introduction : adsorption : adsorbent and adsorbate

RESONANCE Surface Chemistry - 1

INTRODUCTION :

Surface chemistry is that branch of chemistry which deals with study of the phenomena occuring at thesurface or interface, i.e. at the boundary separating two bulk phases. In this chapter our main emphasis willbe on three important topics related to surface chemistry, viz., adsorption, colloids and emulsions.

Adsorption :The phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solidresulting into a higher concentration of the molecules on the surface is called adsorption. As a result ofadsorption, there is a decrease of surface energy. The process of removal of an adsorbed substance from thesurface on which it is adsorbed is called desorption. It is the reverse of adsorption and can be brought aboutby heating or by reducing the pressure.

Adsorbent and adsorbate :The substance on the surface of which adsorption occurs is known as adsorbent. The substances that getadsorbed on the solid surface due to intermolecular attractions are called adsorbate.Charcoal, silica, gel, alumina gel are good adsorbents because they have highly porous structures and havelarge surface area. Colloids on account of their extremely small dimensions possess enoromous surfacearea per unit mass and are, therefore, also good adsorbents.

Examples of adsorption :

Adsorption of a gas by charcoal :Finely divided activated charcoal has a tendency to adsorb a number of gases like ammonia, sulphur dioxide,chlorine, phosgene, etc. In this case, charcoal acts as an adsorbent while gas molecules act as adsorbate.

Adsorption of a dye by charcoal :Animal charcoal is used for decolourising a number of organic substances in the form of their solutions. Thedischarge of the colour is due to the fact that the coloured component (generally an organic dye) getsadsorbed on the surface of the adsorbent (animal charcoal).

Sorption : When both adsorption and absorption take place simultaneously. Eg : Dyes get adsorbed as well as absorbed in the cotton fibre i.e. sorption takes place.

Difference between adsorption and absorption :The terms adsorption and absorption are different. Adsorption is a phenomenon in which there is higherconcentration of another substance on the surface than in the bulk. On the other hand, absorption is aphenomenon in which the molecules of a substance are uniformly distributed throughout the body of othersubstance. For example, when silica gel is placed in the environment of water, it adsorbs the water vapour.The water vapour are present in high concentration at the surface of silica gel. On the other hand, whencalcium chloride is placed in the environment of water, it absorbs water. The water vapour uniformly getdistributed throughout the body of calcium chloride. Thus, silica gel adsorbs water vapour while anhydrouscalcium chloride absorbs water.

The important points of distinction between adsorption and absorption are :

Absorption Adsorption It is the phenomenon in which the particles It is the phenomenon of higher concentration

of gas or liquid get uniformly distributed of gas or liquid on the surface than in the bulkthroughout the body of the solid. of the solid.

The concentration is the same throughout The concentration on the surface of thethe material. Therefore, it is a bulk phenomenon. adsorbent is different (has higher concentration) from that

in the bulk. Therefore, it is a surface phenomenon.

Absorption occurs at uniform rate. Adsorption is rapid in the beginning and its rate slowlydecreases.


Thermodynamics of adsorption : Adsorption is an exothermic process. Therefore H of adsorption is always negative.When a gas is adsorbed the entropy of the gas decreases i.e. S is negative. Adsorption is thus accompanied bydecrease in enthalpy as well as entropy of the system, for a process to be spontaneous requirement is that G must benegative. On the basis of equation, G = H - TS, G can be negative if H has sufficiently high negative value as - TSis positive. Thus, in an adsorption process, which is spontaneous, S is negative, and H is also sufficiently negative andas a combination of these two factors, G is negative.H becomes less and less negative as adsorption proceeds further and further. Ultimately H becomes equal toTS and G becomes zero. This is the state at which equilibrium is attained.

Enthalpy of adsorption Hadsorption : It is the amount of the heat released when 1 mole of an adsorbate gets adsorbedon a particular adsorbent at adsorption equilibrium. It depends upon the nature of both the adsorbate as well asadsorbent.

Types of adsorption : The adsorption is classified into two types :

(i) Physical adsorption (i.e. physisorption) :When the particles of the adsorbate are held to the surface of the adsorbent by the physical forces such asvan der Waal�s forces, the adsorption is called physical adsorption or vanderwaals adsorption.

(ii) Chemical adsorption (i.e. chemisorption) :When the molecules of the adsorbate are held to the surface of the adsorbent by the chemical forces, theadsorption is called chemical adsorption.

Difference between physical adsorption and chemical adsorption

Physical Adsorption Chemical Adsorption

The forces between the adsorbate molecules The forces between the adsorbate moleculesand the adsorbent are weak van der Waal�s and the adsorbent are strong chemical forces.forces.

Low heat of adsorption of the order of High heat of adsorption of the order20-40 kJ mol�1 200-400 kJ mol�1

Usually occurs at low temperature and It occurs at high temperaturedecreases with increasing temperature.

It is reversible. It is irreversible.

The extent of adsorption depends upon There is no correlation between extent ofthe ease of liquefication of the gas. adsorption and the ease of liquefication of gas.

It is less specific in nature, all gases are It is highly specific in nature and occurs onlyadsorbed on the surface of a solid to some when there is bond formation betweenextent. adsorbent and adsorbate molecules.

It forms multimolecular layers. It forms mono-molecular layer.

extent of adsorption (x/m) increases with extent of adsorption (x/m) increases with pressurepressure

Competitive adsorption : When an adsorbent is in contact with more than one species (adsorbate). Therewill be competition among them to get adsorbed on to the surface of the adsorbent. The one that is morestrongly adsorbed gets deposited first in preference to the others. Further a strongly adsorbed substancemay displace a weakly adsorbed substance. Ex.

* NH3 can displace O

2 or N

2 from the surface of charcoal.

(i) Adsorption chromatography. For example, column chromatography, gas chromatography etc. iswidely used in chemical analysis.

(ii) Gas masks function because easily liquefiable poisonous gases like CI2, SO

2, SO

3 etc. get adsorbed

on to the activated charcoal in preference to non poisonous gases even if these are present in verysmall amounts in atmosphere.

(iii) Mixture of Noble gases can be separated by using coconut charcoal by maintaining differenttemperatures.


Adsorption of gases on solids :The extent of adsorption of a gas on a solid surface is affected by the following factors:

The nature of the gas (i.e. nature of the adsorbate). The easily liquefiable gases such as HCl, NH3, Cl

2

etc. are adsorbed more than the permanent gases such as H2, N

2 and O

2. The ease with which a gas can be

liquefied is primarily determined by its critical temperature. Higher the critical temperature (Tc) of a gas, the

more easily it will be liquefied and, therefore, more readily it will be adsorbed on the solid.Gas SO

2CH

4H

2

TC

330K 190 K 33 K

Nature of adsorbent. The extent of adsorption of a gas depends upon the nature of adsorbent. Activatedcharcoal (i.e. activated carbon), metal oxides (silica gel and aluminium oxide) and clay can adsorb gaseswhich are easily liquified. Gases such as H

2, N

2 and O

2 are generally adsorbed on finely divided transition

metals Ni and Co.

Activation of adsorbent.

(a) Metallic adsorbents are activated by mechanical rubbing or by subjecting it to some chemicalreactions.

(b) To increase the adsorbing power of adsorbents, they are sub-divided into smaller pieces. As aresults, the surface area is increased and therefore, the adsorbing power increases.

Effect of temperature.Mostly the process of adsorption is exothermic and the reverse process or desorption is endothermic. If theabove equilibrium is subjected to increase in temperature, then according to Le-Chaterlier�s principle, with

increase in temperature, the desorption will be favoured. Physical adsorption decreases continuously withincrease in temperature whereas chemisorption increases initially, shows a maximum in the curve and thenit decreases continuously.

xm

Temperature

xm

Temperature

The initial increase in chemisorption with increase in temperature is because of activation energy required.This is why the chemical adsorption is also known as �Activated adsorption�.A graph between degree of adsorption (x/m) and temperature �t� at a constant pressure of adsorbate gas is

known as adsorption isobar.

Effect of pressure. The extent of adsorption of a gas per unit mass of adsorbent depends upon the pressure of the gas. The variation of extent of adsorptionexpressed as x/m (where x is the mole of adsorbate and m is the mass of theadsorbent) and the pressure is given as below. A graph between the amount ofadsorption and gas pressure keeping the temperature constant is called anadsorption isotherm.It is clear from the figure-1 that extent of adsorption (x/m) increases with pressure and becomes maximumcorresponding to pressure P

s called equilibrium pressure. Since adsorption is a reversible process, the

desorption also takes place simultaneously. At this pressure (Ps) the amount of gas adsorbed becomes

equal to the amount of gas desorbed.

Freundlich Adsorption isotherm :The variation of extent of adsorption (x/m) with pressure (P) was given mathematically by Freundlich.

At low pressure the graph is almost straight line which indicates that x/m is directly proportional to thepressure. This may be expressed as :

(x/m) p or (x/m) = kp where k is constant. At high pressure the graph becomes almost constant which means that x/m becomes independent of

pressure. This may be expressed as :(x/m) = constant or (x/m) p0 (since p0 = 1) or (x/m) = kp0.

Thus, in the intermediate range of pressure, x/m will depend upon the power of pressure which lies between0 to 1, fractional power of pressure. This may be expressed as

(x/m) p1/n or (x/m) = kp1/n

Where n can take any whole number value which depends upon the nature of adsorbate and adsorbent. Theabove relationship is also called Freundlich�s adsorption isotherm.


The constant k and n can be determined as explained below : Taking logarithms on both sides ofEq. (x/m) = kp1/n we getlog (x/m) = logk + (1/n) log p.

intercept = log k

Slope=(1/n)log(x/m)

log p

One of the drawbacks of Freundlich isotherm is that it fails at high pressure of the gas.

This equation applicable only when adsorbate substance form unimolecular layer on adsorbentsurface. i.e. chemical adsorption.

Langmuir Adsorption Isotherm :Langmuir derived it on theoretical considerations based on kinetic theory of gases.

It is based on the assumption (a) that every adsorption site is equivalent and (b) that the ability of a particle

to bind there is independent of whether or not nearby sites are occupied.He considered adsorption to consist of the following two opposing processes :(i) Adsorption of the gas molecules on the surface of the solid(ii) Desorption of the adsorbed molecules from the surface of the solidHe believed eventually a dynamic equilibrium is established between the above two opposing processes.He also assumed that the layer of the adsorbed gas was unimolecular. This isotherm works particularly wellfor chemisorption.

It is represented by the relation : bp1ap

mx

............ (a)

where �a� and �b� are Langmuir parameter.

At very high pressure x/m =a/b ..........(b)At very low pressure : x/m = ap ........ (c)For determination of the parameters �a� and �b�, Eq. (a) may be written in its inverse form.

ap1

ab

apbp1

xm

............ (d)

A plot of m/x against 1/p gives a straight line with slope and intercept equal to 1/a and b/a respectively.At low pressure according to Eq. (c) x/m increases linearly with p. At high pressure according to Eq. (b)x/m becomes constant i.e. the surface is fully covered and change in pressure has no effect and no furtheradsorption takes place which is clear from the Figure-1.

