 10-5

of
910
2.
of magnitude 2x 10°C each and separated by a (0.2) 8
distance of 3 cm. It is placed in an electric field of
-1.125 x 10 NC 1
2x 10° NC-.Determine the maximum torque on
the dipole. (Ans. 1.2 x 10 Nm)
+ 0.2u uC and - 0.2 u uC are 1.28 ELECTRIC FIELD LINES
3/ Two point charges of
separated by 10m. Determine the electric field at 43. What are electric lines of force ? Give their
an axial point at a distance of 0.1 m from their important properties.
midpoint. Use the standard value of E
Electric lines of force. Michael Faraday (1791-1867)
(Ans. 3.6 x 10 NC) introduced the concept to visualize of lines of force
Calculate the field due to an electric dipole ofthe nature of electric (and magnetic) ields. A smal
length 10 cm and consisting of charges of ± 100 uC positive charge placed in an electric field experiences
at a point 20 cm from each charge. a force in a definite direction and if it is free to move
(Ans.L125 x 10' NCl) it will start moving in that direction.The path along
which this charge would move will be a line of force
HINTS
An electric line of force may be defined as the curv=
1
t= pE sin 9 along which a small positive charge would tend to mov
when free to do so inan electric field and the
=4x 103 x10 x sin 30° tangent
which at any point gives the direction of the electric field
-2x10 Nm.
that point.
 ELECTRIC CHARGES AND FIELD 1.45
In Fig. 1.72, the curve POR is an electric line of
force. The tangent drawn to this curve at the
point P
gives the direction of the
field Ep at the point P.
Similarly,the tangent at the
point Qgivesthe direction
of the field Ep at the
point and so on. Q p

E
R

Fig. 1.73

5. The lines of force are always normal to the surface

E
ofa conductor on which the charges are in

equilibrium.

Reason. If the lines of force are not normal to the

conductor, the component of the field E parallel

Fig. 1.72 An electric line of force.

to the surface would cause the electrons to move
and would set up a currenton the surface. But no
The lines of force do not really exist, they are current flows in the equilibrium condition.

imaginary curves. Yet the concept of lines of force is 6. The lines of force have a tendency to contract
very useful. Michael Faraday gave simple explana lengthwise. This explains attraction between two
tions for many of his discoveries (in electricity and unlike charges.
magnetism) in terms of such lines of force.
The lines of force have a tendency to expand
7.
laterally so as to exert a lateral pressure on neigh
For Your Knowledge bouring ines of force. This explains repulsion
between two similar charges.
The lines of force are imaginary curves, but the field
8. The relative closeness of the lines of force gives a
which they represent is real.
measure of the strength of the electric field in any
The term 'lines of force is misleading. It will be more region. The lines of force are
appropriate
lines'.
to call them electric (or magnetic) field
() close together in a strong field.
(i) far apart in a weak field.
A field line is a space curve i.e., a curve in three
(ii) parallel and equallyspaced in a uniform field.
dimensions.
9. The lines of force do not pass through a conductor
because the electric field inside a charged
conductor is zero.
Properties of Electric Lines of Force

1. The lines of force are continuous smooth curves 1.29 ELECTRIC FIELD LINES FOR DIFFERENT
without any breaks. CHARGED CONDUCTORS
The lines of force start at positive charges and end
2. 44. Sketch and explain the field lines of (i) a positive
they cannot form
-
cge ) a
closed
at negative charges point charge, (iii) two eaual
is a single charge, then the lines of Pot negatrve
loops. If there and opposite charges, (iv) two equal positive charges and
force will start or end at infinity.
(o)a positively charged plane conductor.

3. The tangent to a line of force at any pointgives the Electric field lines for different charge systems :
direction ofthe electricfield atthat point.
(0) Field lines of a positive point charge. Fig.1.74
4. No twolines of force can cross each other. shows the lines of force of an isolated positive point
charge. They are directed radially outwards because a
Reason. If they intersect, then there will be two small positive charge would be accelerated in the
(Fig. 1.73) and
tangents at the point ofintersection outward direction. They extend to infinity. The field is
the electric field at the
hence two directions of
spherically symmetric i.e,, it looks same in all
same point, which is not possible. directions, as seen from the point charge.
 I.46 PHYSICSx ELECTRICC

46 Shane that the Ve dependence of electric fielde

pont chrge is cosistent uith the concept of the elect

field es

Consisteney of the inverse square law with

olectric feld lines AsshowninFig 180, the numbet
radial linesof force originating from a point charge
a givern solid angle AS is constant Consider fwo poi
and P, at distancesr and
P. from the charge 4
Fio. 1.74 Field lines of a a same number of lines fsay n) cut an element of a
&at P
Fig.1.75 Field lines of

positive point charge. negative point charge. Crk AQat P and an element of area

ig. 1.77 lines of two equal positive charges.

() Field lines of a negative point charge. Like that
Field OP

of a positive point charge, the electric field of a (o) Field lines of a positivcly charged plane
negative point charge is also spherically symmetric but conductor. Fig, 1.78 shows the pattern of lines of force
OP
the lines of force point radially inwards as shown in of charged plane conductor. A small positive
positively
Fig. 1.75. They start from infinity.
charge would tend to move normally away from the
() Field teo equal and opposite point
lines of plane conductor. Thus the lines of force are parallel
charges. Fig. 1.76 showsthe electric lines of force of an and normal to the surface of the conductor. They are

electric dipole i.c., a system of two equal and opposite that electric field E is uniform
a small distance They equispaced, indicating
pont charges (tq)separated by at all points near the plane conductor.
start from the positive charge and end on the negative Fig. L80
charge. The lines of force seem to contract lengthwise
as if the two charges are being pulled together. This Number of lines of force cutting unit

explains attraction between two unlike charges. The elerment at P

field is cylindrically symmetric about the dipole axis
ie., the field pattern is same in all planes passing Number of lines of force cutting unit area ele
through the dipole axis. Clearly, the electric field at all
at P,
points on the equatorial line is parallel to the axis of the
dipole.
As electric field strengths Densitv of lines of
Fig. 1.78 Field pattern of a positively charged

plane conductor.
E
45. What is the relation between the density of lines of

force and the electric field strength ? llustrate it in a

diagram. E
Relationbetween electric field strengthand density
of lines of force. Electric field strength is proportional

field strength
130 AREA VECTOR
to the density of lines of force i.e., electric

at a point is proportional to the number of lines of force 47. What is an area vector? How do we speci
cutting a unit area element placed normal to the field at direction of a planar area vector ow do we asso

that point. As illustrated in Fig. 1.79, the electric field at vector to the area of a curved surface ?
Pis stronger than at Q. Area vector. We come across many situ
Fig. 1.76 Field lines of an electric dipole.
Region of where we need to know not only the magitud
weak field
surface area but also its direction The directie
(iv) Field lines of two equal and positive point
charges. Fig, 1.77 showsthe lines of force of two equal planar area vector s specifed by the noralto the pl

and positive point charges. They seem to exert a lateral Fig 1.81(a), a planar area element dS has been

pressure as if the two charges are being pushed away sented by a normal vector dS. The length of ve
from each other. This explains repulsion between two
represents the magnitude dS of the area element.
like charges. The field É is zero at the middle point N -Region of
a unit vector along the normal to the planar area,
of the join of two charges. This point is called neutral strong field

point from which no line of force passes.This field also

has cylindrical symmetry. Fig. 1.79 Density of lines of force is proportional

to the electric field strength.

 ELECTRIC CHARGES AND FIELD 1.47
46. Show that thel/r dependence of electric field of a
pointcharge is consistent witlh
A dS-ds h
the conceptof the clectric Area - dS
field lines.

Consistency of the inverse square

law with the
electric field lines. As shown in Fig. 1.80, the number of
radial of force originating from a
lines
point charge q in
a given A2
P, and
angle
solid

P
is constant.

at distances r and
Consider two points
from the charge q. The
same number of lines (say n) cut an
element of area
, (a) (b)
AQ at P and an element of area AN at Py. Fig. 1.81 (a) A planar area element. (b) An area
element of a curved surface.

T;=OP,
In a curved surface,we can imagine it to be
case of

divided a large number of very small area

into

elements. Each small area element of the curved

r;
=OP, P.
surface can be treated as a planar area. By convention,
the direction of the vector associated with every area
element of a closed surface is along the outward drawn
2
ríAQ normal. As shown in Fig. 1.81(b), the area element ds at
any point on the closed surface is equal to dS n,where
Fig. 1.80

dS is the magnitude of the area element andn is a unit

Number of lines of force cutting unit area vector in the direction of outward normal.
element at P=
131 ELECTRIC FLUX
Number of lines of force cutting unit area element 48. Define the term electric flux. How is it related to
n
at P, = electric field intensity ? What is its SI unit ?
Electric flux. The term flux implies some kind of
As electric field strength cDensity of lines of force low. Fux is the property of any vector field. The
electric flux is a property of electric field.

E_ The electric flux through a given area held inside an

E, electric field is themeasure of the total number of electric
1 lines of force passing normally through that area.
L.e. Ec
As shown in Fig. 1.82, if an electric field E passes
normally through an area element AS, then the
1.30 AREA VECTOR electric
flux through this area is

47. What is an area vector ? How do we specify the =E AS

A.
direction ofa planar area vector ? How do we associate a
Vector to thearea ofacurved surface ? Area = AS
Area vector. We come across many situations

where we need to know not only the magnitude of a

0=0°
surface area but also its direction. The direction of a

planar area vector is specified by the normal to the plane. In

Fig. 1.81(a), a plarnar area element dS has been repre

sented by a normal vector d. The length of vector dS

represents the magnitude dS of the area element. If n is Fig. 1.82 Electric flux through normal area.

a unit vector along the normal to the planar area, then

As shown in Fig. 1.83, if the normal drawn to the
ds = dS n area element AS makes an angle with the uniform
 T48 PHYSICS-X1

field

E cos

A
,
E, then the component
so that the electric

Normal component
of

flux
E
is

of
normal

Ex
to

Surface area
AS will be
Electric

Unit of
flux is

dUnit of Ex
St unit of electric
ascalar

flux
quantity.
unit of S

-E cosx AS -NCm'- Nm'c!

or - EAS cos -As
0 Equivalently, SIunit of electric flux

AreaAS - Vmm'Vm.
1.32 GAUSS'S THEOREM
49.State and prove Gauss'stheorem.
0 Gauss's theorem. This theorem
Ecos gives a relationshin
between the total flux
passing through any
closed
surfaceand the net charge
enclosed within the surface
Gauss theorem states that the total
flux throuoh a
closed surface is 1/ E, times the net charge
enclosed by he
closed surface.
Fig. 1.83 Flux through an inclined area. Mathematically,it can be expressed as
In case the field
E is
non-uniform, we consider a
closed surface S lying
inside the field, as
Fig. 1.84. We
can divide the
shown in
surface S into small area
Proof.For the sake of
elements :AS, AS,, A,e.., simplicity,we prove Gauss's
AS.
Let the corresponding theorem for an isolated positive point charge 4. As
electric fields at these elements
shown in Fig. 1.85, suppose the surface S is a sphere of
be E, E, ,.,EN: radius r centred on q. Then surface S is a
Closed surface S surface.
Gaussian

Fig. 1.84 Electric flux through

a closed surface S.

Then the electric flux through the surface S

will be
-Spherical
Gaussian
surface

i=1 Fig. 1.85 Flux through a sphere enclosing a point charge.

When the number of area elements becomes Electric at any point
infinitely large (N o) and AS0, the above sum
field
on S is
approaches
surface. Thus
a surface integral taken over the closed E= 1

This field points

radially outward at all points on S.
N i=1 Also, any area element
points radially
AS 0 S outwards, so it
is parallel to E, ie., 0=0.
Thus the electric flux through any surface S,
open
or closed, is equal to the surface integral :. Flux through area dS
of the electric is

field É taken over the surface S.

d: = .ds =E dS cos =EdS 0°
 ELECTRIC CHARGES AND FIELD 149
Total flux through surface S is a single fixed value at every point on the surface, we
can easily calculate the electric field of certain

S symmetriccharge distributions by applying Gauss's

theorem.
=Ex Total area of sphere
1.34 COULOMB'S LAW FROM
.4y2 GAUSS'S THEOREM
51. Deduce Coulomb's law from Gauss's theorem.
Deduction of Coulomb's law from Gauss's
theorem. As shown in Fig. 1.86, consider an isolated
This proves Gauss's theorem. positive point charge q. We select a spherical surface S
of radius r centred at charge q as the Gaussian surface.

For Your Knowledge

Spherical

Gauss's theorem -Gaussian

is valid for a closed surface of any
surface
shape and for any general charge distribution.

Ifthe net charge enclosed by a closed surface is zero

(9 = 0), then flux through it is also zero.

=1=0 ds

> The net flux through a closed surface due to a charge

lying outside the closed surface is zero.

The charge q appearing in the Gauss's theorem Fig. 1.86 Applying Gauss's theorem to a
includes the sum of all the charges located anywhere point charge.
inside the closed surface.

The electric field E appearing in Gauss's theorem is By symmetry,E has same magnitude at all points

due to all the charges, both inside and outside the on S. Also E and d at any point on S are directed
closed However, the charge q appearing in
surface.

the theorem is only contained within the closed radially outward. Hence flux through area d is

surface.

