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ELECTRIC CHARGES

AND FIELDS

1.1 INTRODUCTION

All of us have the experience of seeing a spark or hearing a crackle when

we take off our synthetic clothes or sweater, particularly in dry weather.

This is almost inevitable with ladies garments like a polyester saree. Have

you ever tried to find any explanation for this phenomenon? Another

common example of electric discharge is the lightning that we see in the

sky during thunderstorms. We also experience a sensation of an electric

shock either while opening the door of a car or holding the iron bar of a

bus after sliding from our seat. The reason for these experiences is

discharge of electric charges through our body, which were accumulated

due to rubbing of insulating surfaces. You might have also heard that

this is due to generation of static electricity. This is precisely the topic we

are going to discuss in this and the next chapter. Static means anything

that does not move or change with time. Electrostatics deals with the

study of forces, fields and potentials arising from static charges.

Historically the credit of discovery of the fact that amber rubbed with

wool or silk cloth attracts light objects goes to Thales of Miletus, Greece,

around 600 BC. The name electricity is coined from the Greek word

elektron meaning amber. Many such pairs of materials were known which

Physics

on rubbing could attract light objects

like straw, pith balls and bits of papers.

You can perform the following activity

at home to experience such an effect.

Cut out long thin strips of white paper

and lightly iron them. Take them near a

TV screen or computer monitor. You will

see that the strips get attracted to the

screen. In fact they remain stuck to the

screen for a while.

It was observed that if two glass rods

rubbed with wool or silk cloth are

brought close to each other, they repel

each other [Fig. 1.1(a)]. The two strands

FIGURE 1.1 Rods and pith balls: like charges repel and of wool or two pieces of silk cloth, with

unlike charges attract each other.

which the rods were rubbed, also repel

each other. However, the glass rod and

wool attracted each other. Similarly, two plastic rods rubbed with cat’s

fur repelled each other [Fig. 1.1(b)] but attracted the fur. On the other

hand, the plastic rod attracts the glass rod [Fig. 1.1(c)] and repel the silk

or wool with which the glass rod is rubbed. The glass rod repels the fur.

Interactive animation on simple electrostatic experiments:

If a plastic rod rubbed with fur is made to touch two small pith balls

(now-a-days we can use polystyrene balls) suspended by silk or nylon

thread, then the balls repel each other [Fig. 1.1(d)] and are also repelled

by the rod. A similar effect is found if the pith balls are touched with a

http://ephysics.physics.ucla.edu/travoltage/HTML/

glass rod rubbed with silk [Fig. 1.1(e)]. A dramatic observation is that a

pith ball touched with glass rod attracts another pith ball touched with

plastic rod [Fig. 1.1(f )].

These seemingly simple facts were established from years of efforts

and careful experiments and their analyses. It was concluded, after many

careful studies by different scientists, that there were only two kinds of

an entity which is called the electric charge. We say that the bodies like

glass or plastic rods, silk, fur and pith balls are electrified. They acquire

an electric charge on rubbing. The experiments on pith balls suggested

that there are two kinds of electrification and we find that (i) like charges

repel and (ii) unlike charges attract each other. The experiments also

demonstrated that the charges are transferred from the rods to the pith

balls on contact. It is said that the pith balls are electrified or are charged

by contact. The property which differentiates the two kinds of charges is

called the polarity of charge.

When a glass rod is rubbed with silk, the rod acquires one kind of

charge and the silk acquires the second kind of charge. This is true for

any pair of objects that are rubbed to be electrified. Now if the electrified

glass rod is brought in contact with silk, with which it was rubbed, they

no longer attract each other. They also do not attract or repel other light

objects as they did on being electrified.

Thus, the charges acquired after rubbing are lost when the charged

bodies are brought in contact. What can you conclude from these

2 observations? It just tells us that unlike charges acquired by the objects

Electric Charges

and Fields

neutralise or nullify each other’s effect. Therefore the charges were named

as positive and negative by the American scientist Benjamin Franklin.

We know that when we add a positive number to a negative number of

the same magnitude, the sum is zero. This might have been the

philosophy in naming the charges as positive and negative. By convention,

the charge on glass rod or cat’s fur is called positive and that on plastic

rod or silk is termed negative. If an object possesses an electric charge, it

is said to be electrified or charged. When it has no charge it is said to be

neutral.

In olden days, electricity and magnetism were treated as separate subjects. Electricity

dealt with charges on glass rods, cat’s fur, batteries, lightning, etc., while magnetism

described interactions of magnets, iron filings, compass needles, etc. In 1820 Danish

scientist Oersted found that a compass needle is deflected by passing an electric current

through a wire placed near the needle. Ampere and Faraday supported this observation

by saying that electric charges in motion produce magnetic fields and moving magnets

generate electricity. The unification was achieved when the Scottish physicist Maxwell

and the Dutch physicist Lorentz put forward a theory where they showed the

interdependence of these two subjects. This field is called electromagnetism. Most of the

phenomena occurring around us can be described under electromagnetism. Virtually

every force that we can think of like friction, chemical force between atoms holding the

matter together, and even the forces describing processes occurring in cells of living

organisms, have its origin in electromagnetic force. Electromagnetic force is one of the

fundamental forces of nature.

Maxwell put forth four equations that play the same role in classical electromagnetism

as Newton’s equations of motion and gravitation law play in mechanics. He also argued

that light is electromagnetic in nature and its speed can be found by making purely

electric and magnetic measurements. He claimed that the science of optics is intimately

related to that of electricity and magnetism.

The science of electricity and magnetism is the foundation for the modern technological

civilisation. Electric power, telecommunication, radio and television, and a wide variety

of the practical appliances used in daily life are based on the principles of this science.

Although charged particles in motion exert both electric and magnetic forces, in the

frame of reference where all the charges are at rest, the forces are purely electrical. You

know that gravitational force is a long-range force. Its effect is felt even when the distance

between the interacting particles is very large because the force decreases inversely as

the square of the distance between the interacting bodies. We will learn in this chapter

that electric force is also as pervasive and is in fact stronger than the gravitational force

by several orders of magnitude (refer to Chapter 1 of Class XI Physics Textbook).

electroscope [Fig. 1.2(a)]. It consists of a vertical metal rod housed in a

box, with two thin gold leaves attached to its bottom end. When a charged

object touches the metal knob at the top of the rod, charge flows on to

the leaves and they diverge. The degree of divergance is an indicator of

the amount of charge. 3

Physics

Students can make a simple electroscope as

follows [Fig. 1.2(b)]: Take a thin aluminium curtain

rod with ball ends fitted for hanging the curtain. Cut

out a piece of length about 20 cm with the ball at

one end and flatten the cut end. Take a large bottle

that can hold this rod and a cork which will fit in the

opening of the bottle. Make a hole in the cork

sufficient to hold the curtain rod snugly. Slide the

rod through the hole in the cork with the cut end on

the lower side and ball end projecting above the cork.

Fold a small, thin aluminium foil (about 6 cm in

length) in the middle and attach it to the flattened

end of the rod by cellulose tape. This forms the leaves

of your electroscope. Fit the cork in the bottle with

about 5 cm of the ball end projecting above the cork.

A paper scale may be put inside the bottle in advance

to measure the separation of leaves. The separation

is a rough measure of the amount of charge on the

electroscope.

To understand how the electroscope works, use

the white paper strips we used for seeing the

attraction of charged bodies. Fold the strips into half

so that you make a mark of fold. Open the strip and

FIGURE 1.2 Electroscopes: (a) The gold leaf

electroscope, (b) Schematics of a simple iron it lightly with the mountain fold up, as shown

electroscope. in Fig. 1.3. Hold the strip by pinching it at the fold.

You would notice that the two halves move apart.

This shows that the strip has acquired charge on ironing. When you fold

it into half, both the halves have the same charge. Hence they repel each

other. The same effect is seen in the leaf electroscope. On charging the

curtain rod by touching the ball end with an electrified body, charge is

transferred to the curtain rod and the attached aluminium foil. Both the

halves of the foil get similar charge and therefore repel each other. The

divergence in the leaves depends on the amount of charge on them. Let

us first try to understand why material bodies acquire charge.

You know that all matter is made up of atoms and/or molecules.

Although normally the materials are electrically neutral, they do contain

charges; but their charges are exactly balanced. Forces that hold the

molecules together, forces that hold atoms together in a solid, the adhesive

force of glue, forces associated with surface tension, all are basically

electrical in nature, arising from the forces between charged particles.

Thus the electric force is all pervasive and it encompasses almost each

and every field associated with our life. It is therefore essential that we

learn more about such a force.

To electrify a neutral body, we need to add or remove one kind of

FIGURE 1.3 Paper strip charge. When we say that a body is charged, we always refer to this

experiment. excess charge or deficit of charge. In solids, some of the electrons, being

less tightly bound in the atom, are the charges which are transferred

from one body to the other. A body can thus be charged positively by

4 losing some of its electrons. Similarly, a body can be charged negatively

Electric Charges

and Fields

by gaining electrons. When we rub a glass rod with silk, some of the

electrons from the rod are transferred to the silk cloth. Thus the rod gets

positively charged and the silk gets negatively charged. No new charge is

created in the process of rubbing. Also the number of electrons, that are

transferred, is a very small fraction of the total number of electrons in the

material body. Also only the less tightly bound electrons in a material

body can be transferred from it to another by rubbing. Therefore, when

a body is rubbed with another, the bodies get charged and that is why

we have to stick to certain pairs of materials to notice charging on rubbing

the bodies.

A metal rod held in hand and rubbed with wool will not show any sign of

being charged. However, if a metal rod with a wooden or plastic handle is

rubbed without touching its metal part, it shows signs of charging.

Suppose we connect one end of a copper wire to a neutral pith ball and

the other end to a negatively charged plastic rod. We will find that the

pith ball acquires a negative charge. If a similar experiment is repeated

with a nylon thread or a rubber band, no transfer of charge will take

place from the plastic rod to the pith ball. Why does the transfer of charge

not take place from the rod to the ball?

Some substances readily allow passage of electricity through them,

others do not. Those which allow electricity to pass through them easily

are called conductors. They have electric charges (electrons) that are

comparatively free to move inside the material. Metals, human and animal

bodies and earth are conductors. Most of the non-metals like glass,

porcelain, plastic, nylon, wood offer high resistance to the passage of

electricity through them. They are called insulators. Most substances

fall into one of the two classes stated above*.

When some charge is transferred to a conductor, it readily gets

distributed over the entire surface of the conductor. In contrast, if some

charge is put on an insulator, it stays at the same place. You will learn

why this happens in the next chapter.

This property of the materials tells you why a nylon or plastic comb

gets electrified on combing dry hair or on rubbing, but a metal article

like spoon does not. The charges on metal leak through our body to the

ground as both are conductors of electricity.

When we bring a charged body in contact with the earth, all the

excess charge on the body disappears by causing a momentary current

to pass to the ground through the connecting conductor (such as our

body). This process of sharing the charges with the earth is called

grounding or earthing. Earthing provides a safety measure for electrical

circuits and appliances. A thick metal plate is buried deep into the earth

and thick wires are drawn from this plate; these are used in buildings

for the purpose of earthing near the mains supply. The electric wiring in

our houses has three wires: live, neutral and earth. The first two carry

movement of charges which is intermediate between the conductors and

insulators.

5

Physics

electric current from the power station and the third is earthed by

connecting it to the buried metal plate. Metallic bodies of the electric

appliances such as electric iron, refrigerator, TV are connected to the

earth wire. When any fault occurs or live wire touches the metallic body,

the charge flows to the earth without damaging the appliance and without

causing any injury to the humans; this would have otherwise been

unavoidable since the human body is a conductor of electricity.

When we touch a pith ball with an electrified plastic rod, some of the

negative charges on the rod are transferred to the pith ball and it also

gets charged. Thus the pith ball is charged by contact. It is then repelled

by the plastic rod but is attracted by a glass rod which is oppositely

charged. However, why a electrified rod attracts light objects, is a question

we have still left unanswered. Let us try to understand what could be

happening by performing the following experiment.

(i) Bring two metal spheres, A and B, supported on insulating stands,

in contact as shown in Fig. 1.4(a).

(ii) Bring a positively charged rod near one of the spheres, say A, taking

care that it does not touch the sphere. The free electrons in the spheres

are attracted towards the rod. This leaves an excess of positive charge

on the rear surface of sphere B. Both kinds of charges are bound in

the metal spheres and cannot escape. They, therefore, reside on the

surfaces, as shown in Fig. 1.4(b). The left surface of sphere A, has an

excess of negative charge and the right surface of sphere B, has an

excess of positive charge. However, not all of the electrons in the spheres

have accumulated on the left surface of A. As the negative charge

starts building up at the left surface of A, other electrons are repelled

by these. In a short time, equilibrium is reached under the action of

force of attraction of the rod and the force of repulsion due to the

accumulated charges. Fig. 1.4(b) shows the equilibrium situation.