Adsorption from solutions :The process of adsorption can take place from solutions also. It is observed that solid adsorbents adsorbcertain solutes from solution in preference to other solutes and solvents. For example, animal charcoaldecolouries impure sugar solution by adsorbing colouring dye in preference to sugar molecules.The extent of adsorption from solution depends upon the concentration of solute in the solution as given by

Freundlich isotherm : (x/m) = k(c)1/n (n 1)where c is the equilibrium concentration of the solute in solution.

Temperature dependence here also is similar to that for adsorption of gases and in place of equilibriumpressure, we use equilibrium concentrations of the adsorbates in the solution.

Applications of adsorption : In gas masks : Activated charcoal is generally used in gas masks to adsorb poisonous and toxic gases

from air. These masks are commonly used by the miners because there are poisonous gases like CO, CH4

etc. in the atmosphere in coal mines. In dyeing of cloths : Mordants such as alums are used in dyeing of cloths. They adsorb the dye particles

which, otherwise, do not stick to the cloths. In dehumidizers : Silica gel is commonly used to adsorb humidity or moisture from air.


Removal of colouring matter : Many substances such as sugar, juice and vegetable oils are coloured dueto the presence of impurities. They can be decolourised by placing them in contact with adsorbents likeactivated charcoal or fuller�s earth.

Heterogeneous catalysis : The phenomenon of adsorption is useful in the heterogeneous catalysis. Themetals such as Fe, Ni, Pt, Pd, etc, are used in the manufacturing processes such as Contact process,Haber process and the hydrogenation of oils. Their use is based upon the phenomenon of adsorption.

Refining Petroleum : Silica gel is used as adsorbent in petroleum refining.

Chromatography : It is a method for separation of component and is based on preferential adsorptioncolumn is very common device used.

Creating vacuum : High vacuum can be created by removing gas by adsorption.

Adsorption Indicators : In volumetric analysis, adsorption indicator is used Surface of certain precipitatessuch as silver halide have the property of adsorbing some dye like eosin, fluorescein, etc In the case ofprecipitation titration (AgNO

3 vs NaCI) of the indicator is adsorbed at the end point producing a characteristic

colour on the precipitate.

In froth floatation process : Refer metallurgy.

Softening of hard water : Ion exchange resins used for softening of hard water is based upon selective andcompetive adsorption of ions on resins.

Na2Z + Ca+2 CaZ + 2 Na+

The organic polymers containing groups like �COOH, �SO3H and �NH

2 etc. possess the property of selective

adsorption of ions from solution. These are quite useful in the softening of water.

Example-1 A sample of charcoal weighing 6 g was brought into contact with a gas contained in a vessel of onelitre capacity at 27ºC. The pressure of the gas was found to fall from 700 to 400 mm. Calculate the

volume of the gas (reduced to STP) that is adsorbed per gram of the adsorbent under the conditionof the experiment (density of charcoal sample is 1.5 g cm3).

Solution. The adsorption is taking place in a closed vessel hence if pressure falls there is correspondinglyincrease in volume constant, excess of the volume of the gas would be adsorbed.

P1V

1 = P

2V

2

V2 = P

12

1

PV

= 700 × 40

1000 = 1750 mL.

Actual volume of the flask = 1000 � volume of charcoal = 1000 � 50.100.6

= 996 mL.

Volume of the gas adsorbed = 1750 � 996 = 754 mL.

Volume of the gas adsorbed per gram at STP

2

22

1

11

TVP

TVP

gsinU

= 760300

27340067.125

= 60.19 mL.

Catalysts : Berzillus in 1835 used the word catalyst first time for some substance which alter rate of chemicalreaction and themselves remain chemically and quantitatively unchanged after the reaction and the phenomenonis known as catalysis.Eg : Potassium chlorate when heated at 653K to 873K, it gives O

2, When MnO

2 is used in this reaction the

O2 is quickly at the low temperature hence MnO

2 is a catalyst

2KCIO3 2KCI + 3O

2

Homogeneous Catalysis : When catalysts and reactants are in same phase then the process is said to behomogeneous catalysis and

Eg : (i) )g(SO2)g(O)g(SO2 3)g(NO

22

(ii) )aq(COOHCH)(COOCHCH 3)(HCl

33

(iii) .)aq(OHC.)aq(OHC)(OH.)aq(OHC 61266126)(SOH

211221242

Glucose Fructose

Heterogenous Catalysis : When catalysts and reactants are in different phases, then process a know asheterogenous catalysis and catalyst is called heterogeneous catalyst


Eg : (i) )g(SO2)g(O)g(SO2 3)s(Pt

23

(ii) )g(NH2)g(H3)g(N 3)s(Fe

22

(iii) )g(OH6)g(NO4)g(O5)g(NH4 2)s(Pt

23

(iv) Vegetable oils )s(Ni

2 )g(H)( Vegetable ghee (s).

Types of Catalysis(a) Positive Catalysis : A substance which increase the rate of chemical reaction is called positive

catalyst and this process called positive catalysis.

(i) )g(SO2)g(O)g(SO2 3catalystpositive

)s(Pt22 (ii) )g(NH2)g(H3)g(N 3

catalystpositive

)s(Fe22

(b) Negative Catalysis : A substance which decrease the rate of chemical reaction is called negativecatalyst and this process called negative catalysis.

Eg : (i) 22catalystnegative

glycerolorPOH22 OOH2OH2 43

(ii) Decomposition of chloroform reduces in the presence of 1% ethyl alcohol.

HClCOClO21

CHCl 2catalystnegative

alcoholethyl%123

(iii) T.E.L. is used as negative catalyst in petrol which reduce knocking.

(iv) 42catalystnegative

phenol232 SOH2OSOH2

(v) COOHHC2OCHOHC2 56catalystnegative

quinol256

(c) Auto Catalysis : When one of the reaction product behave as catalyst for that reaction and increasethe rate of reaction then the phenomenon is called autocatalysis.�Auto catalytic reactions are slow in the beginning but become increasingly rapid as the reaction

proceeds.

Eg : (i) OHHC)catalystAuto(COOHCHHOHHCOOCCH 523523

(ii)

COOH|

CO10OH8)catalystAuto(MnSOSOKSOH3OOHC5KMnO2 22442424

(iii) 2 23 H3)catalystAuto(As2AsH

(iv) OH)catalystAuto(HNO)NO(CuHNO3Cu 22233

(d) Induced Catalyst : When one reaction catalyse an other reaction than the phenomenon is calledinduced catalysis and that reaction is called induced catalyst. Eg :(i) Sodium sulphite (Na

2SO

3) is oxidised to Na

2SO

4 in atmosphere but sodium arsenite (Na

3AsO

3)

does not oxidises to Na3AsO

4 in air. When both are kept together both are oxidised in air.

Here oxidation of Na2SO

3 catalyse the oxidation of Na

3AsO

3. Hence oxidation reaction of

Na2SO

3 act as an induced catalyst for oxidation reaction of Na

3AsO

3.

(ii) Similarly reduction of HgCl2 to Hg

2 Cl

2 by oxalic acid is very slow while that of KMnO

4 is fast

but mixture of HgCl2 and KMnO

4 is reduced rapidly by oxalic acid.

Promoters/Activators : Substance which themselves are not catalyst but its presence can increase thecatalytic activity of catalyst. A promoters increase the number of active sites on the surface Eg :

(i) 3)promoter(Mo

)catalyst(Fe22 NH2H3N

(ii) Vegetable Oil + H2

)promoter(Cu

)catalyst(Ni Vegetable ghee.

(iii) OHCHH2CO 3)promoter(OCr

)catalyst(ZnO2

32


Catalytic Poisons/ Anti catalysts/ Catalyst Inhibitor :Substance which themselves are not catalyst but whose presence decrease the activity of the catalyst.Poisoning is due to preferential adsorption of poison on the surface of the catalyst.

(i) 3)poisonscatalytic(SH/CO

)catalyst(Fe22 NH2H3N

2

(ii) 3)poisonscatalytic(SAs

)catalyst(asbestosPlatinised22 SO2OSO2

32

(iii) Rosunmund Reactions : HCIRCHOHRCOCl)catalytstpoisons(BaSO

)catalyst(Pd2

4

(iv) C2H

4 + H

2

)catalystpoisons(vapoursAgorCO

)catalyst(Ni C

2H

6

(v) 2H2O

2

)catalystpoisons(HCN

)catalyst(Ptcolloidal 2H

2O +

21

O2.

Characteristics of Catalysis :

(i) A Catalyst remains unchanged in mass and chemical compositions at the end of reactions. However

its physical state can be change. Eg :Granular MnO

2 during decomposition of KClO

3 is left as powder at the end of the reaction.

(ii) Finely devided state of catalyst is more efficient for the reactions because surface area increasesand more adsorption take place.

(iii) A small amount of catalyst is generally sufficient to catalyse almost unlimited reaction but in somecases the rate of reaction depends on amount of catalyst.

Exception : (a) In Friedal Craft reaction more amount of catalyst is required. (b) Hydrolysis of ester in acidic and alkaline medium its rate of reaction is proportional to

concentration of H+ or OH� ions.(iv) A catalyst cannot initiate reaction. But some times the activation energy is so large that practically

a reaction may not start until a catalyst lowers the activation energy significantly. For example,mixture of hydrogen and oxygen do not react at room temperature but the reaction occurs very rapidin presence of Pt black.

H2 + O

2

etemperaturroom No reaction

H2 + O

2

blackPt H2O.

(v) Catalyst are generally specific in nature. A substance which act as a catalyst in a particular reaction,fails to catalyse other reaction.

(vi) Catalyst cannot change equilibrium state but it help to attain equilibrium quickly.

(vii) A catalyst does not change the enthalpy, entropy and free energy of a reaction.

(viii) Optimum temperature : There is a particular temperature at which the efficiency of a catalyst amaximum this temperature is known as optimum temperature. On either side the optimumtemperature, the activity of catalyst decreases.

Adsorption Theory of Heterogeneous Catalyst :This theory explain the mechanism of heterogeneous catalyst. This theory is combination of two theory,intermediate compound formation theory and the old adsorption theory, the catalytic activity is localised onthe surface on the catalyst. The mechanism involve 5 steps.(i) Diffusion of reactant to the surface of the catalyst.(ii) Adsorption of reactant molecules on the surface of the catalyst.(iii) Formation of activated intermediate.(iv) Formation of reactions product on the catalyst surface.(v) Diffusion of reactions product from the catalyst surface or desorption.


Examples :Let us consider addition of H

2 gas to ethelene in presence of Ni catalyst, the reaction take places as follows.

Factors Supporting Theory :

(i) This theory explain the role of active centre, more free valency which provide the more space for themore adsorption and concentration increases as a result increase in rate of reaction.

(ii) Rough surface has more active and pores there will be more free valency so more will be rate ofreaction.

(iii) The theory explain centre action of promoters which occupiedinterstial void as a result surface area for the adsorption increasestherefore rate of reaction increases.