Gauss's theorem is based on the inverse square

do, =É.d = EdS cos 0° =EdS
dependence on distance contained in the coulomb's Net flux through closed surface S is
law. In fact, it is applicable any to field obeying
inverse square law. It will not hold in case of any 4-fE.dš = fEds = E fds
S S
departure from inverse square law.
For a medium of absolute permittivity or dielectric
=Ex total surface area of S- Ex 4r
constant K, the Gauss's theorem can be expressed as Using Gauss's theorem,

9
133 GAUSSIAN SURFACE 1
or Ex 4

50. What isa Gaussian surface ? Give its importance.

Gaussian
enclosing a charge
surface.

is
Any
called
hypothetical

the Gaussian
closed

surface
surface

of that
or E= 1
9
charge. It is chosen to evaluate the surface integral of

the electric field produced by the charge enclosed by it, The force on the point charge 4, if placed on surface
which, in turn, gives the total flux through the surface.
S will be
1 49%
Importance. By choosing a closed surface F=4, E =

(Gaussiarn surface) in such a way that the electric field

E has a normal component which is either zero or has This proves the Coulomb's law.
 I,50

PHYSICS- XI
Examples based on
Electric Flux
and Gauss's
Theorem Example59.A
Formulae Used cylinder is
E with
its
placed in a uniform
axis electricfield
1. parallel to the
Electric flux electric flux ficld.
through aplane surface through
Show that the
the total
a unifomm area Sheld
in Solution. The
cylinder is zero.
electric field E is situation is
shown in Fig. 1.87.

where 0 is the angle E

which the
outward normal to the
drawn normal
to
surface area S nmakes
with the field E ds'
2.
According to
Gauss's
theorem, the
through aclosed total
electric flux
surface S enlosing Fig. 1.87
charge q is Flux through
the entire
cylinder,

3. Flux
density = Total flux
left plane
Area right plane
face Curved
Units Used face
Surface
-|E dS
Electric
NC
flux o- is in
NmC and
cos 180° +EdS cos 0°+ E dS cos 90°
flux density
in =-E[ds+ E [dS +0
Constant Used =-Ex T + Ex =0.
Permittivity
constant of free
space is
Example 60.Calculate the
number of electric
originating
from a charge of 1 C. lines offorce
=8.85x 10-12
4x9x 10-9 c² N-m-2 Solution.The
numberof lines of
from a charge of
1C force
originating

Éxample 57.IfE =6i +3

+4k ,calculate
= Electric flux
through a closed
flux through the electric
asurface surface
of area 20units enclosing a charge of
in Y-Z plane. 1
1C

[Haryana 10- 12 =1.129 x 1011

971] 8.85 x
Solution.Electric field vector, É =6í4+3 +4k Example 61. A positive
As the area centre of a hollow
charge of 17.7 uCis
vector S in the Y-Z sphere of radius 0.5 m. Calculate
placed at the
outward plane points along
drawnnormal i.e., along density through the surface
of the
the flux
positive sphere.
X-direction, so Solution. From Gauss's
S -20 theorem,

Flux, 4 = É.S=(6f+3j +4k),20

Flux, h. =117.7 x 10-6

Nm²
8.85 x 10-122x 10°
= 120 units. Total flux
Flux density
Example 58. A circular Area
plane sheet of radius
placed in a uniform 10 cm is
electric field of5 x 10
NC-, making an 2x 106
angle of60° with the field.
Calculate electric flux 4n (0.5)?
=6.4 x 10 NC,
sheet. through the

Solution. Here r =10 cm

Example62.Calculate the electric flux
= 0.1m, E=5x 10 NC- through each of the
Six faces
As the angle between
of a closed cube of length l, if a
charge g
the plane sheet and the (a) at
is placed
electric
its Centre and (b) at one of its vertices.
field is 60°, angle made by the normal to the Solution. (a) By symmetry,
plane sheet and the electric field the flux through
is0-90°-60°=30° the six faces of the
cube will be samewhen charge
each of

Flux, = ES cos =Ex COs 0IX placed at its centre. q is

-5x 10x3.14 (0.1 x cos 30°

=1.36 x 10
x

Nm2c1.
19
6 E0
 ELECTRIC CHARGES AND FIELD 151
(b) When charge q is placed at one vertex, the flux charge inside the
(ii) By Gauss's theorem, the total
through each of the three faces meeting at this vertex cube is

be 1
will zero, as E is parallel to these faces. As only 9=En =4nx9x 10 x1.05 =9.27 10 12c.
one-eighth of the flux emerging from the charge g
passes through the remaining three faces of the cube, Exomple (64.)An electric feld is uniform, and in the
so the flux through each such face is
positive x direction for positine x and uniform ith the same
11 4 1 9 z
3 8 24 &o
magnitude in the
given that E=
negative

200
x

NC
direction for negatie It is

for x >0
Lample63. The clectric fieldcomponents in Fig. 1.88 are
and E=-200 NC- for x <0.
E, =ax,E, = E, =0, in which a =800 N/ CHf.Calcu
flux through
late (i) the the cube and (i) the charge within A right circular cylinderof length20 cm and radius 5cm
Assume
the cube. a= that 0.1 m has its centre at the origin and its axis along the I-xis so
[NCERT]
that onefaceis at x =+10 cmand the other is at x =-10 cmt

()What is the net outuard flux through each flat face?

(in) What is the flux through the side of the cylinder ?

What is the net outard fiux
(iin) through the cylinder ?
(i) What is the net charge inside the cylinder ?
(CBSE 19C:NCERT)

5cm
Z
Fig. 1.88

Solution. () The AS
electric field is acting only in

X-direction and its Y-and Z-components are zero. For 20 m

X=-10 cm I=10 m

the four non-shaded

A is

these faces.
+n/2. So flux =faces, the angle between

É. AS is zero through each of

E and

Solution. () On the
ig.

left face
1.89

: E =-200 NC,
The magnitude of the at the left face is
electric field
AS =-AS i=-(0.05i m²
E, =r?=a a? x=a at the left face]
The outward flux through the left face is

Flux, 4 = E,.AS= E, AS cos 0

=E, a cos 180° =-E, a =+ 200 x r(0.05). Nm² c

[0 =180°for the left face]

=+1.57 Nm² c

The magnitude of the electric field at the right face is

On the right face :
Eg =ax? =a (2a)l/2
E -200 NC1
[x=2a at the right face]

AŠ = AS =r(0.05)i
Flux,
S =Ep AS cos 0° =E, a
m

[0 =0° for the right face) The outward flux through the right face is

Net flux through the cube

-E.AS =+ L57 Nm²c-!,
(i)For any point on the side of the cylinder EL AS.
-a' (Eg - E) = aa' [(2a)l/2 -ae1
=aav2 -1]=800(0.1²(2-1)
:. Flux through the side of the cylinder,

=1.05 Nm 'c-!. 4-E.AS -EAS cos90°= o.

 152 PHYSICS-XI1

(i)Net outward flux through the cylinder, Example 66. Figure 191 shos five charged lumps of

6,-157 4 L57 +0 -3.14 Nm? C!. plastic and an clectrically neutral coin The cross-section ofa

By
Gaussian surface S is indicated What is the net electric flu
(io) Gauss's theorem, the net charge inside the
through the surface if
cvlinder

Example 65. You

- is

=8,854 x 10x 3.14 - 2.78 x 10 c.

and
9, 94

9,--3.1nC?
+3.1 n C, 29%-5.9nC
are given a charge + Qat the origin O
(Refer to Fig. 1.90). Considera sphere S with centre (2, 0, 0)
of radius v2 m. Consider another sphere of radius v2 m
centered at the origin. Consider the spherical caps (i) PSQ 19
(0) PRQ (in) PWO, with normals outward tothe respective

spheres, and (iv)the flat circle PTQ with normal along the
A-4X1S.

(a) What is the sign of clectric flux through each of the

surfaces ()-(i0) ?

(b)What is the relation between the magnitudes of

fluxes through surfaces (i)-(io) ?

() Calculate the flux through the surface (i) directly. Fig. 1.91
Assume that the area of the cap (ii)is A. [NCERT]
Solution. The neutral coin and the outside charges
q, and q5 make no contribution towards the net charge
enclosed by surface S. Applying Gauss's theorem, we
get

1TR W
(2, 0, 0) +3.1x 10-5.9x 10-9-3.1x 10-9
(0, 0, 0)
8.85x 10-12
=-666.67 Nm² C-l,
Example 67.S, and S, are two concentric spheres enclosing
Fig. 1.90
charges Q and 2Q respectively as shown in Fig. 1.92.
Solution.For the charge + Qsituated O, the
at origin
() What is the ratio of the

field E points along +vex-direction i.e., towards right. electric flux through S,
2Q
(a) The outward drawn normal on cap PSQ points
and S, ?
towards left while it points towards right for ()How will the electric

caps PRQ PWQ and circlePTQ.So the flux is flux through the sphere

negative for (i) and positive for the rest. S, change, a medium
if

(b) The same electric field lines crossing dielectric constantk is

of
(i) also
introduced in the space
cross (ii), (ii). Also, by Gauss's law, the fluxes
through (ii) and (iv) add upto zero. Hence, all inside S, in place of air ?
Fig. 1.92
magnitudes of fluxes are equal. (n)) How the electric flux through sphere S, change.
will
(c) Given area of the cap (ü) =A if a medium of dielectric constant x is
introduced in
Electric field through cap (i) is the space inside S, in place of air?
1 OD
E= Q-9x10' x
|CBSE 02, 14, 14C]

(/2) Solution. ()By Gauss's theorem,

-4.5x 10' O NC-!

Flux through S, is =
Electric flux through the cap (ii) is

4=EA
-4.5 x 10 QA NC'm?. Flux through S, is =2Q+Q_3Q
 ELECTRICCHARGES AND FIELD 1.53

Ratio of electric flux through S, and S, is 7. A charge q is situated at the centre of an imaginary

Q/13 =1:3
hemispherical as shown in Fig, 1.93. Using
surface,

Gauss's theorem and symmnetry considerations, deter

mine the electric flux due to this charge through the

(i) If a mediumn of dielectric constant k is intro hemispherical surface.
Ans. 9
duced in the space inside S,, then flux through S,
becomes

1 Q
K En

(ii) The flux through S, does not change with the

introduction of dielectric medium inside the sphere S,.

Fig. 1.93

Problems For Practice 8A hollow cylindrical of length 1 m and area of

box
placed in a three dimen
25 cm is
cross-section

sional coordinate system as shown in Fig. 1.94. The

Lifthe electric field is givern by
given by E=50x i,
electric field in the region is

=8í+
E 4+ 3k NC-, calculate the electric flux
where E is in NC and x is in metres.

through a surface of area 100 m lying in the X-Y

plane. (Ans. 300 Nm² C-) 1 m
2The electric field in a ertain region of space is

(5í+4i-4k)x10 NC-.Calculate electric flux

1 m
due to this field over an area (2 -î)x 10 m?.
of i

(Ans. 6x 10° Nm' C-ly

Fig. 1.94

Find
uniform field É =3x10° i NC-1,
3Consider a electric
(i) net flux through the cylinder,
through a square
Calculate the flux of this field
(i) charge enclosed by the cylinder. [CBSE D 13]
surface of area 10 cm when
() its plane is parallel to the y-z plane, and 9. The electric field in a region is given by E =b
makes a 60° angle with
(ii) the normal to its plane
Find the charge contained in the cubical volume
[CBSE D 13C]
the x-axis.
bounded by the surfaces x 0, x =4, y=0 =
30 Nm c- (ii) 15Nmc-'] E, =5x 10 NC
[Ans. (i) y= a, z =0
and z=a. Take

field E =5x10 NC-, a= lcmn and b cm. =2(Ans. 2.2 x10-l2 C)

uniform electric
4. Given a

10 cm Je. The electric field components due to a charge inside

of this field through a square of
find the flux the cube of side 0.1 m are as shown.
to the Y-Z plane.
on a side whose plane
What would be the flux
is parallel

through the same square

if
E, = r, where a =400 N/C-m
?
angle with the X-axis =
the plane makes a30° [CBSE D 14] E =0,E 0.

(ii)25 Nm c-')
[Ans.(i) 50 Nm'c-
of a
located at the centre
A point charge of 17.7uCis
flux through
Find the electric
cube of side 0.03 m. [Himachal 931
each face of the cube.
(Ans. 3.3 x 10°
Nmc-)
a charge of

6. A spherical Gaussian surface encloses

the electric flux passing
10 (i) Calculate
8.85 x
C.
(ii) If the radius of the
through the
surface.
how would the flux 0.1 m
is doubled, 0.1 m
Gaussian surface [CBSE D 01, F 07]
change ?
(Ans. ()10 Nm²C (1) No change] Fig. 1.95
154 PHYSICS-X1

Calculate () the flux through the cube,and ()the 8. (0 Flux through the curved surface of the cylinder
ICBSE18C;SP 20) zero.

,
is
charge inside the cube.

|Ans (0 -01 Nm c'()4-8854 x10C Magnitude of the electric field at the left face

E-50x 1-50NC
11. Auniform electric field E E,i N/Cor x > 0and
, EScos 50x 25 x10cos 180
E--E,iN/Chor x <0are given. A right circular 10Nm
--1250
cvlinder of length /cm and radius r cm has its centre
Magnitude of the electric field atthe right face
at the oigin and its axis along the -axis. Find out
the net outward flux. Using Gauss's law write the E= 50x2= 100NC
expression for the net charge within the cylinder. 100×25x 10cos 0

HINTS
|CBSE D 08C]
- 2500x 10Nm!
o =t p-(2500 - 1250) ×10Nm'c
1.E -8i+4j+ 3k

Flux, , -300 Nm?c.

NC,S =100km²
=E.S -(8i+ 4j+ 3Á). 100
(in) Total
- 1250

- 1.250x 10
10Nmc!
charge enclosed by the

9-0r = 8.854
Nm

x
c1

101250 x10c
cylinder,

2. è, - E.s -(5i-4j-4k)x 10°.(2i-j) x 102

= 11067.5x 10°C =1.107 pC
-[5 x 2 + 4 x(- 1) - 0) x 10° Nm² c-l,
-6x10° Nm² c-.
3. () Normal to the area points in the direction of
=-E, a + E, a ---0-42
0= 0°.

