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Unit 1
DIVERSITY IN THE LIVING WORLD
Chapter 1 The Living World Chapter 2 Biological Classification Chapter 3 Plant Kingdom Chapter 4 Animal Kingdom
Biology is the science of life forms and living processes
The living world comprises an amazing diversity of living organisms
Early man could easily perceive the difference between inanimate matter and living organisms
Early man deified some of the inanimate matter (wind, sea, fire etc.) and some among the animals and plants
A common feature of all such forms of inanimate and animate objects was the sense of awe or fear that they evoked
The description of living organisms including human beings began much later in human history
Societies which indulged in anthropocentric view of biology could register limited progress in biological knowledge
Systematic and monumental description of life forms brought in, out of necessity, detailed systems of identification, nomenclature and classification
The biggest spin off of such studies was the recognition of the sharing of similarities among living organisms both horizontally and vertically
That all present day living organisms are related to each other and also to all organisms that ever lived on this earth, was a revelation which humbled man and led to cultural movements for conservation of biodiversity
In the following chapters of this unit, you will get a description, including classification, of animals and plants from a taxonomist’s perspective.
Ernst Mayr (1904 – 2004) Born on 5 July 1904, in Kempten, Germany, Ernst Mayr, the Harvard University evolutionary biologist who has been called ‘The Darwin of the 20th century’, was one of the 100 greatest scientists of all time
Mayr joined Harvard’s Faculty of Arts and Sciences in 1953 and retired in 1975, assuming the title Alexander Agassiz Professor of Zoology Emeritus
Throughout his nearly 80-year career, his research spanned ornithology, taxonomy, zoogeography, evolution, systematics, and the history and philosophy of biology
He almost single-handedly made the origin of species diversity the central question of evolutionary biology that it is today
He also pioneered the currently accepted definition of a biological species
Mayr was awarded the three prizes widely regarded as the triple crown of biology: the Balzan Prize in 1983, the International Prize for Biology in 1994, and the Crafoord Prize in 1999
Mayr died at the age of 100 in the year 2004.
Chapter 1
The Living World
1.1 What is ‘Living’? 1.2 Diversity in the Living World 1.3 Taxonomic Categories 1.4 Taxonomical Aids
How wonderful is the living world ! The wide range of living types is amazing
The extraordinary habitats in which we find living organisms, be it cold mountains, deciduous forests, oceans, fresh water lakes, deserts or hot springs, leave us speechless
The beauty of a galloping horse, of the migrating birds, the valley of flowers or the attacking shark evokes awe and a deep sense of wonder
The ecological conflict and cooperation among members of a population and among populations of a community or even the molecular traffic inside a cell make us deeply reflect on – what indeed is life? This question has two implicit questions within it
The first is a technical one and seeks answer to what living is as opposed to the non-living, and the second is a philosophical one, and seeks answer to what the purpose of life is
As scientists, we shall not attempt answering the second question
We will try to reflect on – what is living?
1.1 What is ‘Living’? When we try to define ‘living’, we conventionally look for distinctive characteristics exhibited by living organisms
Growth, reproduction, ability to sense environment and mount a suitable response come to our mind immediately as unique features of living organisms
One can add a few more features like metabolism, ability to self-replicate, self-organise, interact and emergence to this list
Let us try to understand each of these
All living organisms grow
Increase in mass and increase in number of individuals are twin characteristics of growth
A multicellular organism grows by cell division
In plants, this growth by cell division occurs continuously throughout their life span
In animals, this growth is seen only up to a certain age
However, cell division occurs in certain tissues to replace lost cells
Unicellular organisms grow by cell division
One can easily observe this in in vitro cultures by simply counting the number of cells under the microscope
In majority of higher animals and plants, growth and reproduction are mutually exclusive events
One must remember that increase in body mass is considered as growth
Non-living objects also grow if we take increase in body mass as a criterion for growth
Mountains, boulders and sand mounds do grow
However, this kind of growth exhibited by non-living objects is by accumulation of material on the surface
In living organisms, growth is from inside
Growth, therefore, cannot be taken as a defining property of living organisms
Conditions under which it can be observed in all living organisms have to be explained and then we understand that it is a characteristic of living systems
A dead organism does not grow
Reproduction, likewise, is a characteristic of living organisms
In multicellular organisms, reproduction refers to the production of progeny possessing features more or less similar to those of parents
Invariably and implicitly we refer to sexual reproduction
Organisms reproduce by asexual means also
Fungi multiply and spread easily due to the millions of asexual spores they produce
In lower organisms like yeast and hydra, we observe budding
In Planaria (flat worms), we observe true regeneration, i.e., a fragmented organism regenerates the lost part of its body and becomes, a new organism
The fungi, the filamentous algae, the protonema of mosses, all easily multiply by fragmentation
When it comes to unicellular organisms like bacteria, unicellular algae or Amoeba, reproduction is synonymous with growth, i.e., increase in number of cells
We have already defined growth as equivalent to increase in cell number or mass
Hence, we notice that in single-celled organisms, we are not very clear about the usage of these two terms – growth and reproduction
Further, there are many organisms which do not reproduce (mules, sterile worker bees, infertile human couples, etc)
Hence, reproduction also cannot be an all-inclusive defining characteristic of living organisms
Of course, no non-living object is capable of reproducing or replicating by itself
Another characteristic of life is metabolism
All living organisms are made of chemicals
These chemicals, small and big, belonging to various classes, sizes, functions, etc., are constantly being made and changed into some other biomolecules
These conversions are chemical reactions or metabolic reactions
There are thousands of metabolic reactions occurring simultaneously inside all living organisms, be they unicellular or multicellular
All plants, animals, fungi and microbes exhibit metabolism
The sum total of all the chemical reactions occurring in our body is metabolism
No non­-living object exhibits metabolism
Metabolic reactions can be demonstrated outside the body in cell-free systems
An isolated metabolic reaction(s) outside the body of an organism, performed in a test tube is neither living nor non-living
Hence, while metabolism is a defining feature of all living organisms without exception, isolated metabolic reactions in vitro are not living things but surely living reactions
Hence, cellular organisation of the body is the defining feature of life forms
Perhaps, the most obvious and technically complicated feature of all living organisms is this ability to sense their surroundings or environment and respond to these environmental stimuli which could be physical, chemical or biological
We sense our environment through our sense organs
Plants respond to external factors like light, water, temperature, other organisms, pollutants, etc
All organisms, from the prokaryotes to the most complex eukaryotes can sense and respond to environmental cues
Photoperiod affects reproduction in seasonal breeders, both plants and animals
All organisms handle chemicals entering their bodies
All organisms therefore, are ‘aware’ of their surroundings
Human being is the only organism who is aware of himself, i.e., has self-consciousness
Consciousness therefore, becomes the defining property of living organisms
When it comes to human beings, it is all the more difficult to define the living state
We observe patients lying in coma in hospitals virtually supported by machines which replace heart and lungs
The patient is otherwise brain-dead
The patient has no self-­consciousness
Are such patients who never come back to normal life, living or non-living? In higher classes, you will come to know that all living phenomena are due to underlying interactions
Properties of tissues are not present in the constituent cells but arise as a result of interactions among the constituent cells
Similarly, properties of cellular organelles are not present in the molecular constituents of the organelle but arise as a result of interactions among the molecular components comprising the organelle
These interactions result in emergent properties at a higher level of organisation
This phenomenon is true in the hierarchy of organisational complexity at all levels
Therefore, we can say that living organisms are self-replicating, evolving and self-regulating interactive systems capable of responding to external stimuli
Biology is the story of life on earth
Biology is the story of evolution of living organisms on earth
All living organisms – present, past and future, are linked to one another by the sharing of the common genetic material, but to varying degrees.
1.2 Diversity in the Living World If you look around you will see a large variety of living organisms, be it potted plants, insects, birds, your pets or other animals and plants
There are also several organisms that you cannot see with your naked eye but they are all around you
If you were to increase the area that you make observations in, the range and variety of organisms that you see would increase
Obviously, if you were to visit a dense forest, you would probably see a much greater number and kinds of living organisms in it
Each different kind of plant, animal or organism that you see, represents a species
The number of species that are known and described range between 1.7-1.8 million
This refers to biodiversity or the number and types of organisms present on earth
We should remember here that as we explore new areas, and even old ones, new organisms are continuously being identified
As stated earlier, there are millions of plants and animals in the world; we know the plants and animals in our own area by their local names
These local names would vary from place to place, even within a country
Probably you would recognise the confusion that would be created if we did not find ways and means to talk to each other, to refer to organisms we are talking about
Hence, there is a need to standardise the naming of living organisms such that a particular organism is known by the same name all over the world
This process is called nomenclature
Obviously, nomenclature or naming is only possible when the organism is described correctly and we know to what organism the name is attached to
This is identification
In order to facilitate the study, number of scientists have established procedures to assign a scientific name to each known organism
This is acceptable to biologists all over the world
For plants, scientific names are based on agreed principles and criteria, which are provided in International Code for Botanical Nomenclature (ICBN)
You may ask, how are animals named? Animal taxonomists have evolved International Code of Zoological Nomenclature (ICZN)
The scientific names ensure that each organism has only one name
Description of any organism should enable the people (in any part of the world) to arrive at the same name
They also ensure that such a name has not been used for any other known organism
Biologists follow universally accepted principles to provide scientific names to known organisms
Each name has two components – the Generic name and the specific epithet
This system of providing a name with two components is called Binomial nomenclature
This naming system given by Carolus Linnaeus is being practised by biologists all over the world
This naming system using a two word format was found convenient
Let us take the example of mango to understand the way of providing scientific names better
The scientific name of mango is written as Mangifera indica
Let us see how it is a binomial name
In this name Mangifera represents the genus while indica, is a particular species, or a specific epithet
Other universal rules of nomenclature are as follows: 1
Biological names are generally in Latin and written in italics
They are Latinised or derived from Latin irrespective of their origin
2
The first word in a biological name represents the genus while the second component denotes the specific epithet
3
Both the words in a biological name, when handwritten, are separately underlined, or printed in italics to indicate their Latin origin
4
The first word denoting the genus starts with a capital letter while the specific epithet starts with a small letter
It can be illustrated with the example of Mangifera indica
Name of the author appears after the specific epithet, i.e., at the end of the biological name and is written in an abbreviated form, e.g., Mangifera indica Linn
It indicates that this species was first described by Linnaeus
Since it is nearly impossible to study all the living organisms, it is necessary to devise some means to make this possible
This process is classification
Classification is the process by which anything is grouped into convenient categories based on some easily observable characters
For example, we easily recognise groups such as plants or animals or dogs, cats or insects
The moment we use any of these terms, we associate certain characters with the organism in that group
What image do you see when you think of a dog ? Obviously, each one of us will see ‘dogs’ and not ‘cats’
Now, if we were to think of ‘Alsatians’ we know what we are talking about
Similarly, suppose we were to say ‘mammals’, you would, of course, think of animals with external ears and body hair
Likewise, in plants, if we try to talk of ‘Wheat’, the picture in each of our minds will be of wheat plants, not of rice or any other plant
Hence, all these - ‘Dogs’, ‘Cats’, ‘Mammals’, ‘Wheat’, ‘Rice’, ‘Plants’, ‘Animals’, etc., are convenient categories we use to study organisms
The scientific term for these categories is taxa
Here you must recognise that taxa can indicate categories at very different levels
‘Plants’ – also form a taxa
‘Wheat’ is also a taxa
Similarly, ‘animals’, ‘mammals’, ‘dogs’ are all taxa – but you know that a dog is a mammal and mammals are animals
Therefore, ‘animals’, ‘mammals’ and ‘dogs’ represent taxa at different levels
Hence, based on characteristics, all living organisms can be classified into different taxa
This process of classification is taxonomy
External and internal structure, along with the structure of cell, development process and ecological information of organisms are essential and form the basis of modern taxonomic studies
Hence, characterisation, identification, classification and nomenclature are the processes that are basic to taxonomy
Taxonomy is not something new
Human beings have always been interested in knowing more and more about the various kinds of organisms, particularly with reference to their own use
In early days, human beings needed to find sources for their basic needs of food, clothing and shelter
Hence, the earliest classifications were based on the ‘uses’ of various organisms
Human beings were, since long, not only interested in knowing more about different kinds of organisms and their diversities, but also the relationships among them
This branch of study was referred to as systematics
The word systematics is derived from the Latin word ‘systema’ which means systematic arrangement of organisms
Linnaeus used Systema Naturae as the title of his publication
The scope of systematics was later enlarged to include identification, nomenclature and classification
Systematics takes into account evolutionary relationships between organisms.
1.3 Taxonomic Categories Classification is not a single step process but involves hierarchy of steps in which each step represents a rank or category
Since the category is a part of overall taxonomic arrangement, it is called the taxonomic category and all categories together constitute the taxonomic hierarchy
Each category, referred to as a unit of classification, in fact, represents a rank and is commonly termed as taxon (pl.: taxa)
Taxonomic categories and hierarchy can be illustrated by an example
Insects represent a group of organisms sharing common features like three pairs of jointed legs
It means insects are recognisable concrete objects which can be classified, and thus were given a rank or category
Can you name other such groups of organisms? Remember, groups represent category
Category further denotes rank
Each rank or taxon, in fact, represents a unit of classification
These taxonomic groups/categories are distinct biological entities and not merely morphological aggregates
Taxonomical studies of all known organisms have led to the development of common categories such as kingdom, phylum or division (for plants), class, order, family, genus and species
All organisms, including those in the plant and animal kingdoms have species as the lowest category
Now the question you may ask is, how to place an organism in various categories? The basic requirement is the knowledge of characters of an individual or group of organisms
This helps in identifying similarities and dissimilarities among the individuals of the same kind of organisms as well as of other kinds of organisms.