Ni Ni

H H

more strain

promoters

surface area increase

(v) The theory explain of function of poisons or inhibitors. In poisoning preferentialadsorption of poisons take place on the catalyst, surface area for theadsorption on the catalyst decrease hence rate of reaction decreases.

inhibitors

catalyst


Some Industrial Catalytic reactions

Colloid Solution :Colloid State : A substance is said to be in colloidal state when the size of the particle of disperse phaseis greater than particle of true solution and less than that of suspension solution particle, their range ofdiameters between 1 and 1000 nm (10�9 to 10�6 m).

Colloid solution : It is a heterogeneous system consisting of 2 phase :(1) Disperse Phase (D.P) : The phase which is dispersed through the medium is called dispersed phase ordiscontinuous phase or internal phase.(2) Dispersion Medium (D.M) : A medium in which colloidal particles are dispersed is called dispersionmedium. It is also known as continuous phase or outer phase or external phase.Colloidal solution = D.P. + D.M.Ex. In Gold sol, Gold is D.P and water is D.M.


Colloids :A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as very fineparticles in another substance called dispersion medium. The solution and colloid essentially differ from oneanother by particle size.

In a solution, the particles are ions or small molecules.

In a colloid, the dispersed phase may consist of particles of a single macromolecule (such asprotein or synthetic polymer) or an aggregate of many atoms, ions or molecules. Colloidal particles are largerthan simple molecules but small enough to remain suspended. They have a range of diameters between 1and 1000 nm (10�9 to 10�6 m).

Classification of colloids :

1. On the basis of physical state of D.P. and D.M.On the bases of physical state of D.P. and D.M. colloidal solution may be divided into eight system.

Table : Type of Colloidal Systems

D P D M Type of colloid Examples

Solid Solid Solid Sol Some coloured glasses,and gem stones

Solid Liquid Sol Paints, cell fluidsSolid Gas Aerosol Smoke, dustLiquid Solid Gel Cheese, butter, jelliesLiquid Liquid Emulsion Milk, hair creamLiquid Gas Liquid Aerosol Fog, mist, cloud,

insecticide spraysGas Solid Solid Sol Pumice stone, foam rubberGas Liquid Foam Froath, whipped cream, soap lather.

* Solution of gas in gas is not a colloidal system because it form homogeneous mixture.

2. On the basis of D.M. : Colloidal solution are classified asD.M. Name of colloidal systemWater Hydro sol or aqua solAlcohol Alco solsBenzene Benzo solsAir Aero sols

* Aquadag & oildag are colloidal solution of graphite in water & oil respectively. * Colloidal solution are often termed as sol.

3. On the Basis of interaction of D.P. for D.M. : There are two types-

(i) Lyophilic colloids / liquid loving sols / intrinsic colloid. The colloidal solution in which the particles ofthe dispersed phase have a great affinity (or love) for the dispersion medium, are called lyophilic colloids.These solutions are easily formed and the lyophilic colloids are reversible in nature. In case water acts as thedispersion medium, the lyophilic colloid is called hydrophilic colloid. The common examples of lyophiliccolloids are glue, gelatin, starch, proteins, egg albumin, rubber, etc.

(ii) Lyophobic colloids / solvent hating colloid / extrinsic colloid. The colloidal solutions in which thereis no affinity between particles of the dispersed phase and the dispersion medium are called lyophobiccolloids. Such solutions are formed with difficulty only by special methods. These sols are readily precipitated(or coagulated) on the addition of small amounts of electrolytes, by heating or by shaking and hence are notstable. Further, once precipitated, they do not give back the colloidal sol by simple addition of the dispersionmedium. Hence these sols are also called irreversible sols. They need stabilising agents for their preservation.In case the dispersion medium is water, the lyophobic sol is called hydrophobic colloid. For example, thesolution of metals like Ag and Au, hydroxides like Al (OH)

3, Fe(OH)

3, metal sulphides like As

2S

3 etc.

* Lyophilic sols are more stable than lyophobic sols, the additional stability is due presence of an envelope ofthe solvent layer (say water) around the colloidal particle, the process is known as hydration, To coagulate ahydrophilic sols we have to add a dehydrating agent in addition to electrolyte.


4. On the basis of chemical composition:Inorganic Colloids(i) Metal sols : Cu, Ag, Au, Pt, Sols.(ii) Non Metal sols : S, I

2, Graphite

(iii) Sol of oxide and hydroxide : SnO2, TiO

2, Fe

2O

3, Fe(OH)

3, AI(OH)

3, Cr(OH)

3

(iv) Salt Sol - AgBr, AgI, As2 S

3, etc.

Organic Colloids(i) Homopolar sol - In this type of colloid, particles carry similar type of charge. eg. Sol of rubber in benzenewhich contain - ve charge colloidal particle of latex.(ii) Hydroxy Sol- Starch sol

5. On the basis of charge of on particles(i) positive Sol(a) Metal Oxide & Hydroxide - SnO

2, TiO

2, Fe

2O

3, AI(OH)

3, Fe(OH)

3, Cr(OH)

3.

(b) Basic Dyes Methylene blue, vismark brown.

(ii) negative Sol -(a) Metal sol - Ag, Au, Pt, Cu(b) Acidic dye - congo red, eosin(c) Sulphide Sol- CdS, HgS, As

2S

3, Sb

2S

3.

(d) Natural sol - Blood, clay, charcoal, latex rubber, dust particle in water, starch carbon particle in smoke, gum.

DISTINCTION BETWEEN LYOPHILIC AND LYOPHOBIC COLLOIDS

S.No. Property Lyophilic colloids Lyophobic colloids

1 Ease of preparationThere are easily formed by direct mixing.

These are formed only by special methods

2Reversible or irreversible nature

These are reversible in nature These are irreversible in nature.

3 Particles natureThe particles of colloids are true molecules but are big in size

The particles are aggregates of many molecules

4 Stability These are very stableThese are unstable and require traces of stabilizers

5 Action of electrolytes No effect

The addition of small amount of electrolytes causes precipitation (called coagulation) of colloidal solution.

6 Charge on particlesThe particles do not carry any charge.

The particles move in a specific direction either towards anode or cathode depending upon their charge in an electric field

7 HydrationThe particles of colloids are heavily hydrated due to the attraction for the solvent.

The particles of colloids are not appreciably hydrated.

8 ViscosityThe viscosity and surface tension of the sols are less than that of the dispersion medium

The viscosity and surface tension are nearly the same as that of the dispersion medium.

9 Tyndall effect They do no show tyndall effect They show tyndall effect.

Multimolecular, macromolecular and associated colloids :

Multimolecular colloids : In this type, the particles consist of an aggregate of atoms or small moleculessize less than 1 nm. For example, sols of gold atoms and sulphur (S

8) molecules. In these colloids, the

particles are held together by van der Waal�s forces.

Macromolecular colloids : In this types, the particles of the dispersed phase are sufficiently big in size(macro) to be of colloidal dimensions. These macromolecules forming the dispersed phase are generallypolymers having very high molecular masses. These colloids are quite stable and resemble true solutions inmany respects. Naturally occuring macromolecules are starch, cellulose, proteins, enzymes, gelatin, etc.


Associated colloids (Micelles): These are the substances which behave as normal strong electrolytes atlow concentration but behave as colloidal particles at higher concentration. These associated particles arealso called micelles. Ex. Soap.

[C] Associated colloids (Micelles) :There are some substances which at low concentrations behave as normal strong electrolytes but at higherconcentrations exhibit colloidal behaviour due to the formation of aggregated particles. The aggregated particlesthus formed are called micelles. These are also known as associated colloids. The formation of micellestakes place only above a particular temperature called Kraft Temperature (T

k) and above a particular

concentration called Critical Micelle Concentration (CMC). On dilution, these colloids revert back to individualions. Surface active agents such as soaps and synthetic detergents belong to this class. For soaps, theCMC is ~ 10�4 to 10�3 mol L�1. These colloids have both lyophobic and lyophilic parts. Micelles may containas many as 100 molecules or more. Mechanism of micelle formation: Let us take the example of soapsolutions. Soap is sodium salt of a higher fatty acid and may be represented as RCOO�Na+ e.g., sodiumstearate viz.CH

3(CH

2)

16COO� Na+ which is a major component of many bar soaps. When dissolved in water,

it dissociates into RCOO- and Na+ ions. The RCOO� ions, however, consist of two parts i.e., long hydrocarbonchain R (also called non-polar 'tail') which is hydrophobic (water repelling) and the polar group COO� (alsocalled polar-ionic 'head') which is hydrophilic (water loving). The RCOO� lons are, therefore, present on thesurface with their COO� groups in water and the hydrocarbon chains R staying away from it, and remain atthe surface, but at higher concentration these are pulled into the bulk of the solution and aggregate in aspherical form with their hydrocarbon chains pointing towards the centre with COO� part remaining outwardon the surface. An aggregate thus formed is known as �Ionic micelle'. These micelles may contain as manyas upto 100 such ions.

Aggregation of RCOO� ions to form an ionic micelle.

Similarly, in case of detergents, e.g., sodium lauryl sulphate viz. CH3(CH

2)

11SO

4� Na+, the polar group is �

SO4� along with the long hydrocarbon chain. Hence, the mechanism of micelle formation is same as that of

soaps.

Critical micell concentration [CMC] : The minimum concentration required for micell formation is calledcritical micell concentration. Its value depends upon the nature of D.P. and D.M. For eg. Surface active agent(surfactants, which decrease the surface tension) like soaps and detergents form micelle beyond CMC(~10�3 mol/litre for soaps).* Usually longer the hydrophobic chain, smaller is its CMC.* Also CMC increase with decreasing polarity of the D.M.* The micelles �formation takes place only above a particular temperature called as Kraft Temperature (T

k ).

*At CMC, the micelles are spherical in shape, but that start flattening with increase in concentration andultimately form sheet or film like structures which have a thickness of two molecules. These are calledlamelar micelles or McBain Micelles.

Example of micell :(i) Sodium stearate C

17H

35COO�Na+(Soap).

(ii) Sodium lauryl sulphate CH3 [CH

2]11

SO4� Na+ (Detergent).

(iii) Cetyl trimethyl ammonium bromide (Detergent). CH3(CH

2)

15N+(CH

3)

3Br�.

(iv) Sodium p-dodecylbenzenesulphonate (Detergent).

C H12 25 SO Na3� +

(v) Acidic (negative colloids) and basic (positive colloids) dyes.


[D] The Cleansing Action of Soaps :It has been mentioned earlier that a micelle consists of a hydrophobic hydrocarbon like central core. Thecleansing action of soap is due to these micelles, because oil and grease can be solubilised in theirhydrocarbon, like centres which are not otherwise soluble in water. This is shown diagrammatically in Fig.The dirt goes out along with the soap micelles.