,
the electric field,
a E 5x 10° x(0.01
o = ES cos @ = 3x 10° x(0.10)cos 0° = 0.25 Nm²C
b 0.02
- 30 Nmc-1
q= =8.85 10x0.25 - 2.2 x 1O 12 C.
() =3x 1o x(0.10) xcos 60° = 15 Nm²c-1.
x

YA
4. () = EScos

-5x10° x(0.10) cos O0°= 50 Nm?c-.

(i) o, =5x10° x(0.10° cos(90°– 30°)

=5×10°x(0.10)x=25 Nm²c1,

5. Flux through each phase of the cube

1
17.7×106
6 e 6 8.85x 10-12

= 3.3 x 10 Nm? C1
8.85 x 10C
6. () = x 10-12 C² Nm2 Fig. 1.96
Eg 8.85

() , =10 Nm² c-1

= 10"NmC, because the
is the same as in the case ().

From Gauss's theorem, total flux

charge enclosed

through entire
=[E,(at x = 2a)-
=[a(20)-
aa' = 400
a(a)] a

x(0.1)
E, (atx =

= 0.4
a))

Nm²c
a
7.
spherical surface is 9 = 8.854 x10x0.4 = 3.542 x10C.
11, Proceed as in Example 64 on page 151.

From symmetryconsiderations, flux through the (0 =E,

hemispherical surface is

14 - 2r E, (10) Nmcl
2
()q6,E=2ur'E, (10)c
 ELECTRIC CHARGES AND FIELD
155
135 FIELD DUE TO AN INFINITELY Vectorially,
LONG CHARGED WIRE

52. Apply Gauss's theorem to calculate the electric whereî is the radial unit vector in plane normal to the
fcld ofa thin long straiglht line
infinitely of charge,with a
wire passing through the observation point.
uniform charge density of à Cm

Electric field due to an infinitely long straight charged 1.36 ELECTRIC FIELD DUE TO A UNIFORMLY
wire. Consider a thin infinitely long straight wire having CHARGED INFINITE PLANE SHEET
auniform linear charge density 2 Cm .By symmetry, 53. Apply Gauss's theorem to calculate the electric

the field E of the line charge is directed radially outwards field due to an infinite plane sheet of charge.

and magnitude is same at all points equidistant

its Electric field due to a uniformly charged infinite

from the line charge. To determine the field at a plane sheet. As shown in Fig. 1.98, consider a thiun,

distance rfrom the line charge, we choose a cylindrical infinite plane sheet of charge with uniform surface
Gaussian surface of radius r, length l and with its axis charge density G. We wish to calculate its electric field

along the line charge. As shown in Fig. 1.97(), it has P at distance r from it.
at a point
Curved surface S, and flat circular ends S, and S,:
Plane sheet,

Obviously, ds,||E, ds,IE and ds, 1 Ë .So only the charge density

Curved surface contributes towards the total flux.

Cross-sectional
area A

Fig. 1.98 Gaussian surface for a uniformly

charged infinite plane sheet.

By symmetry, electric E points outwards field

normal to the must have same magni

sheet. Also, it

at two points Pand P

tude and opposite direction
equidistant from the sheet and
on opposite sides. We
Gaussian surface of cross- sectional
choose cylindrical
the
area A and length 2r with its axis perpendicular to
(b) sheet.
(a)
force are parallel to the curved
for line charge. As the lines of
Fig. 1.97 (a) Cylindrical Gaussian surface the curved
through
surface of the cylinder,the
flux
(b) Graph of E vs. r.
plane-end faces of
surface is zero. The flux through the
the cylinder is

S
=EA+ EA =2EA
q =GA
Charge enclosed by the Gaussian surface,
cos 90°
90°+|EdS,
-Eds,cos 0° + EdS, cos
According to Gauss's theorem,
S

= E[ds, +0+0
=Ex area thecurved of
surface or -=Ex 2 rrl

oA
=l
Charge enclosed

Using Gauss's theorem,

by
,
the Gaussian

=q/ E
surface, q

We get
2 EA=

E =°
or E=
28

Or E.2n rl = or E=
2T eg
Vectorially,
2 E

unit vector normal the plane and

field ofa line charge is IM0erseny propor where 9 is the to

Thus the electric

shows
Figure 1.97(b)
the distance from the line charge.
going away from it.

O to
r from the line charge.
of E with distance
the variatíon
1,56 PHYSICS-XI|

Clearly, E is independent of r, the distance from the 55. Two infinite parallel planes have uniform charge
plane sheet. densities to. Determine the electric
the sheet is posilively
field in ()the region
(0) If charged (a >0), the field belween the planes, and (ii) outside it.
is directed away from it.
Electric field of two oppositely
(ii) If the sheet is negatively charged plane
charged (o <0), the field parallel plates. As shown in Fig. 1.100, consider twe
is directed towards it.
plane parallel sheets having uniform
surface charge
For a large planar sheet, the above formula
finite
densities of t G. Supposer be a unit vector pointing
will be approximately valid in
the middle regions of from left to right.
the sheet, away from its edges.
54. Two infinite parallel
planes have uniform charge E
densities of o,and G,.
Determine the electric field at
points (i) to the left of the sheets,
(ii) between them, and
(ii)to the right of the sheets.

Electric field of two positively

charged parallel Eg
plates. Fig. 1.99 showstwo thin plane parallel sheets of
charge having uniform charge densities o, and o,
with o, >o, >0. Suppose r is a unit vector pointing Fig. 1.100
Sheet 1 Sheet 2
from left to right.
In the region :Fields due to
O1 O2
I the two sheets are
E E.

2E0 G,
Total field, E-E +E,= G =0
E 2 E 2eo
In the region II :Fields due to the two sheets are

Sheet 1
2
Sheet
2Eo 2E
Fig. 1.99
Total field,
In the region I:Fields due to the two sheets are 2E 2Eo
In the region III:Fields due
2Eo 2E
to the two sheets are

From the of
principle superposition, the total E =2&o
electric field at any point of region Iis
Ez2&o
r Total field, Em =0.
-+E--; 2&0-(o,+ o,)
Thus the
electric field between two oppositely
In theregion II: Fields due to the two sheets are charged plates of equal charge density
is uniform
which is equal to and is directed from the
positive to
2Eo 2E0 the negative plate, while the field
is zero on the outside
of the two sheets. This arrangement
E, is used
:. Total field,

260
(o-,) producing uniform electric field.
for

In the region III: Fields due to the two sheets are 1.37 FIELDDUE TO A UNIFORMLY
CHARGED THIN SPHERICAL SHELL
2E „20 56. Apply Gauss's theorem to show that for a
spherical shell,
the electric field inside the shell vanishes,
Em=
2Eo (o, t o,)
.. Total field,
whereas outside it, the field is as if all the
charge had been
concentrated at the centre.
 ELECTRIC CHARGES AND FIELD 1.57

Electric

spherical
field

shell.
due
Consider
to a uniformly charged thin
a thin spherical shell of
or E= 9 [For r= R]
4n E, R
aroeof radius Rwith uniform surface charge density
E=
o. From symmetry,we see that the electric field E at

any noint is radial and has same magnitude at points

eOuidistant from the centre (c) When point P lies inside the spherical shell. As is
of the shell i.e., the field is

symmetric. To determine electric field at any

Splherically clear from Fig. 1.102(a), the charge enclosed by the
Doint P at a
distance r from 0, we choose a concentric Gaussian surface is zero, i.e.,
sphere of radius r as the Gaussian surface. q =0
Gaussian

surface
Gaussian
surface

1 4
E
Spherical shell,

R charge density =o
Spherical shell, Fig. 1.102 (a) Gaussian surface for inside points

charge density =g of a thin spherical shell of charge.

Flux through the Gaussian surface,

4- =Ex 4y2

(a)
Fig. 1.101

When point
Gaussian

a thin
surface

spherical

P lies
for outside

shell of charge.

outside the spherical

points of

shell. The
Applying

Ex 4=0
;
Gauss's theorem,

=9
total charge qinside the Gaussian surface is the charge
on the shell of radius R and area 4R. E=0 [For r< R]

q=4Ro Hence electric field due to a uniformly charged splherical

Flux through the Gaussian surface, ¢: = Ex 4r shell is zero at all points inside the shell.

Figure 1.102(b) shows how E varies with distance

r
By Gauss's theorem,
from the centre of the shell of radius R. E is zero from
r=0 to r=R; and beyond r= R, we have
1

= E c
Ex 4n2
1 E
[For r> RJ E= 1 9
4 rE,R*

Vectorially, E=4nr
E«
This field is thesame as that produced by a charge 4

the shell,
placed at the centre O. Hence for points outside
is as if the entire
the field due toa uniformly charged shell
centre.
E =0
at its
Charge of the shell is concentrated R
The Gaussian
(b) When point P lies on the spherical
shell.

the charged spherical shell.

Surface just encloses

Applying Gauss's theorem, Fig. 1.102 (b) Variation of E with r for a

Ex 4nR29 spherical shell of charge.

 1.58 PHYSICS-XIl

1.38 FIELD DUE TO A UNIFORMLY This field is same as that produced by a charge q
placed at the centre 0. Hence forpoints outside the sphere
CHARGED INSULATING SPHERE
the field de to the wnifornly charged sphere is as if the

57. A charge q is uniformly distributed within an entire harge of the sphere is concentrated at its centre.

insulating sphere of radius R. Apply Gauss's theorem to (b) When the point Plies on the sphere. The Gaussian

find the electrnc feld due to this charge distribution at a surface just encloses the charged sphere. Applying
pointdistant rfrom the centre the sphere, where (a)r> R Gauss's theorem,

of
(b) r=R (c) 0 <r<R Shoo the variation of E with

r
graphically.
Ex4nR?
field due to uniformly charged
Electric
insulating
sphere. Consider an insulating sphere (For example, a [For r= R|
nucleus with protons almost uniformly distributed
inside it) of radius A with uniform volume charge
density p. From symmetry, we see that the
(c) When point P lies inside the sphere. The charge
electric field enclosed by the Gaussian surface is
E at any point is radial and has same magnitude at
4
points equidistant from the centre O of the sphere i.e., 4
the field is spherically symmetric. To determine electric

fieldat any point Pat distance r from O, we choose a

concentric sphere of radius r as the Gaussian surface.

(a) When point P lies outside the sphere. The total

charge q inside the Gaussian surface is the charge

inside the sphere of radius R. E

Flux through the Gaussian surface,

E
d, =Ex4ny2 Fig. 1.103 (b) Gaussian surface for inside points

of an insulating sphere of charge.

By Gauss's theorem,

Ex 4u =
E
or Ex 4ny2

Or E= 1
qr _ pr
4nE, R 3en

Fig. 1.103 (a) Gaussian surface for outside points

Atthe centre, r =0
and hence E =0. The variation of
of an insulating sphere of charge.
E with distance r from the centre of a sphere of uni
By Gauss's theorem, formly charged sphere is shown in the figure 1.103(c).

EA

=1 Emax
19
Ex 4nr 4 rEG R

1
1

E
Or 1
[For r> RJ
r=R

Vectorially, Fig. 1.103

4T (c) Variation of E with r for a

uniformly charged sphere.

 1.59
ELECTRIC CHARGES AND FIELD

Examples based
on
Applications of Gauss's Theorem

Formulae Used
charges
their
,,
Example 68. Two
and per

axesis d. Find the magnitude of

long straight

unit length.
the
The
parallel

separation

force
wires

exerted
carry

between
on

, nit length ofone due to the charge on the other.

field at the location of wire 2 due

Electric field of a long straight wire of uniform Solution. Electric
1.
linear charge density to charge on 1 is

E= E=
2tE
27e, d
where r is the perpendicular distance of the 2 due to the above
Force per unit length of wire
observation point from the wire.
field
2, Electric field of an infinite plane sheet of uniform wire2 = Ei,
surface charge density o,
f=Ex charge on unit length of

E= Or
2Eo f=;
27E, d
3. Electric field of two positively charged parallel
Example 69. An electric dipole consists of charges
2mm
plates with charge densities o, and o, such that
+2 x 10C separated by a distance of Itis placed

near a long line charge of density 4.0x 10 C,as shown

the negative charge isat a distance of
1
such that
E=+ (G, + o,) (Outside the plates)
in Fig. 1.104,

2Eo 2 cm fron the line charge. Calculate the force acting on the

1 dipole.
(o-o,) (Inside the plates)
2Eo
4. Electric field of two equally and oppositely

charged parallel plates,

E=0 (For outside points)

-4
E= (For inside points)

2 cm -bk-2 mm
5. Electric field of a thin spherical shell of charge

density G and radius R,

1
E= For r> R (Outside points)

E=0 For r <R (Inside points)

Fig. 1.104

.
1
For r= R (At the
R surface)

Solution. Electric field due to a line charge at

Here q=4n R distance r from it,

uniform charge
1 22
6. Electric field of a solid sphere of
E=
density p and radius R :
1 Force exerted by this field on charge q.
E= For r> R(Outside points)
1 2q
F= qE =
r
E= For r<R (Insidepoints)
4T E R3 Force exerted on negative charge (r=0.02 m),

E=
1

For r= R (Atthe surface) 9x 10 x 2x2 x 10 x 4x 10 N

R
F= 0.02

4
Here =7.2 N, acting towards the line charge
3 (r =2.2 x 10 m),
Force exerted on positive charge
Units Used 9x 10 x2x2 x 10-x4x10-4
Here chargesare in coulomb,
r and Rin metre, A in E, = 2.2 x 10-2
Cmo in Cm,
pin Cm and electric field E in

=6.5 N,acting away from the line charge

NC or Vm!.
 I.60 PHYSICS-X1

Net force on the dipole. Exomple 72.A charge of 17.7 10Cs dstribulot

F- F--72 -65 uniformly oner a large sheet of area 200 n.