1.3.1 Species Taxonomic studies consider a group of individual organisms with fundamental similarities as a species
One should be able to distinguish one species from the other closely related species based on the distinct morphological differences
Let us consider Mangifera indica, Solanum tuberosum (potato) and Panthera leo (lion)
All the three names, indica, tuberosum and leo, represent the specific epithets, while the first words Mangifera, Solanum and Panthera are genera and represents another higher level of taxon or category
Each genus may have one or more than one specific epithets representing different organisms, but having morphological similarities
For example, Panthera has another specific epithet called tigris and Solanum includes species like nigrum and melongena
Human beings belong to the species sapiens which is grouped in the genus Homo
The scientific name thus, for human being, is written as Homo sapiens.
1.3.2 Genus Genus comprises a group of related species which has more characters in common in comparison to species of other genera
We can say that genera are aggregates of closely related species
For example, potato and brinjal are two different species but both belong to the genus Solanum
Lion (Panthera leo), leopard (P
pardus) and tiger (P
tigris) with several common features, are all species of the genus Panthera
This genus differs from another genus Felis which includes cats.
1.3.3 Family The next category, Family, has a group of related genera with still less number of similarities as compared to genus and species
Families are characterised on the basis of both vegetative and reproductive features of plant species
Among plants for example, three different genera Solanum, Petunia and Datura are placed in the family Solanaceae
Among animals for example, genus Panthera, comprising lion, tiger, leopard is put along with genus, Felis (cats) in the family Felidae
Similarly, if you observe the features of a cat and a dog, you will find some similarities and some differences as well
They are separated into two different families – Felidae and Canidae, respectively.
1.3.4 Order You have seen earlier that categories like species, genus and families are based on a number of similar characters
Generally, order and other higher taxonomic categories are identified based on the aggregates of characters
Order being a higher category, is the assemblage of families which exhibit a few similar characters
The similar characters are less in number as compared to different genera included in a family
Plant families like Convolvulaceae, Solanaceae are included in the order Polymoniales mainly based on the floral characters
The animal order, Carnivora, includes families like Felidae and Canidae.
Figure 1.1 Taxonomic categories showing hierarchial arrangement in ascending order
1.3.5 Class This category includes related orders
For example, order Primata comprising monkey, gorilla and gibbon is placed in class Mammalia along with order Carnivora that includes animals like tiger, cat and dog
Class Mammalia has other orders also.
1.3.6 Phylum Classes comprising animals like fishes, amphibians, reptiles, birds along with mammals constitute the next higher category called Phylum
All these, based on the common features like presence of notochord and dorsal hollow neural system, are included in phylum Chordata
In case of plants, classes with a few similar characters are assigned to a higher category called Division.
1.3.7 Kingdom All animals belonging to various phyla are assigned to the highest category called Kingdom Animalia in the classification system of animals
The Kingdom Plantae, on the other hand, is distinct, and comprises all plants from various divisions
Henceforth, we will refer to these two groups as animal and plant kingdoms
The taxonomic categories from species to kingdom have been shown in ascending order starting with species in Figure 1.1
These are broad categories
However, taxonomists have also developed sub-categories in this hierarchy to facilitate more sound and scientific placement of various taxa
Look at the hierarchy in Figure 1.1
Can you recall the basis of arrangement? Say, for example, as we go higher from species to kingdom, the number of common characteristics goes on decreasing
Lower the taxa, more are the characteristics that the members within the taxon share
Higher the category, greater is the difficulty of determining the relationship to other taxa at the same level
Hence, the problem of classification becomes more complex.
Table 1.1 indicates the taxonomic categories to which some common organisms like housefly, man, mango and wheat belong.
TABLE 1.1 Organisms with their Taxonomic Categories
1.4 Taxonomical Aids
Taxonomic studies of various species of plants, animals and other organisms are useful in agriculture, forestry, industry and in general in knowing our bio-resources and their diversity
These studies would require correct classification and identification of organisms
Identification of organisms requires intensive laboratory and field studies
The collection of actual specimens of plant and animal species is essential and is the prime source of taxonomic studies
These are also fundamental to studies and essential for training in systematics
It is used for classification of an organism, and the information gathered is also stored along with the specimens
In some cases the specimen is preserved for future studies
Biologists have established certain procedures and techniques to store and preserve the information as well as the specimens
Some of these are explained to help you understand the usage of these aids.
1.4.1 Herbarium Herbarium is a store house of collected plant specimens that are dried, pressed and preserved on sheets
Further, these sheets are arranged according to a universally accepted system of classification
These specimens, along with their descriptions on herbarium sheets, become a store house or repository for future use
The herbarium sheets also carry a label providing information about date and place of collection, English, local and botanical names, family, collector’s name, etc
Herbaria also serve as quick referral systems in taxonomical studies.
Figure 1.2 Herbarium showing stored specimens
1.4.2 Botanical Gardens These specialised gardens have collections of living plants for reference
Plant species in these gardens are grown for identification purposes and each plant is labelled indicating its botanical/scientific name and its family
The famous botanical gardens are at Kew (England), Indian Botanical Garden, Howrah (India) and at National Botanical Research Institute, Lucknow (India).
1.4.3 Museum Biological museums are generally set up in educational institutes such as schools and colleges
Museums have collections of preserved plant and animal specimens for study and reference
Specimens are preserved in the containers or jars in preservative solutions
Plant and animal specimens may also be preserved as dry specimens
Insects are preserved in insect boxes after collecting, killing and pinning
Larger animals like birds and mammals are usually stuffed and preserved
Museums often have collections of skeletons of animals too.
1.4.4 Zoological Parks These are the places where wild animals are kept in protected environments under human care and which enable us to learn about their food habits and behaviour
All animals in a zoo are provided, as far as possible, the conditions similar to their natural habitats
Children love visiting these parks, commonly called Zoos .
Figure 1.3 Pictures showing animals in different zoological parks of India
1.4.5 Key Key is another taxonomical aid used for identification of plants and animals based on the similarities and dissimilarities
The keys are based on the contrasting characters generally in a pair called couplet
It represents the choice made between two opposite options
This results in acceptance of only one and rejection of the other
Each statement in the key is called a lead
Separate taxonomic keys are required for each taxonomic category such as family, genus and species for identification purposes
Keys are generally analytical in nature
Flora, manuals, monographs and catalogues are some other means of recording descriptions
They also help in correct identification
Flora contains the actual account of habitat and distribution of plants of a given area
These provide the index to the plant species found in a particular area
Manuals are useful in providing information for identification of names of species found in an area
Monographs contain information on any one taxon.
Summary The living world is rich in variety
Millions of plants and animals have been identified and described but a large number still remains unknown
The very range of organisms in terms of size, colour, habitat, physiological and morphological features make us seek the defining characteristics of living organisms
In order to facilitate the study of kinds and diversity of organisms, biologists have evolved certain rules and principles for identification, nomenclature and classification of organisms
The branch of knowledge dealing with these aspects is referred to as taxonomy
The taxonomic studies of various species of plants and animals are useful in agriculture, forestry, industry and in general for knowing our bio-resources and their diversity
The basics of taxonomy like identification, naming and classification of organisms are universally evolved under international codes
Based on the resemblances and distinct differences, each organism is identified and assigned a correct scientific/biological name comprising two words as per the binomial system of nomenclature
An organism represents/occupies a place or position in the system of classification
There are many categories/ranks and are generally referred to as taxonomic categories or taxa
All the categories constitute a taxonomic hierarchy
Taxonomists have developed a variety of taxonomic aids to facilitate identification, naming and classification of organisms
These studies are carried out from the actual specimens which are collected from the field and preserved as referrals in the form of herbaria, museums and in botanical gardens and zoological parks
It requires special techniques for collection and preservation of specimens in herbaria and museums
Live specimens, on the other hand, of plants and animals, are found in botanical gardens or in zoological parks
Taxonomists also prepare and disseminate information through manuals and monographs for further taxonomic studies
Taxonomic keys are tools that help in identification based on characteristics.
Exercises 1
Why are living organisms classified? 2
Why are the classification systems changing every now and then? 3
What different criteria would you choose to classify people that you meet often? 4
What do we learn from identification of individuals and populations? 5
Given below is the scientific name of Mango
Identify the correctly written name
Mangifera Indica Mangifera indica 6
Define a taxon
Give some examples of taxa at different hierarchical levels
7
Can you identify the correct sequence of taxonomical categories? (a) Species Order Phylum Kingdom (b) Genus Species Order Kingdom (c) Species Genus Order Phylum 8
Try to collect all the currently accepted meanings for the word ‘species’
Discuss with your teacher the meaning of species in case of higher plants and animals on one hand, and bacteria on the other hand
9
Define and understand the following terms: Phylum Class Family (iv) Order (v) Genus 10
How is a key helpful in the identification and classification of an organism? 11
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
The Living World
Table of Contents
Unit 1
DIVERSITY IN THE LIVING WORLD
Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 1
The Living World
1.1 What is ‘Living’? 1.2 Diversity in the Living World 1.3 Taxonomic Categories 1.4 Taxonomical Aids 1.1 What is ‘Living’? 1.2 Diversity in the Living World 1.3 Taxonomic Categories 1.3.1 Species 1.3.2 Genus 1.3.3 Family 1.3.4 Order 1.3.5 Class 1.3.6 Phylum 1.3.7 Kingdom 1.4 Taxonomical Aids 1.4.1 Herbarium 1.4.2 Botanical Gardens 1.4.3 Museum 1.4.4 Zoological Parks 1.4.5 Key
Summary Exercises
Landmarks
Table of Contents
Unit 12 Aldehydes, Ketones and Carboxylic Acids
Objectives
After studying this Unit, you will be able to • write the common and IUPAC names of aldehydes, ketones and carboxylic acids; • write the structures of the compounds containing functional groups namely carbonyl and carboxyl groups; • describe the important methods of preparation and reactions of these classes of compounds; • correlate physical properties and chemical reactions of aldehydes, ketones and carboxylic acids, with their structures; • explain the mechanism of a few selected reactions of aldehydes and ketones; • understand various factors affecting the acidity of carboxylic acids and their reactions; • describe the uses of aldehydes, ketones and carboxylic acids.
Carbonyl compounds are of utmost importance to organic chemistry
They are constituents of fabrics, flavourings, plastics and drugs.
In the previous Unit, you have studied organic compounds with functional groups containing carbon-oxygen single bond
In this Unit, we will study about the organic compounds containing carbon-oxygen double bond (>C=O) called carbonyl group, which is one of the most important functional groups in organic chemistry
In aldehydes, the carbonyl group is bonded to a carbon and hydrogen while in the ketones, it is bonded to two carbon atoms
The carbonyl compounds in which carbon of carbonyl group is bonded to carbon or hydrogen and oxygen of hydroxyl moiety (-OH) are known as carboxylic acids, while in compounds where carbon is attached to carbon or hydrogen and nitrogen of -NH2 moiety or to halogens are called amides and acyl halides respectively
Esters and anhydrides are derivatives of carboxylic acids
The general formulas of these classes of compounds are given below:
Aldehydes, ketones and carboxylic acids are widespread in plants and animal kingdom
They play an important role in biochemical processes of life
They add fragrance and flavour to nature, for example, vanillin (from vanilla beans), salicylaldehyde (from meadow sweet) and cinnamaldehyde (from cinnamon) have very pleasant fragrances.
They are used in many food products and pharmaceuticals to add flavours
Some of these families are manufactured for use as solvents (i.e., acetone) and for preparing materials like adhesives, paints, resins, perfumes, plastics, fabrics, etc.
12.1 Nomenclature and Structure of Carbonyl Group
12.1.1 Nomenclature
I
Aldehydes and ketones Aldehydes and ketones are the simplest and most important carbonyl compounds
There are two systems of nomenclature of aldehydes and ketones
(a) Common names Aldehydes and ketones are often called by their common names instead of IUPAC names
The common names of most aldehydes are derived from the common names of the corresponding carboxylic acids [Section 12.6.1] by replacing the ending –ic of acid with aldehyde
At the same time, the names reflect the Latin or Greek term for the original source of the acid or aldehyde
The location of the substituent in the carbon chain is indicated by Greek letters α, β, γ, δ, etc
The α-carbon being the one directly linked to the aldehyde group, β-carbon the next, and so on
For example
The common names of ketones are derived by naming two alkyl or aryl groups bonded to the carbonyl group
The locations of substituents are indicated by Greek letters, α α′, β β′ and so on beginning with the carbon atoms next to the carbonyl group, indicated as αα′
Some ketones have historical common names, the simplest dimethyl ketone is called acetone
Alkyl phenyl ketones are usually named by adding the name of acyl group as prefix to the word phenone
For example
(b) IUPAC names The IUPAC names of open chain aliphatic aldehydes and ketones are derived from the names of the corresponding alkanes by replacing the ending –e with –al and –one respectively
In case of aldehydes the longest carbon chain is numbered starting from the carbon of the aldehyde group while in case of ketones the numbering begins from the end nearer to the carbonyl group
The substituents are prefixed in alphabetical order along with numerals indicating their positions in the carbon chain
The same applies to cyclic ketones, where the carbonyl carbon is numbered one
When the aldehyde group is attached to a ring, the suffix carbaldehyde is added after the full name of the cycloalkane
The numbering of the ring carbon atoms start from the carbon atom attached to the aldehyde group
The name of the simplest aromatic aldehyde carrying the aldehyde group on a benzene ring is benzenecarbaldehyde
However, the common name benzaldehyde is also accepted by IUPAC
Other aromatic aldehydes are hence named as substituted benzaldehydes.