Sodium stearate (C17

H35

COO�

Na+)

(a)

�

Hydrophilic head

Hydrophobic tail(b)

(c) (d) (e)

Fig. : The cleansing action of soap. (a) A sodium stearate molecule (b) The simplified representation of themolecule that shows a hydrophilic head and a hydrophobic tail (c) Grease (oily substance) is not soluble inwater (d) When soap is added to water, the non-polar tails of soap molecules dissolve in grease (e) Finally,the grease is removed in the form of micelles containing grease.

Surfactants : They can be ionic as well as non-ionic. The ionic are soaps and detergent. Thesurfactant gets adsorbed at the interface between the dispersed droplets and dispersion medium ina form of mono molecular layer and lowers the interfacial tension between oil and water so as tofacilitate the mixing of two liquids.

Preparation of lyophobic colloidal sols :

[A] Condensation methods :In these methods particles of atomic or molecular size are induced to combine to form aggregates havingcolloidal dimensions. For this purpose chemical as well as physical methods can be applied.(a) Chemical methods. Colloidal solutions can be prepared by chemical reactions leading to formation ofmolecules by double decomposition, oxidation, reduction or hydrolysis. These molecules then aggregateleading to formation of sols.

(i) Double decomposition : When a hot aqueous dilute solution of arsenous oxide (As2O

3) is mixed with a

saturated solution of H2S in water, a colloidal sol of arsenous sulphide (As

2S

3) is obtained.

As2O

3(in hot water)

+ 3H

2S (saturated solution in H

2O)

iondecompositDouble As2S

3(sol) + 3H

2O

SbOCOOOHCH|

COOKOHCH+ 3H

2S

COOHOHCH|

COOKOHCH+ Sb

2S

3 (orange sol) + 2H

2O

(ii) Oxidation : A colloidal sol of sulphur is obtained by passing H2S into a solution of sulphur dioxide.

SO2 + 2H

2S(saturated solution in H

2O)

Oxidation 3S(sol) + 2H2O

Sulphur sol can also be obtained when H2S is bubbled through Br

2 water or nitric acid (oxidizing agent).

2H2S (aq.) + Br

2 (aq.) 2HBr (aq.) + S (sol).

or by bubbling O2 (g) through a solution of H

2S :

2H2S (aq.) + O

2 (g) 2H

2O (l) + 2S (sol).


(iii) Reduction : Colloidal sol of metals like gold, silver solution are obtained by following method.

2 AuCl3 + 3 HCHO + 3H

2O 2Au(sol) + 3HCOOH + 6HCl.

(purple of cassius)

)sol(Au2SnCl3SnCl3AuCl2 4ductionRe

23

AgNO3 + tannic acid

ductionRe silver sol.

NH2NH

2 can also be used as reducing agent.

*Sol of gold is also known as purple of cassius.

(iv) Hydrolysis : A colloidal sol of metal hydroxides like Al(OH)3 or Cr(OH)

3 is obtained by boiling a dilute

solution of FeCl3 , AlCl

3 or CrCl

3 .

FeCl3 + 3H

2O Fe(OH)

3 (sol) + 3HCl ; AlCl

3 + 3H

2O Al(OH)

3 (sol) + 3HCl

The colloidal sol of sillicic acid is also obtained by hydrolysis of dilute solution of sodium silicate with

hydrochloric acid. Na4SiO

4 + 4HCl Si(OH)

4 (sol) + 4NaCl.

(b) Physical methods : The following physical methods are used to prepare the colloidal solutions.

(i) By Exchange of solvent : When a true solution is mixed with an excess of the other solvent in which thesolute is insoluble but solvent is miscible, a colloidal sol is obtained. For example, when a solution of sulphur in alcohol is poured in excess of water, a colloidal sol of sulphur is obtained. when a solution of phenolphthalein in alcohol is poured in excess of water a white sol of phenolphthalein isfound.Phenolphthalein, I

2 , sulphur sol can be prepared by this methods.

(ii) Excessive cooling : Molecules of certain substance condense together on excess cooling to formcolloidal size particle. The colloidal sol of ice in an organic solvent such as CHCl

3 or ether can be obtained by

freezing a solution of water in the solvent. The molecules of water which can no longer be held in solutionseparately combine to form particles of colloidal size.

(iii) By condensing vapours of a substance into solvent : Substance like sulphur and Hg in water areprepared by passing their vapours in cold water containing small amount of stabilising agent like ammoniumnitrate.

[B] Dispersion Methods :

In these methods large particles of the substance are broken into particles of colloidal dimensions in thepresence of dispersion medium. These are stabilized by adding some suitable stabilizer. Some of themethods employed are given below :

(a) Mechanical dispersion (By colloidal mill) : Here the substance is first finely powdered. It is shakenwith the D.M. to form a suspension. This suspension is passed through a colloidal mill. The simplest type ofcolloidal mill is disc mill which consists of two metal discs nearly touching each other & rotating inopposite. Direction at a high speed (7,000 revolutions per min.). The suspended particles are broken toproduce colloidal size particle.

* This method is used to prepare printing ink.

Driving belt

Discharge

Metal disc

Discharge

SuspensionHollow shaft

(Fig. : A colloid mill)


(b) Electrical disintegration or Bredig's Arc method : This process involves dispersion as well ascondensation. Colloidal sols of less reactive metals such as gold, silver, platinum, copper, lead etc., can beprepared by this method. In this method, electric arc is struck between electrodes of the metal immersed inthe dispersion medium as shown in fig. The intense heat produced vaporises the metal, which then condensesto form particles of colloidal size by surrounding cooling mixture (ice).*A slight trace of KOH is added in water to stabilized colloidal solutions.

(Fig. : Bredig's Arc method)

(c) Ultrasonic dispersion : Ultrasonic vibration (having frequency larger than audible range) can bring aboutthe transformation of coarse suspension or liquids like oil, mercury etc. into colloidal range.*This is the latest method for preparation of metal oxides and metal sulphide sols from their coarse suspension.*It is a suitable technique for oils also. This method also comprises both dispersion and condensation.

(d) Peptization: The term has originated from the digestion of proteins by the enzyme pepsin. Peptizationmay be defined as (the process of converting a precipitate into colloidal sol by shaking it with dispersionmedium in the presence of a small amount of electrolyte). The electrolyte used for this purpose is calledpeptizing agent. This method is applied, generally, to convert a freshly prepared precipitate into a colloidalsol. During peptization, the precipitate adsorbs one of the ions of the electrolyte on its surface. The ionadsorbed on the surface is common either with the anion or cation of the electrolyte. This causes thedevelopment of positive or negative charge on precipitates which ultimately break up into smaller particleshaving the dimensions of colloids.

For example :

(i) When freshly precipitated Fe(OH)3 is shaken with aqueous solution of FeCl

3 (peptizing agent) it adsorbs

Fe3+ ions and thereby breaks up into small-sized particles.

FeCl3 Fe3+ + 3Cl� ; Fe(OH)

3 + Fe3+ Fe(OH)

3 | Fe3+

+ Fe3+

Peptizing agent

ppt of Fe (OH)3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

+3

+3+3

+3

Charge Colloidal particals of Fe (OH)3

(ii) Freshly prepared stannic oxide on treatment with a small amount of dilute hydrochloric acid forms astable colloidal sol of stannic oxide, SnO

2 ; Sn4+ .

SnO2 + 4HCl Sn4+ + 2H

2O + 4Cl� ; SnO

2 + Sn4+ SnO

2 / Sn4+ .

(iii) Freshly precipitated silver chloride can be converted into a colloidal sol by adding a small amount ofhydrochloric acid, AgCl : Cl� .(iv) Cadmium sulphide can be peptised with the help of hydrogen sulphide, CdS : S2� .


Purification of Colloidal SolsThe colloidal sols obtained by various methods are impure and contain impurities of electrolytes and othersoluble substances. These impurities may destabilise the sol. Hence, they have to be removed. A veryimportant method of removal of soluble impurities from sols by a semipermeable membrane is known asdialysis.

A. Dialysis : It is a process of removing a dissolved substance from a colloidal solution by means diffusionthrough suitable membrane. Since particles in true solution (ions or smaller molecules) can pass throughanimal membrane or parchment paper or cellophane sheet but colloidal particle do not, the appratus used forthis purpose is called Dialyser.A bag of suitable membrane containing the colloidal solutions is suspended in a vessel through which freshwater continously flow. The molecules and ions (crystalloids) diffuse through membrane into the outer water& pure colloidal solution is left behind.Movement of ions across the membrane can be expedited by applying electric potential through two electrodesas shown in fig. This method is faster than simple dialysis and is known as Electrodialysis.

Fig. : An apparatus for electrodialysis.

*The most important applications of dialysis is in the purification of blood in the artificial kidney machine. Incase of kidney failure, blood cannot be purified. under such condition, the blood is separated from dissolvedtoxic impurities by dialysis and re-introduced in the bloods stream.*Dialysis is not applicable for non-electrolytes like glucose, sugar, etc.

B. Ultra Filtration : In this method, colloidal sols are purified by carrying out filtration through special type

of graded filters called ultra-filters. These filter papers allow only the electrolytes to pass through. These filterpapers are made of particular pore size by impregnating with colloidal solution and subsequently hardenedby soaking in formaldehyde collodion. In order to accelerate the filtration through such filter papers, increasedpressure or section is employed.

C. Ultra Centrifugation : In this method, the colloidal sol is taken in a tube which is placed in an ultra-

centrifuge. On rotation of the tube at high speeds, the colloidal particles settle down at the bottom of the tubeand the impurities remain in the solution called the centrifugate. The settled colloidal particles are mixed withan appropriate dispersing medium to regenerate the sol.

Example-2 When SO2 is bubbled into H

2S gas, colloidal sol is formed. What type of colloidal sol is it ?

Solution. 2H2S + SO

2 2H

2O + 3S (colloidal).

Lyophobic colloidal sol of sulphur is formed.

Example-3 A reddish brown positively charged sol is obtained by adding small quantity of FeCl3 solution to

freshly prepared and well washed Fe(OH)3 precipitate. How does it take place ?

Solution. It is due to adsorptions of Fe3+ ions on the surface of Fe(OH)3 which gives colloidal sol.

Fe(OH)3 (ppt.) + Fe3+ (ions adsorbed) [Fe(OH)

3]Fe3+ (colloidal sol).

Example-4 Suppose we have a cube of 1.00 cm length. It is cut in all three directions, so as to produce eightcubes, each 0.50 cm on edge length. Then suppose these 0.50 cm cubes are each subdivided intoeight cubes 0.25 cm on edge length, and so on. How many of these successive subdivisions arerequired before the cubes are reduced in size to colloidal dimensions of 100 nm.


Solution. We find that every division in two equal halves also reduces the size of edge lengths to one half.

In first subdivision 1 cm is reduceds to 0.5 cm = 21

cm.

In second subdivision 0.5 cm is reduced to 0.25 cm = 41

cm = 2

2

1

cm

In n subdivision 1 cm is reduced to n

2

1

. Size of colloidal particles lies between 1 to 1000 mm.

Thus to make n subdivision required particle size may be attained.

n

2

1

= 100 nm = 100 × 10�9 m = 100 × 10�7 cm.

n log 2 = 5n × 0.3010 = 5.

n = 3010.05

= 16.61 = 17 subdivisions are required for dimension of 100 nm.