Caleulate the

clectric feld intensity at a disfance of 20cm from if in air

-0.7 N, acting towards the line charge
(CBSE OD 0C)
Example 70.(a)An infinitely long posttioely chargeduire Solution. Surface charge density of the sheet
has a lincar charge densiy ACmAn clectron is reoluing
around the eire as tts contre with a constant elocity in a
177 10c -8,85 x 10 Cm

.
circular plane perpendicular to the wire. Deduce the exyre
200 m

ssion for its kinetic energy. (b) Plot a graph of the kinetic Electric field at a distance of 20 cm from it in air,
energy as a fnctiot of charge density [CBSE F 13; OD 23]
8.85x 1o
Solution. The electrostatic force exerted by the line -5x 10° NC!
2Eo 2x8.85 x 10
charge on the electron provides the centripetal force
for the revolution of electron.
Example 73.Two large, thin metal plates are parallel and
Force exerted by electric field = Centripetal force close to cach other. On their inner faces, the plates have

surface charge densities of opposite stgns and ofmagnitude

cE = 17.7×10 C/nt. What iselectric field intensity E

(a) in the outer region of the first plate, and

Here v is the orbital velocity of the electron
(6) between the plates ? (CBSE SP 23]
But E
2TE, T Solution. (a) In the outer region of the first plate,

the fields due to the two plates are equal and opposite.
or So, E =0.
2TE 2TEg m
(b) Electric field in the region between the two
Kineticenergy of the electron will be
plates,
1 e
17.7x 10 22
= 2.0 x 10-10 Nc!
Eg 8.85x 10-12
(b) As E i,the graph of
1

E
kinetic energy E Ds. charge
Example74. A large plane sheet ofcharge having surface

density 2. will be a straight line

charge density 5.0x 10-16 Clies
in the X-Y plane. Find

as shown in Fig. 1.105.

the electric flux through a circular area radius 0.1 mm if the

of
Fig. 1.105 normal to the circular area makes an angle of 60° with the
Example 71. A long cylindrical volume contains a
Z-axis.

uniformly distributed charge of density p. The radius of Given that : e =8.85 x 10-12 c? N'm2
cylindrical volume is R.A charge particle (q)revolves around
the cylinder in a circular path. Find the kinetic energy of the
Solution. Here g=5.0x 10 Cm, r=0.1 m,
particle. JEE Main June 22) 6=60°
Solution. Charge per unit length of the cylinder,
Field due to a plane sheet of charge,

E=
2-1_pxnR²I =prR? 2E0
Flux through
E= prk' pR? circular area,

2rE 2E 4 = EAS cos 0= cOs

mu' 2&
5.0 x 10x3.14 x (0.1) cos 60°
9pR? 2 x 8.85x 10-12

- 4.44 x 10 Nm²c1.
my = Example 75. A spherical conductor of radius 12 cm has a
2E charge of 1.6x 10C distributed uniformly over its
Surface. What is the electric field (i) nside the sphere,
K=mu²_pgR?
2 (ii)just outside the sphere, (H)at a point 18 cm from the
centre of the sphere ? (NCERT|
 ELECTRIC CHARGES AND FIELD 1.61
Solution. Here q =1.6 x 107 C 7. A spherical shell of metal has aradius of 0.25 m and
R=12 cm =0.12m carries a charge of0.2 uC. Calculate the electric field
at a point (i) inside the shell, (i1) just
() Inside the sphere, E =0. This is because the charge
intensity

on the outer outside the shell and (ii) 3.0m from the centre of
resides surface of the spherical conductor.
the shell. [Ans. (i) 0 (i) 2.88 x 10 NC!
(i) Iust outside the sphere, r=R=0.12 m.Here the (ii) 200 NC-)
charge may be assumed to be concentrated at the HINTS
centre of the sphere.
Er 9x 10 x 0.04

E= 1
1. 2=2rns, Er =4ne, 2 9x10 x2
4ne R?
=2 x10 Cml,
9x 10 x 1.6 x 107
2. Here h=2x10Cm,r= 0.2 m
(0.12)²
1 22 2x2x 10-8
= 10 NC1 ..E=
1
=9x 109 0.2

(iü) At a point 18 cm from the centre,

=1800 Vm-1
r=18cm =0.18m.
1 22 9x10x2x1o-4 Vm
1 9x 10 x 1.6 x 10-7
3. E= = 4.5 x10 1.

E= 4

=4.44 x 10 NC-,
(0.18)
4. E=9
q=&, AS E= 8.85 x 102 x 1x 100

Droblems For Practice =8.85 x 10-10C

5. From Example 70,

An line charge produces a field of
1. infinite en
cm. Calculate the E= =9x10 x1.6x10 x2.0 x 108
9x10*NC at a distance of 4

linear charge density. (Ans. 2 x 10 Cm-ly

= 2.88 x 1o- J.
a charge of
2. A cylinder of large length carries
6. Upward electric force on particle
2x10Cm-. Find the electric field at a distance of
Vm-) = Weight of the particle
0.2 m from it. (Ans. 1800

long wire is stretched horizontally mg = qE =9.

3. An infinitely
surface of the earth. It carries a
4 metre above the
2€,M8
cm of its length. Calculate its
charge l uC per
on the earth's surface
at a point
electric field
2x8.85 >x 102x5x 10 x 9.8
(Ans. 4.5x 10°
Vm-l)y
vertically below the wire.
4x 106
plates each of area 1 m
are placed
Two large metal
= x 10-13 C.
4.
a distance of 10 cm and carry 2.16
facing each other at
If the Number of required to be removed,
charges on their faces. electrons
equal and opposite
the plates is 100
find NC, 2.16 x 10-13
electric field between 1.355 x106
x 10-10 C) e 1019
the charge on each plate. (Ans. 8.85 l.6 x

a long line charge at any point inside the shell = 0.

3. An electron is revolving around 7. (i) Electric field

having charge density 2x 10Cm. Find the kinetic

1

energy of the electron, assuming that it is (i) E= R

orbit.
of theradius of electron's
independent 9x10x0.2 x 10-6 = 2.88
x 10-|) 10+ NC-1,
(Ans. 2.88 x

(0.25)
mass 5x 10g is kept over a large
of 1
0. A particle
density 4x 10°Cm. E=
horizontal
sheet of charge (rii)

to this particle,
so that
What charge should
be given
does not fall down. How many 9x10x0.2 x 106 =200 NC-1
released. it
if
to give this charge
? (3.0)
electrons should be removed
1.355 x 109)
2.16 x 10"C,
(Ans.
 T62 PHYSICSX1

VERY SHORT ANsWER CONCEPTUAL PROBLEMS

Problem 1. The electric charge of any body is actually farther end of the paper. The rod exerts greater
a surplus or deficit attraele
of electrons. Why not protons ? thanrepulsion on the paper because negative
charge
Solution. Electrons are loosely bound to atoms and closer to the rod than the
positive charge. Hernce the od
can be readily exchanged attracts the piece of paper
during rubbing Protons are
firmly bound inside the nucleus.They cannot be easily Problerm Can two like
8.
charges attract each ofher 21[
detached. Hence electric charge of any
body is just a yes, how ?
surplus or deficit of electrons
and not protons. Solution.Yes. If one charge is larger than
the other, the
Problem 2. When a glass rod is rubbed with silk charge induces equal and opposite charge on fhe
larger
both acquire charges. What is the source oftheir nearer end of the body with smaller charge. The
electri-
fication2 opposite
induced charge is larger than the small
charge initially
Solution. For the present on it.
clectrification of a body, only
electrons are responsible. During rubbing Problem 9. Why do the
electrons are gramophone records get
transferred from glass rod covered with dust easily ?
to silk. The glass rod acquires a
positive dharge andsilk acquires an Solution. The gramophonerecordsget charged due
equal negative charge.
to
Problem the mass of a body the rubbing action of the needle.
3.Is
affected on charging ? Sothey attract the dust
Solution. Yes. Electrons have a particles from the air.
definite mass. The
mass of a body slightly Problemn 10. An ebonite rod held in
hand can be
increases if it gains
the mass decreases if the body electrons while charged by
loses electrons. rubbing with flannel but a copper rod
Problem 4.Two identical cannot be charged like this. Why ?
metallic spheres of
exactly
equal masses are taken. One is Solution. Ebonite rod is insulating.
given a positive charge q Whatever charge
coulombs and other an equal appears on it due to rubbing,
negative charge. Are their stays on it. Copper is good
masses after charging conductor.Any charge developed
equal ? [IIT] on it flows to the earth
Solution. No. The positive charge of a through our body. Socopper rod cannot
body is due to be charged like this
deficit of electrons Itcan be charged by
while the negative charge is providing it a plastic or rubber
due to handle.
surplus of electrons. Problem 11. Electrostatic
Hence the mass of the experiments do not work
charged sphere will be slightly more negatively well on humid days. Give
than that of the reason.
positively charged spheres. Solution. Electrostatic experiments require accumu
Problemn 5. A positively charged rod repels a sus lation of charges. Whatever charges appear
during the
pended object. Can we conclude that the object is posi-
experimnentation, they are drained away through humid
tively charged ? air which is more
conducting than dry air due to the
Solution Yes, the object is positively charged Presence ot a larger number of charged particles in

it.
Repulsion is the surest test of electrification. Problem 12 A comb run
through one's dry hair
Problem 6. A positively charged rod attracts a attracts small bits of paper. Why ? What happens if the
hair is wet or if it is a
suspended object. Can we conclude that the object rainy day ?
is NCERT]
negatively charged ? Solution When the comb runs
through dry hair, it gets
Solution. No. A positively charged rod can
charged by friction. The
attract molecules in the paper
get
both a neutral polarized by the charged comb,
object

Problem 7.How does a

and a less positively charged object.
attraction. If the hair is wet,
resulting in a net force
or if it is rainy day,
t
positively charged glass rod friction
between hair and the
attract a neutral piece of
paper ? comb reduces. The comb does
not
get charged and thus it will
Solution. The positively charged rod induces not attract small bits of paper.
negative
charge on the coser end Problem 13. Ordinary rubber
and positive charge on the is an insulator.
special But the
rubber tyres of
aircrafts are made slightly
conducting. Why is this
necessary? NCERTI
Solution During landing, the
tyres of aircraft may get
highly charged due to friction
between tyres and the air
strip. If the tyres are
made slightly conducting they will
lose the charge to the earth
otherwise too much of static
electricity accumulated may
produce spark and result in
fire.
Fig. 1.106
 ELECTRIC CHARGES AND FIELD 1.63
Problem 14.Vehicles carrying
inflammable materials
usually have metallic ropes touching the
Solution. The charge on any body is always an integral
ground during ultiple of Here
motion. Why ?

e.
INCERT]
4 0.8 10 1"C
Solution. Moving vehicle gets charged due to friction. 0.5
The inflammable material may catch fire e 1.6x1019 c
due to the spark
produced by charged vehicle. When metallic rope is
used, This is not an integer. So a body cannot have a charge
the charge developed on the vehicle is transferred to tho of 0.8 x 101"C.
ground and so the fire is prevernted.
Problem 22. If the distance between two equal point
Problem 15. An inflated balloon is charoed by charges is doubled and their individual charges are also
nubbing with fur.Will it stick readily to a conducting doubled, what would happen to the force between them ?
wall or to an insulating wall ? Give reason.
[Roorkee] Solution. The original force between the two charges is
Solution. It will
stick readily to the conducting wall. It
1

induces an equal amount of charge on the F=

conducting
wall and much smaller charge on
insulating wall. So a
large force of attraction acts between the balloon and the When the individual charges and the distance
conducting wall. between them are doubled, the force becomes
Problem A metal sphere is fixed on a smooth
16. 1 1
horizontal insulating plate. Another metal sphere is
F'= 2q x 24
9x9-F
4ne (2r)²
placed a small distance away. If the fixed
sphere is given Hence the
a charge, how will the other sphere react ? force will remain same.

Solution. The charge on Problemn 23. The electrostatic force between two
the fixed sphere induces
unlikecharge at the closer end and like charee on the far Charges is a central force. Why ?
end of the free sphere. Net attraction acts on the free Solution. The electrostatic force between two charges
sphere and so it gets accelerated towards the fixed sphere. acts along the line joining the two charges.So it is a central
Problem 17. Is there some way of producing high force.

voltage on your body without getting a shock ? Problemn 24. How the Coulomb force between two
is

Solution. Ifwe stand on an insulating surface and charges affected by the presence of a third charge ?
touch the live wire of a high power supply, a high poten Solution. The Coulomb force between two charges
tial is developed on our body,without causing any shock.
does not depend on the presence of a third charge.
Problem 18. A charged rod attracts bits of dry cork Problem 25. Two equal balls having equal positive
which after touching the rod, often jump away
trom it charge 'g coulombs are suspended by two
insulating
violently. Why ?
strings of equal length. What would be the effect on the
Solution. The charged rod attracts the bits of dry cork
force when a plastic sheet is inserted between the two?
by inducing unlike charge at their near ends and like

charge at their far ends. When the cork bits touch the rod,
[CBSE OD 14]

they share the charge of the rod of the same sign and so
Solution. The force between the two balls decreases
get strongly repelled away. because k(Plastic) > l and Foc 1/ K.

Problem 19. What does 4 t 9, =0 signify in

Problem Force between two point charges kept at
26.

electrostatics ? ICBSE OD 01CI a distant d apart in air is F. Ifthese charges are kept at
the samne distance in water, how does the electric force
Or
between them change ? [CBSE OD 11]
Iwocharges 4, and q,, separated by a small distance
Solution. Dielectric constant for water, K= 80
Satisfy the equation 4, +4, = 0. What does it tell about

the charges ? (CBSE F O03| Fvater = Fair

K 80
Solution The equation signifies that the electric
Thus the force in water is 1/ 80 times the original force
charges are algebraically additive and here 4 and 42 are
in air.
equaland opposite.
Problemn 27. The dielectric constant of water is 80.
Problem 20. Name the experiment which established
the quantum nature of electric charge. What is its permittivity ?