The process is called induction of charge and happens almost

instantly. The accumulated charges remain on the surface, as shown,

till the glass rod is held near the sphere. If the rod is removed, the

charges are not acted by any outside force and they redistribute to

their original neutral state.

(iii) Separate the spheres by a small distance while the glass rod is still

held near sphere A, as shown in Fig. 1.4(c). The two spheres are found

to be oppositely charged and attract each other.

(iv) Remove the rod. The charges on spheres rearrange themselves as

shown in Fig. 1.4(d). Now, separate the spheres quite apart. The

charges on them get uniformly distributed over them, as shown in

Fig. 1.4(e).

In this process, the metal spheres will each be equal and oppositely

charged. This is charging by induction. The positively charged glass rod

does not lose any of its charge, contrary to the process of charging by

FIGURE 1.4 Charging

by induction.

contact.

When electrified rods are brought near light objects, a similar effect

takes place. The rods induce opposite charges on the near surfaces of

6 the objects and similar charges move to the farther side of the object.

Electric Charges

and Fields

[This happens even when the light object is not a conductor. The

mechanism for how this happens is explained later in Sections 1.10 and

2.10.] The centres of the two types of charges are slightly separated. We

know that opposite charges attract while similar charges repel. However,

the magnitude of force depends on the distance between the charges

and in this case the force of attraction overweighs the force of repulsion.

As a result the particles like bits of paper or pith balls, being light, are

pulled towards the rods.

Example 1.1 How can you charge a metal sphere positively without

touching it?

Solution Figure 1.5(a) shows an uncharged metallic sphere on an

http://www.physicsclassroom.com/mmedia/estatics/estaticTOC.html

Interactive animation on charging a two-sphere system by induction:

insulating metal stand. Bring a negatively charged rod close to the

metallic sphere, as shown in Fig. 1.5(b). As the rod is brought close

to the sphere, the free electrons in the sphere move away due to

repulsion and start piling up at the farther end. The near end becomes

positively charged due to deficit of electrons. This process of charge

distribution stops when the net force on the free electrons inside the

metal is zero. Connect the sphere to the ground by a conducting

wire. The electrons will flow to the ground while the positive charges

at the near end will remain held there due to the attractive force of

the negative charges on the rod, as shown in Fig. 1.5(c). Disconnect

the sphere from the ground. The positive charge continues to be

held at the near end [Fig. 1.5(d)]. Remove the electrified rod. The

positive charge will spread uniformly over the sphere as shown in

Fig. 1.5(e).

FIGURE 1.5

of induction and the rod does not lose any of its charge.

EXAMPLE 1.1

by induction, by bringing a positively charged rod near it. In this

case the electrons will flow from the ground to the sphere when the

sphere is connected to the ground with a wire. Can you explain why?

7

Physics

1.5 BASIC PROPERTIES OF ELECTRIC CHARGE

We have seen that there are two types of charges, namely positive and

negative and their effects tend to cancel each other. Here, we shall now

describe some other properties of the electric charge.

If the sizes of charged bodies are very small as compared to the

distances between them, we treat them as point charges. All the

charge content of the body is assumed to be concentrated at one point

in space.

We have not as yet given a quantitative definition of a charge; we shall

follow it up in the next section. We shall tentatively assume that this can

be done and proceed. If a system contains two point charges q1 and q2,

the total charge of the system is obtained simply by adding algebraically

q1 and q2 , i.e., charges add up like real numbers or they are scalars like

the mass of a body. If a system contains n charges q1, q2, q3, …, qn, then

the total charge of the system is q1 + q2 + q3 + … + qn . Charge has

magnitude but no direction, similar to the mass. However, there is one

difference between mass and charge. Mass of a body is always positive

whereas a charge can be either positive or negative. Proper signs have to

be used while adding the charges in a system. For example, the

total charge of a system containing five charges +1, +2, –3, +4 and –5,

in some arbitrary unit, is (+1) + (+2) + (–3) + (+4) + (–5) = –1 in the

same unit.

We have already hinted to the fact that when bodies are charged by

rubbing, there is transfer of electrons from one body to the other; no new

charges are either created or destroyed. A picture of particles of electric

charge enables us to understand the idea of conservation of charge. When

we rub two bodies, what one body gains in charge the other body loses.

Within an isolated system consisting of many charged bodies, due to

interactions among the bodies, charges may get redistributed but it is

found that the total charge of the isolated system is always conserved.

Conservation of charge has been established experimentally.

It is not possible to create or destroy net charge carried by any isolated

system although the charge carrying particles may be created or destroyed

in a process. Sometimes nature creates charged particles: a neutron turns

into a proton and an electron. The proton and electron thus created have

equal and opposite charges and the total charge is zero before and after

the creation.

Experimentally it is established that all free charges are integral multiples

of a basic unit of charge denoted by e. Thus charge q on a body is always

given by

8 q = ne

Electric Charges

and Fields

where n is any integer, positive or negative. This basic unit of charge is

the charge that an electron or proton carries. By convention, the charge

on an electron is taken to be negative; therefore charge on an electron is

written as –e and that on a proton as +e.

The fact that electric charge is always an integral multiple of e is termed

as quantisation of charge. There are a large number of situations in physics

where certain physical quantities are quantised. The quantisation of charge

was first suggested by the experimental laws of electrolysis discovered by

English experimentalist Faraday. It was experimentally demonstrated by

Millikan in 1912.

In the International System (SI) of Units, a unit of charge is called a

coulomb and is denoted by the symbol C. A coulomb is defined in terms

the unit of the electric current which you are going to learn in a

subsequent chapter. In terms of this definition, one coulomb is the charge

flowing through a wire in 1 s if the current is 1 A (ampere), (see Chapter 2

of Class XI, Physics Textbook , Part I). In this system, the value of the

basic unit of charge is

e = 1.602192 × 10–19 C

Thus, there are about 6 × 1018 electrons in a charge of –1C. In

electrostatics, charges of this large magnitude are seldom encountered

and hence we use smaller units 1 μC (micro coulomb) = 10–6 C or 1 mC

(milli coulomb) = 10–3 C.

If the protons and electrons are the only basic charges in the universe,

all the observable charges have to be integral multiples of e. Thus, if a

body contains n1 electrons and n 2 protons, the total amount of charge

on the body is n 2 × e + n1 × (–e) = (n 2 – n1) e. Since n1 and n2 are integers,

their difference is also an integer. Thus the charge on any body is always

an integral multiple of e and can be increased or decreased also in steps

of e.

The step size e is, however, very small because at the macroscopic

level, we deal with charges of a few μC. At this scale the fact that charge of

a body can increase or decrease in units of e is not visible. The grainy

nature of the charge is lost and it appears to be continuous.

This situation can be compared with the geometrical concepts of points

and lines. A dotted line viewed from a distance appears continuous to

us but is not continuous in reality. As many points very close to

each other normally give an impression of a continuous line, many

small charges taken together appear as a continuous charge

distribution.

At the macroscopic level, one deals with charges that are enormous

compared to the magnitude of charge e. Since e = 1.6 × 10–19 C, a charge

of magnitude, say 1 μC, contains something like 1013 times the electronic

charge. At this scale, the fact that charge can increase or decrease only in

units of e is not very different from saying that charge can take continuous

values. Thus, at the macroscopic level, the quantisation of charge has no

practical consequence and can be ignored. At the microscopic level, where

the charges involved are of the order of a few tens or hundreds of e, i.e., 9

Physics

they can be counted, they appear in discrete lumps and quantisation of

charge cannot be ignored. It is the scale involved that is very important.

every second, how much time is required to get a total charge of 1 C

on the other body?

Solution In one second 109 electrons move out of the body. Therefore

the charge given out in one second is 1.6 × 10–19 × 109 C = 1.6 × 10–10 C.

The time required to accumulate a charge of 1 C can then be estimated

to be 1 C ÷ (1.6 × 10–10 C/s) = 6.25 × 109 s = 6.25 × 109 ÷ (365 × 24 ×

3600) years = 198 years. Thus to collect a charge of one coulomb,

from a body from which 109 electrons move out every second, we will

need approximately 200 years. One coulomb is, therefore, a very large

EXAMPLE 1.2

It is, however, also important to know what is roughly the number of

electrons contained in a piece of one cubic centimetre of a material.

A cubic piece of copper of side 1 cm contains about 2.5 × 10 24

electrons.

cup of water?

Solution Let us assume that the mass of one cup of water is

250 g. The molecular mass of water is 18g. Thus, one mole

(= 6.02 × 1023 molecules) of water is 18 g. Therefore the number of

EXAMPLE 1.3

Each molecule of water contains two hydrogen atoms and one oxygen

atom, i.e., 10 electrons and 10 protons. Hence the total positive and

total negative charge has the same magnitude. It is equal to

(250/18) × 6.02 × 1023 × 10 × 1.6 × 10–19 C = 1.34 × 107 C.

Coulomb’s law is a quantitative statement about the force between two

point charges. When the linear size of charged bodies are much smaller

than the distance separating them, the size may be ignored and the

charged bodies are treated as point charges. Coulomb measured the

force between two point charges and found that it varied inversely as

the square of the distance between the charges and was directly

proportional to the product of the magnitude of the two charges and

acted along the line joining the two charges. Thus, if two point charges

q1, q2 are separated by a distance r in vacuum, the magnitude of the

force (F) between them is given by

q1 q 2

F =k (1.1)

r2

How did Coulomb arrive at this law from his experiments? Coulomb

used a torsion balance* for measuring the force between two charged metallic

* A torsion balance is a sensitive device to measure force. It was also used later

by Cavendish to measure the very feeble gravitational force between two objects,

10 to verify Newton’s Law of Gravitation.

Electric Charges

and Fields

spheres. When the separation between two spheres is much

larger than the radius of each sphere, the charged spheres

may be regarded as point charges. However, the charges

on the spheres were unknown, to begin with. How then

could he discover a relation like Eq. (1.1)? Coulomb

thought of the following simple way: Suppose the charge

on a metallic sphere is q. If the sphere is put in contact

with an identical uncharged sphere, the charge will spread

over the two spheres. By symmetry, the charge on each

sphere will be q/2*. Repeating this process, we can get

charges q/2, q/4, etc. Coulomb varied the distance for a

fixed pair of charges and measured the force for different

separations. He then varied the charges in pairs, keeping

the distance fixed for each pair. Comparing forces for Charles Augustin de

different pairs of charges at different distances, Coulomb Coulomb (1736 – 1806)

arrived at the relation, Eq. (1.1). Coulomb, a French

Coulomb’s law, a simple mathematical statement, physicist, began his career

as a military engineer in

was initially experimentally arrived at in the manner

the West Indies. In 1776, he

described above. While the original experiments returned to Paris and

established it at a macroscopic scale, it has also been retired to a small estate to

established down to subatomic level (r ~ 10–10 m). do his scientific research.

Coulomb discovered his law without knowing the He invented a torsion

explicit magnitude of the charge. In fact, it is the other balance to measure the

way round: Coulomb’s law can now be employed to quantity of a force and used

furnish a definition for a unit of charge. In the relation, it for determination of

Eq. (1.1), k is so far arbitrary. We can choose any positive forces of electric attraction

value of k. The choice of k determines the size of the unit or repulsion between small

9 charged spheres. He thus

of charge. In SI units, the value of k is about 9 × 10 .

arrived in 1785 at the

The unit of charge that results from this choice is called inverse square law relation,

a coulomb which we defined earlier in Section 1.4. now known as Coulomb’s

Putting this value of k in Eq. (1.1), we see that for law. The law had been

q1 = q2 = 1 C, r = 1 m anticipated by Priestley and

F = 9 × 109 N also by Cavendish earlier,

though Cavendish never

That is, 1 C is the charge that when placed at a

published his results.

distance of 1 m from another charge of the same Coulomb also found the

magnitude in vacuum experiences an electrical force of inverse square law of force

repulsion of magnitude 9 × 109 N. One coulomb is between unlike and like

evidently too big a unit to be used. In practice, in magnetic poles.

electrostatics, one uses smaller units like 1 mC or 1 μC.