Important properties of colloidal sols :

Heterogeneous character :Colloidal sols are heterogeneous in character as they consist of two phases.(a) dispersed phase and (b) dispersion medium.

Visibility : Due to scattering caused by the colloidal particles, it will appear as a bright spotmoving randomly.

Filterability : Colloidal particles pass through an ordinary filter paper. However, the particle do notpass through other fine membranes.

Colligative Properties :Colloidal sols show the colligative properties viz. relative lowering of vapour pressure, elevation in boilingpoint, depression in freezing point and osmotic pressure. However, due to high average molecular masses ofcolloidal particles, mole fraction of the dispersed phase is very low. Hence, the values of the colligativeproperties observed experimentally are very small. Only osmotic pressure measurements are used indetermining the molecular mass of polymers.

Optical Properties-Tyndall effect :Tyndall, in 1869, observed that if a strong beam of light is passed through a colloidal sol placed in a darkplace, the path of the beam gets illuminated. This phenomenon is called Tyndall effect, which is due to thescattering of light by the colloidal particles. The illuminated path of beam is called Tyndall cone. Thisphenomenon is due to scattering of light from the surface of colloidal particles. In a true solution there are noparticles of sufficiently large diameter to scatter light & hence the beam is invisible.

Light source

True solution

Tyndall cone

Scattered light

microscopeEye

Colloidal solution

*The intensity of scattered light depends on the difference between the refractive indice of the D.P and D.M.,In lyophobic colloids, this difference is appreciable and therefore the tyndal effect is quite well defined but inlyophilic sols the difference is very small and the tyndal effect is very weak. Thus in sols of silicic acid, bloodserum, albumin, etc. there is little or no tyndal effect.


Example of Tyndall Effect Blue colour of sky and sea water. Visibility of tail of comets. Light thrown from a projector in cinema hall. Appearance of dust particle in a semi darked room.

Application of Tyndall Effect :(i) In making ultramicroscopes.(ii) In finding heterogenity of solution.

Example-5 Under what conditions is Tyndal effect observed ?Solution. Tydalls effect is applicable when :

(a) The diameter of the dispersed particles is not much smaller than the wavelength of the light used.(b) The refractive indices of the dispersed phase and the dispersion medium must differ greatly inmagnitude.

Example-6 In the lower layer of the atmosphere, there is a great deal of dust. When the weather is fine, it ispossible to see the magnificent red colour of the setting sun. What have these observation to do withcolloids ?

Solution. Dust in the atmosphere is often colloidal. When the sun is low down on the horizon, light from it hasto pass through a great deal of dust to reach your eyes. The blue part of the light is scattered awayfrom your eyes. You see the red part of the spectrum, which remains. Red sunsets are the Tyndalleffect on a large scale.

Mechanical Properties :Brownian movement: Robert Brown, a botanist, discovered in 1827 that pollen grains placed in water donot remain at rest but move about continuously and randomly. Later on, this phenomenon was observed incase of colloidal particles when they were seen under an ultramicroscope. The particles were seen to be inconstant zig-zag motion as shown in fig. This zig-zag motion is called Brownian movement.

(Fig. Brownian movement)

Brownian movement arises because of the impact of the molecules of the dispersion medium with thecolloidal particles. It has been postulated that the impact of the molecules of the dispersion medium on thecolloidal particle are unequal leading to zig-zag motion. However, as the size of the particle increases, theeffect of the impacts average out and the Brownian movement becomes slow. Ultimately, when the dispersedparticle becomes big enough to acquire the dimensions of suspension, no Brownian movement is observed.

Factors Affecting Brownian Movement :(i) If particles is large then brownian movement becomes less.(ii) Brownian movement increases with increasing temperature.(iii) The brownian movement does not change with time & remains same for months or even for a year.

Important :(i) In confirmation of kinetic energy.(ii) Determination of Avogadro numbers.(iii) Stability of colloidal solution : Brownian movement does not allow the colloidal particles to settle

down to gravity & thus is responsible for their stability.


Electrical Properties (Electrophoresis) :The particles of the colloids are electrically charged and carry positive or negative charge. The dispersionmedium has an equal and opposite charge making the system neutral as a whole. Due to similar nature ofthe charge carried by the particles, they repel each other and do not combine to form bigger particles. Thatis why, a sol is stable and particles do not settle down. Arsenious sulphide, gold, silver and platinum particlesin their respective colloidal sols are negatively charged while particles of ferric hydroxide, aluminium hydroxideare positively charged. The existence of the electric charge is shown by the phenomenon of electrophoresis.It involves the 'movement of colloidal particles either towards the cathode or anode, under the influence of theelectric field'. The apparatus used for electrophoresis as shown in fig.

(Fig. : A set up for electrophoresis.)

The colloidal solution is placed in a U-tube fitted with platinum electrodes. On passing an electric current, thecharged colloidal particles move towards the oppositely charged electrode. Thus, if arsenic sulphide sol istaken in the U-tube, in which negatively charge particle of arsenic sulphide move towards the anode.*Earlier this process was called cataphoresis because most of the colloidal sols studied at that time werepositively charged and moved towards cathode.

Electrosmosis :When movement of colloidal particles is prevented by some suitable means (porous diaphragm or semipermeable members), it is observed that the D.M. begins to move in an electric field. This phenomenon istermed as electromosis.

Sedimentation potential or Dorn potential :When the charged colloidal particles are made to settle down under centrifugal field, there occurs a chargeseparation and a potential difference is developed. This effect is called Dorn effect and the potential differencethus developed is called Dorn potential or sedimentation potential.This process is reverse of electrophoresis.

Isoelectric point :The H+ concentration at which the colloidal particles have no charge is known as the isoelectric point. At thispoint stability of colloidal particles becomes very less & do not move under influence of electric field.

Streaming potential : A potential difference is developed across a porous partition when the dispersionmedium of a charged colloid is forced through it. This is called Streaming potential. This process is reverseof electro-osmosis.

Charge on colloidal particles :Colloidal particles are either positively charged or negatively charged. This charge is due to preferentialadsorption of either positive or negative ions on their surface. There is adsorption of common ion present inexcess.Fe(OH)

3 sol prepared by the hydrolysis of FeCl

3 solution adsorbs Fe3+ and this is positively charged.

FeCl3 + 3H

2O Fe(OH

3) + 3HCl ; Fe(OH)

3 + FeCl

3 Fe(OH)

3 Fe3+ : 3Cl�

Fixed part Diffused part. Positive charge on colloidal sol is due to adsorption of Fe3+ ion (common ion between Fe(OH)

3 and FeCl

3).

As2S

3 colloidal sol is obtained when As

2O

3 is saturated with H

2S :

As2O

3 + 3H

2S As

2S

3 + 3H

2O.

As2S

3 adsorbs S2� ions (common between H

2S and As

2S

3 and thus is negatively charged).

As2S

3 + H

2S As

2S

3 S2� : 2H+.

AgI in contact with AgNO3 forms positively charged colloidal sol due to adsorption of Ag+ ion.

AgI + AgNO3 [AgI]Ag+ : NO

3� , AgI in contact with KI forms negatively charged colloidal sol due to

adsorption of I� ion AgI + KI AgI I� : K+.


SnO2 in acidic medium forms positively charged colloidal sol due to adsorption of Sn4+ formed.

SnO2

+ 4H+ Sn4+ + 2H2O SnO

2 + Sn4+ SnO

2 Sn4+

SnO2 in alkaline medium forms negatively charged colloidal sol due to adsorption of SnO

32� formed.

SnO2 + 2OH� SnO

32� + H

2O SnO

2 + SnO

32� SnO

2 SnO

32�

Electric Double Layer Theory or Helm-holtz Electric double layer :The surface of colloid particles acquire a positive or negative charge by the selective (preferential) adsorptionsof common ions carrying positive or negative charge respectively to form first layer. This layer attract counterions from D.M. form a second layer. The combination of two layers of opposite charge around the colloidalparticle is called Helm-holtz electric double layer. The first layer of ions is firmely held and is termed as fixedlayer while the second layer is mobile which is termed as diffused layer. The charge of opposite ions of fixedand diffused layer double layer results in a difference in potential between two opposite charge layer is calledthe electrokinetic potential or zetapotential which can be given by

Z = D

u4where = viscosity coefficient, D = dielectric constant of medium.

u = velocity of colloidal particles when an electric field is applied.Example :

When silver nitrate solution is added to KI solution, the precipitation of AgI adsorb iodide ions fromthe D.M with the formation of fixed layer and negatively charged colloidal solution form, however when KIsolution is added to AgNO

3 solution positive charge sol result due to the adsorbs of Ag+ ions from D.M.

AgI/I� AgI/Ag+

Negative charged Positively charged.This fixed layer attracts counter ions from the medium forming a second layer.

Fixed layerK+

K+

K+

K+

K+

K+

K+

K+

K+

AgI

I�I�

I�

I�

I�I�I�I�I�

I�

Zetapotential

Diffused layer

Example-7 Classify the following sols according to theirs charges :(a) gold sol (b) ferric hydroxide sol (c) gelatine (d) blood(e) sulphur (f) arsenious sulphide (g) titanium oxide.

Solution. Negatively charged colloidal sol : (a), (c), (d), (e), (f).positively charged colloidal sol : (b), (g).

Example-8 SnO2 forms positively charged colloidal sol in acidic medium and negatively charged colloidal sol in

basic medium. Explain ?Solution. SnO

2 is amphoteric reacting with acid and base both. In acidic medium (say HCl) Sn4+ ion is formed

which is preferentially adsorbed on SnO2 giving positively charged colloidal sol :

SnO2 + 4HCl SnCl

4 + 2H

2O ;

SnO2 + SnCl

4 [SnO

2]Sn4+(positively charged) + 4Cl�.

Coagulation/Flocculation :The presence of small amounts of appropriate electrolytes is necessary for the stability of the colloids.However, when an electrolyte is added in larger concentration; the particles of the sol take up the ions whichare oppositely charged and thus get neutralised. The neutral particles then start aggregating giving particlesof larger size which are then precipitated. This process of aggregation of colloidal particles into aninsoluble precipitate by the addition of some suitable electrolyte is known as coagulation. At lowerconcentration of electrolytes, the aggregation of particles is called flocculation that can be reversed onshaking while at higher concentration of electrolyte, coagulation takes place and the same cannot be reversed


simply by shaking. The stability of the lyophobic colloids is due to presence of charge on colloidal particles.If, somehow, the charge is removed, the particles will come near to each other to form aggregates and settleddown under the force of gravity.