Solution. Millikan's oil drop experiment for deter

Solution. Dielectric constant, K=
mining electronic charge.
Poblem 21. Can a body have a charge of 0.8 x 10C? .. Permittivity, = KE, =8.854 x 1012 x 80
Justify your answer by comment ? = 7.083 x 10 10 c²N'm2,
 T64 PHYSICS-XI1

Problem 28,Give an example to illustrate that electro Problem 35. An electron and a proton are
static forces are much stronger than gravitational forces. same electric field. Will they experience same
keptin
the
Solution A charged rod can force and
glass lift a piece of paper have same acceleration ?
against the gravitational pull of the earth on this piece. Solution. Both and proton
This shows that the clectrostatic force on
electron will exerien
the piece of
paper is much greater
force of same
magnitude, F eE Since a proton has
than the gravitational force on it 1e34
fimes more mass than an electron, so
its acceleration
Problem 29, Two electrically charged be
w

particles, 1/1836 times thatof the electron.

having charges of different magnitude, when placed at a
distance from each other, experience a force of
Problem 36. Why direction of an electric field is
attraction F. These fwo particles are put in contact and
taken outward (away) for a positive charge and
inward
again placed at the same distance from each (towards) for a negative charge ?
other.
What is the nature of new force between By convention, the
Solution,
them ? direction of electric field

Is the magnitude is thesame asthat of force on

the force
of of interaction between
a unit positive charge. As
them now more or less than F ? this force is outward in the field of a
[CBSE SP 11) inward
positive charge,and
in the field of a negative
Solution When the fwoparticles are put in
charge,so the directions
contact, are taken accordingly.
they share the difference of charge identically. Hence the
two particles repel, with a force less than F. Problem 37. A charged particle is free to move in an
Problem 30. An electron moves electric field. Will it always move along an
along a metal tube field ? electric
with variable cross-section, as shown in Fig. 1.107. How
will its velocity change when it approaches the Solution. The tangent at any point to the line of force
neck of
the tube ? gives the direction of electric field and hence of force on a
charge at that point. the charged particle
If
starts from
rest, it will move along the line of force. If it is in motion
and moves initially at an angle with the line of force, then
resultant path is not along the line of force.

Fig. 1.107 Problem 38. A small test charge is

released at rest at a
point in an electrostatic field
Solution. The positive charge induced on the neck configuration. Will it
of travel along the line of
the tube will accelerate the electron force ?
towards the neck. [NCERT)
Problem 31. Why should a Solution. Not necessarily. The
test charge be of negli- along the
test charge will move
gibly small magnitude ? line of force only if it is a straight line. This is
because a line of force gives the
Solution. The magnitude of the test charge must direction of acceleration
small enough so that be and not that of velocity.
it does not disturb the distribution of
the charges whose electric we Problemn 39. Why do charges
field wish to measure reside on the surface of
otherwise the measured field will be different from the the conductor ?
actual field.
Solution. Charges lie at the ends of lines of force.
Problem 32. In defining electric field due to a point These lines of force have a tendency to
contract in length.
charge,the test charge has tobe vanishingly smnall. How The lines of force pull charges from inside a conductor to
this conditioncan be justified, when we
know that charge its outer surface.
less than that on an

charge
situations,

charge on an
,
Solution.
electron

the
or a proton is not
Because of charge quantisation,

source
the test
cannot go below e. However, in macroscopic
charge is much larger than the
possible ?

of a
Problemn

charged conductor
Solution.

conductor and
40.

This is
Why
?
is electric

because charges reside on the

not inside
field zero inside

surface
a
it.
electron or proton,so the limit g, -> 0for the
test charge is justified.
Problem 41. Why do the electrostatic field lines not
form closed loops?
Problem 33.What is the advantage of
introducingthe
[CBSE OD 14, 15;D20]
concept of electric field ? Solution, Electrostatic field lines start from a positive
By knowing the electrical field at a point, the
Solution. arge and end on a negative charge or they fade out at

force on a charge placed at that point can be infinity in case of isolated charges without forming
determined. any
closed
Problem 34. How docharges interact ? loop.

Solution. The electric field of one charge exerts a force

Alternatively, electrostatic field is a conservative field.
on the other charge and vice versa.
The work done in movng a charge along a closed path
must be zero. Hence, electrostatic field lines
ChargeElectric field cannot torm
Charge.
closed loops.
 ELECTRIC CHARGES AND FIELD
L.65

Problem 42. Why do the electric field lines never: Solution. If the field lines are not normal, then the field
each other? (CbSt OD A;D 20)
cross F would bave a langential component which will make
Solution. If two lines of torce intersect, then there electrons move alongthe surface creating surface currents
would be two tangents and ttwo directions of electric field be
and the conductor will not in equilibrium.

at the point of intersection, which is impossible.

Problem 47. Fig. 1.111 shows two p

Proble
43. Draw
large metal plates, P, and P,, tightly held
lines of force to represent
field.
against each other and placed between
auniform electric
two equal and unlike point charges
[CBSE OD 05)
perpendicular to the line joining them. +0
Solution. The lines of
(i) What will happen to the plates
force of a uniform electric ?
when they are released
field are equidistant Fig. 1.108 Uniform electric field.
(ii) Draw the pattern of the electric
parallel lines as shown in
field lines for the system.
1.108.
Fig. [CBSE F 09] Fig. 1.111
Problem 44. Fig. 1.109 shows electric lines of force
Solution.
due point charges 4, and q, placed at points A and B
to
(i) When released, the two plates tend to move
Write the nature of charge on them. the charges induced in them.
respectively.
apart slightly due to

[CBSE F 03] lines for the

(1) The pattern of the electric field

system is shown in Fig. 1.112.

P P

A B

+Q0
Fig. 1.109

pointing towards 4
Solution.As the lines of force are
as wellas q9, so both q,
and q, must be negative charges.

(+Q) is kept in
Problem 45. A positive point
charge Fiq. 1.112
Sketch
conducting plate.
the vicinity of an uncharged shown in Fig. 1.113,
charge on 48. In the electric field
originating from the point
Problemn
electric field lines lines on the lefttwice the
have
[CBSE OD 09, 17C the electric field
tothe surface of the plate. those on the right. If the
separation as that between
the charge +2, the lines of
Solution. Starting from point A is 40 NC-1, calculate
plate, inducing negative
magnitude of the field at
force will terminate at
the metal
of force will be proton placed at point A Also
by a
positions, the lines the force experienced
on it. At all B.
charge
as shown in Fig. 1.110.
find the magnitude of electric field at point
metal surface,
perpendicular tothe
+Q
A
B

Induced

-ve charge Fig. 1.113

point A,
Solution. Force on proton at

F= eE, 1.6 x 10x40 = 6.4 x10 N 18

at point B
between the lines of
force
1.110
Fig.
lines
As the separation
that the field that at point A, so
twice

=x
of
Problem 46. Why is it necessary is
of a conductor
from a point charge placed in the
vicinity
- E 40 =20 NC!.
every point. Ey
conductor at
be normal to the [CBSE F 091
st
 PHYSICS-XIl

P'roblem 49 The electric lines of force tend to surface 'S. What is the electric
flux due
contract lengthwise and expand laterally. What do they to
indicate?
figuration through the surface 'S? this
con-
Net charge
Solution The lengthwise contraction indicates
enclosed by the
surface s
attraction between unlike charges while
lateral expansion
indicates repulsionbetween like
charges. +24-44
TroblenmS A point charge
placed atany point on the
axis of an electric
dipole at some large
distance Problem
55. Two
experiences a force F. What will charges of
be the force acting on magnitudes-20 andsn
the point charge are located at
when its distance from the points (a,0)and (a,0)
doubled ? dipole is respectively. Whas
is the electric
flux due to
ICBSE OD 231 thesecharges
through a sphere of
Solution At any radius'3a' with its
axial point of a dipole, centre at the origin ?
varics as electric field (CBSE OD 1
Solution. Only the
charge -20 is enclosed
by the
on sphere of radius 3a. By Gauss's
theorem.

.Whenthe distance 20
of the point charge is
the force reduces to doubled,
F/8
Problem 51 Consider the Problem 56. A point charge +Qis placed at
situation shownin Fig, the centre
What are the signs ofq, 1.114. O of an uncharged
andg,2 If the lines are drawnin hollow spherical
conductor of inner
radius 'a' and outer
proportion to the charge, radius "b. Find the
foliowing :
what is the 4, lq,
(a) The magnitude
and sign of
ratio ?
the charge induced on
the
Solution. is a Here q inner and outer surfaces of
negative charge and q, is the conducting shell.
a positive charge.
(b) The magnitude of
electric
field vector at a distance
92 18 ()r Gand (i) r = 25,

=1:3. from the

Fig. 1.114 centre of the shell. Fig. 1.116

Problem 52. An arbitrary surface encloses a CBSE SP 181

What is the dipole.
electric fiux through this surface ? Solution. (a) The charge +Q at the centre induces
charge-Qonthe innersurface
As the total
(Exemplar Problemn) on the outer surface of the shell.
of the shell and charge +Q
Solution.
charge of a dipole is zero,so
Gauss's theorem, the by (b) ()Imagine
electric a concentric
flux through the closed spherical surface of radius
Surface is zero. r=as
2 the Gaussian surface. By symmetry, E will have
Problem 53. The force on an same magnitude
electron kept in an

;
at all points on this surface
electric field in a particular
direction is F. What will be and will
Point radially outwards.
the magnitude and
direction of the force experienced
proton at fhe same point in the field
bya Flux through the Gaussian
surface,
? Mass of the proton
is 1836 times the mass of the
electron. = Ex4n2
(CBSE FO7)
Solution.A proton has charge equal Charge enclosed by the
that of an electron. Hence
and opposite to
the proton will experience a
Gaussian surface =+Q.
By Gauss's theorem,
force equal and opposite to that of
F.

Problem 54Figure 1.115 showsthree charges +24, -4 Ex4ny²Q

and +34. Two charges +24and -g are enclosed within a
1
E= 1 4Q
+24
()
For outside points liker=
2b, the field is similar to
that of a point charge.

Fig. 1.115 1 1

4nE (2b) 4ze,4b?

 ELECTRIC CHARGES AND FIELD 1.67

SHORT ANSWER CONCEPTUAL PROBLEMS

Problem 1 Compare electrostatic and gravitational P'roblem 4. Two point eiectric charges of unknown

interactions.
magnitude and sign are placed a distance d' apart. The
Solution. Similarities between electrostatic and electricfield intensity is zero at a point,not between the

gravitational interactions : charges but on the line joining them. Write the essential
1. Both forces act according conditions for this to happen.
to similar laws :
1
F= The two charges must have opposite signs.
and F=G",2My Solution. (i)

(i)The magnitude of the charge lying near the point of

2. Both are conservative zero electric field intensity must be smaller than the
forces.

3. Both are central

magnitude of the other charge.
forces.
Problem 5. The electric field E due to a point charge
Dissimilarities between and
tational interactions :
electrostatic gravi

at any point is defined as lim

F
,where q is the test

1. Electrostatic interactions may be attractive or

F the force acting on it. What is the
charge and is

repulsive while gravitational interactions are

physical significance of lim in this field expression ?
always attractive. 40

2. Electrostatic interactions depend on the nature Draw the electric lines of a point chargeQ when ()Q>0
of the medium while gravitational interactions and (ii) Q <0. [CBSE D 071
do not depend on the nature of the medium.
The lim indicates that the test charge q is
Solution.
3. Electrostatic interactions are much stronger q0
than gravitational interactions.
small enough so that its presence does not affect the

Problem 2.Distinguish betweenelectric charge and distribution source charge and hence does not change
of

mass. the value of electric field which we wish to measure.

Solution. For electric lines of force of point charge Q, see

Fig. 1.74 and Fig. 1.75.

Mass
Electric Charge
A charge Q located at a point r is in
can be Mass is always positive. Problem 6.
1. Electric charge
equilibrium under the combined electric field of three
positive, negative or zero.
2. Electric chargeis always Quantisation of mass is
charges 14243 If the charges 4,,4, are located at

yet not established.

quantised.
points r and r, respectively, find the direction of the
3. Charge on a body does
not change with its
Mass of a body
increases with its speed. force on e, due to 4,in terms of 4,,4, r, and r.

speed. (CBSE SP 0S]

charge is always Mass is not conserved

4. Electric
exered by 4, and 4, on Q are
by itself as it can be The
Solution. forces
Conserved.
changed into energy shown in Fig. 1.118.

and vice versa.

distribution of
Represent the surface
Problem 3.

plate by using dashes in such

a
Charge for a square metal
density of charge, the
surface
Way that the greaterthe
tarther away are the dashes from the plate.
Solution. The distribution

of surface charge density on a

Square metal plate is as shown

in Fig. 1.117. As the surface
Charged
charge density is proportional plate
metal
to the curvature and curvature
is maximumat the corners and
Zero at planesurface, so dashes
are equidistant from the

straightportion and far-away Fig. 1.117

Fig. 1.118

from the corners.