The common and IUPAC names of some aldehydes and ketones are given in Table 12.1.
Table 12.1: Common and IUPAC Names of Some Aldehydes and Ketones
12.1.2 Structure of the Carbonyl Group
The carbonyl carbon atom is sp2-hybridised and forms three sigma (σ) bonds
The fourth valence electron of carbon remains in its p-orbital and forms a π-bond with oxygen by overlap with p-orbital of an oxygen
In addition, the oxygen atom also has two non bonding electron pairs
Thus, the carbonyl carbon and the three atoms attached to it lie in the same plane and the π-electron cloud is above and below this plane
The bond angles are approximately 120° as expected of a trigonal coplanar structure .
The carbon-oxygen double bond is polarised due to higher electronegativity of oxygen relative to carbon
Hence, the carbonyl carbon is an electrophilic (Lewis acid), and carbonyl oxygen, a nucleophilic (Lewis base) centre
Carbonyl compounds have substantial dipole moments and are polar than ethers
The high polarity of the carbonyl group is explained on the basis of resonance involving a neutral (A) and a dipolar (B) structures as shown.
Intext Questions 12.1 Write the structures of the following compounds
α-Methoxypropionaldehyde 3-Hydroxybutanal 2-Hydroxycyclopentane carbaldehyde (iv) 4-Oxopentanal (v) Di-sec
butyl ketone (vi) 4-Fluoroacetophenone
12.2 Preparation of Aldehydes and Ketones Some important methods for the preparation of aldehydes and ketones are as follows:
12.2.1 Preparation of Aldehydes and Ketones 1
By oxidation of alcohols
Aldehydes and ketones are generally prepared by oxidation of primary and secondary alcohols, respectively (Unit 11, Class XII).
By dehydrogenation of alcohols This method is suitable for volatile alcohols and is of industrial application
In this method alcohol vapours are passed over heavy metal catalysts (Ag or Cu)
Primary and secondary alcohols give aldehydes and ketones, respectively (Unit 11, Class XII)
3
From hydrocarbons By ozonolysis of alkenes: As we know, ozonolysis of alkenes followed by reaction with zinc dust and water gives aldehydes, ketones or a mixture of both depending on the substitution pattern of the alkene (Unit 13, Class XI)
By hydration of alkynes: Addition of water to ethyne in the presence of H2SO4 and HgSO4 gives acetaldehyde
All other alkynes give ketones in this reaction (Unit 13, Class XI).
12.2.2 Preparation of Aldehydes 1
From acyl chloride (acid chloride) Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium sulphate
This reaction is called Rosenmund reduction.
From nitriles and esters Nitriles are reduced to corresponding imine with stannous chloride in the presence of hydrochloric acid, which on hydrolysis give corresponding aldehyde.
This reaction is called Stephen reaction
Alternatively, nitriles are selectively reduced by diisobutylaluminium hydride, (DIBAL-H) to imines followed by hydrolysis to aldehydes:
Similarly, esters are also reduced to aldehydes with DIBAL-H.
From hydrocarbons
Aromatic aldehydes (benzaldehyde and its derivatives) are prepared from aromatic hydrocarbons by the following methods:
By oxidation of methylbenzene Strong oxidising agents oxidise toluene and its derivatives to benzoic acids
However, it is possible to stop the oxidation at the aldehyde stage with suitable reagents that convert the methyl group to an intermediate that is difficult to oxidise further
The following methods are used for this purpose
(a) Use of chromyl chloride (CrO2Cl2): Chromyl chloride oxidises methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde.
This reaction is called Etard reaction
(b) Use of chromic oxide (CrO3): Toluene or substituted toluene is converted to benzylidene diacetate on treating with chromic oxide in acetic anhydride
The benzylidene diacetate can be hydrolysed to corresponding benzaldehyde with aqueous acid.
By side chain chlorination followed by hydrolysis Side chain chlorination of toluene gives benzal chloride, which on hydrolysis gives benzaldehyde
This is a commercial method of manufacture of benzaldehyde.
By Gatterman – Koch reaction
When benzene or its derivative is treated with carbon monoxide and hydrogen chloride in the presence of anhydrous aluminium chloride or cuprous chloride, it gives benzaldehyde or substituted benzaldehyde.
This reaction is known as Gatterman-Koch reaction
12.2.3 Preparation of Ketones 1
From acyl chlorides
Treatment of acyl chlorides with dialkylcadmium, prepared by the reaction of cadmium chloride with Grignard reagent, gives ketones.
From nitriles
Treating a nitrile with Grignard reagent followed by hydrolysis yields a ketone.
3
From benzene or substituted benzenes When benzene or substituted benzene is treated with acid chloride in the presence of anhydrous aluminium chloride, it affords the corresponding ketone
This reaction is known as Friedel-Crafts acylation reaction.
Example 12.1 Give names of the reagents to bring about the following transformations: Hexan-1-ol to hexanal Cyclohexanol to cyclohexanone p-Fluorotoluene to p-fluorobenzaldehyde (iv) Ethanenitrile to ethanal
(v) Allyl alcohol to propenal (vi) But-2-ene to ethanal Solution
C5H5NH+CrO3Cl-(PCC) Anhydrous CrO3 CrO3 in the presence of acetic anhydride/1
CrO2Cl2 2
HOH (iv) (Diisobutyl)aluminiumhydride (DIBAL-H) (v) PCC (vi) O3/H2O-Zn dust
Intext Question
12.2 Write the structures of products of the following reactions;
(iv)
12.3 Physical Properties The physical properties of aldehydes and ketones are described as follows
Methanal is a gas at room temperature
Ethanal is a volatile liquid
Other aldehydes and ketones are liquid or solid at room temperature
The boiling points of aldehydes and ketones are higher than hydrocarbons and ethers of comparable molecular masses
It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions
Also, their boiling points are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding
The following compounds of molecular masses 58 and 60 are ranked in order of increasing boiling points.
The lower members of aldehydes and ketones such as methanal, ethanal and propanone are miscible with water in all proportions, because they form hydrogen bond with water.
However, the solubility of aldehydes and ketones decreases rapidly on increasing the length of alkyl chain
All aldehydes and ketones are fairly soluble in organic solvents like benzene, ether, methanol, chloroform, etc
The lower aldehydes have sharp pungent odours
As the size of the molecule increases, the odour becomes less pungent and more fragrant
In fact, many naturally occurring aldehydes and ketones are used in the blending of perfumes and flavouring agents.
Example 12.2
Arrange the following compounds in the increasing order of their boiling points:
CH3CH2CH2CHO, CH3CH2CH2CH2OH, H5C2-O-C2H5, CH3CH2CH2CH3 Solution The molecular masses of these compounds are in the range of 72 to 74
Since only butan-1-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan-1-ol would be the highest
Butanal is more polar than ethoxyethane
Therefore, the intermolecular dipole-dipole attraction is stronger in the former
n-Pentane molecules have only weak van der Waals forces
Hence increasing order of boiling points of the given compounds is as follows: CH3CH2CH2CH3 < H5C2-O-C2H5 < CH3CH2CH2CHO < CH3CH2CH2CH2OH
Intext Question 12.3 Arrange the following compounds in increasing order of their boiling points
CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3
12.4 Chemical Reactions
Since aldehydes and ketones both possess the carbonyl functional group, they undergo similar chemical reactions
1
Nucleophilic addition reactions Contrary to electrophilic addition reactions observed in alkenes (refer Unit 13, Class XI), the aldehydes and ketones undergo nucleophilic addition reactions.
Mechanism of nucleophilic addition reactions A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp2 hybridised orbitals of carbonyl carbon
The hybridisation of carbon changes from sp2 to sp3 in this process, and a tetrahedral alkoxide intermediate is produced
This intermediate captures a proton from the reaction medium to give the electrically neutral product
The net result is addition of Nu– and H+ across the carbon oxygen double bond as shown in Fig
12.2.
Reactivity Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons
Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent
Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former.
Example 12.3
Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal? Explain your answer.
Solution The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than carbon atom of the carbonyl group present in propanal
The polarity of the carbonyl group is reduced in benzaldehyde due to resonance as shown below and hence it is less reactive than propanal.
Some important examples of nucleophilic addition and nucleophilic addition-elimination reactions: (a) Addition of hydrogen cyanide (HCN): Aldehydes and ketones react with hydrogen cyanide (HCN) to yield cyanohydrins
This reaction occurs very slowly with pure HCN
Therefore, it is catalysed by a base and the generated cyanide ion (CN-) being a stronger nucleophile readily adds to carbonyl compounds to yield corresponding cyanohydrin
Cyanohydrins are useful synthetic intermediates
(b) Addition of sodium hydrogensulphite: Sodium hydrogensulphite adds to aldehydes and ketones to form the addition products.
The position of the equilibrium lies largely to the right hand side for most aldehydes and to the left for most ketones due to steric reasons
The hydrogensulphite addition compound is water soluble and can be converted back to the original carbonyl compound by treating it with dilute mineral acid or alkali
Therefore, these are useful for separation and purification of aldehydes.
(c) Addition of Grignard reagents: (refer Unit 11, Class XII)
(d) Addition of alcohols: Aldehydes react with one equivalent of monohydric alcohol in the presence of dry hydrogen chloride to yield alkoxyalcohol intermediate, known as hemiacetals, which further react with one more molecule of alcohol to give a gem-dialkoxy compound known as acetal as shown in the reaction.
Ketones react with ethylene glycol under similar conditions to form cyclic products known as ethylene glycol ketals
Dry hydrogen chloride protonates the oxygen of the carbonyl compounds and therefore, increases the electrophilicity of the carbonyl carbon facilitating the nucleophilic attack of ethylene glycol
Acetals and ketals are hydrolysed with aqueous mineral acids to yield corresponding aldehydes and ketones respectively
(e) Addition of ammonia and its derivatives: Nucleophiles, such as ammonia and its derivatives H2N-Z add to the carbonyl group of aldehydes and ketones
The reaction is reversible and catalysed by acid
The equilibrium favours the product formation due to rapid dehydration of the intermediate to form >C=N-Z.
Z = Alkyl, aryl, OH, NH2, C6H5NH, NHCONH2, etc.
Table 12.2: Some N-Substituted Derivatives of Aldehydes and Ketones (>C=N-Z)
* 2,4-DNP-derivatives are yellow, orange or red solids, useful for characterisation of aldehydes and ketones.
2
Reduction
Reduction to alcohols: Aldehydes and ketones are reduced to primary and secondary alcohols respectively by sodium borohydride (NaBH4) or lithium aluminium hydride (LiAlH4) as well as by catalytic hydrogenation (Unit 11, Class XII)
Reduction to hydrocarbons: The carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with zinc-amalgam and concentrated hydrochloric acid [Clemmensen reduction] or with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol (Wolff-Kishner reduction).
Bernhard Tollens (1841-1918) was a Professor of Chemistry at the University of Gottingen, Germany.
3
Oxidation Aldehydes differ from ketones in their oxidation reactions
Aldehydes are easily oxidised to carboxylic acids on treatment with common oxidising agents like nitric acid, potassium permanganate, potassium dichromate, etc
Even mild oxidising agents, mainly Tollens’ reagent and Fehlings’ reagent also oxidise aldehydes.
Ketones are generally oxidised under vigorous conditions, i.e., strong oxidising agents and at elevated temperatures
Their oxidation involves carbon-carbon bond cleavage to afford a mixture of carboxylic acids having lesser number of carbon atoms than the parent ketone.
The mild oxidising agents given below are used to distinguish aldehydes from ketones: Tollens’ test: On warming an aldehyde with freshly prepared ammoniacal silver nitrate solution (Tollens’ reagent), a bright silver mirror is produced due to the formation of silver metal
The aldehydes are oxidised to corresponding carboxylate anion
The reaction occurs in alkaline medium.
Fehling’s test: Fehling reagent comprises of two solutions, Fehling solution A and Fehling solution B
Fehling solution A is aqueous copper sulphate and Fehling solution B is alkaline sodium potassium tartarate (Rochelle salt)
These two solutions are mixed in equal amounts before test
On heating an aldehyde with Fehling’s reagent, a reddish brown precipitate is obtained
Aldehydes are oxidised to corresponding carboxylate anion
Aromatic aldehydes do not respond to this test.
Oxidation of methyl ketones by haloform reaction: Aldehydes and ketones having at least one methyl group linked to the carbonyl carbon atom (methyl ketones) are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound
The methyl group is converted to haloform
This oxidation does not affect a carbon-carbon double bond, if present in the molecule.