The constant k in Eq. (1.1) is usually put as

k = 1/4πε0 for later convenience, so that Coulomb’s law is written as

1 q1 q2

F = (1.2)

4 π ε0 r2

ε0 is called the permittivity of free space . The value of ε0 in SI units is

ε 0 = 8.854 × 10–12 C2 N–1m–2

two charges (q/2 each) add up to make a total charge q. 11

Physics

Since force is a vector, it is better to write

Coulomb’s law in the vector notation. Let the

position vectors of charges q1 and q2 be r1 and r2

respectively [see Fig.1.6(a)]. We denote force on

q1 due to q2 by F12 and force on q2 due to q1 by

F21. The two point charges q1 and q2 have been

numbered 1 and 2 for convenience and the vector

leading from 1 to 2 is denoted by r21:

r21 = r2 – r1

In the same way, the vector leading from 2 to

1 is denoted by r12:

r12 = r1 – r2 = – r21

The magnitude of the vectors r21 and r12 is

denoted by r21 and r12 , respectively (r12 = r21). The

direction of a vector is specified by a unit vector

along the vector. To denote the direction from 1

to 2 (or from 2 to 1), we define the unit vectors:

FIGURE 1.6 (a) Geometry and r21 r

(b) Forces between charges. rˆ21 = , rˆ12 = 12 , rˆ21 = rˆ12

r21 r12

Coulomb’s force law between two point charges q1 and q2 located at

r1 and r2 is then expressed as

1 q1 q 2

F21 = rˆ21 (1.3)

4 π εo 2

r21

• Equation (1.3) is valid for any sign of q1 and q2 whether positive or

negative. If q1 and q2 are of the same sign (either both positive or both

negative), F21 is along r̂ 21, which denotes repulsion, as it should be for

like charges. If q1 and q2 are of opposite signs, F21 is along – r̂ 21(= r̂ 12),

which denotes attraction, as expected for unlike charges. Thus, we do

not have to write separate equations for the cases of like and unlike

charges. Equation (1.3) takes care of both cases correctly [Fig. 1.6(b)].

• The force F12 on charge q1 due to charge q2, is obtained from Eq. (1.3),

by simply interchanging 1 and 2, i.e.,

1 q1 q 2

F12 = rˆ12 = −F21

4 π ε0 2

r12

• Coulomb’s law [Eq. (1.3)] gives the force between two charges q1 and

q2 in vacuum. If the charges are placed in matter or the intervening

space has matter, the situation gets complicated due to the presence

of charged constituents of matter. We shall consider electrostatics in

12 matter in the next chapter.

Electric Charges

and Fields

Example 1.4 Coulomb’s law for electrostatic force between two point

charges and Newton’s law for gravitational force between two

stationary point masses, both have inverse-square dependence on

the distance between the charges/masses. (a) Compare the strength

of these forces by determining the ratio of their magnitudes (i) for an

electron and a proton and (ii) for two protons. (b) Estimate the

accelerations of electron and proton due to the electrical force of their

mutual attraction when they are 1 Å (= 10-10 m) apart? (mp = 1.67 ×

10–27 kg, me = 9.11 × 10–31 kg)

Solution

(a) (i) The electric force between an electron and a proton at a distance

http://webphysics.davidson.edu/physlet_resources/bu_semester2/co1_coulomb.html

Interactive animation on Coulomb’s law:

r apart is:

1 e2

Fe = −

4 πε 0 r 2

where the negative sign indicates that the force is attractive. The

corresponding gravitational force (always attractive) is:

m p me

FG = −G

r2

where mp and me are the masses of a proton and an electron

respectively.

Fe e2

= = 2.4 × 1039

FG 4 πε 0Gm pm e

(ii) On similar lines, the ratio of the magnitudes of electric force

to the gravitational force between two protons at a distance r

apart is :

Fe e2

= = 1.3 × 1036

FG 4 πε 0Gm p m p

However, it may be mentioned here that the signs of the two forces

are different. For two protons, the gravitational force is attractive

in nature and the Coulomb force is repulsive . The actual values

of these forces between two protons inside a nucleus (distance

between two protons is ~ 10-15 m inside a nucleus) are Fe ~ 230 N

whereas FG ~ 1.9 × 10–34 N.

The (dimensionless) ratio of the two forces shows that electrical

forces are enormously stronger than the gravitational forces.

(b) The electric force F exerted by a proton on an electron is same in

magnitude to the force exerted by an electron on a proton; however

the masses of an electron and a proton are different. Thus, the

magnitude of force is

1 e2

|F| = = 8.987 × 109 Nm2/C2 × (1.6 ×10–19C)2 / (10–10m)2

4 πε 0 r 2

= 2.3 × 10–8 N

Using Newton’s second law of motion, F = ma, the acceleration

that an electron will undergo is

a = 2.3×10–8 N / 9.11 ×10–31 kg = 2.5 × 1022 m/s2

Comparing this with the value of acceleration due to gravity, we

EXAMPLE 1.4

the motion of electron and it undergoes very large accelerations

under the action of Coulomb force due to a proton.

The value for acceleration of the proton is

2.3 × 10–8 N / 1.67 × 10–27 kg = 1.4 × 1019 m/s2 13

Physics

Example 1.5 A charged metallic sphere A is suspended by a nylon

thread. Another charged metallic sphere B held by an insulating

handle is brought close to A such that the distance between their

centres is 10 cm, as shown in Fig. 1.7(a). The resulting repulsion of A

is noted (for example, by shining a beam of light and measuring the

deflection of its shadow on a screen). Spheres A and B are touched

by uncharged spheres C and D respectively, as shown in Fig. 1.7(b).

C and D are then removed and B is brought closer to A to a

distance of 5.0 cm between their centres, as shown in Fig. 1.7(c).

What is the expected repulsion of A on the basis of Coulomb’s law?

Spheres A and C and spheres B and D have identical sizes. Ignore

the sizes of A and B in comparison to the separation between their

centres.

EXAMPLE 1.5

14 FIGURE 1.7

Electric Charges

and Fields

q′. At a distance r between their centres, the magnitude of the

electrostatic force on each is given by

1 qq ′

F =

4 πε 0 r 2

neglecting the sizes of spheres A and B in comparison to r. When an

identical but uncharged sphere C touches A, the charges redistribute

on A and C and, by symmetry, each sphere carries a charge q/2.

Similarly, after D touches B, the redistributed charge on each is

q′/2. Now, if the separation between A and B is halved, the magnitude

of the electrostatic force on each is

EXAMPLE 1.5

1 (q / 2)(q ′ /2) 1 (qq ′ )

F′ = = =F

4 πε 0 (r /2) 2

4πε 0 r 2

The mutual electric force between two charges is given

by Coulomb’s law. How to calculate the force on a

charge where there are not one but several charges

around? Consider a system of n stationary charges

q1, q2, q3, ..., qn in vacuum. What is the force on q1 due

to q2, q3, ..., qn? Coulomb’s law is not enough to answer

this question. Recall that forces of mechanical origin

add according to the parallelogram law of addition. Is

the same true for forces of electrostatic origin?

Experimentally it is verified that force on any

charge due to a number of other charges is the vector

sum of all the forces on that charge due to the other

charges, taken one at a time. The individual forces

are unaffected due to the presence of other charges.

This is termed as the principle of superposition.

To better understand the concept, consider a

system of three charges q1, q2 and q3, as shown in

Fig. 1.8(a). The force on one charge, say q1, due to two

other charges q2, q3 can therefore be obtained by

performing a vector addition of the forces due to each

one of these charges. Thus, if the force on q1 due to q2

is denoted by F12, F12 is given by Eq. (1.3) even though

other charges are present.

1 q1q 2

Thus, F12 = rˆ12

4πε 0 r12

2

In the same way, the force on q1 due to q3, denoted FIGURE 1.8 A system of (a) three

by F13, is given by charges (b) multiple charges.

1 q1q3

F13 = rˆ13

4 πε 0 r13

2 15

Physics

which again is the Coulomb force on q1 due to q3, even though other

charge q2 is present.

Thus the total force F1 on q1 due to the two charges q2 and q3 is

given as

1 q1q 2 1 q1q 3

F1 = F12 + F13 = rˆ12 + rˆ13 (1.4)

4πε 0 r12

2

4 πε 0 r13

2

charges more than three, as shown in Fig. 1.8(b).

The principle of superposition says that in a system of charges q1,

q2, ..., qn, the force on q1 due to q2 is the same as given by Coulomb’s law,

i.e., it is unaffected by the presence of the other charges q3, q4, ..., qn. The

total force F1 on the charge q1, due to all other charges, is then given by

the vector sum of the forces F12, F13, ..., F1n:

i.e.,

F1 = F12 + F13 + ...+ F1n = ⎢ 2 rˆ12 + 2 rˆ13 + ... + 2 rˆ1n ⎥

4 πε 0 ⎣ r12 r13 r1n ⎦

q1 n qi

= ∑ rˆ1i

4 πε 0 i = 2 r12i

(1.5)

The vector sum is obtained as usual by the parallelogram law of

addition of vectors. All of electrostatics is basically a consequence of

Coulomb’s law and the superposition principle.

Example 1.6 Consider three charges q1, q2, q3 each equal to q at the

vertices of an equilateral triangle of side l. What is the force on a

charge Q (with the same sign as q) placed at the centroid of the

triangle, as shown in Fig. 1.9?

FIGURE 1.9

EXAMPLE 1.6

we draw a perpendicular AD to the side BC,

AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O

16 from A is (2/3) AD = ( 1/ 3 ) l. By symmatry AO = BO = CO.

Electric Charges

and Fields

Thus,

3 Qq

Force F1 on Q due to charge q at A = along AO

4 πε 0 l2

3 Qq

Force F2 on Q due to charge q at B = 4 πε along BO

0 l2

3 Qq

Force F3 on Q due to charge q at C = 4 πε 2 along CO

0 l

3 Qq

The resultant of forces F 2 and F 3 is 4 πε 2 along OA, by the

0 l

3 Qq

parallelogram law. Therefore, the total force on Q = 4 πε 2 ( rˆ − rˆ )

0 l

EXAMPLE 1.6

= 0, where r̂ is the unit vector along OA.

It is clear also by symmetry that the three forces will sum to zero.

Suppose that the resultant force was non-zero but in some direction.

Consider what would happen if the system was rotated through 60º

about O.

of an equilateral triangle, as shown in Fig. 1.10. What is the force on

each charge?

FIGURE 1.10

and –q at C are F12 along BA and F13 along AC respectively, as shown

in Fig. 1.10. By the parallelogram law, the total force F1 on the charge

q at A is given by

F1 = F r̂1 where r̂1 is a unit vector along BC.

The force of attraction or repulsion for each pair of charges has the

EXAMPLE 1.7

q2

same magnitude F =

4 π ε0 l 2

unit vector along AC. 17

Physics

Similarly the total force on charge –q at C is F3 = 3 F n̂ , where n̂ is

the unit vector along the direction bisecting the ∠BCA.

It is interesting to see that the sum of the forces on the three charges

EXAMPLE 1.7

is zero, i.e.,

F1 + F2 + F3 = 0

The result is not at all surprising. It follows straight from the fact

that Coulomb’s law is consistent with Newton’s third law. The proof

is left to you as an exercise.

Let us consider a point charge Q placed in vacuum, at the origin O. If we

place another point charge q at a point P, where OP = r, then the charge Q

will exert a force on q as per Coulomb’s law. We may ask the question: If

charge q is removed, then what is left in the surrounding? Is there

nothing? If there is nothing at the point P, then how does a force act

when we place the charge q at P. In order to answer such questions, the

early scientists introduced the concept of field. According to this, we say

that the charge Q produces an electric field everywhere in the surrounding.

When another charge q is brought at some point P, the field there acts on

it and produces a force. The electric field produced by the charge Q at a

point r is given as

1 Q 1 Q

E ( r) = rˆ = rˆ (1.6)

4 πε 0 r 2

4 πε 0 r 2

where rˆ = r/r, is a unit vector from the origin to the point r. Thus, Eq.(1.6)

specifies the value of the electric field for each value of the position

vector r. The word “field” signifies how some distributed quantity (which

could be a scalar or a vector) varies with position. The effect of the charge

has been incorporated in the existence of the electric field. We obtain the

force F exerted by a charge Q on a charge q, as

1 Qq

F= rˆ (1.7)

4 πε 0 r 2

Note that the charge q also exerts an equal and opposite force on the

charge Q. The electrostatic force between the charges Q and q can be

looked upon as an interaction between charge q and the electric field of

Q and vice versa. If we denote the position of charge q by the vector r, it

experiences a force F equal to the charge q multiplied by the electric

field E at the location of q. Thus,

F(r) = q E(r) (1.8)

Equation (1.8) defines the SI unit of electric field as N/C*.

Some important remarks may be made here:

(i) From Eq. (1.8), we can infer that if q is unity, the electric field due to

FIGURE 1.11 Electric a charge Q is numerically equal to the force exerted by it. Thus, the

field (a) due to a electric field due to a charge Q at a point in space may be defined

charge Q, (b) due to a as the force that a unit positive charge would experience if placed

charge –Q.

18 * An alternate unit V/m will be introduced in the next chapter.

Electric Charges

and Fields

at that point. The charge Q, which is producing the electric field, is

called a source charge and the charge q, which tests the effect of a

source charge, is called a test charge. Note that the source charge Q

must remain at its original location. However, if a charge q is brought

at any point around Q, Q itself is bound to experience an electrical

force due to q and will tend to move. A way out of this difficulty is to

make q negligibly small. The force F is then negligibly small but the

ratio F/q is finite and defines the electric field:

⎛ F⎞

E = lim ⎜ ⎟ (1.9)

q →0 ⎝ q ⎠

in the presence of q) is to hold Q to its location by unspecified forces!