 PIYSICS-X1

According to Coulomb's
Ois
law, force exerted by q on Importance :
(0 The tangent any
at point onthe
direction of the Curvegives the
electric field at that point
i) The relative
closeness of the
indicates the
lines of fo
Force exerted by e, on Qis relative strength of electric fela
atdifferent points
1 Probleo What ismeant by the
statementthat s
electric field of a point
charge has spherical
whereas that of an symmet
electric dipole is cylindricall
Acconding to the principle symmetric ?
of superposition,
exerted by total force
and q, on Qis
Goltion The electric field due to a point charge a
distance r fromit is given by: E

Clearly, the magnitude

The charge Owill bein equilibrium of field E will be same
if theforce exerted points on the surface of
at all
by q, isequal and a sphere of radius rdrawn around
oppositeto
by qand 4z
the combined force
exerted the point charge and
does not depend on the
direction of
Hence the field due to apointcharge
Force exerted by 4, on
Q=-F Electric
is spherically
symmetric.
field at distance r on
the equatorial
electric dipole of dipole
line of an
moment p is given by
1
E=

Theelectric field Eis

same at all points which
cylinder of lie on a
+ 42 -
radiusrwith its axis on the
dipole axis and the
field pattern
I7-P dipole
looks same in all
axis. We planes passing through
the
The above say that the electric
vector gives the field of an
direction of the dipole is electric
dueto q force on Q cylindrically symmetric.
Problem 10An electric
Problem7Twopoint charges +q and in a uniform
dipole free to move is
placed
'a' -q are placed a electric field.
distance apart. Draw Explain alongwith
the line on which its motion
when it is placed, diagram
field is parallel the resultant
to the line
joining the two (a)parallel to the
charges. (b)perpendicular
Solution As shown in Fig, field,
to the field.
1.119, the resultant at Solution
points lying on the perperndicular all (a) Since the line
of action of the two
bisector (dotted line) forces
parallel to theline is PaSses through the same
point, the net force
joining the charges + q torque acting on the and the net
and-4. dipole is zero. So no
duced when a dipole is motion is pro
p placed parallel to the
electric field.
(b)The two equal and
opposite forces -4E
and +gE
constitute a couple and
hence a
torque acts on the
dipole, given by
t =pE sin 90° = q. 2aE [: p=.2a)
This torque
rotates the dipole about
an axis
perpendicular to the electric
E field and
passingthrough the
midpoint of the dipole.

Fig. 1.119

Peoble sWhatisan electric of force ? What is

itsimportance ?
line
O-4
|Punjab 98C)
Solution An electric line
mnay be defined as
of force
the path
straight or curved,
along which a unit
charge would tend to positive (by
move if free to do so.
Fig. 1.12o Electric dipole (a) parallel
(b) perpendicular to uniform
electrie fieldE.
 ELECTRIC CHARGES AND FIELD
1.69

Problem 11. An electric dipole is a pair of

equal and () When 0 -90°, torque is maximum [Fig. 1.121(a)]
opposite charges, separated by a small fixed distance
aween them. The dipole is free to move. TmaxPEsin 90°mpE
What is the
action on it, when it is placed in (ii) When 0-30° or 150°, torque is half the maximum
(i) a uniform electric field,and value [Fig. 1.121(6)1.

(ii) a non-uniform electric field ? t=p Esin (30° or 150°) tnax

Solution. (1) In a uniform electric field, an electric
(iii) When =0° or 180°, torque is minimum
dipole experiences two equal, opposite and parallel forces
t its two ends. The net force on it is zero but it [Fig. 1.121()

experiences a torque due to which it rotates about an axis Tmin P Esin (0° or 180°) =0
nerpendicular to the electric field and passing through its
Problem 13. Two small identical electrical dipoles
mid-point. p are kept at an
AB and CD, each of dipole moment
(in) In anon-uniform electric field, an electric dipole
forces at its two angle of 120° to each other in an external electric fieid
experiencestwo unequal and non-parallel
ends. The two forces add up to give a resultant force E pointing along the x-axis as shown in Fig. 1.122.

and a torque. So the dipole will acceleratelinearly in the Y!

direction of the resultant force and also accelerate
angularily in the direction of the resultant torque.
120°
Problem 12. An electric dipole of dipole moment p is
X

placed in a uniformn electric field É. Write the expression

for the torque experienced by the dipole.Identify two 9B

pairs of perpendicular vectors in the expression. Show Y
the dipole in the
diagrammatically the orientation of Fig. 1.122
which the torque is (i) maximunm (ii) half the
field for

maximum value (iii) zero. [CBSE SP08) Find the

()dipole moment of the arrangement, and

Torque experienced by the electric dipole of
Solution.
magnitude and direction of the net torque
moment p in a uniform electric field E is given by (i)
dipole
acting on it. [CBSE D 11;OD 20)

Solution. (i) The directions of the

two dipole moments and their

are :
The pairs of perpendicular vectors resultant are shown in Fig. 1.123.

T andE X
1.7 and p 2.
Given Pa Pe=P
Resultant dipole moment,

Fig. 1.123
+4 PR =NP p+2xpxp cos 120°

J30° moment acts along the bisector of ZAOC

This dipole
i.e., at an angle of 30° with
+X direction.

(a)
(i) Torque, r pEsin 30°=pE

the torque t acts into the plane of

By right hand rule,

Z-direction.
paper along
flux. Write SI units.
its
Problem 14. (a) Define electric
[CBSE 18C]

spherical balloon carries a charge that is

(b) A
As the balloon is
uniformly distributedover
its surface.

in how does the total

blown up and increases size,
? Give
surface change
flux coming out of the
(c)
electric
|CBSE D 071
reason,
Fig. 1.121
1.70 PHYSICS-XII

Solution. (a) The electric flux through a given surface Solution. ()For 0<x<a, the
charge
area is the total number of electric lines of force passing Gaussian surface Iis zero.
enclosed by

normally through that area.

It is given by

Ao, - E.AS = E AS cos 0

SI unit of electric flux Nm'
(b) As the balloon is blown up and increases in
size,
the total charge on its surface remains constant.
Hence, by
Gauss's theorem, the total electric flux coming out of

it
surface remains unchanged.

Problem 15. A thin metallic spherical shell of radius

R carries a charge Q on its surface. A point charge
2 is Fig. 1.125

placed at its centre C and ano ther charge +20 is placed By Gauss's theorem,
outside the shell at a distance x from the centre as
shown in Fig. 1.124.

Find A Ex4rx =0 .. E=0

(i) the force on the 20 (ii) For asx<b, the net charge enclosed by the
charge at the centre
Gaussian surface II is q.
of shell and at the
point A, Ex4ry? or E=
(ii) the electric flux

through the shell.

(iii) For b<x<o, the net charge enclosed by the

[CBSE D 15] Fig. 1.124

Gaussiarn surface Il is (4+).

centre
Solution.
of the
Net force
shell is zero.
on the charge Q/2, placed at the
Ex4ry2 9+Q E= 1 q+2
Force on the charge 20 kept at point A at distance r
Problem A spherical rubber balloon carries a
17.
from the centre is charge that uniformly distributed
is over its surface. As

F=Ex20=
4ren
1
3Q/2 x2Q=
2 130 the balloon is blown up; how does E
vary for points
() inside the balloon,(i) on the surface of the
balloon
and (iii) outside the balloon ?
(i) Electric fiux through the shell, = Q2 Solution. (i) For points inside the balloon, E = 0.
(ii) As the balloon is blown up, surface charge density
Problem 16. Two thin concentric and coplanar sphe- G decreases and so the field, (E=o/E) on its surface
rical shells, of radii a and b (b>a)carry charges, q and Q. decreases.

respectively. Find the magnitude of the electric field, at

a point distant x,from their common centre for (ii) For points outside the balloon, E=
()0<x <a (ii) a <x <b (iii) b<x<o As the balloon is blown up, the charge
enclosed by
[CBSE F 15, D 16C] the Gaussian surface remains same, so
E does notchange.

1HOTS Problems on Higher Order Thinking Skills

Problemn 1.The dimensionsof anatomn are of the order Solution. The electric fields of protons and
of an Angstrom. Thus there must be large electric fields electrons bind the atoms to a neutral entity. Fields are
between the protons and electrons. Why, there is the caused by excess charges, there can be no excess
electrostatic field inside a conductor zero charg
on the inner surface of an isolated conductor.
[ExemplarProblem)
 ELECTRIC CHARGES AND FIELD

Problem 2. Twodipoles,
made from charges tqand +O
esDectively. kave equal dipole moments. Give the
(i) ratio
letpeen the 'separations' of these

betoeen he dipole axes ofthese two

tuo pairs of charges angle ()
dipoles. [CBSE SP 13]
Solution. As the two dipoles have equal dipole
moments, so

() qa =Qd

++ ++ (a)
must have

+
(ii) their dipole axes
same direction i.e., 0=0°
Problem 3.Sketch the electric field

lines for a uniformly charged hollow

cylinder as shown in Fig. 1.126(a).

Fig. 1.126 (6)

Solution.

(C

(a) (e)

()
Top view (ii) Side view
Fig. 1.127

Solution. Only Fig. 1.127(c) is right and the remaining

Fig. 1.126 (b)
figures cannot represent the electrostatic field lines.

Problem 4. Two point charges placed at a distance r in Figure 1.127(a) is wrong because field lines must be
a force F on each ofher. At what distance
exert will these normal to a conductor.
air

charges experience the same force F in a medium of dielectric Figure 1.127 (b) is wrong because lines of force cannot
constantK? start from a negative charge.

1 Figure 1.127(c) is right because it satisfies all the

Solution.
4192 F ...)
properties of lines of force.

Figure 1.127(d) is wrong because lines of force cannot

med =
1
9192 =F ..(ii) intersecteach other.

Figure 1.127(e) is wrong because electrostatic field

lines cannot form closed loops.

Dividing ()by (i), we get
Problem 7.Figure 1.128 shows the electric lines around
=1 or threecharges A, Band C.

(a) Which charge is positive ?

Problem 5. A force F isacting between two charges
(b) Which charge has the largest magnitude ? Why ?
placed sone distance apárt in vncuum. Ifa brass rod is placed
(c) Inwhich region or regions of the picture could the
betocen these two charges, howdoes the force change ?
electric field be zero ? B

Solution, For any metal, k =0 (i)near A

=0 (ii) near B
'brass
K
(iii) near C
.e,, in the presence of brass rod, the force between the
(iv) nowhere.
two charges becomes zero.
Justify your answer.
Troblem 6. Which amongthe curves shown in Fig 1.127,

electrostatic field lines ? Fig. 1.128

tahot possibly represent
 L72 PHYSICS-XI|

Solution. (a) Charges A and C are positive because

the lines of force are emerging out from them. Therefore, deflect
(6) Chas the largest
Charge magnitude because the y-directionis
maximum numberof field lines are associated with it. 1
h =0 xt+
(c) 0) Near A. No neutral point can exist between 2
unlike charges A and B or between B and C. The
Fig. 1.131 As the particle
neutral point exists between like charges A and Also,
so it has

C.
the neutral y-direction,
point will be closer to the charge with smaller Force on thecharge- q in the upward direction is
magnitude. Hence, electric (a) Negative cha
field is zero near charge A.
ma = qE (b) (ii) and (iv).
Problem 8. A glass rod rubbed with silk brought close
is
qE
same sign ar
to fwo uncharged spheres in contact with each other, inducing
:. Acceleration, a=
charges on thenm as
(c) To find e/
ment h as the
shown in Fig. 1.129. Time taken to cross the field, t =
Describe what happen Time taken by a
when
Vertical at the far edge of the plate
deflection
will be t=
(i) the spheres are slightly 1
separated, and Fig. 1.129 y= ut + at?
2 =0+
2 Problem 11. Fi
1m U
2mu
(i) the glass rod is subsequently of electric field line
removed, and finally
Like the motion of a projectile in gravitational from rest at pointP
(iii) the spheres are separated field,
far apart ? the path of a charged particle in an electric field is by the electric field.
Solution.
parabolic.
momentum of the
A B A B A B
Problem 10. Figure 1.132 shows -tracks of
three charged
particlesin a uniform
electrostatic field. Give the signs of the
three charges. Which particle has the highest charge to mass
ratio ?
[NCERT;CBSE D 01C]
(a) (6) (c)

Fig. 1.130

()) When the spheres are slightly

separated with
the glass rod undisturbed, there is
little change
in the distribution of charges, as shown in Fig. 1.132

Fig. 1.130(a). (a) Suppose that a particle is attracted towards the

(ii) When the glass rod is removed, there is positive plate :what must the charge on it be ?
redistribution of charge on each sphere. Their (6) two particles have identical
Suppose,
positive and negative charges will face each curved trajectories.
Which of the following are
necessarily ?
other, as shown in Fig. 1.130(6).
(ii) The charge on each
sphere gets uniformly
(i)They have same charge
mass; (ii) The charges have
(ii) They have
same ; true

thesamesign; (iv) They

distributed as shown in Fig. have the same elm ratio. Solution. Th
1.130(c).
to each other
Problem 9. A particle of nass nand charge (-q) (c) You are given the
enters the initial velocity v of a beam particle pattern (a). Co
region between the two charged and the length of
plates initially moving along the capacitor l. What other strongest in ca
X-axis with speed v, (like particle 1 in Fig. measurement would one
1.152). The length of enable to find elm ? Momer
plate is L and a uniform electric field E is
the plates. Show
maintained between Soluion. Particles 1 and 2 have negative charges >Momer
that the vertical deflection of the particle at the because they are being deflected
edge of he plate towards the positive > Momer
far is qEE/(2m o). plate of the electrostatic
field.
Compare this motion with Problemn
motion of a projectile in Particle 3 has
positive charge because
gravitational field. it is being
deflected towards the
negative plate. coulomb force
Solution.The motion of the charge
-qin the region
of the electric field E between Acceleration acting on charge
the two charged plates is qin y-direction, between the tu
shown in Fig. 1.131. a==
F_qE and (2uC, -3
 ELECTRIC CHARGES AND
Thorefore,
FIELD 173
deflection of
charged partice in
is timetin 1 1
y-direction
Solution. (i) As F so F versus graph is a

h=0 2 a²19E2
xt+- 2 m straight line for both pairs of charges.

As the particle
()For the pair (1 uC, 2uC), Fis repulsive,the graph
suffers maximum defection in OA has a +ve slope. For the pair (2uC,
y-direction,
so ) highest charge
ithas
to mass (g/ m)
-3uC),F is
ratio. attractive, the graph OB has -ve slope. a

(a) Negative charge.