Iodoform reaction with sodium hypoiodite is also used for detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.
Example 12.4
An organic compound (A) with molecular formula C8H8O forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide
It neither reduces Tollens’ or Fehlings’ reagent, nor does it decolourise bromine water or Baeyer’s reagent
On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C7H6O2
Identify the compounds (A) and (B) and explain the reactions involved.
Solution (A) forms 2,4-DNP derivative
Therefore, it is an aldehyde or a ketone
Since it does not reduce Tollens’ or Fehling reagent, (A) must be a ketone
(A) responds to iodoform test
Therefore, it should be a methyl ketone
The molecular formula of (A) indicates high degree of unsaturation, yet it does not decolourise bromine water or Baeyer’s reagent
This indicates the presence of unsaturation due to an aromatic ring
Compound (B), being an oxidation product of a ketone should be a carboxylic acid
The molecular formula of (B) indicates that it should be benzoic acid and compound (A) should, therefore, be a monosubstituted aromatic methyl ketone
The molecular formula of (A) indicates that it should be phenyl methyl ketone (acetophenone)
Reactions are as follows:
Reactions due to a-hydrogen Acidity of α-hydrogens of aldehydes and ketones: The aldehydes and ketones undergo a number of reactions due to the acidic nature of α-hydrogen. The acidity of α-hydrogen atoms of carbonyl compounds is due to the strong electron withdrawing effect of the carbonyl group and resonance stabilisation of the conjugate base.
Aldol condensation: Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively
This is known as Aldol reaction.
The name aldol is derived from the names of the two functional groups, aldehyde and alcohol, present in the products
The aldol and ketol readily lose water to give α,β-unsaturated carbonyl compounds which are aldol condensation products and the reaction is called Aldol condensation
Though ketones give ketols (compounds containing a keto and alcohol groups), the general name aldol condensation still applies to the reactions of ketones due to their similarity with aldehydes.
Cross aldol condensation: When aldol condensation is carried out between two different aldehydes and / or ketones, it is called cross aldol condensation
If both of them contain α-hydrogen atoms, it gives a mixture of four products
This is illustrated below by aldol reaction of a mixture of ethanal and propanal.
Ketones can also be used as one component in the cross aldol reactions.
Other reactions
Cannizzaro reaction: Aldehydes which do not have an α-hydrogen atom, undergo self oxidation and reduction (disproportionation) reaction on heating with concentrated alkali
In this reaction, one molecule of the aldehyde is reduced to alcohol while another is oxidised to carboxylic acid salt.
Electrophilic substitution reaction: Aromatic aldehydes and ketones undergo electrophilic substitution at the ring in which the carbonyl group acts as a deactivating and meta-directing group.
Intext Questions 12.4 Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions
Ethanal, Propanal, Propanone, Butanone
Benzaldehyde, p-Tolualdehyde, p-Nitrobenzaldehyde, Acetophenone
Hint: Consider steric effect and electronic effect
12.5 Predict the products of the following reactions:
(iv)
12.5 Uses of Aldehydes and Ketones In chemical industry aldehydes and ketones are used as solvents, starting materials and reagents for the synthesis of other products
Formaldehyde is well known as formalin (40%) solution used to preserve biological specimens and to prepare bakelite (a phenol-formaldehyde resin), urea-formaldehyde glues and other polymeric products
Acetaldehyde is used primarily as a starting material in the manufacture of acetic acid, ethyl acetate, vinyl acetate, polymers and drugs
Benzaldehyde is used in perfumery and in dye industries
Acetone and ethyl methyl ketone are common industrial solvents
Many aldehydes and ketones, e.g., butyraldehyde, vanillin, acetophenone, camphor, etc
are well known for their odours and flavours.
Carboxylic Acids
Carbon compounds containing a carboxyl functional group, –COOH are called carboxylic acids
The carboxyl group, consists of a carbonyl group attached to a hydroxyl group, hence its name carboxyl
Carboxylic acids may be aliphatic (RCOOH) or aromatic (ArCOOH) depending on the group, alkyl or aryl, attached to carboxylic carbon
Large number of carboxylic acids are found in nature
Some higher members of aliphatic carboxylic acids (C12 – C18) known as fatty acids, occur in natural fats as esters of glycerol
Carboxylic acids serve as starting material for several other important organic compounds such as anhydrides, esters, acid chlorides, amides, etc.
12.6 Nomenclature and Structure of Carboxyl Group
12.6.1 Nomenclature
Since carboxylic acids are amongst the earliest organic compounds to be isolated from nature, a large number of them are known by their common names
The common names end with the suffix –ic acid and have been derived from Latin or Greek names of their natural sources
For example, formic acid (HCOOH) was first obtained from red ants (Latin: formica means ant), acetic acid (CH3COOH) from vinegar (Latin: acetum, means vinegar), butyric acid (CH3CH2CH2COOH) from rancid butter (Latin: butyrum, means butter).
In the IUPAC system, aliphatic carboxylic acids are named by replacing the ending –e in the name of the corresponding alkane with – oic acid
In numbering the carbon chain, the carboxylic carbon is numbered one
For naming compounds containing more than one carboxyl group, the alkyl chain leaving carboxyl groups is numbered and the number of carboxyl groups is indicated by adding the multiplicative prefix, dicarboxylic acid, tricarboxylic acid, etc
to the name of parent alkyl chain
The position of –COOH groups are indicated by the arabic numeral before the multiplicative prefix
Some of the carboxylic acids along with their common and IUPAC names are listed in Table 12.3.
Table 12.3 Names and Structures of Some Carboxylic Acids
12.6.2 Structure of Carboxyl Group
In carboxylic acids, the bonds to the carboxyl carbon lie in one plane and are separated by about 120°
The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible resonance structure shown below:
Intext Question 12.6 Give the IUPAC names of the following compounds: Ph CH2CH2COOH (CH3)2C=CHCOOH (iv)
12.7 Methods of Preparation of Carboxylic Acids
Some important methods of preparation of carboxylic acids are as follows
1
From primary alcohols and aldehydes Primary alcohols are readily oxidised to carboxylic acids with common oxidising agents such as potassium permanganate (KMnO4) in neutral, acidic or alkaline media or by potassium dichromate (K2Cr2O7) and chromium trioxide (CrO3) in acidic media (Jones reagent).
Carboxylic acids are also prepared from aldehydes by the use of mild oxidising agents (Section 12.4).
2
From alkylbenzenes Aromatic carboxylic acids can be prepared by vigorous oxidation of alkyl benzenes with chromic acid or acidic or alkaline potassium permanganate
The entire side chain is oxidised to the carboxyl group irrespective of length of the side chain
Primary and secondary alkyl groups are oxidised in this manner while tertiary group is not affected
Suitably substituted alkenes are also oxidised to carboxylic acids with these oxidising reagents (refer Unit 13, Class XI).
3
From nitriles and amides Nitriles are hydrolysed to amides and then to acids in the presence of H+ or as catalyst
Mild reaction conditions are used to stop the reaction at the amide stage.
4
From Grignard reagents Grignard reagents react with carbon dioxide (dry ice) to form salts of carboxylic acids which in turn give corresponding carboxylic acids after acidification with mineral acid.
As we know, the Grignard reagents and nitriles can be prepared from alkyl halides (refer Unit 10, Class XII)
The above methods (3 and 4) are useful for converting alkyl halides into corresponding carboxylic acids having one carbon atom more than that present in alkyl halides (ascending the series).
5
From acyl halides and anhydrides Acid chlorides when hydrolysed with water give carboxylic acids or more readily hydrolysed with aqueous base to give carboxylate ions which on acidification provide corresponding carboxylic acids
Anhydrides on the other hand are hydrolysed to corresponding acid(s) with water.
6
From esters
Acidic hydrolysis of esters gives directly carboxylic acids while basic hydrolysis gives carboxylates, which on acidification give corresponding carboxylic acids.
Example 12.5 Write chemical reactions to affect the following transformations: Butan-1-ol to butanoic acid Benzyl alcohol to phenylethanoic acid 3-Nitrobromobenzene to 3-nitrobenzoic acid (iv) 4-Methylacetophenone to benzene-1,4-dicarboxylic acid (v) Cyclohexene to hexane-1,6-dioic acid
(vi) Butanal to butanoic acid
Solution
(iv)
(v)
(vi)
Intext Question 12.7 Show how each of the following compounds can be converted to benzoic acid
Ethylbenzene Acetophenone Bromobenzene (iv) Phenylethene (Styrene)
12.8 Physical Properties Aliphatic carboxylic acids upto nine carbon atoms are colourless liquids at room temperature with unpleasant odours
The higher acids are wax like solids and are practically odourless due to their low volatility
Carboxylic acids are higher boiling liquids than aldehydes, ketones and even alcohols of comparable molecular masses
This is due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding
The hydrogen bonds are not broken completely even in the vapour phase
In fact, most carboxylic acids exist as dimer in the vapour phase or in the aprotic solvents.
In vapour state or in aprotic solvent
Simple aliphatic carboxylic acids having upto four carbon atoms are miscible in water due to the formation of hydrogen bonds with water
The solubility decreases with increasing number of carbon atoms
Higher carboxylic acids are practically insoluble in water due to the increased hydrophobic interaction of hydrocarbon part
Benzoic acid, the simplest aromatic carboxylic acid is nearly insoluble in cold water
Carboxylic acids are also soluble in less polar organic solvents like benzene, ether, alcohol, chloroform, etc.
Hydrogen bonding of RCOOH with H2O
12.9 Chemical Reactions The reaction of carboxylic acids are classified as follows:
12.9.1 Reactions Involving Cleavage of O–H Bond
Acidity
Reactions with metals and alkalies
The carboxylic acids like alcohols evolve hydrogen with electropositive metals and form salts with alkalies similar to phenols
However, unlike phenols they react with weaker bases such as carbonates and hydrogencarbonates to evolve carbon dioxide
This reaction is used to detect the presence of carboxyl group in an organic compound.
Carboxylic acids dissociate in water to give resonance stabilised carboxylate anions and hydronium ion.
For the above reaction:
where Keq, is equilibrium constant and Ka is the acid dissociation constant
For convenience, the strength of an acid is generally indicated by its pka value rather than its Ka value
pKa = – log Ka The pKa of hydrochloric acid is –7.0, where as pKa of trifluoroacetic acid (the strongest carboxylic acid), benzoic acid and acetic acid are 0.23, 4.19 and 4.76, respectively
Smaller the pKa, the stronger the acid ( the better it is as a proton donor)
Strong acids have pKa values < 1, the acids with pKa values between 1 and 5 are considered to be moderately strong acids, weak acids have pKa values between 5 and 15, and extremely weak acids have pKa values >15.
Carboxylic acids are weaker than mineral acids, but they are stronger acids than alcohols and many simple phenols (pKa is ~16 for ethanol and 10 for phenol)
In fact, carboxylic acids are amongst the most acidic organic compounds you have studied so far
You already know why phenols are more acidic than alcohols
The higher acidity of carboxylic acids as compared to phenols can be understood similarly
The conjugate base of carboxylic acid, a carboxylate ion, is stabilised by two equivalent resonance structures in which the negative charge is at the more electronegative oxygen atom
The conjugate base of phenol, a phenoxide ion, has non-equivalent resonance structures in which the negative charge is at the less electronegative carbon atom
Therefore, resonance in phenoxide ion is not as important as it is in carboxylate ion
Further, the negative charge is delocalised over two electronegative oxygen atoms in carboxylate ion whereas it is less effectively delocalised over one oxygen atom and less electronegative carbon atoms in phenoxide ion (Unit 11, Class XII)
Thus, the carboxylate ion is more stabilised than phenoxide ion, so carboxylic acids are more acidic than phenols.
Effect of substituents on the acidity of carboxylic acids: Substituents may affect the stability of the conjugate base and thus, also affect the acidity of the carboxylic acids
Electron withdrawing groups increase the acidity of carboxylic acids by stabilising the conjugate base through delocalisation of the negative charge by inductive and/or resonance effects
Conversely, electron donating groups decrease the acidity by destabilising the conjugate base.
Electron withdrawing group (EWG) stabilises the carboxylate anion and strengthens the acid
Electron donating group (EDG) destabilises the carboxylate anion and weakens the acid
The effect of the following groups in increasing acidity order is Ph < I < Br < Cl < F < CN < NO2 < CF3 Thus, the following acids are arranged in order of increasing acidity (based on pKa values): CF3COOH > CCl3COOH > CHCl2COOH > NO2CH2COOH > NC-CH2COOH >
FCH2COOH > ClCH2COOH > BrCH2COOH > HCOOH > ClCH2CH2COOH > (continue) C6H5COOH > C6H5CH2COOH > CH3COOH > CH3CH2COOH (continue ) Direct attachment of groups such as phenyl or vinyl to the carboxylic acid, increases the acidity of corresponding carboxylic acid, contrary to the decrease expected due to resonance effect shown below:
This is because of greater electronegativity of sp2 hybridised carbon to which carboxyl carbon is attached
The presence of electron withdrawing group on the phenyl of aromatic carboxylic acid increases their acidity while electron donating groups decrease their acidity.
12.9.2 Reactions Involving Cleavage of C–OH Bond 1
Formation of anhydride Carboxylic acids on heating with mineral acids such as H2SO4 or with P2O5 give corresponding anhydride.