++

++

+

++

++

++

+ ++

++

+

++

++

++

++

++

++

+

++

++

++

+++

++

+

++

++

++

+++

++

++ +

++

++

+

++

++

+

++

++

+

++

++

++

+

++

++

+

++

++

++

+ ++

++

+

++

++

++

+++

++ ++

++

+

++

++

++

+

++

++

+

++

++

+

++

++

++

++

++

++

++

++

++

++ +

+

++ +

+++

++

+++

++

+

++

++

++

+ ++

++

+

++

++

++

++

++

++

+

++

++

++

+ ++

++

+

++

++

++

++

Neutralisedsol particles Coagulated sol

Coagulation of lyophobic sols can be carried out by the following methods.(i) By electrophoresis

(ii) By mutual precipitation : It is a process in which oppositely charged sol are mixed in proper proportionto neutralise the charge of each other causing coagulation of both the sol.

Example : Positively charged Fe(OH)3 and negatively charged As

2S

3 colloidal particle containing sol on

mixing get coagulated.

(iiii) By Prolonged Dialysis : On prolonged dialysis, traces of the electrolyte present in the sol are removedalmost completely and the colloidal unstable and ultimately coagulate.

(iv) By Boiling : Sols such as sulphur and silver halides dispersed in water may be coagulated by boilingbecause increased collisions between sol particle and the water molecule removed the adsorbed electrolytes.This takes away the charge from the particles and helps them to coagulate.

(v) By cooling : Certain sol can also be coagulated by lowering temperature. For example, accumulation ofcream on the surface of milk on cooling. This is because at lower temperature the dispersion mediummolecules do not exert sufficient force on to the dispersed particles and hence the Brownian motion becomesless effective.

(vi) By the addition of electrolyte : When excess of an electrolyte is added, the colloidal particles areprecipitated.

Coagulation value or Flocculation value :It needs to be noted that the coagulation of a colloidal solution by an electrolyte does not take place until theadded electrolyte has certain minimum concentration in the solution. The minimum concentration ofelectrolyte in millimoles required to cause coagulation of one litre of colloidal solution is calledcoagulation value. It is express in terms of millimoles/litre.

Coagulation value = litreinsolofvolumeeelectrolytofoleslimmil

Comparision of relative coagulating power of two electrolyte for the same colloidal solution : The coagulation value decrease with increase in charge of the coagulating ion.

Coagulating power valuencoagulatio1

.

BeelectrolytofpowergcoagulatinAeelectrolytofpowergcoagulatin

= AvaluencoagulatioBvaluencoagulatio

.

Factor-Affecting Coagulations :(i) Nature of sols : The lyophobic colloid can easily coagulate because it is a less stable colloid, butlyophilic colloids coagulate hardly by the addition of electrolyte due to protective layer of D.M. surroundingthe colloidal particle.

(ii) Nature of electrolyte : In equimolar electrolyte, strong electrolyte have greater coagulating power thanweak electrolyte. Example :

0.1M NaCl > 0.1M CH3COOH.

Hardy-Schulze Rule :According to this rule greater is the valency of coagulating ion, greater its power to cause precipitation. Thisis known as Hardy-Schulze.


In case of positive charged sol, the coagulating power of anion is in the order of.[Fe(CN)

6]4� > PO

43� > SO

42� > Cl�.

In case of negative charged sol, the coagulating power of cation is in the order of.Al3+ > Ba2+ > Na+.

The coagulating power of bivalent ion is 20-80 times higher than monovalent ion and coagulating power oftrivalents is many times more than bivalent.

Example-9 The particles of a particular colloidal solution of arsenic trisulphide (As2S

3) are negatively charged.

Which 0.0005 M solution would be most effective in coagulating this colloidal solution. KCl, MgCl2,

AlCl3 or Na

3PO

4? Explain.

Solution. Since As2S

3 is a negatively charged colloidal sol hence positively charged ion will cause its coagulation.

By Hardy-Schulze rule �greater the charge on ion, greater the coagulating power to coagulate

oppositely charged colloidal sol�, hence out of K+, Mg2+, Al3+ and Na+, Al3+ would be most effective.

Protective colloidal sols :Lyophilic colloidal sols are much more stable than lyophobic colloidal sols. This is due to the extensivesolvation of lyophilic colloidal sols, which forms a protective layer outside it and thus prevents it from formingassociated colloids. Lyophobic sols can easily precipitate by addition of small amount of an electrolyte.They can be prevented from coagulation by previous addition of some lyophilic colloid. This is due to formationof a protective layer by lyophilic sols outside lyophobic sols. Process of protecting the lyophobic colloidsolution from precipitation by an electrolyte due to previous addition of some lyophilic colloid is calledprotection of colloid and lyophilic colloidal sols are called protective sols.Eg : Gelatin, Sodium caseinate, Egg albumin, Gum arabic, Potato starch etc.,Gelatin (lyophilic) protects gold sol (lyophobic) colloids is expressed in terms of gold number.

Gold Number : Zpsigmondy (1901) introduce a term called gold number it is defined as ��the minimum

amount of the protective colloid in milligrams which when added to 10 ml of a standard gold sol is justsufficient to prevent a colour change from red to blue on the addition of 1 ml of 10% sodium chloride solution.It may be noted that smaller of the gold number, greater will be protecting power of the protective colloid.

Protecting power numbergold1 .

Gold Number = mLinsolgoldofvolume10mginsollyophilicofweight

The gold numbers of a few protective colloids are as follows :

Gelatin Haemoglobin Egg albumin Gum arabic Dextrin Starch0.005 - 0.01 0.03 - 0.07 0.1 - 0.2 0.15 - 0.25 6 � 6.2 20 - 25

Uses of protective action :(i) Gelatin is added in the preparation of ice cream to protect the particle of ice.(ii) Protargol and Argyrol, is a silver sol protected by organic material used as eye drop.

Applications of Colloids :Colloids including emulsions find a number of uses in our daily life and industry. Some of the uses are givenbelow.

In medicines : A wide variety of medicinal and pharmaceutical preparations are emulsions. Colloidialmedicines are easily adsorbed by the body tissue because of large surface area.* Colloidal antimony is used in curing kalaazar.* Milk of magnesia, an emulsion, is used for stomach disorder.* Colloidal gold is used for intramuscular injection.* Colloidal sulphur are used as Germicides.* Argyrol is a silver sol used as an eye lotion.* Colloidal Fe(OH)3 is given to arsenic poisoning patients as it adsorbs arsenic and then gets omited out.


Tanning :Animal hides are colloidal in nature. Which contain positive charge colloidal particles of protein. This hide iskept in a tank containing tannic acid, which contains negatively charge colloidal particle. Therefore, mutualcoagulation takes place this results in hardening of leather, this process is termed as tanning of leather.Chromium salts are also used in place of tannic acid.

Photographic plate & Film :Photographic plate or films are prepared by coating an emulsion of the light sensitive silver bromide in gelatinover glass plates or celluloid films. Gelatin prevent the coagulation of colloidal particle of AgBr.

Rubber plating :The negatively charged rubber particles from rubber sol are deposited on wares and handles of different tools.Rubber gloves are formed by rubber plating on suitable templates.

Sewage disposal :Sewage water contains charged colloidal particles of dirt, rubbish, etc., and these do not settle down easily.The particles can be removed by discharging them at electrodes. Dirty water is passed through a tunnel fittedwith metallic electrodes which are maintained at high potential difference. The particles migrate to the oppositelycharged electrode, lose their charge and get coagulated. The deposited matter is used as a manure and thewater left behind is used for irrigation.

Cottrell smoke precipitator :Smoke is a dispersion of negatively charged colloidal particles of carbon in air and can be made free of thesecolloidal particles by passing it through cottrell precipitator as shown in fig. installed in the chimney of anindustrial plant. It consists of two metal discs charged to a high potential. The carbon particles get dischargedand precipitate, while gases come out from the chimney.

Fig. : Cottrell smoke precipitator.

Formation of deltas : The river water contains colloidal particles of sand and clay which carry negativecharge. The sea water contains +ve ions such as Na+, Mg2+, Ca2+, etc. As the river water meets sea water,these ions discharge the sand or clay particle which are precipitated in the form of delta.

Artificial rain : Cloud consists of charge particle of water disperse in air. Rain is caused by aggregation ofthese minute particles, artificial rain can be done by throwing electrified sand or Agl from aeroplanes, colloidalH

2O

particle present in cloud will get coagulated by these sand or Agl particles to form bigger water drops

causing rain.

Stop bleeding from a cut :Blood is a colloidal solution containing a �ve charge colloidal particle (Albuminoid), bleeding can be stopped

by use of alum or FeCl3 solution. The addition of Al3+ or Fe3+ causes coagulation of blood, so bleeding stops.

Stop Screen :It is used warfare for concenalment and camouflage by spraying very fine particles of titanium oxide fromaeroplane. As titanium oxide is very heavy, the smoke screen drops down rapidly as a curtain of dazzlingwhiteness. This is due to scattering of light by colloidal particles.

Preparation of nano-materials : These materials are prepared for use as catalyst by using reversemicelles.

In disinfectants : The disinfectants such as dettol and lysol give emulsions of the oil-in-water type whenmixed with water.


In metallurgical operations : Emulsions play an important role in industry. The metal ores areconcentrated by froth-floatation process which involves the treatment of the pulverised ore in emulsion of pineoil.

Building roads : Asphalt emulsified in water is used for building roads without the necessity of melting theasphalt.

STEM TECHNOLOGY :The size and shape of the colloidal particles is determined with the help of an electron microscope which hasmuch more resolving power (of the order of 10�12m.) The different techniques used to study the colloidalparticles are :(i) Scanning electron microscope (SEM)(ii) Transmission electron microscope (TEM) and(iii) Scanning transmission electron microscope (STEM).

Emulsions :Pair of immiscible liquid is called emulsion. Emulsion are unstable and some time they are separated intotwo layers on keeping still, for the stabilising of an emulsion, a third component is added called emulsifying-Agent form an interfacial film between D.P. and D.M.Emulsion droplets are bigger than sol particles and can be seen under an ordinary microscope and sometimeseven with a magnifying glass.Example :Milk is an emulsion in which liquid fat is D.P. and liquid water is D.M. and casein is emulsifying agent.

Demulsification : The separation of an emulsion into its constituent liquids is called demulsification.Various techniques employed for this are freezing, boiling, centrifugation, electrostatic precipitation or chemicalmethods which destroys the emulsifying agents.Demulsification can be brought about by :(i) Freezing(ii) Heating(iii) Centrifugal action (Separation of cream of milk done by centrifugation).(iv) Removal of emulsifiers by adding a better solvent for them like alcohol, phenol etc, called demulsifiers.

Types of emulsions :Depending on the nature of the dispersed phase, the emulsions are classified as :(a) Oil in water emulsions (b) Water in oil emulsions

(a) Oil in water emulsions (o/w) : This type of emulsions is formed when oil D.P. and water D.M.

Ex. :- Milk and vanishing cream are oil - in - water type emulsions.

(b) Water in oil emulsions (w/o) : This type of emulsions is formed when water is D.P. and oil is D.M.Ex. : - Cold cream and cod liver oil.

Inversion of phase : The conversion of emulsion of oil in water (o/w) into water in oil (w/o) or vice versa iscalled the inversion of phase.