(b) arnd (i0). The particles must have
(ii)
(tn) Clearly, Etraction =3Fulšion the slope of OBis
charges of three times the slope of
same sign and same e/m ratio. OA.
(c) To find e/m we measure
the vertical displace
ment h as the particle crosses
the capacitorplates.
Time taken by a particle to cross the capacitor Repulsion
plates,

:. h=E2
2 m -0.5
Problem 11. Figure 1.133 shous three different patterns
of electric field lines. In each pattern, a proton is released
from rest at point P and then accelerated towards the point Q
by the electric field. Rank the patterns according to the linear
momentum the proton when it reaches Qgreatest
of
first. Fig. 1.134

Problem 13. Two pointcharges, q, and q,are located at

P points (a, 0, 0) and (0, b, 0) respectively. Find the electric

field, due to both these charges, at the point (0,0,c)
[CBSE SP13]
(a)
Solution. Net electric field at the point (0, 0, c) due to
the charges 4, and q, is

19
(b)
AY

P 92 (0, b, 0)

(c)

Fig. 1.133 (a, 0, 0)

Solution. The lines of force near point P are closest

to each other in pattern (c) and farthest apart in

70, 0, c)
pattern (a). Consequently, electric field near point P is
strongest in case (c) and weakest in case (a).

Momentum of proton at point Q in pattern (c)

Fig. 1.135
> Momentumof proton at point Q in pattem ()

> Momentum of proton at point Q in pattem (a) But i--ai+ck

Problem 12. Plot a graph showing the variation of

COulomb
1

zehere r is the distance

bj+ ck ;=(b+dn
force (F) versus

vetoeen thetuo charges of each pair of charges :(1 uC, 2uC) Enet = 14-ai+ ck).-bj- cky

nd (2 uC,-3uC) Interpret the graphs obtained.

[CBSE OD 11]
 PHYSICS-XI

Problem 14. A peint datge is pland ot thecentre of 6elds and are equal and opposite and also E
splherical Giussian surteHoe oilleiectri fuxh chnge feld a
E, are equal and opposite. Hence
the net
and

() the sphere is replaced by a cube of same or diferent the centre is zero

and (i), the potentiala
olumc, (i) In both figures 1.137(i)
positive charges are +ve and those due to
the due to
()a secoa charge is placed near, and outside,

are equally negative.

Hence the net

original sphere, negative charges

at the centre iszero.
() a second charge is placed inside the sphere, and
potential

16. Thrce charges + q. tqand-2qare placed

by an ? Problem
(i) the oiginal charge is replaced electric dipole
the dipole
triangle. What is
the verices of an equilateral
-E.aš - I at
Solution By Gauss's law, moment of the system
?
24 pC

(0 & does not change because it depends only on

the total charge enclosed by the Gaussian 3030

surface and not its shape or size.

the total fiux is

(m) does not change because
by the charge inside the surface arnd B
determined A
not on the charge outside. +q
charge inside
(in) change because the
will
total

the surface has changed.

() becomes zero, because a dipole consists

so the net
of

two equal and opposite charges and

is zero.
charge inside the surface
are placed at the four
Problem 15. Four point charges Fig. 1.138
as shownin
Coriersofa square in the fuo wavs (i)and (ü) Solution. Let each side of the equilateral triangle be d.
at
the (i) electric ficld (ii) electric potential,
Fig 1.136. Will

in the twoo Dipole moment along CA =qd

the centre thesquare, be the same or different
CB = qd
of
onfigratians and oy ? ICBSE SP 081 Dipole moment along

By parallelogram law, the net dipole moment acts

-Q +Q
along diagonal CD. Its magnitude is
D C D C

p=N(gdy + (qdy' + 2qd x qd x cos 60° = V3 qd

This dipole moment acts along the bisector of the

B B
angle at charge -24.
+0
(i) Problem 17. Two similar balls each having mass m and
charge q are hung from a silk thread of length Prove that
I.
Fig. 1.136 equilibrium separation,
1/3

Solution ) In Fig 1.137(0), the electric fields E

and E- get added and also E, and Ep get added. Hence

makes a angle with the vertical.

field at the centre O.In Fig. 1.137i), the hei eachi thread small
there is a net JEE Main July 21]

C Solution. Refer to Fig. 1.139. According to

D
Lp
D
Eo
Coulomb's law,the electrostatic repulsion between the
two balls will be

F=
1
() (i) For smallangular displacement 0, we have
arc x/2 x
sin 0-0(rad)
Fig. 1.137 radius 1 21
 ELECTRIC CHARGES AND FELD L75
Problem 19. A simple pendulum consists of a small

Sphere of mass m suspended by a thread of length l The

sphere carries aposititve charge q The pendulum is placedin
a uniforn electric ficld of strength E directed vertically

upoards. Witlh what period will the pendulum oscillate if the

electrostatic force acting on the sphere is less than the
gravitational force ? Assume oscillations to be small.

Solution The sphere experiences an upward pull

qE.
x/2
0
sin ..Net force on the sphere
tmg
=mg - qE, in downward direction

g
Acceleration,

Fig. 1.139 Force mg -qt

Mass 111

17
.Restoringforce on each ball

= 1Ig sin -
T=2n-2r
m
For equilibrium,
Problem 20. Four particles, each having a charge q, are
1
placed on the vertices of a regular pentagon. The distance of
corner fromthe centre is 'a'. Deternnine the electric field
each

71/3 at the centre of the pentagon.

O Or X= Solution. Suppose the four charges are placed at the

2rE,mg corners A, B, C and D of the pentagon ABCDE. If we

+ coulomb Place a charge q at the corner
each ofvalue a B

Froblem 18.Five point charges, E also, then by symmetry,

of a regularhexagon L metres.
of the total electric
side field at O
re placed onffve vertices
Find the magnitude of force on a charge -q coulomb placed must be zero. Thus, the field
[IIT 92 ;CBSE OD 19] at the centre O due to the
at the centre of the hexagon. D
charges A, B C and D
A
is

Solution. The situation is shown in Fig. 1.140.

the
charge-qatO due to charges at A
on
and D are equal and opposite to

Forces
field due to the charge q at E
equal and opposite and
hence cancel out.
alone.

Electric field at O due to

Fig. 1.141

charge q at E
1 4
along EO
C

. Electric field at O due to the charges at A, B, C

and D
1

along OE.

number of charges, each equal

Problem 21. An ifinite
Fig. 1.140
along X-axis at x =1x -2, x=4,x =8, ..
to g are placed
and E
to charges at B
and so on.
due
the forces
at the point x =0due to this set

| ()Find
Similarly,
the electricfield
cancel out.
by
O is exerted up ofchages.
The on charge -4 at above set up,
only force be the electric field, if in the
(i) What will
charge +q at C. It is given by charges have opposite signs.
the consecutive
kq newion,along OC. [IIT 95]

F=k194
(Oc)
 176
PHYSICS-XI

Solution. () Atthe point x =0, the electric fields due Solution

to all the
2F cos 9
charges are in the same x-direction and hence kOg
get added up.
1

Net force on charge 4,

2F

1
Fpet cos 9

AnE, 1
2kQgx
(ii)
Electric field at a =0 3/2
is

4 d/2 d2
42
For maximum value of t Fig. 1.143
dFnet
1
=0
4 16 64 dx
1/2
Problem 22. Two identical positive charges Qeach are x2r=0
fixed at a distance of 2a'apart from each other. Another
point charge g,with mass 'm'is placed atmidpoint between d
fuofixed charges. For a small displacement along the line
joining the fixed carges, the charge 2/2
q, execiutes SHM.Find the
time period of oscillation Problem 24. Eight identical point charges of q
of charge go: JEE Main July 22] each are placed at the coulomb
corners of a cube of each side 0.1
Solution. m.
Calculate the electric field at the centre of the cube. G
Calculate the field at the centre when one
m of the corner
charges is removed.
a-x Solution.Length of each side, I =0.1 m
Distanceof each corner from the
Fig. 1.142 centre of the cube is

Net force on the charge q, in the r= V31 N3 x0.1

=5V3x 10-2 m
displaced position, 2 2
F= When the eight point charges
all
are placed at the
(a+x (a-x corners,
fields electric
of the diagonally
charges cancel out in pairs. opposite

(a-x-(a+ x Net field at the centre =0.

.:.
(a-x') When a charge is removed from one corner, the
electricfields due to three
4x pairs of charges cancel out.
[x<<a] However, the charge at the
remaining comer creates field,
E= 1
qx 1

x=-kx
9x 10 x q
=1.2 x g NC
(5/3x 102 10

T-2n7 The field points towards the

comer having nocharge.
Problem 25. If the total charge enclosed by a surfacce is
zero, does it imply that the electric ield
evervelere on the
surface is zero ? Conversely, if the electric fieldeeryehere
on the surface is zero, does it imply
Problem 23. Two point charges O each that the charge inside is
are placed at a zero ?
distance d apart. A third point charge q is placed at a distance
[ExemplarProbleml
from the mid-point on Solution.As q=0, so from the Gauss's theorenn,we
the perpendicular
bisector. Find the have
value of xat which charge g uoill experience the maximum
Coulomb'sforce. |JEE Main June 221 4-fE.as -0
 ELECTRIC
CHARGES AND FIELD L77
Clearly, the above equation does not
imply that E
necessarily
zero at all points on the Problem 27. A sall metal splere carrying charge +Q
is Gaussian surface.
1s located at the centre of a
also be possible that spherical cavity in a large
It may E is non-zero but it is tcharged metal sphere as shown in Fig. 1.145. Use Gauss's
to d S at all points on the
perpendicular surface, even theorem to find electric field at points P and P
then the integral (CBSE D 05, OD 14C]
6E.dS would be zero. Metal

However, the converse is true. If E

is zero at
Free space

on Gaussian surface,then from

every +Q
point Gauss's theorem

we get, q=0
ie. no net charge is enclosed by the Gaussian surface.
Fig. 1.145

Problem 26. Figure 1.144 shows a cylindrical Gaussian Solution.(i) To determine the electric field at point

surface jor an infinitely long thin straight eire of uniform P, consider a concentric spherical surface of radius r,

linear charge density. as the Gaussian surface. By symmetry, the field E will
have same magnitude at all points on this surface and

will point radially outward.

:. Flux through the Gaussian surface,

= Ex 4r
Charge enclosed by Gaussian surface = +Q
By Gauss's theorem,
1
= Ex 4ri or E=

)
Fig. 1.144
(i) As the electric field insidea conductor is zero, so

Answer the following: field at point =0.

()
For which surface is the

Over which surface is E

electric

zero ?
flux zero ?
radius
placed
Problem 28.
r, and outer
at the centre
A
radiuS

of the
,
splherical conducting
has a charge
shell.
shell

Q.A charge g
of inner

()inner
is

(ii) Over which surface is| E| constant ? (a) What is the surface charge density on the

(i0) Over which surface does| E |change ?

(b)
surface,

Write
(i)outer surface

the expression
of the shell ?

for the electric field at a point

[CBSE OD 10]
X>, from the centre the shell.
of
Solution. Electric field of a line charge,
Solution.The charge q at the centre induces charge
E=
2
,
acting radially
outward
-qon the inner surface of the shell and charge +q on the
E of the shell.
outer surface
() At thetwo plane end faces,

4-[E.aš =0.

) For any finite distance from the line charge,

(ü)
E

At
cannot

all
be

points
zero.

of the curved
surface, | E| is

Constant.

decreases with the

At the plane end faces,IE| Fig. 1.146

increase in distance r.
 PHYSICS-XII
L78 1.148 shows the
Solution. Figure
inner surface, charge
charge density on the model of the atom. distr

(a) ()Surface bution for the given

(i)Surface charge density on the outer surface,

(6) For x >,the field is similar to that of a point

charge.

Q+
E= 1

Problem 29. Electric field in Fig. 1.146 is directed along Fig. 1.148 An early model of atom.

+X direction and given by E, =5 Ar +2 B, where E is in

As the atom is neutral,the total negative charge in a
NC and x is in netre, A and
-
sphere radius R must be
of Ze. If p is the negatiye
B are constants with dimen
charge density, then we must have
sions. TakingA = 10 NC m

B-5 NC 4TR 3Ze

and calculate : Ze+ p=0 or p=
() the clectric flux through the
3
cube. (i) net charge cnclosed
|= 10cm By spherical symmetry of the charge distribution,
within the cube.
the electric field É depends only on radial distance r
[CBSE Sample Paper 08]
Fig. 1.147
and not on the direction of r. It should point radially
Solution. (i) The electric field is acting only in inwards or outwards. So we imagine a spherical
X-direction and its Y and Z-components are zero. So Gaussian surface of radius r centered at the nucleus.
the flux passes only through faces Mand N.
()For r <R.Flux through the Gaussian

;=
surface,
The magnitude of the electric field at the face
M(I =0)is Ex 4n?
E =5 Ax +2 B=5x 10 x +2x5=10 NC-1
0 Charge enclosed by the Gaussian surface,

Flux, q= Positive nuclear charge

M=Ey cos =10x (0.10)x cos 1 80°

+ Negative charge in a sphere of radius r

=-0.1 Nm-l =Ze 4 4 3Ze

+
3 PZe+
The magnitude of the electric field at the face
N (r=10cm) is
Ey =5 Ar + 2B=5x 10 x0.10 +2 x 5
o=15 NC-1 Applying Gauss's theorem, =q/Ey we get
=Ey 4n=

E)
Flux, cos Ze
Ex
=15 x (0.10)x cos 0°=0.15 Nm 2c1
Net flux through the cube,
or
= t y =-0.1+0.15 = 0.05 Nm ?c1. (r<R)

(ii)Total charge enclosed within the cube, The field Epoints radially
outward.
4=e =8.854 x 10x0.05 =4.43 x 10-13 (i) For r>R. As the atom is neutral, the total
C.
Problem 30. An
charge enclosed by the Gaussian surface is zero. By
early model ofan atom
haoe npositiely
considered it to
ualiis s theorem,
charged point nucleus charge Ze, surrounded
of
byauniform density of negative charge up toa radius R. The Ex 4r=0
atom as a whole is neutral. For this model, what is the E- 0. (r>R)
electric ata distance r from the nucleus
field
? (i)Atr= R. Both of the above cases give the same
[NCERT;AlIMS 18) result: E=0.
 ELECTRIC CHARGES AND FIELD 79

GUIDELINES TO NCERT EXERCISES

11 What is the forcebetwcen two small

2x
charged spheres
The factor ke| Gmn, represents the ratio of
kaing charges of T0 Cand 3 x 10Cplaced 30cm apart clectrostatic force to the gravitational force betweer an
air
in
electron and a protor. Also, the large value of the ratio
Ans. Here 4 =2x 10 C, q, =3x107C. signifies that the electrostatic force is much stronger thar
r=30cm =0.30m the gravitational force.