2
Esterification Carboxylic acids are esterified with alcohols or phenols in the presence of a mineral acid such as concentrated H2SO4 or HCl gas as a catalyst.
Mechanism of esterification of carboxylic acids: The esterification of carboxylic acids with alcohols is a kind of nucleophilic acyl substitution
Protonation of the carbonyl oxygen activates the carbonyl group towards nucleophilic addition of the alcohol
Proton transfer in the tetrahedral intermediate converts the hydroxyl group into –+OH2 group, which, being a better leaving group, is eliminated as neutral water molecule
The protonated ester so formed finally loses a proton to give the ester.
Reactions with PCl5, PCl3 and SOCl2 The hydroxyl group of carboxylic acids, behaves like that of alcohols and is easily replaced by chlorine atom on treating with PCl5, PCl3 or SOCl2
Thionyl chloride (SOCl2) is preferred because the other two products are gaseous and escape the reaction mixture making the purification of the products easier.
4
Reaction with ammonia Carboxylic acids react with ammonia to give ammonium salt which on further heating at high temperature give amides
For example:
12.9.3 Reactions Involving –COOH Group 1
Reduction Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride or better with diborane
Diborane does not easily reduce functional groups such as ester, nitro, halo, etc
Sodium borohydride does not reduce the carboxyl group.
Decarboxylation Carboxylic acids lose carbon dioxide to form hydrocarbons when their sodium salts are heated with sodalime (NaOH and CaO in the ratio of 3 : 1)
The reaction is known as decarboxylation.
Alkali metal salts of carboxylic acids also undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid
The reaction is known as Kolbe electrolysis (Unit 13, Class XI).
12.9.4
Substitution Reactions in the Hydrocarbon Part 1
Halogenation
Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids
The reaction is known as Hell-Volhard-Zelinsky reaction.
Ring substitution
Aromatic carboxylic acids undergo electrophilic substitution reactions in which the carboxyl group acts as a deactivating and meta-directing group
They however, do not undergo Friedel-Crafts reaction (because the carboxyl group is deactivating and the catalyst aluminium chloride (Lewis acid) gets bonded to the carboxyl group).
Intext Question 12.8 Which acid of each pair shown here would you expect to be stronger? CH3CO2H or CH2FCO2H CH2FCO2H or CH2ClCO2H CH2FCH2CH2CO2H or CH3CHFCH2CO2H
(iv)
12.10 Uses of Carboxylic Acids
Methanoic acid is used in rubber, textile, dyeing, leather and electroplating industries
Ethanoic acid is used as solvent and as vinegar in food industry
Hexanedioic acid is used in the manufacture of nylon-6, 6
Esters of benzoic acid are used in perfumery
Sodium benzoate is used as a food preservative
Higher fatty acids are used for the manufacture of soaps and detergents.
Summary Aldehydes, ketones and carboxylic acids are some of the important classes of organic compounds containing carbonyl group
These are highly polar molecules
Therefore, they boil at higher temperatures than the hydrocarbons and weakly polar compounds such as ethers of comparable molecular masses
The lower members are more soluble in water because they form hydrogen bonds with water
The higher members, because of large size of hydrophobic chain of carbon atoms, are insoluble in water but soluble in common organic solvents
Aldehydes are prepared by dehydrogenation or controlled oxidation of primary alcohols and controlled or selective reduction of acyl halides
Aromatic aldehydes may also be prepared by oxidation of methylbenzene with chromyl chloride or CrO3 in the presence of acetic anhydride, formylation of arenes with carbon monoxide and hydrochloric acid in the presence of anhydrous aluminium chloride, and cuprous chloride or by hydrolysis of benzal chloride
Ketones are prepared by oxidation of secondary alcohols and hydration of alkynes
Ketones are also prepared by reaction of acyl chloride with dialkylcadmium
A good method for the preparation of aromatic ketones is the Friedel-Crafts acylation of aromatic hydrocarbons with acyl chlorides or anhydrides
Both aldehydes and ketones can be prepared by ozonolysis of alkenes
Aldehydes and ketones undergo nucleophilic addition reactions onto the carbonyl group with a number of nucleophiles such as, HCN, NaHSO3, alcohols (or diols), ammonia derivatives, and Grignard reagents
The α-hydrogens in aldehydes and ketones are acidic
Therefore, aldehydes and ketones having at least one α-hydrogen, undergo Aldol condensation in the presence of a base to give α-hydroxyaldehydes (aldol) and α-hydroxyketones(ketol), respectively
Aldehydes having no α-hydrogen undergo Cannizzaro reaction in the presence of concentrated alkali
Aldehydes and ketones are reduced to alcohols with NaBH4, LiAlH4, or by catalytic hydrogenation
The carbonyl group of aldehydes and ketones can be reduced to a methylene group by Clemmensen reduction or Wolff-Kishner reduction
Aldehydes are easily oxidised to carboxylic acids by mild oxidising reagents such as Tollens’ reagent and Fehling’s reagent
These oxidation reactions are used to distinguish aldehydes from ketones
Carboxylic acids are prepared by the oxidation of primary alcohols, aldehydes and alkenes by hydrolysis of nitriles, and by treatment of Grignard reagents with carbon dioxide
Aromatic carboxylic acids are also prepared by side-chain oxidation of alkylbenzenes
Carboxylic acids are considerably more acidic than alcohols and most of simple phenols
Carboxylic acids are reduced to primary alcohols with LiAlH4, or better with diborane in ether solution and also undergo α-halogenation with Cl2 and Br2 in the presence of red phosphorus (Hell-Volhard Zelinsky reaction)
Methanal, ethanal, propanone, benzaldehyde, formic acid, acetic acid and benzoic acid are highly useful compounds in industry.
Exercises
12.1 What is meant by the following terms ? Give an example of the reaction in each case
Cyanohydrin Acetal Semicarbazone (iv) Aldol (v) Hemiacetal (vi) Oxime (vii) Ketal (vii) Imine (ix) 2,4-DNP-derivative (x) Schiff’s base 12.2 Name the following compounds according to IUPAC system of nomenclature: CH3CH(CH3)CH2CH2CHO CH3CH2COCH(C2H5)CH2CH2Cl CH3CH=CHCHO (iv) CH3COCH2COCH3 (v) CH3CH(CH3)CH2C(CH3)2COCH3 (vi) (CH3)3CCH2COOH (vii) OHCC6H4CHO-p 12.3 Draw the structures of the following compounds
3-Methylbutanal p-Nitropropiophenone p-Methylbenzaldehyde (iv) 4-Methylpent-3-en-2-one (v) 4-Chloropentan-2-one (vi) 3-Bromo-4-phenylpentanoic acid (vii) p,p’-Dihydroxybenzophenone (viii) Hex-2-en-4-ynoic acid 12.4 Write the IUPAC names of the following ketones and aldehydes
Wherever possible, give also common names
CH3CO(CH2)4CH3 CH3CH2CHBrCH2CH(CH3)CHO CH3(CH2)5CHO (iv) Ph-CH=CH-CHO (v) (vi) PhCOPh 12.5 Draw structures of the following derivatives
The 2,4-dinitrophenylhydrazone of benzaldehyde Cyclopropanone oxime Acetaldehydedimethylacetal (iv) The semicarbazone of cyclobutanone (v) The ethylene ketal of hexan-3-one (vi) The methyl hemiacetal of formaldehyde 12.6 Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents
PhMgBr and then H3O+ Tollens’ reagent Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid 12.7 Which of the following compounds would undergo aldol condensation, which the Cannizzaro reaction and which neither? Write the structures of the expected products of aldol condensation and Cannizzaro reaction
Methanal 2-Methylpentanal Benzaldehyde (iv) Benzophenone (v) Cyclohexanone (vi) 1-Phenylpropanone (vii) Phenylacetaldehyde (viii) Butan-1-ol (ix) 2,2-Dimethylbutanal 12.8 How will you convert ethanal into the following compounds? Butane-1,3-diol But-2-enal But-2-enoic acid 12.9 Write structural formulas and names of four possible aldol condensation products from propanal and butanal
In each case, indicate which aldehyde acts as nucleophile and which as electrophile
12.10 An organic compound with the molecular formula C9H10O forms 2,4-DNP derivative, reduces Tollens’ reagent and undergoes Cannizzaro reaction
On vigorous oxidation, it gives 1,2-benzenedicarboxylic acid
Identify the compound
12.11 An organic compound (A) (molecular formula C8H16O2) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C)
Oxidation of (C) with chromic acid produced (B)
(C) on dehydration gives but-1-ene
Write equations for the reactions involved
12.12 Arrange the following compounds in increasing order of their property as indicated: Acetaldehyde, Acetone, Di-tert-butyl ketone, Methyl tert-butyl ketone (reactivity towards HCN) CH3CH2CH(Br)COOH, CH3CH(Br)CH2COOH, (CH3)2CHCOOH, CH3CH2CH2COOH (acid strength) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength) 12.13 Give simple chemical tests to distinguish between the following pairs of compounds
Propanal and Propanone Acetophenone and Benzophenone Phenol and Benzoic acid (iv) Benzoic acid and Ethyl benzoate (v) Pentan-2-one and Pentan-3-one (vi) Benzaldehyde and Acetophenone (vii) Ethanal and Propanal 12.14 How will you prepare the following compounds from benzene? You may use any inorganic reagent and any organic reagent having not more than one carbon atom Methyl benzoate m-Nitrobenzoic acid p-Nitrobenzoic acid (iv) Phenylacetic acid (v) p-Nitrobenzaldehyde
12.15 How will you bring about the following conversions in not more than two steps? Propanone to Propene Benzoic acid to Benzaldehyde Ethanol to 3-Hydroxybutanal (iv) Benzene to m-Nitroacetophenone (v) Benzaldehyde to Benzophenone (vi) Bromobenzene to 1-Phenylethanol (vii) Benzaldehyde to 3-Phenylpropan-1-ol (viii) Benazaldehyde to α-Hydroxyphenylacetic acid (ix) Benzoic acid to m- Nitrobenzyl alcohol 12.16 Describe the following: Acetylation Cannizzaro reaction Cross aldol condensation (iv) Decarboxylation 12.17 Complete each synthesis by giving missing starting material, reagent or products
12.18 Give plausible explanation for each of the following: Cyclohexanone forms cyanohydrin in good yield but 2,2,6-trimethylcyclo-hexanone does not
There are two –NH2 groups in semicarbazide
However, only one is involved in the formation of semicarbazones
During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed
12.19 An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen
The molecular mass of the compound is 86
It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test
On vigorous oxidation it gives ethanoic and propanoic acid
Write the possible structure of the compound
12.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol
Why?
Answers to Some Intext Questions 12.1
12.2
12.3 CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH
12.4 Butanone < Propanone < Propanal < Ethanal Acetophenone < p-Tolualdehyde , Benzaldehyde < p-Nitrobenzaldehyde
12.5
12.6 3-Phenylpropanoic acid 3-Methylbut-2-enoic acid 2-Methylcyclopentanecarboxylic acid
(iv) 2,4,6-Trinitrobenzoic acid 12.7
12.8
Table of Contents
Unit 12
Aldehydes, Ketones and Carboxylic Acids
12.1 Nomenclature and Structure of Carbonyl Group
12.1.1 Nomenclature
12.1.2 Structure of the Carbonyl Group
12.2 Preparation of Aldehydes and Ketones
12.2.1 Preparation of Aldehydes and Ketones
12.2.2 Preparation of Aldehydes
12.2.3 Preparation of Ketones
12.3 Physical Properties
12.4 Chemical Reactions
12.5 Uses of Aldehydes and Ketones
12.6 Nomenclature and Structure of Carboxyl Group
12.6.1 Nomenclature
12.6.2 Structure of Carboxyl Group
12.7 Methods of Preparation of Carboxylic Acids
12.9 Chemical Reactions
12.9.1 Reactions Involving Cleavage of O–H Bond
12.9.2 Reactions Involving Cleavage of C–OH Bond
12.9.3 Reactions Involving –COOH Group
12.10 Uses of Carboxylic Acids
Summary
Exercises
Answers to Some Intext Questions
Landmarks
Cover
Chapter 12 Aldehydes, Ketones and Carboxylic Acids
Figure 1.7 Solution Let the original charge on sphere A be q and that on B be q′
At a distance r between their centres, the magnitude of the electrostatic force on each is given by 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
Thus the electrostatic force on A, due to B, remains unaltered
7 Forces between Multiple Charges 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 (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
Thus, F12 In the same way, the force on q1 due to q3, denoted by F13, is given by
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 (1.4) The above calculation of force can be generalised to a system of charges more than three, as shown in (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.,
Figure 1.8 A system of (a) three charges (b) multiple charges.
(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 ?
Figure 1.9 Solution In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = () l and the distance AO of the centroid O from A is (2/3) AD = () l
By symmatry AO = BO = CO
Thus, Force F1 on Q due to charge q at A = along AO Force F2 on Q due to charge q at B = along BO Force F3 on Q due to charge q at C = along CO The resultant of forces F2 and F3 is along OA, by the parallelogram law
Therefore, the total force on Q = = 0, where 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 60° about O
Example 1.7 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in 0
What is the force on each charge?