[A] Identification of the type of emulsion : These two types may be identified by :

Dilution test : An emulsion can be diluted with any amount of the dispersion medium, while the dispersedliquid, if added, forms a separate layer. Thus if a few drops of water added to the emulsion are soluble in it,it is oil in water type and if immiscible, it is water in oil type.

Dye test : If a small amount of oil soluble dye gives a uniform colour to the emulsion, it is water in oil typeotherwise it is oil in water type.

Electrical conductivity test : If conductivity of emulsion increases significantly by adding a very smallamount of electrolyte, it is oil in water type and if there is no significant increase in conductivity, it is water inoil type.

[B] Applications of emulsions :

Disinfactants like phenyl, dettol when mixed with water form emulsion. Digestion of fat in small intestine occurs easily due to emulsion. In metallurgical process the concentration of ore by froath floatation method is based upon emulsion. Milk is an emulsion of liquid fat in water in which casein emulsifying agent.


Cleansing action of soap is due to formation of emulsions. Soaps and detergents emulsify the grease alongwith the adhering dirt and carry them away in the wash water.

For concentrating ores, the finely powdered ore is treated with an oil. Oil forms emulsion with the oreparticles. When air is bubbled into the mixture, emulsion containing the particles of the mineral are carried tothe surface.

Gelation

Ex : Gelatin dissolved in water forming a colloidal. Sol Which when cooled sets into a gelly.*Gel have honey-comb structure :

Ex : Sillicic acid, Gum arabic, Sodium oleate, Gelatin, Solid alcohol, etc.

Types of Gel :

(i) Elastic gel : Those gel which have elastic properties.Ex. Gelatin, Starch, Agar-Agar etc.

(ii) Non- elastic gel : Those gel which are rigid.Eg : Silica gel.

Properties of Gel :

1. Syneresis/weeping of gel : The spontaneous liberation of liquid from a gel is called syneresis or weepingof gels. It is reverse of swelling.Eg : Gelatin, Agar-Agar show syneresis at low concentration while sillicic acid shows it at high concentration.

2. Imbibition or swelling of gel : When gel is kept in a suitable liquid (water) it absorb large volume of liquid.The phenomenon is called, imbibition or swelling of gel.

3. Thixotropic : Some gels when shaken to form a sol, on keeping changes into gel are termed as thixotropicgel and phenomenon is called thixotropy.

Eg : Gelatin and silica liquify on shaking changing into corresponding sol and the sol on keeping changesback into gel.


1. Which of the following statements about physical adsorption is not correct ?(A) It is usually monolayer(B) It is reversible in nature(C) It involves van der Waals interactions between adsorbent and adsorbate(D) It involves small enthalpy of adsorption as compared to chemisorption.

2. Which of the following statements about chemisorption is not applicable?(A) It involves chemical forces between adsorbent and absorbate(B) It is irreversible in nature(C) It involves high heat of adsorption(D) It does not require activation energy

3. Which of the following statements regarding adsorption is not correct ?(A) Extent of adsorption of gases on charcoal increases with increase in pressure of the gas(B) Extent of adsorption is independent of temperature(C) Extent of chemisorption by a given mass of adsorbent is limited(D) Extent of adsorption is dependent on the nature of adsorbent

4. Following is the variation of physical adsorption with temperature:

(A) (B) (C) (D)

5. Adsorption is the phenomenon in which a substance:(A) accumulates on the surface of the other substance(B) goes into the body of the other substances(C) remains close to the other substance(D) none of these

6. There is desorption of physical adsorption when:(A) temperature is increased (B) temperature is decreased(C) pressure is increased (D) concentration is increased

7. The rate of chemisorption :(A) decreases with increase of pressure (B) increases with increase of pressure(C) is independent of pressure (D) is independent of temperature

8. Which of the following is not characteristic of chemisorption?(A) it is irreversible (B) it is specific(C) it is multilayer phenomenon (D) heat of adsorption of about � 400 kJ

9. Softening of hard water is done using sodium aluminium silicate (zeolite). The causes :(A) adsorption of Ca2+ and Mg2+ ions of hard water replacing Na+ ions.(B) adsorption of Ca2+ and Mg2+ ions of hard water replacing Al3+ ions(C) both (A) and (B)(D) none of these

10. Which one is false in the following statement ?(A) A catalyst is specific in its action(B) A very small amount of the catalyst alters the rate of a reaction(C) The number of free vacancies on the surface of the catalyst increases on sub-division(D) Ni is used as a catalyst in the manufacture of ammonia

]11. A catalyst increases rate of reaction by :(A) Decreasing enthalpy (B) Decreasing internal energy(C) Decreasing activation energy (D) Increasing activation energy


12. Colloidal solution of gold prepared by different methods of different colours because of :(A) different diameters of colloidal gold particles (B) variable valency of gold(C) different concentration of gold particles (D) impurities produced by different methods

13. An example of extrinsic colloid (lyophobic colloids) is :(A) As

2S

3 sol (B) Fe(OH)

3 sol (C) Egg albumin (D) (A) & (B) both

14. Which of the following sols is not positively charged?(A) Arsenious sulphide (B) Aluminium hydroxide(C) Ferric hydroxide (D) Silver iodide in silver nitrate solution

15. A colloidal solution can be purified by the following method :(A) dialysis (B) peptization (C) filtration (D) oxidation

16. Peptisation is:(A) conversion of a colloidal into precipitate form(B) conversion of precipitate into colloidal sol(C) conversion of metal into colloidal sol by passage of electric current(D) conversion of colloidal sol into macromolecules

17. Bleeding is stopped by the application of ferric chloride. This is because:(A) the blood starts flowing in opposite direction(B) the blood reacts and forms a solid, which seals the blood vessel(C) the blood is coagulated and thus the blood vessel is sealed(D) the ferric chloride seals the blood vessel.

18. Gold number of a lyophilic sol is such property that:(A) the larger its value, the greater is the peptising power(B) the lower its value, the greater is the peptising power(C) the lower its value, the greater is the protecting power(D) the larger its value, the greater is the protecting power

19. Protective sols are:(A) lyophilic (B) lyophobic (C) both (A) and (B) (D) none of (A) and (B)

20. Which of the following ions is most effective in the coagulation of an arsenious sulphide solution ?(A) K+ (B) Mg2+ (C) Al3+ (D) C

21. Which of the following ions is most effective in the coagulation of ferric hydroxide solution ?(A) Cl¯ (B) Br

� (C) NO

2¯ (D) SO

42�

22. Small liquid droplets dispersd in another liquid is called :(A) Suspension (B) Emulsion (C) Gel (D) True solution

23. At CMC, the surfactant molecules :(A) Decomposes (B) Become completely soluble(C) Associate (D) Dissociate

24. Some type of gels like gelatin loose water slowly. The process is known as :(A) Synerisis (B) Thixotropy (C) Peptisation (D) Imbibition

25. The pressure of the gas was found to decrease from 720 to 480 mm. When 5g of sample of activatedcharcoal was kept in a flask of one litre capacity maintained at 27ºC. If the density of charcoal at 1.25 gm/

mL. The volume of gas adsorbed per gm of charcoal at 480 mm of Hg is(A) 80.03 mL (B) 32.20 mL (C) 100.08 mL (D) None of these

26. Coagulation value of the electrolytes AlCl3 and NaCl for As

2S

3 sol are 0.093 and 52 repectively. How many

times AlCl3 has greater coagulating power than NaCl.

(A) 930 (B) 520 (C) 560 (D) None of these

27. Graph between log x/m and log p is a straight line inclined at an angle of 45º. When pressure is 0.5 atm and

ln k = 0.693, the amount of solute adsorbed per gm of adsorbent will be :(A) 1 (B) 1.5 (C) 0.25 (D) 2.5

28. Which of the following statements is not correct for a lyophobic solution ?(A) It can be easily solvated (B) It carries charges(C) The coagulation of this sol is irreversible in nature (D) It is less stable in a solvent


29. Which of the following statements is correct for a lyophilic solution ?(A) It is not easily solvated (B) It is unstable(C) The coagulation of this sol is irreversible in nature (D) It is quite stable in a solvent

30. Liquid-liquid sol is known as(A) aerosol (B) foam (C) emulsion (D) gel

31. The colloidal system consisting of a liquid adsorbate in a solid adsorbent is termed as :(A) aerosol (B) foam (C) emulsion (D) gel

32. Which of the following statements is not correct ?(A) A colloidal solution is a heterogeneous two-phase system(B) Silver sol in water is an example of lyophilic solution.(C) Metal hydroxides in water are examples of lyophobic solution(D) Liquid-liquid colloidal solution is not a stable system

33. Size of colloidal particles may range from :(A) 1 to 1000 nm (B) 10 to 100 pm (C) 1 to 100 µm (D) 1 to 10 mm

34. Which of the following represents a multimolecular colloidal particles?(A) Starch (B) A sol of gold (C) Proteins (D) Soaps

35. Which of the following anions will have minimum flocculation value for the ferric oxide solution ?(A) Cl¯ (B) Br¯ (C) SO

42� (D) [Fe(CN)

6l3-

36. Which of the following represents a macromolecular colloidal particles ?(A) Solution of gold (B) Cellulose (C) Soaps (D) Synthetic detergents

37. Which of the following statements is not correct?(A) Peptization is the process by which certain substances are converted into the colloidal state.(B) Metal sols of gold, silver and platinum can be prepared by Bredig's arc method.(C) Impurities present in a solution makes it more stable.(D) Dialysis is a process to remove impurities of ions and molecules from a solution.

38. Select correct statement (s):(A) hydrophilic colloid is a colloid in which there is a strong attraction between the dispersed phase andwater(B) hydrophobic colloid is a colloid in which there is a lack of attraction between the dispersed phase and water(C) hydrophobic sols are often formed when a solid crystallises rapidly from a chemical reaction or a supersaturated solution(D) all of the above

39. A reddish brown sol (containing Fe3+) is obtained by:(A) the addition of small amount of FeCl

3 solution to freshly prepared Fe(OH)

3 precipitate

(B) the addition of Fe(OH)3 to freshly prepared FeCl

3 solution

(C) the addition of NH4OH to FeCl

3 solution dropwise

(D) the addition of NaOH to FeCl3 solution dropwise

40. Which is an example of coagulation?(A) curdling of milk (B) purification of water by addition of alum(C) formation of deltas at the river beds (D) All the three are example of coagulation

41. Explain the adsorption of nitrogen on iron.

42. How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent ofadsorption of a gas on a solid ?

43. How is adsorption of a gas is related to its critical temperature ?

44. The volume of nitrogen gas Um (measured at STP) required to cover a sample of silica gel with a mono-

molecular layer is 129 cm3 g�1 of gel. Calculate the surface area per gram of the gel if each nitrogen moleculeoccupies 16.2 × 10�20 m2.

45. What is the volume of collodial particle VC as compared to the volume of a solute particle in a true solution

VS ?


46. Gold number Of haemoglobin is 0.03. Hence, 100 mL of gold sol will require what amount of haemoglobin sothat gold is not coagulated by 1 mL of 10% NaCI solution ?