This may look strange but actually this is what happens in practice.

When we are considering the electric force on a test charge q due to a

charged planar sheet (Section 1.15), the charges on the sheet are held to

their locations by the forces due to the unspecified charged constituents

inside the sheet.

(ii) Note that the electric field E due to Q, though defined operationally

in terms of some test charge q, is independent of q. This is because

F is proportional to q, so the ratio F/q does not depend on q. The

force F on the charge q due to the charge Q depends on the particular

location of charge q which may take any value in the space around

the charge Q. Thus, the electric field E due to Q is also dependent on

the space coordinate r. For different positions of the charge q all over

the space, we get different values of electric field E. The field exists at

every point in three-dimensional space.

(iii) For a positive charge, the electric field will be directed radially

outwards from the charge. On the other hand, if the source charge is

negative, the electric field vector, at each point, points radially inwards.

(iv) Since the magnitude of the force F on charge q due to charge Q

depends only on the distance r of the charge q from charge Q,

the magnitude of the electric field E will also depend only on the

distance r. Thus at equal distances from the charge Q, the magnitude

of its electric field E is same. The magnitude of electric field E due to

a point charge is thus same on a sphere with the point charge at its

centre; in other words, it has a spherical symmetry.

Consider a system of charges q1, q2, ..., qn with position vectors r1,

r2, ..., rn relative to some origin O. Like the electric field at a point in

space due to a single charge, electric field at a point in space due to the

system of charges is defined to be the force experienced by a unit

test charge placed at that point, without disturbing the original

positions of charges q1, q2, ..., qn. We can use Coulomb’s law and the

superposition principle to determine this field at a point P denoted by

position vector r. 19

Physics

Electric field E1 at r due to q1 at r1 is given by

1 q1

E1 = rˆ1P

4 πε 0 r1P2

where r̂1P is a unit vector in the direction from q1 to P,

and r1P is the distance between q1 and P.

In the same manner, electric field E2 at r due to q2 at

r2 is

1 q2

E2 = rˆ2P

4πε 0 r22P

where r̂2P is a unit vector in the direction from q2 to P

FIGURE 1.12 Electric field at a and r 2P is the distance between q 2 and P. Similar

point due to a system of charges is expressions hold good for fields E3, E4, ..., En due to

the vector sum of the electric fields charges q3, q4, ..., qn.

at the point due to individual By the superposition principle, the electric field E at r

charges. due to the system of charges is (as shown in Fig. 1.12)

E(r) = E1 (r) + E2 (r) + … + En(r)

1 q1 1 q2 1 qn

= rˆ + rˆ2P + ... + rˆnP

4 πε 0 r1P

2 1P

4 πε 0 r2P

2

4 πε 0 rn2P

n

1 q

E(r) =

4π ε 0

∑ r 2i rˆi P (1.10)

i =1 i P

E is a vector quantity that varies from one point to another point in space

and is determined from the positions of the source charges.

You may wonder why the notion of electric field has been introduced

here at all. After all, for any system of charges, the measurable quantity

is the force on a charge which can be directly determined using Coulomb’s

law and the superposition principle [Eq. (1.5)]. Why then introduce this

intermediate quantity called the electric field?

For electrostatics, the concept of electric field is convenient, but not

really necessary. Electric field is an elegant way of characterising the

electrical environment of a system of charges. Electric field at a point in

the space around a system of charges tells you the force a unit positive

test charge would experience if placed at that point (without disturbing

the system). Electric field is a characteristic of the system of charges and

is independent of the test charge that you place at a point to determine

the field. The term field in physics generally refers to a quantity that is

defined at every point in space and may vary from point to point. Electric

field is a vector field, since force is a vector quantity.

The true physical significance of the concept of electric field, however,

emerges only when we go beyond electrostatics and deal with time-

dependent electromagnetic phenomena. Suppose we consider the force

between two distant charges q1, q2 in accelerated motion. Now the greatest

speed with which a signal or information can go from one point to another

20 is c, the speed of light. Thus, the effect of any motion of q1 on q2 cannot

Electric Charges

and Fields

arise instantaneously. There will be some time delay between the effect

(force on q2) and the cause (motion of q1). It is precisely here that the

notion of electric field (strictly, electromagnetic field) is natural and very

useful. The field picture is this: the accelerated motion of charge q1

produces electromagnetic waves, which then propagate with the speed

c, reach q2 and cause a force on q2. The notion of field elegantly accounts

for the time delay. Thus, even though electric and magnetic fields can be

detected only by their effects (forces) on charges, they are regarded as

physical entities, not merely mathematical constructs. They have an

independent dynamics of their own, i.e., they evolve according to laws

of their own. They can also transport energy. Thus, a source of time-

dependent electromagnetic fields, turned on briefly and switched off, leaves

behind propagating electromagnetic fields transporting energy. The

concept of field was first introduced by Faraday and is now among the

central concepts in physics.

uniform electric field of magnitude 2.0 × 104 N C–1 [Fig. 1.13(a)]. The

direction of the field is reversed keeping its magnitude unchanged

and a proton falls through the same distance [Fig. 1.13(b)]. Compute

the time of fall in each case. Contrast the situation with that of ‘free

fall under gravity’.

FIGURE 1.13

Solution In Fig. 1.13(a) the field is upward, so the negatively charged

electron experiences a downward force of magnitude eE where E is

the magnitude of the electric field. The acceleration of the electron is

ae = eE/me

where me is the mass of the electron.

Starting from rest, the time required by the electron to fall through a

2h 2h m e

distance h is given by t e = =

ae eE

For e = 1.6 × 10–19C, me = 9.11 × 10–31 kg,

E = 2.0 × 104 N C–1, h = 1.5 × 10–2 m,

te = 2.9 × 10–9s

In Fig. 1.13 (b), the field is downward, and the positively charged

proton experiences a downward force of magnitude eE . The

EXAMPLE 1.8

ap = eE/mp

where mp is the mass of the proton; mp = 1.67 × 10–27 kg. The time of

fall for the proton is

21

Physics

2h 2h m p

tp = = = 1.3 × 10 –7 s

ap eE

Thus, the heavier particle (proton) takes a greater time to fall through

the same distance. This is in basic contrast to the situation of ‘free

fall under gravity’ where the time of fall is independent of the mass of

the body. Note that in this example we have ignored the acceleration

due to gravity in calculating the time of fall. To see if this is justified,

let us calculate the acceleration of the proton in the given electric

field:

eE

ap =

mp

=

1.67 × 10 −27 kg

EXAMPLE 1.8

= 1.9 × 1012 m s –2

which is enormous compared to the value of g (9.8 m s –2), the

acceleration due to gravity. The acceleration of the electron is even

greater. Thus, the effect of acceleration due to gravity can be ignored

in this example.

Example 1.9 Two point charges q1 and q2, of magnitude +10–8 C and

–10–8 C, respectively, are placed 0.1 m apart. Calculate the electric

fields at points A, B and C shown in Fig. 1.14.

FIGURE 1.14

Solution The electric field vector E1A at A due to the positive charge

q1 points towards the right and has a magnitude

(9 × 109 Nm 2C-2 ) × (10 −8 C)

E1A = = 3.6 × 104 N C–1

(0.05 m)2

The electric field vector E2A at A due to the negative charge q2 points

EXAMPLE 1.9

towards the right and has the same magnitude. Hence the magnitude

of the total electric field EA at A is

EA = E1A + E2A = 7.2 × 104 N C–1

EA is directed toward the right.

22

Electric Charges

and Fields

The electric field vector E1B at B due to the positive charge q1 points

towards the left and has a magnitude

(9 × 109 Nm2 C –2 ) × (10 −8 C)

E1B = = 3.6 × 104 N C–1

(0.05 m)2

The electric field vector E2B at B due to the negative charge q2 points

towards the right and has a magnitude

(9 × 109 Nm 2 C –2 ) × (10 −8 C)

E 2B = = 4 × 103 N C–1

(0.15 m)2

The magnitude of the total electric field at B is

EB = E1B – E2B = 3.2 × 104 N C–1

EB is directed towards the left.

The magnitude of each electric field vector at point C, due to charge

q1 and q2 is

(9 × 109 Nm 2C –2 ) × (10−8 C)

E1C = E2C = = 9 × 103 N C–1

(0.10 m)2

The directions in which these two vectors point are indicated in

EXAMPLE 1.9

Fig. 1.14. The resultant of these two vectors is

π π

E C = E1 cos + E 2 cos = 9 × 103 N C–1

3 3

EC points towards the right.

We have studied electric field in the last section. It is a vector quantity

and can be represented as we represent vectors. Let us try to represent E

due to a point charge pictorially. Let the point charge be placed at the

origin. Draw vectors pointing along the direction of the electric field with

their lengths proportional to the strength of the field at

each point. Since the magnitude of electric field at a point

decreases inversely as the square of the distance of that

point from the charge, the vector gets shorter as one goes

away from the origin, always pointing radially outward.

Figure 1.15 shows such a picture. In this figure, each

arrow indicates the electric field, i.e., the force acting on a

unit positive charge, placed at the tail of that arrow.

Connect the arrows pointing in one direction and the

resulting figure represents a field line. We thus get many

field lines, all pointing outwards from the point charge.

Have we lost the information about the strength or

magnitude of the field now, because it was contained in

the length of the arrow? No. Now the magnitude of the

field is represented by the density of field lines. E is strong

near the charge, so the density of field lines is more near

the charge and the lines are closer. Away from the charge, FIGURE 1.15 Field of a point charge.

the field gets weaker and the density of field lines is less,

resulting in well-separated lines.

Another person may draw more lines. But the number of lines is not

important. In fact, an infinite number of lines can be drawn in any region. 23

Physics

It is the relative density of lines in different regions which is

important.

We draw the figure on the plane of paper, i.e., in two-

dimensions but we live in three-dimensions. So if one wishes

to estimate the density of field lines, one has to consider the

number of lines per unit cross-sectional area, perpendicular

to the lines. Since the electric field decreases as the square of

the distance from a point charge and the area enclosing the

charge increases as the square of the distance, the number

of field lines crossing the enclosing area remains constant,

whatever may be the distance of the area from the charge.

We started by saying that the field lines carry information

about the direction of electric field at different points in space.

FIGURE 1.16 Dependence of

Having drawn a certain set of field lines, the relative density

electric field strength on the

distance and its relation to the (i.e., closeness) of the field lines at different points indicates

number of field lines. the relative strength of electric field at those points. The field

lines crowd where the field is strong and are spaced apart

where it is weak. Figure 1.16 shows a set of field lines. We

can imagine two equal and small elements of area placed at points R and

S normal to the field lines there. The number of field lines in our picture

cutting the area elements is proportional to the magnitude of field at

these points. The picture shows that the field at R is stronger than at S.

To understand the dependence of the field lines on the area, or rather

the solid angle subtended by an area element, let us try to relate the

area with the solid angle, a generalization of angle to three dimensions.

Recall how a (plane) angle is defined in two-dimensions. Let a small

transverse line element Δl be placed at a distance r from a point O. Then

the angle subtended by Δl at O can be approximated as Δθ = Δl/r.

Likewise, in three-dimensions the solid angle* subtended by a small

perpendicular plane area ΔS, at a distance r, can be written as

ΔΩ = ΔS/r2. We know that in a given solid angle the number of radial

field lines is the same. In Fig. 1.16, for two points P1 and P2 at distances

r1 and r2 from the charge, the element of area subtending the solid angle

ΔΩ is r12 ΔΩ at P1 and an element of area r22 ΔΩ at P2, respectively. The

number of lines (say n) cutting these area elements are the same. The

number of field lines, cutting unit area element is therefore n/( r12 ΔΩ) at

P1 andn/( r22 ΔΩ) at P2 , respectively. Since n and ΔΩ are common, the

strength of the field clearly has a 1/r 2 dependence.

The picture of field lines was invented by Faraday to develop an

intuitive non- mathematical way of visualizing electric fields around

charged configurations. Faraday called them lines of force. This term is

somewhat misleading, especially in case of magnetic fields. The more

appropriate term is field lines (electric or magnetic) that we have

adopted in this book.

Electric field lines are thus a way of pictorially mapping the electric

field around a configuration of charges. An electric field line is, in general,

* Solid angle is a measure of a cone. Consider the intersection of the given cone

with a sphere of radius R. The solid angle ΔΩ of the cone is defined to be equal

24 2

to ΔS/R , where ΔS is the area on the sphere cut out by the cone.

Electric Charges

and Fields

a curve drawn in such a way that the tangent to it at each

point is in the direction of the net field at that point. An

arrow on the curve is obviously necessary to specify the

direction of electric field from the two possible directions

indicated by a tangent to the curve. A field line is a space

curve, i.e., a curve in three dimensions.