According to Coulomb'slaw, 14.() Explain the meaning of the statement dlectric

charge a body is quantised."

of
x3x107
2T2-9x10,2x
1

F= 10
(i) Why can one ignore quantisation of electric charge
(0.30) when dealing with macroscopic ie., large scale

=6x10N (repulsive). charges ?

Ans. (i)Quantisation of electric charge means that the
12. The electrostatic force on a small sphere of charge toal charge (o of a body is always an
integral multiple of
0.4 uC due to another small sphere ofcharge -0.8uC in air is
a basic charge (o) which is the charge on an electron. Thus
0.2 N. (i) What is the distance between two spheres ? (ii) What
q=ne, where n = 0, ± 1, ± 2, ± 3,.....
is the force on the second sphere due to the first ?

(i)While dealing with macroscopic charges (4 =ne),

Ans. ()Here 4, =0.4 uC = 0,4 x 106 C
we can ignore quantisation of electric charge. This is

4, = -0.8uC=-0.8 x 10 °C F=0.2N, r=? because e is very small and nis very large andsoq behaves
1 as if it were continuous i.e., as if a large amount of charge
As F= 2 is flowing continuously.

1.5. When a glass rod is rubbed with a silk cloth, churges appear
1
492 on both. A similar phenomenon is observed with many other

pairs of bodies. Explain how this observation is consistent with

9x10 106 x0.8x106

x0.4 x
=144 x 104 the law of conservation of charge.

0.2 Ans, Itis observed that the positive charge developed

or r= 12x10 =0.12m =12 cm. on the glass rod has the same magnitude as the negative
charge developed on silk cloth. So total charge
after
opposite
(ü)Thetwo charges mutually exert equal and
forces,
rubbing is zero as before rubbing. Hence the law of
conservation of charge is being obeyed here.
due
.. Force on the second sphere
-0.2 N (attractive).
to the first
1.6. Four point charges qa =2 C qp =-5 C 4c =2 uC

at the corners of a square ABCD of side

ke/Gm, m, is dimensionless. Look 4n=-5Care located

1.3. Checkthat the ratio
uCplaced
10cm. What is the force on a charge of 1 at the centre

value of this
4p atable physical constants and determine the
of the sqare ?
of
ratio. What does this ratio signify ?
Ans, Here OA =OB= 0C= OD= 2
[Nm?c x[C? = nounit
Ans. k
=5/2cm=5/2 x 10m
Gmm, [Nm'kg 1 *(kgl[kgl 10em 4-2
4o-5HC C

As the ratío ke|Gm, m, has no unit, so it is

dimensionless.
Now k=9x 10 Nm'c2
G= 6.67 x 101 Nm'kg
e= 1.6 x 101 kg

=9.1x 10
m, kg
and
=
m, 1.66 x 10 kg

9x 10 x(1.6 x 10 19 B
10 cm
x 10 27
103 x 1.66
x10 x9.1 x
Gm y 6.67

Fig. 1.149

= 2.287 x10*
 PHYSICS-X1

Forces exerted on the charge of 1 C located at the Electric field at the

midpoint 0due
Centre are to4
9x102x10*1x10 910'3x1
(010
- 2.7 x 10 NC
-36N, alomg O along OB
Electric field at the midpoint
R.9105x10 x1x104 O due to q

}
(52×10 Eg
9x10 3 x105

-9N along o -2.7x 10 NC

(0.10
along OB
t.9x10210x1x 104
(2×10 Resultant field at the midpoint Ois

-36 N,along O E= E + Ep =(2.7+ 2.7) x 10

- 54x10 NC, along
F9*10x5x 10 x lxi06
OB.

(5V2 10
(i) Force on a
the midpoint O,
negative charge of 1.5 x 10"C placad .
-9N,along OD F= qE= 1.5 x 10 x5.4 x 10

Clearly, - and Fp-F = 8.1 x10N,along OA

Hence total The force on
force on uC charge is
1
negative charge
a
acts in a direction
opposite to that of the electric field.
1.9. A has two charges
system
4a = 25 x10 C and
qn =-2.5 10 C
1.7. (a)
FF-E -F=
An electrostatic
zero N. B(0,0,
x
+15 cm) respectively.
located at points
What is
A(0,0,

the total
-15om) and

electric charge and

field line is a dipole moment of the
system ?
That is, a field line continuous curve.
cannot have sudden Ans.
(b)Explain
breaks. Why not ? Clearly, the two charges lie on
why tuwo field lines never side of the origin and at 15 cm
Z-axis on either
cross each other at from
point ? any Fig. 1.151.
it, as shown
|CBSE D 05, 03;

in
Ans. (a)
OD14]
Electric lines of 2a =30cm
region of an
electric field.
force
throughout the
The eleciric field of a
exist = 0.30 m, q =2.5 x 10C
charge
decreasesgradually with
increasing
Z
becomes zero at infinity distance from it and
Le.,electric
field cannot
abruptiy.Soa line of force vanish 4,=-2.5 x 10 C
cannot have sudden B (0, 0, + 15 cm)
must be a continuous breaks, it
Curve.
(b) Iftwo lines of
force intersect, then
two tangents and hence two there would be
directions of electric field at
the point of intersection,
which is not possible.
1.8. Two point charges 4, =+3uC and q =-3 uC are
located 20cm apart in vacuum. () Find the at the
midpoint O of the line ABjoining the tuo
electric field

negative
charges. (i) 1f a
test charge magnitude 15 x 10 C A0,0, - 15 cm)
of
is placed at the 4-2.5 x 1o'c
centre, fnd the force experienced by the test charge

(CBSEOD 031
Ans. Thedirections of the fields E, and E, due to the Fig. 1.151
cCharges 4 and 4 at the mnidpoint P are as shown in
Fig 1.150.
Total charge =4 t q =2.5 x 10-2.5x10 =0
Dipole moment,

10 cm p=qx2a 10
10 can = 2.5 x x0.30
B
-0.75 x 10 Cm
The dipole moment acts in the direction from B to A
Fig.1.150
Le, along negative Z-axis.
 ELECTRIC CHARGES AND FIELD L81
L10. Anelectric dipole with dipole moment 4 x 10 Cmis 1.14. Consider a uniform electric feld :
at 30 oith the direction of a uniform through
aligned electric ficld 3 x 10 NC )Wihat is the flux of this field a

of
i

5 x104 NC Calculate the magnitude is parallel to the

the torque
magnitude 10 cm on side whose plane

of
of a
sgquare
on the dipole.
acting Y-Z-plane ?(ii) What is the flur through the same square if the
Ans. Here p=4x10 Cm,9- 30°, E =5x 10' Nc! normal to its plane makes a 60 angle with the X-axis ?
(CBSE D 13C]
.Torque, t=pEsin e
Y-Z plane points
=4x10x5x10' xsin 30°
Ans. ()Normal
in X-direction, so
to a plane parallel to

= 10 Nm. m?
AS = 0.10 x0.10í m² = 0.01

1.11. A polythene piece rubbed with wool is found to have a Electric flux,

neoative charge of 3.2 x 10C.() Estimate the mumber of

clectrons transferred. (i) Is therea transfer of mass from wool to

¢, =E.8s=3 x 10° i .0.01

polythene
? -30í.í = 30 Nm²c.
Ans.()Here =3.2x 10 C, e=1.6x10-19 C
q

(i) Here =60°

=
As q ne, therefore = EAS 60° =3 x 10 cos >x0.01 cos 60°

Number of electrons transferred,

=30x-15 Nmic-1.
3.2 x 10-7
n= 2x1o12 1.15.Consider a uniform electric ficld :
1.6x10-19
È=3x10°i NC-. What the net flux ofthis field througha
Since polythene

transferred
has negative charge, so
from wool to polythene during rubbing.
electrons are
of side 20
cube
coordinate
oriented so that

planes
c
fuces are parallel to the

?
is

ifs

a of mass from wool to Y-Z plane

(ii) Yes, there is transfer
Ans.The flux entering one face parallel to is

has a finite mass of to Y-Z plane.

polytherne because each electron
equal to the flux leaving other face parallel
Flux through other faces is zero. Hence net flux through
9.1x 10 kg.
the cube is zero.
Mass transferred
measurement of the electric field at the surface
1.16.Careful
=m, xn=9.1 x 10-31 x2 x10!2 net outward ux through the
of a black box indicates that the

= x 10 kg
1.82 Surface ofthe box is .0 x 10° NmC.()What is the net charge

(i) If the net outward fluux through the surface of

inside the box ?

small.
Clearly,

1.12. (a)

have their centres

mutual force
the

Two
amount of mass transferred

insulated

separated

of electrostatic repulsion
charged copper spheres A and B
by a distance of 50 cm. What
if
is the
is negligibly

the charge on each is

the

inside

Ans. () ,
box were zero, could you conclude
the box ? Why or why not ?

Using Gauss theorem,

= 8.0× 10' Nm²c2
that there were no charges

to
65 x 10C? The radii of A and Bare negligible compared
of repulsion if
the distance separation. (b)What is the force

9=-p =8.0 10x

of
and the
each sphere is charged double the above amount, Charge, x

distance between them is halved ?

= 0.07 x 10C = 0.07 uC

Ans. Refer tothesolution of Example 9 on page 1.12.

we cannot say that there are no charges at all

in
(ii) No,
showstracks of three charged particles
113.Figure 1.152 the net charge inside
inside the box, We can only say that
field, Give the signs of the three charges.
2 uniform electrostatic
the box is zero.
has the highest charge to mass rato ?
Which particle
1.17. A point charge +10 uC is a distance 5cm directly
1) y

of a square of side 10 cm as shown in

above the centre

the electric flux through

Fig. 1.153(a). What is the magnitude of

?(Hint :Think of the square as one face ofa cube with

the square

edge 10 cL)
with
Ans. We can imagine the square as face of a cube
1.152 edge 10 cm and with the charge of + 10uC placed at its
Fig.

10on page 1.72. centre, as shown in Fig. 1.153(b).

Ans. Refer to thesolution of Problem

 T82 PHYSICS-XI1

910 r15 z10

-6.67 x 10"C= 667 nC

As the field acts inwards, the

charge 4
10
cm cm negative. mk
10
10 cm 10cm 4=-6.67 nC.

(a)
(h) 1.21. A uniformly charged
conducting sphere
Fig. 1.153 diameter has a surface charge density of 80.0 uC/
the charge on the sphere. (ii) What is the total
Symmetry of six faces of a cube about centre dlectric f
ensures
its leaving the surface of the sphere ?
that the flux through each square CSED
when the face is same R= 2.4
charge q is placed at the centre. Ans. Here = 1.2 m
Total
2
flux,

G= 80.0uCm =80 x105 Cm -2

() Charge on the sphere is
OT q= 4n R' G=4x3.14
6
x 10 × 106 x 4nx9 x 10
x (1.2) x 80 10c
=1.45 x10- C
= 1.88 x 10 Nmc-1 (ii) Flux,

1.18.

Gaussian
A

surface
point charge

9.0 cm on
of 2.0

edge.
uC
What
is at the centre

is the net electric

of a cubic e=1= 1.45 x 103 x 4n x9x 10
flux
through the surface ?
= 1.6 x 10 Nm² C

Ans. Here q= 2.0 u C= 2.0 x 10°C, An

1.22. infinite line charge
Eo =8.85x10-12c?N'm2 9x 10* NC- produces a feld of
at a distance of 2 cm.
By Gauss's theorem, Calculate the linear charge
electric flux is density.

2.0 × 10-6
10° Nm2 c-1
Ans. E=9x10* NC, r=2cm =0.02 m
8.85 x 10-l2 2.26 x

Electric field of a line charge, E=

1L19. A point charge causes an -
2E
N C to
electric flux of 1.0 x 10 .:. Linear
charge
pass through a density,
spherical Gaussian
10.0 cm radius centred on the charge.
surface of
2=2renEr 2r x = 1

Gaussian
(i) If the radiusof the x9x10* x0.02
surface were doubled, how much 4n x9x109
flux would pass
through the ?(i1) What
surface is the value of the point charge ?
=0.01 x 10 Cm =0.1 uCm.
Ans. (i) r=-10 NmC, because the 1.23. Two large,
enclosed is the same in both the charge thin metal plates are parallel and close
cases. to
each other. On their
inner faces, the
(i) Charge, plates
have surface charge
densities of opposite signs and of
What is E (a) to the magnitude 17.0 x 102 C
left of the plates, (b) to the right of the plates.
and (c) between the plates ?
1

47 x9x 10 x(-1.0 ×10°) Ans. Here g= 17.0 x 10 cm-2

(a) On the left, the fields
=-8,84 x 10 C=-8,84nC. of the two plates are
equal and opposite, so E= Zero.
L20.Aconducting sphere 10 cm has an unknown
radius
On
of
(b) the right,
charge. If the electric field the fields of the
20 cm from the centre ofthesphere is two plates are
15 x 10° NC and points radially inuward, what is the net
equal and opposite,so E= Zero.
charge on the sphere (c) Between the plates,
? the fields due to both
Ans. Flectric field at the outside points
plates are in same direction. So the resultant
of a conducting field is
sphere
is
E= 104xx9x
28 2E o =17 x 10

-19.2 x1010 NC l