Figure 1.10 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in 0
By the parallelogram law, the total force F1 on the charge q at A is given by F1 = F where is a unit vector along BC
The force of attraction or repulsion for each pair of charges has the same magnitude The total force F2 on charge q at B is thus F2 = F 2, where 2 is a unit vector along AC
Similarly the total force on charge –q at C is F3 = F , where 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 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
1.8 Electric Field 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 (1.6) where 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 (1.7) 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: From Eq
(1.8), we can infer that if q is unity, the electric field due to a charge Q is numerically equal to the force exerted by it
Thus, the electric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed 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: (1.9)
Figure 1.11 Electric field (a) due to a charge Q, (b) due to a charge –Q.
A practical way to get around the problem (of keeping Q undisturbed 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
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
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
1.8.1 Electric field due to a system of charges 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
Electric field E1 at r due to q1 at r1 is given by E1 = where 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 E2 = where is a unit vector in the direction from q2 to P and r2P is the distance between q2 and P
Similar expressions hold good for fields E3, E4, ..., En due to charges q3, q4, ..., qn
By the superposition principle, the electric field E at r due to the system of charges is (as shown in 2) E(r) = E1 (r) + E2 (r) + … + En(r) = E(r) (1.10) 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
8.2 Physical significance of electric field 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.
Figure 1.12 Electric field at a point due to a system of charges is the vector sum of the electric fields at the point due to individual charges.
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 is c, the speed of light
Thus, the effect of any motion of q1 on q2 cannot 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 for a short interval of time and then 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
Example 1.8 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1
The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance
Compute the time of fall in each case
Contrast the situation with that of ‘free fall under gravity’
Figure 1.13
Solution In 3(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 distance h is given by 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 3 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE
The acceleration of the proton is 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
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:
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 4
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 = 3.6 × 104 N C–1 The electric field vector E2A at A due to the negative charge q2 points 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
The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude = 3.6 × 104 N C–1 The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude = 4 × 103 N C–1 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 × 103 N C–1 The directions in which these two vectors point are indicated in 4
The resultant of these two vectors is = 9 × 103 N C–1 EC points towards the right
1.9 Electric Field Lines 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, 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
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
Having drawn a certain set of field lines, the relative density (i.e., closeness) of the field lines at different points indicates 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 generalisation of angle to three dimensions
Recall how a (plane) angle is defined in two-dimensions
Let a small transverse line element ∆l be placed at a distance r from a point O
Then the angle subtended by ∆l at O can be approximated as ∆θ = ∆l/r
Likewise, in three-dimensions the solid angle* subtended by a small perpendicular plane area ∆S, at a distance r, can be written as ∆Ω = ∆S/r2
We know that in a given solid angle the number of radial field lines is the same
In 6, for two points P1 and P2 at distances r1 and r2 from the charge, the element of area subtending the solid angle ∆Ω is ∆Ω at P1 and an element of area ∆Ω 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/(∆Ω) at P1 and n/(∆Ω) at P2, respectively
Since n and ∆Ω are common, the strength of the field clearly has a 1/r2 dependence.
Figure 1.15 Field of a point charge.
The picture of field lines was invented by Faraday to develop an intuitive non-mathematical way of visualising 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, 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 1.16 Dependence of electric field strength on the distance and its relation to the number of field lines.
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: Field lines start from positive charges and end at negative charges
If there is a single charge, they may start or end at infinity
In a charge-free region, electric field lines can be taken to be continuous curves without any breaks
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.)
Figure 1.17 Field lines due to some simple charge configurations.
(iv) Electrostatic field lines do not form any closed loops
This follows from the conservative nature of electric field (Chapter 2)
10 Electric Flux 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.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 ofliquid flow
In the picture of electric field lines described above, we saw that the number of field lines crossing a unit area, placed normal to the field at a point is a measure of the strength of electric field at that point
This means that if we place a small planar element of area ∆S normal to E at a point, the number of field lines crossing it is proportional* to E ∆S
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 ∆S cosθ
Thus, the number of field lines crossing ∆S is proportional to E ∆S cosθ
When θ = 90°, field lines will be parallel to ∆S and will not cross it at all .
Figure 1.18 Dependence of flux on the inclination θ between E and .
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 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 9
Thus, the area element vector ∆S at a point on a closed surface equals ∆S where ∆S is the magnitude of the area element and is a unit vector in the direction of outward normal at that point
We now come to the definition of electric flux
Electric flux ∆φ through an area element ∆S is defined by ∆φ = E.∆S = E ∆S 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 ∆S
For a closed surface, with the convention stated already, θ is the angle between E and the outward normal to the area element
Notice we could look at the expression E ∆S cosθ in two ways: E (∆S cosθ ) i.e., E times the projection of area normal to E, or E⊥ ∆S, 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.∆S (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 ∆S → 0 and the sum in Eq
(1.12) is written as an integral
11 Electric Dipole 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.
Figure 1.19 Convention for defining normal and ∆S.
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/r2 (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: 1.11.1 The field of an electric dipole 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: when the point is on the dipole axis, and 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
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 0(a)
Then [1.13(a)] where is the unit vector along the dipole axis (from –q to q)
Also [1.13(b)] The total field at P is (1.14) For r >> a (r >> a) (1.15) For points on the equatorial plane The magnitudes of the electric fields due to the two charges +q and –q are given by [1.16(a)] [1.16(b)] and are equal
The directions of E+q and E–q are as shown in 0(b)
Clearly, the components normal to the dipole axis cancel away
The components along the dipole axis add up
The total electric field is opposite to
We have E = – (E +q + E –q) cosθ (1.17) At large distances (r >> a), this reduces to (1.18) 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 (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 (r >> a) (1.20)
Figure 1.20 Electric field of a dipole at (a) a point on the axis, (b) a point on the equatorial plane of the dipole
p is the dipole moment vector of magnitude p = q × 2a and directed from –q to q.
At a point on the equatorial plane (r >> a) (1.21) Notice the important point that the dipole field at large distances falls off not as 1/r2 but as1/r3
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
1.11.2 Physical significance of dipoles 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
Example 1.10 Two charges ±10 µC are placed 5.0 mm apart
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 1(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 1(b)
fIGURE 1.21 Solution (a) Field at P due to charge +10 µC = = 4.13 × 106 N C–1 along BP Field at P due to charge –10 µC = 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 (r/a >> 1) 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, E = = 2.6 × 105 N C–1 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 = = 3.99 × 106 N C–1 along BQ Field at Q due to charge –10 µC at A = = 3.99 × 106 N C–1 along QA
Clearly, the components of these two forces with equal magnitudes 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
* Centre of a collection of positive point charges is defined much the same way as the centre of mass: .
= 2 × along BA = 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 the axis of the dipole: (r/a >> 1)
= 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
1.12 Dipole in a Uniform External Field Consider a permanent dipole of dipole moment p in a uniform external field E, as shown in 2
(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 each force multiplied by the arm of the couple (perpendicular 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 paper
The comb, as we know, acquires charge through friction
But the paper is not charged
What then explains the attractive force? Taking the clue from the preceding discussion, the charged comb ‘polarises’ 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! 13 Continuous Charge Distribution 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 ∆S 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 ∆Q on that element
We then define a surface charge density σ at the area element by
Figure 1.22 Dipole in a uniform electric field.
(1.23) 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 ∆S 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 charge distribution
The linear charge density λ of a wire is defined by
Figure 1.23 Electric force on a dipole: (a) E parallel to p, (b) E antiparallel to p.
(1.24) where ∆l is a small line element of wire on the macroscopic scale that, however, includes a large number of microscopic charged constituents, and ∆Q is the charge contained in that line element
The units for λ are C/m
The volume charge density (sometimes simply called charge density) is defined in a similar manner: (1.25) where ∆Q is the charge included in the macroscopically small volume element ∆V 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 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.
Figure 1.24 Definition of linear, surface and volume charge densities
In each case, the element (∆l, ∆S, ∆V) chosen is small on the macroscopic scale but contains a very large number of microscopic constituents.
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 ∆V
The charge in a volume element ∆V is ρ∆V
Now, consider any general point P (inside or outside the distribution) with position vector R
Electric field due to the charge ρ∆V is given by Coulomb’s law: (1.26) where r′ is the distance between the charge element and P, and ′ 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: (1.27) Note that ρ, r′, all can vary from point to point
In a strict mathematical method, we should let ∆V→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
1.14 Gauss’s Law 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 5
The flux through an area element ∆S is (1.28) where we have used Coulomb’s law for the electric field due to a single charge q
The unit vector 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 ∆S and have the same direction
Therefore,
* At the microscopic level, charge distribution is discontinuous, because they are discrete charges separated by intervening space where there is no charge.
(1.29) since the magnitude of a unit vector is 1
The total flux through the sphere is obtained by adding up flux through all the different area elements: Since each area element of the sphere is at the same distance r from the charge,
Now S, the total area of the sphere, equals 4πr2
Thus, (1.30) 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
Figure 1.25 Flux through a sphere enclosing a point charge q at its centre.
= 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 6
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: Gauss’s law is true for any closed surface, no matter what its shape or size
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
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, represents only the total charge inside S
Figure 1.26 Calculation of the flux of uniform electric field through the surface of a cylinder.
(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
Example 1.11 The electric field components in 7 are 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 ∆S is ± π/2
Therefore, the flux φ = E.∆S 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 φL= EL.∆S = =EL ∆S cosθ = –EL ∆S, since θ = 180° = –ELa2 φR= ER.∆S = ER ∆S cosθ = ER ∆S, since θ = 0° = ERa2 Net flux through the cube = φR + φL = ERa2 – ELa2 = a2 (ER – EL) = αa2 [(2a)1/2 – a1/2] = αa5/2 = 800 (0.1)5/2 = 1.05 N m2 C–1 (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. Example 1.12 An electric field is uniform, and in the positive x 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 N/C for x > 0 and E = –200 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
(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 ∆S are parallel
Therefore, the outward flux is φL= E.∆S = – 200 = + 200 ∆S, since = – ∆S = + 200 × π (0.05)2 = + 1.57 N m2 C–1 On the right face, E and ∆S are parallel and therefore φR = E.∆S = + 1.57 N m2 C–1
(b) For any point on the side of the cylinder E is perpendicular to ∆S and hence E.∆S = 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 (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 1.15 Applications of Gauss’s Law 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
1.15.1 Field due to an infinitely 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 9
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 9(b)
Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero
At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r
The surface area of the curved part is 2πrl, where l is the length of the cylinder
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 i.e., E = Vectorially, E at any point is given by (1.32) where 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.
Figure 1.29 (a) Electric field due to an infinitely long thin straight wire is radial, (b) The Gaussian surface for a long thin wire of uniform linear charge density.
Note that when we write a vector A as a scalar multiplied by a unit vector, i.e., as A = A , 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 if A > 0 and opposite to if A < 0
When we want to restrict to non-negative values, we use the symbol and call it the modulus of A
Thus,
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
15.2 Field due to a uniformly charged infinite plane sheet Let σ be the uniform surface charge density of an infinite plane sheet
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.∆S through both the surfaces are equal and add up
Therefore the net flux through the Gaussian surface is 2 EA
The charge enclosed by the closed surface is σA
Therefore by Gauss’s law, 2 EA = σA/ε0 or, E = σ/2ε0 Vectorically, (1.33) where 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
15.3 Field due to a uniformly charged thin spherical shell Let σ be the uniform surface charge density of a thin spherical shell of radius R
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)
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 ∆S at every point are parallel and the flux through each element is E ∆S
Summing over all ∆S, the flux through the Gaussian surface is E × 4 π r2
The charge enclosed is σ × 4 π R2
By Gauss’s law E × 4 π r2 = Or, where q = 4 π R2 σ is the total charge on the spherical shell
Vectorially, (1.34) 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.
Figure 1.30 Gaussian surface for a uniformly charged infinite plane sheet.
Field inside the shell: In 1(b), the point P is inside the shell
The Gaussian surface is again a sphere through P centred at O
The flux through the Gaussian surface, calculated as before, is E × 4 π r2
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
Example 1.13 An early model for an atom considered it to have a 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 1.31 Gaussian surfaces for a point with (a) r > R, (b) r < R.
Figure 1.32 Solution The charge distribution for this model of the atom is as shown in 2
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 or 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 spherical surface centred at the nucleus
We consider two situations, namely, r < R and r > R
r < R : The electric flux φ enclosed by the spherical surface is φ = E (r) × 4 π r2 where E (r) is the magnitude of the electric field at r
This is because 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, i.e., Substituting for the charge density ρ obtained earlier, we have Gauss’s law then gives, The electric field is directed radially outward
r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral
Thus, from Gauss’s law, E (r) × 4 π r2 = 0 or E (r) = 0; r > R At r = R, both cases give the same result: E = 0
Summary 1. Electric and magnetic forces determine the properties of atoms, 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
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 r21 separating them
Mathematically, F21 = force on q2 due to where is a unit vector in the direction from q1 to q2 and k = 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
* Compare this with a uniform mass shell discussed in Section 8.5 of Class XI Textbook of Physics.
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 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|/4πε0r2; 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
On symmetry operations In Physics, we often encounter systems with various symmetries
Consideration of these 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 ofthe sheet
Compare this with the effort needed to arrive at this result by a direct calculation using Coulomb’s law.