Figure 1.17 shows the field lines around some simple

charge configurations. As mentioned earlier, the field lines

are in 3-dimensional space, though the figure shows them

only in a plane. The field lines of a single positive charge

are radially outward while those of a single negative

charge are radially inward. The field lines around a system

of two positive charges (q, q) give a vivid pictorial

description of their mutual repulsion, while those around

the configuration of two equal and opposite charges

(q, –q), a dipole, show clearly the mutual attraction

between the charges. The field lines follow some important

general properties:

(i) Field lines start from positive charges and end at

negative charges. If there is a single charge, they may

start or end at infinity.

(ii) In a charge-free region, electric field lines can be taken

to be continuous curves without any breaks.

(iii) Two field lines can never cross each other. (If they did,

the field at the point of intersection will not have a

unique direction, which is absurd.)

(iv) Electrostatic field lines do not form any closed loops.

This follows from the conservative nature of electric

field (Chapter 2).

Consider flow of a liquid with velocity v, through a small

flat surface dS, in a direction normal to the surface. The

rate of flow of liquid is given by the volume crossing the

area per unit time v dS and represents the flux of liquid

flowing across the plane. If the normal to the surface is

not parallel to the direction of flow of liquid, i.e., to v, but

makes an angle θ with it, the projected area in a plane

perpendicular to v is v dS cos θ. Therefore the flux going

out of the surface dS is v. n̂ dS.

For the case of the electric field, we define an

analogous quantity and call it electric flux.

We should however note that there is no flow of a

physically observable quantity unlike the case of liquid

flow.

In the picture of electric field lines described above, FIGURE 1.17 Field lines due to

we saw that the number of field lines crossing a unit area, some simple charge configurations.

placed normal to the field at a point is a measure of the

strength of electric field at that point. This means that if 25

Physics

we place a small planar element of area ΔS

normal to E at a point, the number of field lines

crossing it is proportional* to E ΔS. Now

suppose we tilt the area element by angle θ.

Clearly, the number of field lines crossing the

area element will be smaller. The projection of

the area element normal to E is ΔS cosθ. Thus,

the number of field lines crossing ΔS is

proportional to E ΔS cosθ. When θ = 90°, field

lines will be parallel to ΔS and will not cross it

at all (Fig. 1.18).

The orientation of area element and not

merely its magnitude is important in many

contexts. For example, in a stream, the amount

of water flowing through a ring will naturally

depend on how you hold the ring. If you hold

it normal to the flow, maximum water will flow

FIGURE 1.18 Dependence of flux on the

inclination θ between E and n̂ . through it than if you hold it with some other

orientation. This shows that an area element

should be treated as a vector. It has a

magnitude and also a direction. How to specify the direction of a planar

area? Clearly, the normal to the plane specifies the orientation of the

plane. Thus the direction of a planar area vector is along its normal.

How to associate a vector to the area of a curved surface? We imagine

dividing the surface into a large number of very small area elements.

Each small area element may be treated as planar and a vector associated

with it, as explained before.

Notice one ambiguity here. The direction of an area element is along

its normal. But a normal can point in two directions. Which direction do

we choose as the direction of the vector associated with the area element?

This problem is resolved by some convention appropriate to the given

context. For the case of a closed surface, this convention is very simple.

The vector associated with every area element of a closed surface is taken

to be in the direction of the outward normal. This is the convention used

in Fig. 1.19. Thus, the area element vector ΔS at a point on a closed

surface equals ΔS n̂ where ΔS is the magnitude of the area element and

n̂ is a unit vector in the direction of outward normal at that point.

We now come to the definition of electric flux. Electric flux Δφ through

an area element ΔS is defined by

Δφ = E.ΔS = E ΔS cosθ (1.11)

which, as seen before, is proportional to the number of field lines cutting

the area element. The angle θ here is the angle between E and ΔS. For a

closed surface, with the convention stated already, θ is the angle between

FIGURE 1.19 E and the outward normal to the area element. Notice we could look at

Convention for the expression E ΔS cosθ in two ways: E (ΔS cosθ ) i.e., E times the

defining normal

n̂ and ΔS. * It will not be proper to say that the number of field lines is equal to EΔS. The

number of field lines is after all, a matter of how many field lines we choose to

draw. What is physically significant is the relative number of field lines crossing

26 a given area at different points.

Electric Charges

and Fields

projection of area normal to E, or E⊥ ΔS, i.e., component of E along the

normal to the area element times the magnitude of the area element. The

unit of electric flux is N C–1 m2.

The basic definition of electric flux given by Eq. (1.11) can be used, in

principle, to calculate the total flux through any given surface. All we

have to do is to divide the surface into small area elements, calculate the

flux at each element and add them up. Thus, the total flux φ through a

surface S is

φ ~ Σ E.ΔS (1.12)

The approximation sign is put because the electric field E is taken to

be constant over the small area element. This is mathematically exact

only when you take the limit ΔS → 0 and the sum in Eq. (1.12) is written

as an integral.

An electric dipole is a pair of equal and opposite point charges q and –q,

separated by a distance 2a. The line connecting the two charges defines

a direction in space. By convention, the direction from –q to q is said to

be the direction of the dipole. The mid-point of locations of –q and q is

called the centre of the dipole.

The total charge of the electric dipole is obviously zero. This does not

mean that the field of the electric dipole is zero. Since the charge q and

–q are separated by some distance, the electric fields due to them, when

added, do not exactly cancel out. However, at distances much larger than

the separation of the two charges forming a dipole (r >> 2a), the fields

due to q and –q nearly cancel out. The electric field due to a dipole

therefore falls off, at large distance, faster than like 1/r 2 (the dependence

on r of the field due to a single charge q). These qualitative ideas are

borne out by the explicit calculation as follows:

The electric field of the pair of charges (–q and q) at any point in space

can be found out from Coulomb’s law and the superposition principle.

The results are simple for the following two cases: (i) when the point is on

the dipole axis, and (ii) when it is in the equatorial plane of the dipole,

i.e., on a plane perpendicular to the dipole axis through its centre. The

electric field at any general point P is obtained by adding the electric

fields E–q due to the charge –q and E+q due to the charge q, by the

parallelogram law of vectors.

(i) For points on the axis

Let the point P be at distance r from the centre of the dipole on the side of

the charge q, as shown in Fig. 1.20(a). Then

q

E −q = − ˆ

p [1.13(a)]

4 πε 0 (r + a )2

where p̂ is the unit vector along the dipole axis (from –q to q). Also

q

E +q = ˆ

p [1.13(b)] 27

4 π ε 0 (r − a )2

Physics

The total field at P is

q ⎡ 1 1 ⎤

E = E +q + E − q = ⎢ − ⎥pˆ

4 π ε 0 ⎣ (r − a )2

(r + a )2 ⎦

q 4a r

= ˆ

p (1.14)

4 π ε o ( r 2 − a 2 )2

For r >> a

4qa

E= ˆ

p (r >> a) (1.15)

4 π ε 0r 3

The magnitudes of the electric fields due to the two

charges +q and –q are given by

q 1

E +q = [1.16(a)]

4 πε 0 r 2 + a 2

q 1

E –q = [1.16(b)]

4 πε 0 r + a 2

2

at (a) a point on the axis, (b) a point The directions of E +q and E –q are as shown in

on the equatorial plane of the dipole. Fig. 1.20(b). Clearly, the components normal to the dipole

p is the dipole moment vector of

axis cancel away. The components along the dipole axis

magnitude p = q × 2a and

directed from –q to q.

add up. The total electric field is opposite to p̂ . We have

E = – (E +q + E –q ) cosθ p̂

2qa

=− ˆ

p (1.17)

4 π ε o (r 2 + a 2 )3 / 2

At large distances (r >> a), this reduces to

2qa

E=− ˆ

p (r >> a ) (1.18)

4 π εo r 3

From Eqs. (1.15) and (1.18), it is clear that the dipole field at large

distances does not involve q and a separately; it depends on the product

qa. This suggests the definition of dipole moment. The dipole moment

vector p of an electric dipole is defined by

p = q × 2a p̂ (1.19)

that is, it is a vector whose magnitude is charge q times the separation

2a (between the pair of charges q, –q) and the direction is along the line

from –q to q. In terms of p, the electric field of a dipole at large distances

takes simple forms:

At a point on the dipole axis

2p

E= (r >> a) (1.20)

4 πε o r 3

At a point on the equatorial plane

p

E=− (r >> a) (1.21)

28 4 πε o r 3

Electric Charges

and Fields

Notice the important point that the dipole field at large distances

falls off not as 1/r 2 but as1/r 3. Further, the magnitude and the direction

of the dipole field depends not only on the distance r but also on the

angle between the position vector r and the dipole moment p.

We can think of the limit when the dipole size 2a approaches zero,

the charge q approaches infinity in such a way that the product

p = q × 2a is finite. Such a dipole is referred to as a point dipole. For a

point dipole, Eqs. (1.20) and (1.21) are exact, true for any r.

In most molecules, the centres of positive charges and of negative charges*

lie at the same place. Therefore, their dipole moment is zero. CO2 and

CH4 are of this type of molecules. However, they develop a dipole moment

when an electric field is applied. But in some molecules, the centres of

negative charges and of positive charges do not coincide. Therefore they

have a permanent electric dipole moment, even in the absence of an electric

field. Such molecules are called polar molecules. Water molecules, H2O,

is an example of this type. Various materials give rise to interesting

properties and important applications in the presence or absence of

electric field.

Determine the electric field at (a) a point P on the axis of the dipole

15 cm away from its centre O on the side of the positive charge, as

shown in Fig. 1.21(a), and (b) a point Q, 15 cm away from O on a line

passing through O and normal to the axis of the dipole, as shown in

Fig. 1.21(b).

EXAMPLE 1.10

FIGURE 1.21

* Centre of a collection of positive point charges is defined much the same way

∑ qi ri

as the centre of mass: rcm = i .

∑ qi 29

i

Physics

Solution (a) Field at P due to charge +10 μC

10 −5 C 1

= ×

4 π (8.854 × 10 −12 2

C N −1

m ) −2

(15 − 0.25)2 × 10 −4 m 2

= 4.13 × 106 N C–1 along BP

Field at P due to charge –10 μC

10 –5 C 1

= ×

4 π (8.854 × 10 −12 C2 N −1 m −2 ) (15 + 0.25)2 × 10 −4 m 2

= 3.86 × 106 N C–1 along PA

The resultant electric field at P due to the two charges at A and B is

= 2.7 × 105 N C–1 along BP.

In this example, the ratio OP/OB is quite large (= 60). Thus, we can

expect to get approximately the same result as above by directly using

the formula for electric field at a far-away point on the axis of a dipole.

For a dipole consisting of charges ± q, 2a distance apart, the electric

field at a distance r from the centre on the axis of the dipole has a

magnitude

2p

E = (r/a >> 1)

4πε 0r 3

where p = 2a q is the magnitude of the dipole moment.

The direction of electric field on the dipole axis is always along the

direction of the dipole moment vector (i.e., from –q to q). Here,

p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m

Therefore,

2 × 5 × 10−8 C m 1

E = × = 2.6 × 105 N C–1

4 π (8.854 × 10 −12 2

C N m ) −1 −2

(15)3 × 10 −6 m 3

along the dipole moment direction AB, which is close to the result

obtained earlier.

(b) Field at Q due to charge + 10 μC at B

10−5 C 1

= 4 π (8.854 × 10 −12 C 2 N −1 m −2 ) × [152 + (0.25)2 ] × 10 −4 m 2

10 −5 C 1

= ×

−12

4 π (8.854 × 10 C N m ) 2

[15 −1 −2 2

+ (0.25)

2

] × 10 −4 m 2

= 3.99 × 106 N C–1 along QA.

cancel along the direction OQ but add up along the direction parallel

to BA. Therefore, the resultant electric field at Q due to the two

charges at A and B is

0.25

EXAMPLE 1.10

15 2

+ (0.25)2

= 1.33 × 105 N C–1 along BA.

As in (a), we can expect to get approximately the same result by

directly using the formula for dipole field at a point on the normal to

30 the axis of the dipole:

Electric Charges

and Fields

p

E = (r/a >> 1)

4 π ε0 r 3

5 × 10−8 C m 1

= ×

EXAMPLE 1.10

4 π (8.854 × 10−12 C2 N –1 m –2 ) (15)3 × 10 −6 m 3

= 1.33 × 105 N C–1.

The direction of electric field in this case is opposite to the direction

of the dipole moment vector. Again the result agrees with that obtained

before.

Consider a permanent dipole of dipole moment p in a uniform

external field E, as shown in Fig. 1.22. (By permanent dipole, we

mean that p exists irrespective of E; it has not been induced by E.)

There is a force qE on q and a force –qE on –q. The net force on

the dipole is zero, since E is uniform. However, the charges are

separated, so the forces act at different points, resulting in a torque

on the dipole. When the net force is zero, the torque (couple) is

independent of the origin. Its magnitude equals the magnitude of FIGURE 1.22 Dipole in a

each force multiplied by the arm of the couple (perpendicular uniform electric field.

distance between the two antiparallel forces).

Magnitude of torque = q E × 2 a sinθ

= 2 q a E sinθ

Its direction is normal to the plane of the paper, coming out of it.

The magnitude of p × E is also p E sinθ and its direction

is normal to the paper, coming out of it. Thus,

τ =p×E (1.22)

This torque will tend to align the dipole with the field

E. When p is aligned with E, the torque is zero.

What happens if the field is not uniform? In that case,

the net force will evidently be non-zero. In addition there

will, in general, be a torque on the system as before. The

general case is involved, so let us consider the simpler

situations when p is parallel to E or antiparallel to E. In

either case, the net torque is zero, but there is a net force

on the dipole if E is not uniform.

Figure 1.23 is self-explanatory. It is easily seen that

when p is parallel to E, the dipole has a net force in the

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

the net force on the dipole is in the direction of decreasing

field. In general, the force depends on the orientation of p

with respect to E.

This brings us to a common observation in frictional

electricity. A comb run through dry hair attracts pieces of FIGURE 1.23 Electric force on a

paper. The comb, as we know, acquires charge through dipole: (a) E parallel to p, (b) E

friction. But the paper is not charged. What then explains antiparallel to p.

the attractive force? Taking the clue from the preceding 31

Physics

discussion, the charged comb ‘polarizes’ the piece of paper, i.e., induces

a net dipole moment in the direction of field. Further, the electric field

due to the comb is not uniform. In this situation, it is easily seen that the

paper should move in the direction of the comb!

We have so far dealt with charge configurations involving discrete charges

q1, q2, ..., qn. One reason why we restricted to discrete charges is that the

mathematical treatment is simpler and does not involve calculus. For

many purposes, however, it is impractical to work in terms of discrete

charges and we need to work with continuous charge distributions. For

example, on the surface of a charged conductor, it is impractical to specify

the charge distribution in terms of the locations of the microscopic charged

constituents. It is more feasible to consider an area element ΔS (Fig. 1.24)

on the surface of the conductor (which is very small on the macroscopic

scale but big enough to include a very large number of electrons) and

specify the charge ΔQ on that element. We then define a surface charge

density σ at the area element by

ΔQ

σ= (1.23)

ΔS

We can do this at different points on the conductor and thus arrive at

a continuous function σ, called the surface charge density. The surface

charge density σ so defined ignores the quantisation of charge and the

discontinuity in charge distribution at the microscopic level*. σ represents

macroscopic surface charge density, which in a sense, is a smoothed out

average of the microscopic charge density over an area element ΔS which,

as said before, is large microscopically but small macroscopically. The

units for σ are C/m2.

Similar considerations apply for a line charge distribution and a volume

FIGURE 1.24 charge distribution. The linear charge density λ of a wire is defined by

Definition of linear,

ΔQ

surface and volume λ = (1.24)

charge densities. Δl

In each case, the where Δl is a small line element of wire on the macroscopic scale that,

element (Δl, ΔS, ΔV ) however, includes a large number of microscopic charged constituents,

chosen is small on and ΔQ is the charge contained in that line element. The units for λ are

the macroscopic C/m. The volume charge density (sometimes simply called charge density)

scale but contains is defined in a similar manner:

a very large number

of microscopic ΔQ

ρ= (1.25)

constituents. ΔV

where ΔQ is the charge included in the macroscopically small volume

element ΔV that includes a large number of microscopic charged

constituents. The units for ρ are C/m3.

The notion of continuous charge distribution is similar to that we

adopt for continuous mass distribution in mechanics. When we refer to

32 discrete charges separated by intervening space where there is no charge.

Electric Charges

and Fields

the density of a liquid, we are referring to its macroscopic density. We

regard it as a continuous fluid and ignore its discrete molecular

constitution.

The field due to a continuous charge distribution can be obtained in

much the same way as for a system of discrete charges, Eq. (1.10). Suppose

a continuous charge distribution in space has a charge density ρ. Choose

any convenient origin O and let the position vector of any point in the

charge distribution be r. The charge density ρ may vary from point to

point, i.e., it is a function of r. Divide the charge distribution into small

volume elements of size ΔV. The charge in a volume element ΔV is ρΔV.

Now, consider any general point P (inside or outside the distribution)

with position vector R (Fig. 1.24). Electric field due to the charge ρΔV is

given by Coulomb’s law:

1 ρ ΔV

ΔE = rˆ' (1.26)

4 πε 0 r' 2

where r′ is the distance between the charge element and P, and r̂ ′ is a

unit vector in the direction from the charge element to P. By the

superposition principle, the total electric field due to the charge

distribution is obtained by summing over electric fields due to different

volume elements:

1 ρ ΔV

E≅ Σ rˆ' (1.27)

4 πε 0 all ΔV r' 2

Note that ρ, r′, rˆ ′ all can vary from point to point. In a strict

mathematical method, we should let ΔV→0 and the sum then becomes

an integral; but we omit that discussion here, for simplicity. In short,

using Coulomb’s law and the superposition principle, electric field can

be determined for any charge distribution, discrete or continuous or part

discrete and part continuous.

As a simple application of the notion of electric flux, let us consider the

total flux through a sphere of radius r, which encloses a point charge q

at its centre. Divide the sphere into small area elements, as shown in

Fig. 1.25.

The flux through an area element ΔS is

q

Δφ = E i Δ S = rˆ i ΔS (1.28)

4 πε 0 r 2

where we have used Coulomb’s law for the electric field due to a single

charge q. The unit vector r̂ is along the radius vector from the centre to

the area element. Now, since the normal to a sphere at every point is

along the radius vector at that point, the area element ΔS and r̂ have

the same direction. Therefore,

q FIGURE 1.25 Flux

Δφ = ΔS (1.29) through a sphere

4 πε 0 r 2 enclosing a point

since the magnitude of a unit vector is 1. charge q at its centre.

The total flux through the sphere is obtained by adding up flux

through all the different area elements: 33

Physics

q

φ= Σ ΔS

all ΔS 4 π ε0 r 2

Since each area element of the sphere is at the same

distance r from the charge,

FIGURE 1.26 Calculation of the q q

φ= Σ ΔS = S

flux of uniform electric field 4 πε o r 2 all ΔS

4 πε 0 r 2

through the surface of a cylinder.

Now S, the total area of the sphere, equals 4πr 2. Thus,

q q

φ= × 4 πr 2 = (1.30)

4 πε 0 r 2

ε0

Equation (1.30) is a simple illustration of a general result of

electrostatics called Gauss’s law.

We state Gauss’s law without proof:

Electric flux through a closed surface S

= q/ε0 (1.31)

q = total charge enclosed by S.

The law implies that the total electric flux through a closed surface is

zero if no charge is enclosed by the surface. We can see that explicitly in

the simple situation of Fig. 1.26.

Here the electric field is uniform and we are considering a closed

cylindrical surface, with its axis parallel to the uniform field E. The total

flux φ through the surface is φ = φ1 + φ2 + φ3, where φ1 and φ2 represent

the flux through the surfaces 1 and 2 (of circular cross-section) of the

cylinder and φ3 is the flux through the curved cylindrical part of the

closed surface. Now the normal to the surface 3 at every point is

perpendicular to E, so by definition of flux, φ3 = 0. Further, the outward

normal to 2 is along E while the outward normal to 1 is opposite to E.

Therefore,

φ1 = –E S1, φ2 = +E S2

S1 = S2 = S

where S is the area of circular cross-section. Thus, the total flux is zero,

as expected by Gauss’s law. Thus, whenever you find that the net electric

flux through a closed surface is zero, we conclude that the total charge

contained in the closed surface is zero.

The great significance of Gauss’s law Eq. (1.31), is that it is true in

general, and not only for the simple cases we have considered above. Let

us note some important points regarding this law:

(i) Gauss’s law is true for any closed surface, no matter what its shape

or size.

(ii) The term q on the right side of Gauss’s law, Eq. (1.31), includes the

sum of all charges enclosed by the surface. The charges may be located

anywhere inside the surface.

(iii) In the situation when the surface is so chosen that there are some

charges inside and some outside, the electric field [whose flux appears

on the left side of Eq. (1.31)] is due to all the charges, both inside and

outside S. The term q on the right side of Gauss’s law, however,

34 represents only the total charge inside S.

Electric Charges

and Fields

(iv) The surface that we choose for the application of Gauss’s law is called

the Gaussian surface. You may choose any Gaussian surface and

apply Gauss’s law. However, take care not to let the Gaussian surface

pass through any discrete charge. This is because electric field due

to a system of discrete charges is not well defined at the location of

any charge. (As you go close to the charge, the field grows without

any bound.) However, the Gaussian surface can pass through a

continuous charge distribution.

(v) Gauss’s law is often useful towards a much easier calculation of the

electrostatic field when the system has some symmetry. This is

facilitated by the choice of a suitable Gaussian surface.

(vi) Finally, Gauss’s law is based on the inverse square dependence on

distance contained in the Coulomb’s law. Any violation of Gauss’s

law will indicate departure from the inverse square law.

Ex = αx1/2, Ey = Ez = 0, in which α = 800 N/C m1/2. Calculate (a) the

flux through the cube, and (b) the charge within the cube. Assume

that a = 0.1 m.

FIGURE 1.27

Solution

(a) Since the electric field has only an x component, for faces

perpendicular to x direction, the angle between E and ΔS is

± π/2. Therefore, the flux φ = E.ΔS is separately zero for each face

of the cube except the two shaded ones. Now the magnitude of

the electric field at the left face is

EL = αx1/2 = αa1/2

(x = a at the left face).

The magnitude of electric field at the right face is

ER = α x1/2 = α (2a)1/2

(x = 2a at the right face).

The corresponding fluxes are

EXAMPLE 1.11

φ = E .ΔS = ΔS E L ⋅ n

L L

ˆ L =E ΔS cosθ = –E ΔS, since θ = 180°

L L

= –ELa2

φR= ER.ΔS = ER ΔS cosθ = ER ΔS, since θ = 0°

= ERa2

Net flux through the cube 35

Physics

= φR + φL = ERa2 – ELa2 = a2 (ER – EL) = αa2 [(2a)1/2 – a1/2]

= αa5/2 ( 2 –1)

= 800 (0.1)5/2 ( )

2 –1

EXAMPLE 1.11

2 –1

= 1.05 N m C

(b) We can use Gauss’s law to find the total charge q inside the cube.

We have φ = q/ε0 or q = φε0. Therefore,

q = 1.05 × 8.854 × 10–12 C = 9.27 × 10–12 C.

direction for positive x, and uniform with the same magnitude but in

the negative x direction for negative x. It is given that E = 200 î N/C

for x > 0 and E = –200 î N/C for x < 0. A right circular cylinder of

length 20 cm and radius 5 cm has its centre at the origin and its axis

along the x-axis so that one face is at x = +10 cm and the other is at

x = –10 cm (Fig. 1.28). (a) What is the net outward flux through each

flat face? (b) What is the flux through the side of the cylinder?

(c) What is the net outward flux through the cylinder? (d) What is the

net charge inside the cylinder?

Solution

(a) We can see from the figure that on the left face E and ΔS are

parallel. Therefore, the outward flux is

φ = E.ΔS = – 200 ˆii ΔS

L

= + 200 × π (0.05)2 = + 1.57 N m2 C–1

On the right face, E and ΔS are parallel and therefore

φR = E.ΔS = + 1.57 N m2 C–1.

(b) For any point on the side of the cylinder E is perpendicular to

ΔS and hence E.ΔS = 0. Therefore, the flux out of the side of the

cylinder is zero.

(c) Net outward flux through the cylinder

φ = 1.57 + 1.57 + 0 = 3.14 N m2 C–1

FIGURE 1.28

EXAMPLE 1.12

(d) The net charge within the cylinder can be found by using Gauss’s

law which gives

q = ε0φ

= 3.14 × 8.854 × 10–12 C

= 2.78 × 10–11 C

36

Electric Charges

and Fields

The electric field due to a general charge distribution is, as seen above,

given by Eq. (1.27). In practice, except for some special cases, the

summation (or integration) involved in this equation cannot be carried

out to give electric field at every point in

space. For some symmetric charge

configurations, however, it is possible to

obtain the electric field in a simple way using

the Gauss’s law. This is best understood by

some examples.

long straight uniformly

charged wire

Consider an infinitely long thin straight wire

with uniform linear charge density λ. The wire

is obviously an axis of symmetry. Suppose we

take the radial vector from O to P and rotate it

around the wire. The points P, P′, P′′ so

obtained are completely equivalent with

respect to the charged wire. This implies that

the electric field must have the same magnitude

at these points. The direction of electric field at

every point must be radial (outward if λ > 0,

inward if λ < 0). This is clear from Fig. 1.29.