10. Some of the important properties of field lines are: Field lines are continuous curves without any breaks
Two field lines cannot cross each other
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:
Dipole electric field on the axis at a distance r from the centre:
The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2 dependence of electric field due to a point charge
13. In a uniform electric field E, a dipole experiences a torque given by = p × E but experiences no net force
. The flux ∆φ of electric field E through a small area element ∆S is given by ∆φ = E.∆S The vector area element ∆S is ∆S = ∆S where ∆S is the magnitude of the area element and is normal to the area element, which can be considered planar for sufficiently small ∆S
For an area element of a closed surface, is taken to be the direction 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: Thin infinitely long straight wire of uniform linear charge density λ
where r is the perpendicular distance of the point from the wire and is the radial unit vector in the plane normal to the wire passing through the point
Infinite thin plane sheet of uniform surface charge density σ
where is a unit vector normal to the plane, outward on either side
Thin spherical shell of uniform surface charge density σ
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
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
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
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/r2, typical of field due to a single charge
An electric dipole is the simplest example of this fact
Exercises
Physical quantity Symbol Dimensions Unit Remarks Vector area element ∆S [L2] m2 ∆S = ∆S Electric field E [MLT–3A–1] V m–1 Electric flux φ [ML3 T–3A–1] V m ∆φ = E.∆S Dipole moment p [LTA] C m Vector directed from negative to positive charge Charge density: linear λ [L–1 TA] C m–1 Charge/length surface σ [L–2 TA] C m–2 Charge/area volume ρ [L–3 TA] C m–3 Charge/volume
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
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? 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 separated by a distance of 50 cm
What is the mutual force of 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 4
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 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 (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 5 cannot possibly represent electrostatic field lines? 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 6(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
(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), where is the unit vector in the outward normal direction, and σ is the surface charge density near the hole
30 Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E 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
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
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 3)
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.)
Interactive animation on simple electrostatic experiments: http://demoweb.physics.ucla.edu/content/100-simple-electrostatic-experiments
Example 1.1
Interactive animation on charging a two-sphere system by induction: http://www.physicsclassroom.com/mmedia/estatics/itsn.cfm
Example 1.2
Example 1.3
Charles Augustin de Coulomb (1736 –1806)
Example 1.4
Interactive animation on Coulomb’s law: http://webphysics.davidson.edu/physlet_resources/bu_semester2/menu_semester2.html
Example 1.5
Example 1.5
Example 1.6
Example 1.6
Example 1.7
Example 1.7
* An alternate unit V/m will be introduced in the next chapter.
Example 1.8
Example 1.8
Example 1.9
Example 1.9
* Solid angle is a measure of a cone
Consider the intersection of the given cone with a sphere of radius R
The solid angle ∆Ω of the cone is defined to be equal to ∆S/R2, where ∆S is the area on the sphere cut out by the cone.
* It will not be proper to say that the number of field lines is equal to E∆S
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 a given area at different points.
Example 1.10
Example 1.10
Example 1.10
Example 1.11
Example 1.11
Example 1.12
Example 1.13
Example 1.13
Contents
Chapter-1
Landmarks
Cover
Chapter One
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
1.2 Electric Charge 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 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
Figure 1.1 Rods and pith balls: like charges repel and unlike charges attract each other.
It was observed that if two glass rods rubbed with wool or silk cloth are brought close to each other, they repel each other
The two strands of wool or two pieces of silk cloth, with 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 but attracted the fur
On the other hand, the plastic rod attracts the glass rod and repel the silk or wool with which the glass rod is rubbed
The glass rod repels the fur
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 and are also repelled by the rod
A similar effect is found if the pith balls are touched with a glass rod rubbed with silk
A dramatic observation is that a pith ball touched with glass rod attracts another pith ball touched with plastic rod
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 like charges repel and 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 observations? It just tells us that unlike charges acquired by the objects 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 electrically neutral
A simple apparatus to detect charge on a body is the gold-leaf electroscope
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
Unification of electricity and magnetism 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).
Students can make a simple electroscope as follows : 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.
Figure 1.2 Electroscopes: (a) The gold leaf electroscope, (b) Schematics of a simple 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 iron it lightly with the mountain fold up, as shown in
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.
Figure 1.3 Paper strip experiment.
To electrify a neutral body, we need to add or remove one kind of charge
When we say that a body is charged, we always refer to this 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 losing some of its electrons
Similarly, a body can be charged negatively 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
1.3 Conductors and Insulators 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 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.
* There is a third category called semiconductors, which offer resistance to the movement of charges which is intermediate between the conductors and insulators.
1.4 Charging by Induction 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.
Figure 1.4 Charging by induction.
Bring two metal spheres, A and B, supported on insulating stands, in contact as shown in (a)
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 (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
(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
Separate the spheres by a small distance while the glass rod is still held near sphere A, as shown in (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 (d)
Now, separate the spheres quite apart
The charges on them get uniformly distributed over them, as shown in (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 contact
When electrified rods are brought near light objects, a similar effect takes place
The rods induce opposite charges on the near surfaces of the objects and similar charges move to the farther side of the object
[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 insulating metal stand
Bring a negatively charged rod close to the metallic sphere, as shown in (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 (c)
Disconnect the sphere from the ground
The positive charge continues to be held at the near end
Remove the electrified rod
The positive charge will spread uniformly over the sphere as shown in (e)
Figure 1.5 In this experiment, the metal sphere gets charged by the process of induction and the rod does not lose any of its charge.
Similar steps are involved in charging a metal sphere negatively 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?
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.
1.5.1 Additivity of charges 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 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.
1.5.2 Charge is conserved 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
1.5.3 Quantisation of charge 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 q = ne 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 n2 protons, the total amount of charge on the body is n2 × e + n1 × (–e) = (n2 – 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
In this respect, 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
However, at the microscopic level, where the charges involved are of the order of a few tens or hundreds of e, i.e., they can be counted, they appear in discrete lumps and quantisation of charge cannot be ignored
It is the magnitude of scale involved that is very important.
Example 1.2 If 109 electrons move out of a body to another body 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 unit for many practical purposes
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 × 1024 electrons
Example 1.3 How much positive and negative charge is there in a 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 molecules in one cup of water is (250/18) × 6.02 × 1023.
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.
1.6 Coulomb’s Law 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 (1.1) How did Coulomb arrive at this law from his experiments? Coulomb used a torsion balance* for measuring the force between two charged metallic 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 different pairs of charges at different distances, Coulomb arrived at the relation, Eq
(1.1)
* 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, to verify Newton’s Law of Gravitation.
Coulomb’s law, a simple mathematical statement, was initially experimentally arrived at in the manner described above
While the original experiments established it at a macroscopic scale, it has also been established down to subatomic level (r ~ 10–10 m)
Coulomb discovered his law without knowing the explicit magnitude of the charge
In fact, it is the other way round: Coulomb’s law can now be employed to furnish a definition for a unit of charge
In the relation, Eq
(1.1), k is so far arbitrary
We can choose any positive value of k
The choice of k determines the size of the unit of charge
In SI units, the value of k is about 9 × 109
The unit of charge that results from this choice is called a coulomb which we defined earlier in Section 1.4
Putting this value of k in Eq
(1.1), we see that for q1 = q2 = 1 C, r = 1 m F = 9 × 109 N
Charles Augustin de Coulomb (1736 – 1806) Coulomb, a French physicist, began his career as a military engineer in the West Indies
In 1776, he returned to Paris and retired to a small estate to do his scientific research
He invented a torsion balance to measure the quantity of a force and used it for determination of forces of electric attraction or repulsion between small charged spheres
He thus arrived in 1785 at the inverse square law relation, now known as Coulomb’s law
The law had been anticipated by Priestley and also by Cavendish earlier, though Cavendish never published his results
Coulomb also found the inverse square law of force between unlike and like magnetic poles.
That is, 1 C is the charge that when placed at a distance of 1 m from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude 9 × 109 N
One coulomb is evidently too big a unit to be used
In practice, in 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 (1.2) ε0 is called the permittivity of free space
The value of ε0 in SI units is = 8.854 × 10–12 C2 N–1m–2
* Implicit in this is the assumption of additivity of charges and conservation: two charges (q/2 each) add up to make a total charge q.
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: , Coulomb’s force law between two point charges q1 and q2 located at r1 and r2 is then expressed as (1.3) Some remarks on Eq
(1.3) are relevant:
Figure 1.6 (a) Geometry and (b) Forces between charges
• 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 21, which denotes repulsion, as it should be for like charges
If q1 and q2 are of opposite signs, F21 is along –21(=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
• The force F12 on charge q1 due to charge q2, is obtained from Eq
(1.3), by simply interchanging 1 and 2, i.e., Thus, Coulomb’s law agrees with the Newton’s third law
• 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 matter in the next chapter.
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 and masses respectively.(a) Compare the strength of these forces by determining the ratio of their magnitudes for an electron and a proton and for two protons
(b) Estimate the accelerations of electron and proton due to the electrical force of their mutual attraction when they are1 Å (= 10-10 m) apart? (mp = 1.67 × 10–27 kg, me = 9.11 × 10–31 kg) Solution (a) The electric force between an electron and a proton at a distance r apart is:
where the negative sign indicates that the force is attractive
The corresponding gravitational force (always attractive) is:
where mp and me are the masses of a proton and an electron respectively.
On similar lines, the ratio of the magnitudes of electric force to the gravitational force between two protons at a distance r apart is: 1.3 × 1036 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 |F| == 8.987 × 109 Nm2/C2 × (1.6 ×10–19C)2 / (10–10m)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 can conclude that the effect of gravitational field is negligible on 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 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 (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 (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 (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.
Figure 1.7 Solution Let the original charge on sphere A be q and that on B be q′
At a distance r between their centres, the magnitude of the electrostatic force on each is given by
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
Thus the electrostatic force on A, due to B, remains unaltered.
1.7 Forces between Multiple Charges 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 nstationary 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?
Figure 1.8 A system of (a) three charges (b) multiple charges.
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 theprinciple of superposition
To better understand the concept, consider a system of three charges q1, q2 and q3, as shown in (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. Thus, F12
In the same way, the force on q1 due to q3, denoted by F13, is given by
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 (1.4) The above calculation of force can be generalised to a system of charges more than three, as shown in (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.,
(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 ?
Figure 1.9 Solution In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos 30º = () l and the distance AO of the centroid O from A is (2/3) AD = () l. By symmatry AO = BO = CO
Thus, Force F1 on Q due to charge q at A =along AO Force F2 on Q due to charge q at B =along BO Force F3 on Q due to charge q at C =along CO The resultant of forces F2 and F3 isalong OA, by the parallelogram law
Therefore, the total force on Q == 0, whereis 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 60° about O
Example 1.7 Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in 0
What is the force on each charge?
Figure 1.10 Solution The forces acting on charge q at A due to charges q at B and –q at C are F12 along BA and F13 along AC respectively, as shown in 0
By the parallelogram law, the total force F1 on the charge q at A is given by F1 = Fwhereis a unit vector along BC
The force of attraction or repulsion for each pair of charges has the same magnitude The total force F2 on charge q at B is thus F2 = F2, where2 is a unit vector along AC
Similarly the total force on charge –q at C is F3 =F, whereis the unit vector along the direction bisecting the ∠BCA
It is interesting to see that the sum of the forces on the three charges 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.
1.8 Electric Field 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 offield
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 (1.6) wherer/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 (1.7) 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: From Eq
(1.8), we can infer that if q is unity, the electric field due to a charge Q is numerically equal to the force exerted by it
Thus, the electric field due to a charge Q at a point in space may be defined as the force that a unit positive charge would experience if placed 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: (1.9)
Figure 1.11 Electric field (a) due to a charge Q, (b) due to a charge –Q.
A practical way to get around the problem (of keeping Q undisturbed 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
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
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.
1.8.1 Electric field due to a system of charges 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
Electric field E1 at r due to q1 at r1 is given by E1 = whereis 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 E2 =
whereis a unit vector in the direction from q2 to P and r2P is the distance between q2 and P
Similar expressions hold good for fields E3, E4, ..., En due to charges q3, q4, ..., qn
By the superposition principle, the electric field E at r due to the system of charges is (as shown in 2)
Figure 1.12 Electric field at a point due to a system of charges is the vector sum of the electric fields at the point due to individual charges.
E(r) = E1 (r) + E2 (r) + … + En(r) = E(r)(1.10) 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.
1.8.2 Physical significance of electric field 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 is c, the speed of light
Thus, the effect of any motion of q1 on q2 cannot 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 for a short interval of time and then 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.
Example 1.8 An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 104 N C–1
The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance
Compute the time of fall in each case
Contrast the situation with that of ‘free fall under gravity’.
Figure 1.13
Solution In 3(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 distance h is given by 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 3 (b), the field is downward, and the positively charged proton experiences a downward force of magnitude eE
The acceleration of the proton is 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
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:
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 4.
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 = 3.6 × 104 N C–1 The electric field vector E2A at A due to the negative charge q2 points 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
The electric field vector E1B at B due to the positive charge q1 points towards the left and has a magnitude = 3.6 × 104 N C–1 The electric field vector E2B at B due to the negative charge q2 points towards the right and has a magnitude = 4 × 103 N C–1 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 × 103 N C–1 The directions in which these two vectors point are indicated in 4
The resultant of these two vectors is = 9 × 103 N C–1
EC points towards the right.