Consider a pair of line elements P1 and P2

of the wire, as shown. The electric fields

produced by the two elements of the pair when

summed give a resultant electric field which

is radial (the components normal to the radial

vector cancel). This is true for any such pair

and hence the total field at any point P is

radial. Finally, since the wire is infinite,

electric field does not depend on the position

of P along the length of the wire. In short, the

electric field is everywhere radial in the plane

cutting the wire normally, and its magnitude

depends only on the radial distance r.

To calculate the field, imagine a cylindrical

Gaussian surface, as shown in the Fig. 1.29(b).

Since the field is everywhere radial, flux

through the two ends of the cylindrical

Gaussian surface is zero. At the cylindrical

FIGURE 1.29 (a) Electric field due to an

part of the surface, E is normal to the surface infinitely long thin straight wire is radial,

at every point, and its magnitude is constant, (b) The Gaussian surface for a long thin

since it depends only on r. The surface area wire of uniform linear charge density.

of the curved part is 2πrl, where l is the length

of the cylinder. 37

Physics

Flux through the Gaussian surface

= flux through the curved cylindrical part of the surface

= E × 2πrl

The surface includes charge equal to λ l. Gauss’s law then gives

E × 2πrl = λl/ε0

λ

i.e., E =

2πε 0r

Vectorially, E at any point is given by

λ

E= ˆ

n (1.32)

2πε0r

where n̂ is the radial unit vector in the plane normal to the wire passing

through the point. E is directed outward if λ is positive and inward if λ is

negative.

Note that when we write a vector A as a scalar multiplied by a unit

vector, i.e., as A = A â , the scalar A is an algebraic number. It can be

negative or positive. The direction of A will be the same as that of the unit

vector â if A > 0 and opposite to â if A < 0. When we want to restrict to

non-negative values, we use the symbol A and call it the modulus of A .

Thus, A ≥ 0 .

Also note that though only the charge enclosed by the surface (λl )

was included above, the electric field E is due to the charge on the entire

wire. Further, the assumption that the wire is infinitely long is crucial.

Without this assumption, we cannot take E to be normal to the curved

part of the cylindrical Gaussian surface. However, Eq. (1.32) is

approximately true for electric field around the central portions of a long

wire, where the end effects may be ignored.

Let σ be the uniform surface charge density of an infinite plane sheet

(Fig. 1.30). We take the x-axis normal to the given plane. By symmetry,

the electric field will not depend on y and z coordinates and its direction

at every point must be parallel to the x-direction.

We can take the Gaussian surface to be a

rectangular parallelepiped of cross sectional area

A, as shown. (A cylindrical surface will also do.) As

seen from the figure, only the two faces 1 and 2 will

contribute to the flux; electric field lines are parallel

to the other faces and they, therefore, do not

contribute to the total flux.

The unit vector normal to surface 1 is in –x

direction while the unit vector normal to surface 2

is in the +x direction. Therefore, flux E.ΔS through

both the surfaces are equal and add up. Therefore

FIGURE 1.30 Gaussian surface for a the net flux through the Gaussian surface is 2 EA.

uniformly charged infinite plane sheet.

The charge enclosed by the closed surface is σA.

38 Therefore by Gauss’s law,

Electric Charges

and Fields

2 EA = σA/ε0

or, E = σ/2ε0

Vectorically,

σ

E= ˆ

n (1.33)

2ε 0

where n̂ is a unit vector normal to the plane and going away from it.

E is directed away from the plate if σ is positive and toward the plate

if σ is negative. Note that the above application of the Gauss’ law has

brought out an additional fact: E is independent of x also.

For a finite large planar sheet, Eq. (1.33) is approximately true in the

middle regions of the planar sheet, away from the ends.

Let σ be the uniform surface charge density of a thin spherical shell of

radius R (Fig. 1.31). The situation has obvious spherical symmetry. The

field at any point P, outside or inside, can depend only on r (the radial

distance from the centre of the shell to the point) and must be radial (i.e.,

along the radius vector).

(i) Field outside the shell: Consider a point P outside the

shell with radius vector r. To calculate E at P, we take the

Gaussian surface to be a sphere of radius r and with centre

O, passing through P. All points on this sphere are equivalent

relative to the given charged configuration. (That is what we

mean by spherical symmetry.) The electric field at each point

of the Gaussian surface, therefore, has the same magnitude

E and is along the radius vector at each point. Thus, E and

ΔS at every point are parallel and the flux through each

element is E ΔS. Summing over all ΔS, the flux through the

Gaussian surface is E × 4 π r 2. The charge enclosed is

σ × 4 π R 2. By Gauss’s law

σ

E × 4 π r2 = 4 π R2

ε0

σ R2 q

Or, E = =

ε 0 r 2 4 π ε0 r 2

where q = 4 π R2 σ is the total charge on the spherical shell.

Vectorially,

q FIGURE 1.31 Gaussian

E= rˆ (1.34)

4 πε 0 r 2 surfaces for a point with

(a) r > R, (b) r < R.

The electric field is directed outward if q > 0 and inward if

q < 0. This, however, is exactly the field produced by a charge

q placed at the centre O. Thus for points outside the shell, the field due

to a uniformly charged shell is as if the entire charge of the shell is

concentrated at its centre.

(ii) Field inside the shell: In Fig. 1.31(b), the point P is inside the

shell. The Gaussian surface is again a sphere through P centred at O. 39

Physics

The flux through the Gaussian surface, calculated as before, is

E × 4 π r 2. However, in this case, the Gaussian surface encloses no

charge. Gauss’s law then gives

E × 4 π r2 = 0

i.e., E = 0 (r < R ) (1.35)

that is, the field due to a uniformly charged thin shell is zero at all points

inside the shell*. This important result is a direct consequence of Gauss’s

law which follows from Coulomb’s law. The experimental verification of

this result confirms the 1/r2 dependence in Coulomb’s law.

positively charged point nucleus of charge Ze, surrounded by a

uniform density of negative charge up to a radius R. The atom as a

whole is neutral. For this model, what is the electric field at a distance

r from the nucleus?

FIGURE 1.32

shown in Fig. 1.32. The total negative charge in the uniform spherical

charge distribution of radius R must be –Z e, since the atom (nucleus

of charge Z e + negative charge) is neutral. This immediately gives us

the negative charge density ρ, since we must have

4 πR3

ρ = 0 – Ze

3

3 Ze

or ρ = −

4 π R3

To find the electric field E(r) at a point P which is a distance r away

from the nucleus, we use Gauss’s law. Because of the spherical

symmetry of the charge distribution, the magnitude of the electric

field E(r) depends only on the radial distance, no matter what the

direction of r. Its direction is along (or opposite to) the radius vector r

from the origin to the point P. The obvious Gaussian surface is a

EXAMPLE 1.13

namely, r < R and r > R.

(i) r < R : The electric flux φ enclosed by the spherical surface is

φ = E (r ) × 4 π r 2

where E (r ) is the magnitude of the electric field at r. This is because

* Compare this with a uniform mass shell discussed in Section 8.5 of Class XI

40 Textbook of Physics.

Electric Charges

and Fields

the field at any point on the spherical Gaussian surface has the

same direction as the normal to the surface there, and has the same

magnitude at all points on the surface.

The charge q enclosed by the Gaussian surface is the positive nuclear

charge and the negative charge within the sphere of radius r,

4 πr3

i.e., q = Z e + ρ

3

Substituting for the charge density ρ obtained earlier, we have

r3

q = Ze−Ze

R3

Gauss’s law then gives,

Z e ⎛1 r ⎞

E (r ) = ⎜ − ⎟; r < R

4 π ε0 ⎝ r 2 R 3 ⎠

The electric field is directed radially outward.

(ii) r > R: In this case, the total charge enclosed by the Gaussian

EXAMPLE 1.13

spherical surface is zero since the atom is neutral. Thus, from Gauss’s

law,

E (r ) × 4 π r 2 = 0 or E (r ) = 0; r > R

At r = R, both cases give the same result: E = 0.

ON SYMMETRY OPERATIONS

symmetries helps one arrive at results much faster than otherwise by a straightforward

calculation. Consider, for example an infinite uniform sheet of charge (surface charge

density σ) along the y-z plane. This system is unchanged if (a) translated parallel to the

y-z plane in any direction, (b) rotated about the x-axis through any angle. As the system

is unchanged under such symmetry operation, so must its properties be. In particular,

in this example, the electric field E must be unchanged.

Translation symmetry along the y-axis shows that the electric field must be the same

at a point (0, y1, 0) as at (0, y2, 0). Similarly translational symmetry along the z-axis

shows that the electric field at two point (0, 0, z1) and (0, 0, z2) must be the same. By

using rotation symmetry around the x-axis, we can conclude that E must be

perpendicular to the y-z plane, that is, it must be parallel to the x-direction.

Try to think of a symmetry now which will tell you that the magnitude of the electric

field is a constant, independent of the x-coordinate. It thus turns out that the magnitude

of the electric field due to a uniformly charged infinite conducting sheet is the same at all

points in space. The direction, however, is opposite of each other on either side of the

sheet.

Compare this with the effort needed to arrive at this result by a direct calculation

using Coulomb’s law.

41

Physics

SUMMARY

molecules and bulk matter.

2. From simple experiments on frictional electricity, one can infer that

there are two types of charges in nature; and that like charges repel

and unlike charges attract. By convention, the charge on a glass rod

rubbed with silk is positive; that on a plastic rod rubbed with fur is

then negative.

3. Conductors allow movement of electric charge through them, insulators

do not. In metals, the mobile charges are electrons; in electrolytes

both positive and negative ions are mobile.

4. Electric charge has three basic properties: quantisation, additivity

and conservation.

Quantisation of electric charge means that total charge (q) of a body

is always an integral multiple of a basic quantum of charge (e) i.e.,

q = n e, where n = 0, ±1, ±2, ±3, .... Proton and electron have charges

+e, –e, respectively. For macroscopic charges for which n is a very large

number, quantisation of charge can be ignored.

Additivity of electric charges means that the total charge of a system

is the algebraic sum (i.e., the sum taking into account proper signs)

of all individual charges in the system.

Conservation of electric charges means that the total charge of an

isolated system remains unchanged with time. This means that when

bodies are charged through friction, there is a transfer of electric charge

from one body to another, but no creation or destruction of charge.

5. Coulomb’s Law: The mutual electrostatic force between two point

charges q1 and q2 is proportional to the product q1q2 and inversely

proportional to the square of the distance r 21 separating them.

Mathematically,

k (q1q2 )

F21 = force on q2 due to q1 = 2

rˆ21

r21

1

where r̂21 is a unit vector in the direction from q1 to q2 and k =

4 πε 0

is the constant of proportionality.

In SI units, the unit of charge is coulomb. The experimental value of

the constant ε0 is

ε0 = 8.854 × 10–12 C2 N–1 m–2

The approximate value of k is

k = 9 × 109 N m2 C–2

6. The ratio of electric force and gravitational force between a proton

and an electron is

k e2

≅ 2.4 × 1039

G m em p

7. Superposition Principle: The principle is based on the property that the

forces with which two charges attract or repel each other are not

affected by the presence of a third (or more) additional charge(s). For

an assembly of charges q1, q2, q3, ..., the force on any charge, say q1, is

42

Electric Charges

and Fields

the vector sum of the force on q1 due to q2, the force on q1 due to q3,

and so on. For each pair, the force is given by the Coulomb’s law for

two charges stated earlier.

8. The electric field E at a point due to a charge configuration is the

force on a small positive test charge q placed at the point divided by

the magnitude of the charge. Electric field due to a point charge q has

a magnitude |q|/4πε0r 2; it is radially outwards from q, if q is positive,

and radially inwards if q is negative. Like Coulomb force, electric field

also satisfies superposition principle.

9. An electric field line is a curve drawn in such a way that the tangent

at each point on the curve gives the direction of electric field at that

point. The relative closeness of field lines indicates the relative strength

of electric field at different points; they crowd near each other in regions

of strong electric field and are far apart where the electric field is

weak. In regions of constant electric field, the field lines are uniformly

spaced parallel straight lines.

10. Some of the important properties of field lines are: (i) Field lines are

continuous curves without any breaks. (ii) Two field lines cannot cross

each other. (iii) Electrostatic field lines start at positive charges and

end at negative charges —they cannot form closed loops.

11. An electric dipole is a pair of equal and opposite charges q and –q

separated by some distance 2a. Its dipole moment vector p has

magnitude 2qa and is in the direction of the dipole axis from –q to q.

12. Field of an electric dipole in its equatorial plane (i.e., the plane

perpendicular to its axis and passing through its centre) at a distance

r from the centre:

−p 1

E=

4 πε o (a 2 + r 2 )3 / 2

−p

≅ , for r >> a

4 πε o r 3

Dipole electric field on the axis at a distance r from the centre:

2 pr

E =

4 πε 0 (r 2 − a 2 )2

2p

≅ for r >> a

4 πε 0r 3

The 1/r 3 dependence of dipole electric fields should be noted in contrast

to the 1/r 2 dependence of electric field due to a point charge.

13. In a uniform electric field E, a dipole experiences a torque τ given by

τ =p×E

but experiences no net force.

14. The flux Δφ of electric field E through a small area element ΔS is

given by

Δφ = E.ΔS

The vector area element ΔS is

ΔS = ΔS n̂

area element, which can be considered planar for sufficiently small ΔS. 43

Physics

of outward normal, by convention.

15. Gauss’s law: The flux of electric field through any closed surface S is

1/ε0 times the total charge enclosed by S. The law is especially useful

in determining electric field E, when the source distribution has simple

symmetry:

(i) Thin infinitely long straight wire of uniform linear charge density λ

λ

E= ˆ

n

2 πε 0 r

where r is the perpendicular distance of the point from the wire and

n̂ is the radial unit vector in the plane normal to the wire passing

through the point.

(ii) Infinite thin plane sheet of uniform surface charge density σ

σ

E= ˆ

n

2 ε0

(iii) Thin spherical shell of uniform surface charge density σ

q

E= rˆ (r ≥ R )

4 πε 0 r 2

E=0 (r < R )

where r is the distance of the point from the centre of the shell and R

the radius of the shell. q is the total charge of the shell: q = 4πR2σ.

The electric field outside the shell is as though the total charge is

concentrated at the centre. The same result is true for a solid sphere

of uniform volume charge density. The field is zero at all points inside

the shell

from negative to

positive charge

Charge density

surface σ [L–2

TA] Cm –2

Charge/area

44

Electric Charges

and Fields

POINTS TO PONDER

1. You might wonder why the protons, all carrying positive charges, are

compactly residing inside the nucleus. Why do they not fly away? You

will learn that there is a third kind of a fundamental force, called the

strong force which holds them together. The range of distance where

this force is effective is, however, very small ~10-14 m. This is precisely

the size of the nucleus. Also the electrons are not allowed to sit on

top of the protons, i.e. inside the nucleus, due to the laws of quantum

mechanics. This gives the atoms their structure as they exist in nature.

2. Coulomb force and gravitational force follow the same inverse-square

law. But gravitational force has only one sign (always attractive), while

Coulomb force can be of both signs (attractive and repulsive), allowing

possibility of cancellation of electric forces. This is how gravity, despite

being a much weaker force, can be a dominating and more pervasive

force in nature.

3. The constant of proportionality k in Coulomb’s law is a matter of

choice if the unit of charge is to be defined using Coulomb’s law. In SI

units, however, what is defined is the unit of current (A) via its magnetic

effect (Ampere’s law) and the unit of charge (coulomb) is simply defined

by (1C = 1 A s). In this case, the value of k is no longer arbitrary; it is

approximately 9 × 109 N m2 C–2.

4. The rather large value of k, i.e., the large size of the unit of charge

(1C) from the point of view of electric effects arises because (as

mentioned in point 3 already) the unit of charge is defined in terms of

magnetic forces (forces on current–carrying wires) which are generally

much weaker than the electric forces. Thus while 1 ampere is a unit

of reasonable size for magnetic effects, 1 C = 1 A s, is too big a unit for

electric effects.

5. The additive property of charge is not an ‘obvious’ property. It is related

to the fact that electric charge has no direction associated with it;

charge is a scalar.

6. Charge is not only a scalar (or invariant) under rotation; it is also

invariant for frames of reference in relative motion. This is not always

true for every scalar. For example, kinetic energy is a scalar under

rotation, but is not invariant for frames of reference in relative

motion.

7. Conservation of total charge of an isolated system is a property

independent of the scalar nature of charge noted in point 6.

Conservation refers to invariance in time in a given frame of reference.

A quantity may be scalar but not conserved (like kinetic energy in an

inelastic collision). On the other hand, one can have conserved vector

quantity (e.g., angular momentum of an isolated system).

8. Quantisation of electric charge is a basic (unexplained) law of nature;

interestingly, there is no analogous law on quantisation of mass.

9. Superposition principle should not be regarded as ‘obvious’, or equated

with the law of addition of vectors. It says two things: force on one

charge due to another charge is unaffected by the presence of other

charges, and there are no additional three-body, four-body, etc., forces

which arise only when there are more than two charges.

10. The electric field due to a discrete charge configuration is not defined

at the locations of the discrete charges. For continuous volume charge

distribution, it is defined at any point in the distribution. For a surface

charge distribution, electric field is discontinuous across the surface. 45

Physics

11. The electric field due to a charge configuration with total charge zero

is not zero; but for distances large compared to the size of

the configuration, its field falls off faster than 1/r 2, typical of field

due to a single charge. An electric dipole is the simplest example of

this fact.

EXERCISES

1.1 What is the force between two small charged spheres having

charges of 2 × 10–7C and 3 × 10–7C placed 30 cm apart in air?

1.2 The electrostatic force on a small sphere of charge 0.4 μC due to

another small sphere of charge – 0.8 μC in air is 0.2 N. (a) What is

the distance between the two spheres? (b) What is the force on the

second sphere due to the first?

1.3 Check that the ratio ke2/G memp is dimensionless. Look up a Table

of Physical Constants and determine the value of this ratio. What

does the ratio signify?

1.4 (a) Explain the meaning of the statement ‘electric charge of a body

is quantised’.

(b) Why can one ignore quantisation of electric charge when dealing

with macroscopic i.e., large scale charges?

1.5 When a glass rod is rubbed with a silk cloth, charges appear on

both. A similar phenomenon is observed with many other pairs of

bodies. Explain how this observation is consistent with the law of

conservation of charge.

1.6 Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC are

located at the corners of a square ABCD of side 10 cm. What is the

force on a charge of 1 μC placed at the centre of the square?

1.7 (a) An electrostatic field line is a continuous curve. That is, a field

line cannot have sudden breaks. Why not?

(b) Explain why two field lines never cross each other at any point?

1.8 Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart

in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining

the two charges?

(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at

this point, what is the force experienced by the test charge?

1.9 A system has two charges qA = 2.5 × 10–7 C and qB = –2.5 × 10–7 C

located at points A: (0, 0, –15 cm) and B: (0,0, +15 cm), respectively.

What are the total charge and electric dipole moment of the system?

1.10 An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30°

with the direction of a uniform electric field of magnitude 5 × 104 NC–1.

Calculate the magnitude of the torque acting on the dipole.

1.11 A polythene piece rubbed with wool is found to have a negative

charge of 3 × 10–7 C.

(a) Estimate the number of electrons transferred (from which to

which?)

(b) Is there a transfer of mass from wool to polythene?

1.12 (a) Two insulated charged copper spheres A and B have their centres

46 separated by a distance of 50 cm. What is the mutual force of

Electric Charges

and Fields

electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The

radii of A and B are negligible compared to the distance of

separation.

(b) What is the force of repulsion if each sphere is charged double

the above amount, and the distance between them is halved?

1.13 Suppose the spheres A and B in Exercise 1.12 have identical sizes.

A third sphere of the same size but uncharged is brought in contact

with the first, then brought in contact with the second, and finally

removed from both. What is the new force of repulsion between A

and B?

1.14 Figure 1.33 shows tracks of three charged particles in a uniform

electrostatic field. Give the signs of the three charges. Which particle

has the highest charge to mass ratio?

FIGURE 1.33

1.15 Consider a uniform electric field E = 3 × 103 î N/C. (a) What is the

flux of this field through a square of 10 cm on a side whose plane is

parallel to the yz plane? (b) What is the flux through the same

square if the normal to its plane makes a 60° angle with the x-axis?

1.16 What is the net flux of the uniform electric field of Exercise 1.15

through a cube of side 20 cm oriented so that its faces are parallel

to the coordinate planes?

1.17 Careful measurement of the electric field at the surface of a black

box indicates that the net outward flux through the surface of the

box is 8.0 × 103 Nm2/C. (a) What is the net charge inside the box?

(b) If the net outward flux through the surface of the box were zero,

could you conclude that there were no charges inside the box? Why

or Why not?

1.18 A point charge +10 μC is a distance 5 cm directly above the centre

of a square of side 10 cm, as shown in Fig. 1.34. What is the

magnitude of the electric flux through the square? (Hint: Think of

the square as one face of a cube with edge 10 cm.)

FIGURE 1.34 47

Physics

1.19 A point charge of 2.0 μC is at the centre of a cubic Gaussian

surface 9.0 cm on edge. What is the net electric flux through the

surface?

1.20 A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass

through a spherical Gaussian surface of 10.0 cm radius centred on

the charge. (a) If the radius of the Gaussian surface were doubled,

how much flux would pass through the surface? (b) What is the

value of the point charge?

1.21 A conducting sphere of radius 10 cm has an unknown charge. If

the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C

and points radially inward, what is the net charge on the sphere?

1.22 A uniformly charged conducting sphere of 2.4 m diameter has a

surface charge density of 80.0 μC/m2. (a) Find the charge on the

sphere. (b) What is the total electric flux leaving the surface of the

sphere?

1.23 An infinite line charge produces a field of 9 × 104 N/C at a distance

of 2 cm. Calculate the linear charge density.

1.24 Two large, thin metal plates are parallel and close to each other. On

their inner faces, the plates have surface charge densities of opposite

signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer

region of the first plate, (b) in the outer region of the second plate,

and (c) between the plates?

ADDITIONAL EXERCISES

1.25 An oil drop of 12 excess electrons is held stationary under a constant

electric field of 2.55 × 104 NC–1 in Millikan’s oil drop experiment. The

density of the oil is 1.26 g cm–3. Estimate the radius of the drop.

(g = 9.81 m s–2; e = 1.60 × 10–19 C).

1.26 Which among the curves shown in Fig. 1.35 cannot possibly

represent electrostatic field lines?

48

Electric Charges

and Fields

FIGURE 1.35

1.27 In a certain region of space, electric field is along the z-direction

throughout. The magnitude of electric field is, however, not constant

but increases uniformly along the positive z-direction, at the rate of

105 NC–1 per metre. What are the force and torque experienced by a

system having a total dipole moment equal to 10–7 Cm in the negative

z-direction ?

1.28 (a) A conductor A with a cavity as shown in Fig. 1.36(a) is given a

charge Q. Show that the entire charge must appear on the outer

surface of the conductor. (b) Another conductor B with charge q is

inserted into the cavity keeping B insulated from A. Show that the

total charge on the outside surface of A is Q + q [Fig. 1.36(b)]. (c) A

sensitive instrument is to be shielded from the strong electrostatic

fields in its environment. Suggest a possible way.

FIGURE 1.36

1.29 A hollow charged conductor has a tiny hole cut into its surface.

Show that the electric field in the hole is (σ/2ε0) n̂ , where n̂ is the

unit vector in the outward normal direction, and σ is the surface

charge density near the hole.

1.30 Obtain the formula for the electric field due to a long thin wire of

uniform linear charge density λ without using Gauss’s law. [Hint:

Use Coulomb’s law directly and evaluate the necessary integral.]

1.31 It is now believed that protons and neutrons (which constitute nuclei

of ordinary matter) are themselves built out of more elementary units

called quarks. A proton and a neutron consist of three quarks each.

Two types of quarks, the so called ‘up’ quark (denoted by u) of charge

+ (2/3) e, and the ‘down’ quark (denoted by d) of charge (–1/3) e,

together with electrons build up ordinary matter. (Quarks of other

types have also been found which give rise to different unusual

varieties of matter.) Suggest a possible quark composition of a

proton and neutron. 49

Physics

1.32 (a) Consider an arbitrary electrostatic field configuration. A small

test charge is placed at a null point (i.e., where E = 0) of the

configuration. Show that the equilibrium of the test charge is

necessarily unstable.

(b) Verify this result for the simple configuration of two charges of

the same magnitude and sign placed a certain distance apart.

1.33 A particle of mass m and charge (–q) enters the region between the

two charged plates initially moving along x-axis with speed vx (like

particle 1 in Fig. 1.33). The length of plate is L and an uniform

electric field E is maintained between the plates. Show that the

vertical deflection of the particle at the far edge of the plate is

qEL2/(2m vx2).

Compare this motion with motion of a projectile in gravitational field

discussed in Section 4.10 of Class XI Textbook of Physics.

1.34 Suppose that the particle in Exercise in 1.33 is an electron projected

with velocity vx = 2.0 × 106 m s–1. If E between the plates separated

by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper

plate? (|e|=1.6 × 10–19 C, me = 9.1 × 10–31 kg.)

50

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