1.9 Electric Field Lines 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, 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
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
Having drawn a certain set of field lines, the relative density (i.e., closeness) of the field lines at different points indicates 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.
Figure 1.15 Field of a point charge
To understand the dependence of the field lines on the area, or rather the solid anglesubtended by an area element, let us try to relate the area with the solid angle, a generalisation of angle to three dimensions
Recall how a (plane) angle is defined in two-dimensions
Let a small transverse line element ∆l be placed at a distance r from a point O
Then the angle subtended by ∆l at O can be approximated as ∆θ = ∆l/r
Likewise, in three-dimensions the solid angle* subtended by a small perpendicular plane area ∆S, at a distance r, can be written as ∆Ω = ∆S/r2
We know that in a given solid angle the number of radial field lines is the same
In 6, for two points P1 and P2 at distances r1 and r2 from the charge, the element of area subtending the solid angle ∆Ω is∆Ω at P1 and an element of area
Figure 1.16 Dependence of electric field strength on the distance and its relation to the number of field lines.
∆Ω 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/(∆Ω) at P1 and n/(∆Ω) at P2, respectively
Since n and ∆Ω are common, the strength of the field clearly has a 1/r2dependence
The picture of field lines was invented by Faraday to develop an intuitive non-mathematical way of visualising 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, 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 Field lines due to some simple charge configurations.
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: Field lines start from positive charges and end at negative charges
If there is a single charge, they may start or end at infinity
In a charge-free region, electric field lines can be taken to be continuous curves without any breaks
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).
1.10 Electric Flux 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 dSand 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 dSis v.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, we saw that the number of field lines crossing a unit area, placed normal to the field at a point is a measure of the strength of electric field at that point
This means that if we place a small planar element of area ∆S normal to E at a point, the number of field lines crossing it is proportional* to E ∆S
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 ∆S cosθ
Thus, the number of field lines crossing ∆S is proportional to E ∆S cosθ
When θ = 90°, field lines will be parallel to ∆Sand will not cross it at all .
Figure 1.18 Dependence of flux on the inclination θ between E and n
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 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 theoutward normal
This is the convention used in 9
Thus, the area element vector ∆S at a point on a closed surface equals ∆Sn where ∆S is the magnitude of the area element and n is a unit vector in the direction of outward normal at that point
We now come to the definition of electric flux
Electric flux ∆φ through an area element ∆S is defined by ∆φ = E.∆S = E ∆S 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 ∆S
For a closed surface, with the convention stated already, θ is the angle between E and the outward normal to the area element
Notice we could look at the expression E ∆S cosθ in two ways: E (∆S cosθ ) i.e., E times the projection of area normal to E, or E⊥ ∆S, 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.∆S (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 ∆S → 0 and the sum in Eq
(1.12) is written as an integral.
Figure 1.19 Convention for defining normal
1.11 Electric Dipole 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/r2 (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:
1.11.1 The field of an electric dipole 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: when the point is on the dipole axis, and 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+qdue to the charge q, by the parallelogram law of vectors
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 0(a)
Then [1.13(a)] whereis the unit vector along the dipole axis (from –q to q)
Also [1.13(b)] The total field at P is
(1.14) For r >> a (r >> a) (1.15)
Figure 1.20 Electric field of a dipole at (a) a point on the axis, (b) a point on the equatorial plane of the dipole. p is the dipole moment vector of magnitude p = q × 2a and directed from –q to q
For points on the equatorial plane The magnitudes of the electric fields due to the two charges +q and –q are given by [1.16(a)] [1.16(b)] and are equal
The directions of E+q and E–q are as shown in 0(b)
Clearly, the components normal to the dipole axis cancel away
The components along the dipole axis add up
The total electric field is opposite to
We have E = – (E +q + E –q) cosθ (1.17) At large distances (r >> a), this reduces to (1.18) From Eqs
(1.15) and (1.18), it is clear that the dipole field at large distances does not involveq 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(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 (r >> a) (1.20)
At a point on the equatorial plane (r >> a) (1.21) Notice the important point that the dipole field at large distances falls off not as 1/r2 but as1/r3
Further, the magnitude and the direction of the dipole field depends not only on the distance rbut 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.
1.11.2 Physical significance of dipoles
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.
Example 1.10 Two charges ±10 µC are placed 5.0 mm apart
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 1(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 1(b).
figure 1.21 Solution (a) Field at P due to charge +10 µC = = 4.13 × 106 N C–1 along BP Field at P due to charge –10 µC
= 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 (r/a >> 1) 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, E == 2.6 × 105 N C–1 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 = = 3.99 × 106 N C–1 along BQ
Field at Q due to charge –10 µC at A = = 3.99 × 106 N C–1 along QA.
Clearly, the components of these two forces with equal magnitudes 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
.
= 2 ×along BA = 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 the axis of the dipole: (r/a >> 1)
= 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.
1.12 Dipole in a Uniform External Field Consider a permanent dipole of dipole moment p in a uniform external field E, as shown in 2
(By permanent dipole, we mean that p exists irrespective of E; it has not been induced byE.) 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 each force multiplied by the arm of the couple (perpendicular distance between the two antiparallel forces).
Figure 1.22 Dipole in a uniform electric field
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 paper
The comb, as we know, acquires charge through friction
But the paper is not charged
What then explains the attractive force? Taking the clue from the preceding discussion, the charged comb ‘polarises’ 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!
1.13 Continuous Charge Distribution 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 ∆S 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 ∆Q on that element
We then define a surface charge density σ at the area element by
(1.23) 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*. σrepresents macroscopic surface charge density, which in a sense, is a smoothed out average of the microscopic charge density over an area element ∆S 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 charge distribution
The linear charge density λ of a wire is defined by Figure 1.23 Electric force on a dipole: (a) E parallel to p, (b) E antiparallel to p.
(1.24) where ∆l is a small line element of wire on the macroscopic scale that, however, includes a large number of microscopic charged constituents, and ∆Q is the charge contained in that line element
The units for λ are C/m
The volume charge density (sometimes simply called charge density) is defined in a similar manner: (1.25) where ∆Q is the charge included in the macroscopically small volume element ∆V 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 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.
Figure 1.24 Definition of linear, surface and volume charge densities. In each case, the element (∆l, ∆S, ∆V) chosen is small on the macroscopic scale but contains a very large number of microscopic constituents.
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 ∆V
The charge in a volume element ∆V is ρ∆V
Now, consider any general point P (inside or outside the distribution) with position vector R
Electric field due to the charge ρ∆V is given by Coulomb’s law: (1.26) where r′ is the distance between the charge element and P, and′ 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: (1.27) Note that ρ, r′,all can vary from point to point
In a strict mathematical method, we should let ∆V→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.
1.14 Gauss’s Law 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 5.
Figure 1.25 Flux through a sphere enclosing a point charge q at its centre.
The flux through an area element ∆S is (1.28) where we have used Coulomb’s law for the electric field due to a single charge q
The unit vectoris 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 ∆S andhave the same direction
Therefore, (1.29) since the magnitude of a unit vector is 1
The total flux through the sphere is obtained by adding up flux through all the different area elements:
Since each area element of the sphere is at the same distance r from the charge,
Now S, the total area of the sphere, equals 4πr2
Thus, (1.30)
Figure 1.26 Calculation of the flux of uniform electric field through the surface of a cylinder.
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 6
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: Gauss’s law is true for any closed surface, no matter what its shape or size
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
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, represents only the total charge inside S
(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.
Example 1.11 The electric field components in 7 are 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 ∆S is ± π/2
Therefore, the flux φ = E.∆S 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 φL= EL.∆S ==EL ∆S cosθ = –EL ∆S, since θ = 180° = –ELa2 φR= ER.∆S = ER ∆S cosθ = ER ∆S, since θ = 0° = ERa2 Net flux through the cube = φR + φL = ERa2 – ELa2 = a2 (ER – EL) = αa2 [(2a)1/2 – a1/2] = αa5/2 = 800 (0.1)5/2
= 1.05 N m2 C–1 (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
Example 1.12 An electric field is uniform, and in the positive x direction for positive x, and uniform with the same magnitude but in the negative x direction for negative x
It is given that E= 200N/C for x > 0 and E = –200N/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
(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 ∆S are parallel
Therefore, the outward flux is φL= E.∆S = – 200 = + 200 ∆S, since= – ∆S = + 200 × π (0.05)2 = + 1.57 N m2 C–1 On the right face, E and ∆S are parallel and therefore φR = E.∆S = + 1.57 N m2 C–1
(b) For any point on the side of the cylinder E is perpendicular to ∆S and hence E.∆S = 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 (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
1.15 Applications of Gauss’s Law 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.
1.15.1 Field due to an infinitely 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 9.
Figure 1.29 (a) Electric field due to an infinitely long thin straight wire is radial, (b) The Gaussian surface for a long thin wire of uniform linear charge density
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 9(b).Since the field is everywhere radial, flux through the two ends of the cylindrical Gaussian surface is zero
At the cylindrical part of the surface, E is normal to the surface at every point, and its magnitude is constant, since it depends only on r
The surface area of the curved part is 2πrl, where l is the length of the cylinder
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 i.e., E = Vectorially, E at any point is given by (1.32) whereis 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, 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 vectorif A > 0 and opposite toif A < 0
When we want to restrict to non-negative values, we use the symboland call it the modulus of A
Thus,
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.
1.15.2 Field due to a uniformly charged infinite plane sheet Let σ be the uniform surface charge density of an infinite plane sheet
We take thex-axis normal to the given plane
By symmetry, the electric field will not depend on y and zcoordinates 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 areaA, 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.
Figure 1.30 Gaussian surface for a uniformly charged infinite plane sheet.
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.∆S through both the surfaces are equal and add up
Therefore the net flux through the Gaussian surface is 2 EA
The charge enclosed by the closed surface is σA
Therefore by Gauss’s law, 2 EA = σA/ε0 or, E = σ/2ε0 Vectorically, (1.33) where n 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 ofx 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.
1.15.3 Field due to a uniformly charged thin spherical shell Let σ be the uniform surface charge density of a thin spherical shell of radius R
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)
Field outside the shell: Consider a point P outside the shell with radius vector r
To calculateE 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 ∆S at every point are parallel and the flux through each element is E ∆S. Summing over all ∆S, the flux through the Gaussian surface is E × 4 π r2
The charge enclosed is σ × 4 π R2
By Gauss’s law E × 4 π r2 =
Or, where q = 4 π R2 σ is the total charge on the spherical shell
Vectorially, (1.34)
Figure 1.31 Gaussian 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.
Field inside the shell: In 1(b), the point P is inside the shell
The Gaussian surface is again a sphere through P centred at O. The flux through the Gaussian surface, calculated as before, is E × 4 π r2
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.
Example 1.13 An early model for an atom considered it to have a 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 Solution The charge distribution for this model of the atom is as shown in 2
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 or 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 spherical surface centred at the nucleus
We consider two situations, namely, r < R and r > R
r < R : The electric flux φ enclosed by the spherical surface is φ = E (r) × 4 π r2 where E (r) is the magnitude of the electric field at r
This is because 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, i.e., Substituting for the charge density ρ obtained earlier, we have Gauss’s law then gives,
The electric field is directed radially outward
r > R: In this case, the total charge enclosed by the Gaussian spherical surface is zero since the atom is neutral
Thus, from Gauss’s law, E (r) × 4 π r2 = 0 or E (r) = 0; r > R
At r = R, both cases give the same result: E = 0.
On symmetry operations In Physics, we often encounter systems with various symmetries
Consideration of these 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 densityσ) 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 ofthe sheet
Compare this with the effort needed to arrive at this result by a direct calculation using Coulomb’s law.
Summary 1. Electric and magnetic forces determine the properties of atoms, 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 r21separating them
Mathematically, F21 = force on q2 due to whereis a unit vector in the direction from q1 to q2 and k =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
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, sayq1, is 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|/4πε0r2; 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: Field lines are continuous curves without any breaks
Two field lines cannot cross each other
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:
Dipole electric field on the axis at a distance r from the centre:
The 1/r3 dependence of dipole electric fields should be noted in contrast to the 1/r2dependence of electric field due to a point charge
13. In a uniform electric field E, a dipole experiences a torquegiven by = p × E but experiences no net force
14. The flux ∆φ of electric field E through a small area element ∆S is given by ∆φ = E.∆S The vector area element ∆S is ∆S = ∆S where ∆S is the magnitude of the area element andis normal to the area element, which can be considered planar for sufficiently small ∆S
For an area element of a closed surface,is taken to be the direction 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: Thin infinitely long straight wire of uniform linear charge density λ
where r is the perpendicular distance of the point from the wire andis the radial unit vector in the plane normal to the wire passing through the point
Infinite thin plane sheet of uniform surface charge density σ
whereis a unit vector normal to the plane, outward on either side
Thin spherical shell of uniform surface charge density σ
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.
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 m2C–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.
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/r2, 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 separated by a distance of 50 cm
What is the mutual force of 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 4
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 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 (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 5 cannot possibly represent electrostatic field lines?
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 6(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
(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), whereis 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 E 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
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 3)
The length of plate is Land 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.)
Chapter 1 ELECTRIC CHARGES AND FIELDS