47. Finely divided catalyst has greater surface area and has greater catalytic activity than the compact solid. Ifa total surface area of 6291456 cm2 is required for adsorption in a catalysed gaseous reaction, then howmany splits should be made to a cube of exactly 1 cm in length to achieve required surface area.(Given : One split of a cube gives eight cubes of same size) ?

48. Volume of N2 at NTP required to form a mono layer on the surface of iron catalyst is 8.15 ml/gram of the

adsorbent. What will be the surface area of the adsorbent per gm if each nitrogen molecule occupies16 × 10�22 m2 ?

49. For the coagulation of 200 mL of As2S

3 solution, 10 mL of 1 M NaCl is required. What is the coagulating value

(number of milli moles of solute needed for coagulation of 1 liter of solution) of NaCl ?

50. Size of colloidal particles may range from_____________to_____________ ?


KVPY PROBLEMS (PREVIOUS YEARS)

1. A gel toothpaste is a mixture of a : [KVPY_ 2008_SA](A) liquid in a solid (B) solid in a gas(C) liquid in a liquid (D) gas in a solid

2. A water droplet spreads on a clean gold surface. The droplet, however, does not spread onto a gold surface

which is pre-treated with a dilute solution of hexanethiol [CH3 � (CH

2)

5-SH] and washed with excess water.

This is because - [KVPY_2007_SB]

(A) chemisorption of hexanethiol on gold surface renders it hydrophilic

(B) chemisorption of hexanethiol onto gold surface renders it hydrophobic

(C) physisorption of hexanethiol onto gold surface renders it hydrophobic

(D) physisorption of hexanethiol onto gold surface renders it hydrophilic.

3. The adsorption isothermal for a gas is given by the relation x = ap/(1 + bp) where x is moles of gas adsorbedper gram of the adsorbent, p is the pressure of the gas, and a and b are constants. Then x :

[KVPY_2011_SB](A) increases with p(B) remains unchanged with p(C) decreases with p(D) increases with p at low pressure and the same at high pressure.

4. Emulsification of 10 ml of oil in water produces 2.4 × 1018 droplets. If the surfaces tension at the oil-waterinterface is 0.03 Jm�2 and the area of each droplet is 12.5 × 10�16 m2, the energy spent in the formation of oildroplets is : [KVPY_2011_SB](A) 90 J (B) 30 J (C) 900 J (D) 10 J


EXERCISE

1. (A) 2. (D) 3. (B) 4. (B) 5. (A) 6. (A) 7. (B)

8. (C) 9. (A) 10. (D) 11. (C) 12. (A) 13. (D) 14. (A)

15. (A) 16. (B) 17. (C) 18. (C) 19. (A) 20. (C) 21. (D)

22. (B) 23. (C) 24. (A) 25. (C) 26. (C) 27. (A) 28. (A)

29. (D) 30. (C) 31. (D) 32. (B) 33. (A) 34. (B) 35. (D)

36. (B) 37. (C) 38. (D) 39. (A) 40. (D)

41. When nitrogen gas is brought in contact with iron at 83 K, it is physisorbed on iron surface as nitrogenmolecules, N

2.As the temperature is increased the amount of nitrogen adsorbed decreases rapidly and at

room temperature, practically there is no adsorption of nitrogen on iron. At 773 K and above, nitrogen ischemisorbed on the iron surface as nitrogen atoms.

42. (a) Smaller the size of the particles of the adsorbent, greater is the surface area and hence greater isthe adsorption

(b) At constant temperature, adsorption first increases with increase of pressure and then attainsequilibrium.

(c) In physical adsorption, it decreases with increase of temperature bu in chemisorption, first it increasesand then decreases.

43. Higher is the critical temperature of a gas, greater the van der Waal�s forces of attraction and hence greater

is the adsorption.

44. 561.8 cm3 45. ~ 103 46. 0.30 mg 47. 20 48. 0.35 m2/g

49. 50 50. 1 to 1000 nm

KVPY PROBLEMS (PREVIOUS YEARS)

1. (B) 2. (B) 3. (D) 4. (A)

RESONANCE ALGEBRA -1

Matrices & DeterminantAs for everything else, so for a mathematical theory,

beauty can be perceived but not explained..... Cayley Arthur

Any rectangular arrangement of numbers (real or complex) (or of real valued or complex valuedexpressions) is called a matrix. If a matrix has m rows and n columns then the order of matrix iswritten as m × n and we call it as order m by nThe general m × n matrix is

A =

mnmj3m2m1m

inij3i2i1i

n2j2232221

n1j1131211

a.....a.....aaa...................................a......a......aaa...................................a.....a......aaaa.....a......aaa

where aij denote the element of ith row & jth column. The above matrix is usually denoted as [aij]m × n .

Notes :(i) The elements a11, a22, a33,........ are called as diagonal elements. Their sum is called as

trace of A denoted as tr(A)

(ii) Capital letters of English alphabets are used to denote matrices.

(iii) Order of a matrix : If a matrix has m rows and n columns, then we say that its order is "m by n",written as "m × n".

Row matrix :A matrix having only one row is called as row matrix (or row vector).General form of row matrixis A = [a11, a12, a13, ...., a1n]

This is a matrix of order "1 × n" (or a row matrix of order n)

Column matrix :A matrix having only one column is called as column matrix (or column vector).

Column matrix is in the form A =

1m

21

11

a...

aa

This is a matrix of order "m × 1" (or a column matrix of order m)Square matrix :

A matrix in which number of rows & columns are equal is called a square matrix. The generalform of a square matrix is

A =

nn2n1n

n22221

n11211

a.......aa............................a........aaa.......aa

which we denote as A = [aij]n.

This is a matrix of order "n × n" (or a square matrix of order n)

Zero matrix :A = [aij]m × n is called a zero matrix, if aij = 0 i & j.

ALGEBRA


e.g. : (i)

000000

(ii)

000000000

Upper triangular matrix :A = [aij]m × n is said to be upper triangular, if aij = 0 for i > j (i.e., all the elements below the

diagonal elements are zero).

e.g. : (i)

vu00zyx0dcba

(ii)

z00yx0cba

Lower triangular matrix :A = [aij]m × n is said to be a lower triangular matrix, if aij = 0 for i < j. (i.e., all the elements abovethe diagonal elements are zero.)

e.g. : (i)

zyx0cb00a

(ii)

0zyx00cb000a

Diagonal matrix :A square matrix [aij]n is said to be a diagonal matrix if aij = 0 for i j. (i.e., all the elements ofthe square matrix other than diagonal elements are zero)Note : Diagonal matrix of order n is denoted as Diag (a11, a22, ......ann).

e.g. : (i)

c000b000a

(ii)

c000000000b0000a

Scalar matrix :Scalar matrix is a diagonal matrix in which all the diagonal elements are same. A = [aij]n is ascalar matrix, if (i) aij = 0 for i j and (ii) aij = k for i = j.

e.g. : (i)

a00a

(ii)

a000a000a

Unit matrix (identity matrix) :Unit matrix is a diagonal matrix in which all the diagonal elements are unity. Unit matrix oforder 'n' is denoted by n (or ).i.e. A = [aij]n is a unit matrix when aij = 0 for i j & aii = 1

eg. 2 =

1001

, 3 =

100010001

.

Comparable matrices : Two matrices A & B are said to be comparable, if they have the same order(i.e., number of rows of A & B are same and also the number of columns).

e.g. : (i) A =

213

432& B =

310243

are comparable

e.g. : (ii) C =

213

432& D =

321403

are not comparable


Equality of matrices :Two matrices A and B are said to be equal if they are comparable and all the correspondingelements are equal.

Let A = [aij] m × n & B = [bij]p × qA = B iff (i) m = p, n = q

(ii) aij = bij i & j.

Multiplication of matrix by scalar :Let be a scalar (real or complex number) & A = [aij]m × n be a matrix. Thus the product A isdefined as A = [bij]m × n where bij = aij i & j.

e.g. : A =

21003120

5312& – 3A (–3) A =

6300936015936

Note : If A is a scalar matrix, then A = , where is a diagonal entry of A

Addition of matrices :Let A and B be two matrices of same order (i.e. comparable matrices). Then A + B is defined tobe.

A + B = [aij]m × n + [bij]m × n.= [cij]m × n where cij = aij + bij i & j.

e.g. : A =

013211

, B =

7532

21

, A + B =

760010

Substraction of matrices :Let A & B be two matrices of same order. Then A – B is defined as A + (– B) where – B is (– 1)B.

Properties of addition & scalar multiplication :Consider all matrices of order m × n, whose elements are from a set F (F denote Q, R or C).Let Mm × n (F) denote the set of all such matrices.Then(a) A Mm × n (F) & B Mm × n (F) A + B Mm × n(F)(b) A + B = B + A(c) (A + B) + C = A + (B + C)(d) O = [o]m × n is the additive identity.(e) For every A Mm × n(F), – A is the additive inverse.(f) (A + B) = A + B(g) A = A(h) (1 + 2) A = 1A + 2A

Multiplication of matrices :Let A and B be two matrices such that the number of columns of A is same as number of rowsof B. i.e., A = [aij]m × p & B = [bij]p × n.

Then AB = [cij]m × n where cij =

p

1kkjikba , which is the dot product of ith row vector of A and jth

column vector of B.

e.g. : A =

132321

, B =

021101001110

, AB =

27311943


Notes :(1) The product AB is defined iff the number of columns of A is equal to the number ofrows of B. A is called as premultiplier & B is called as post multiplier. AB is defined

/ BA is defined.(2) In general AB BA, even when both the products are defined.(3) A (BC) = (AB) C, whenever it is defined.

Properties of matrix multiplication :Consider all square matrices of order 'n'. Let Mn (F) denote the set of all square matrices oforder n. (where F is Q, R or C). Then(a) A, B Mn (F) AB Mn (F)(b) In general AB BA(c) (AB) C = A(BC)(d) n, the identity matrix of order n, is the multiplicative identity.

An = A = n A A Mn (F)(e) For every non singular matrix A (i.e., |A| 0) of Mn (F) there exist a unique (particular)

matrix B Mn (F) so that AB = n = BA. In this case we say that A & B are multiplicativeinverse of one another. In notations, we write B = A–1 or A = B–1.

(f) If is a scalar (A) B = (AB) = A(B).(g) A(B + C) = AB + AC A, B, C Mn (F)(h) (A + B) C = AC + BC A, B, C Mn (F).

Notes : (1) Let A = [aij]m × n. Then An = A & m A = A, where n & m are identity matrices of ordern & m respectively.

(2) For a square matrix A, A2 denotes AA, A3 denotes AAA etc.

Transpose of a matrix :

Let A =[aij]m × n. Then the transpose of A is denoted by A( or AT) and is defined as

A = [bij]n × m where bij = aji i & j.

i.e. A is obtained by rewriting all the rows of A as columns (or by rewriting all the columns of A asrows).

e.g. : A =

wzyxdcba4321

, AA =

wd4zc3yb2xa1

Results : (i) For any matrix A = [aij]m × n, (A) = A

(ii) Let be a scalar & A be a matrix. Then (A) = A

(iii) (A + B) = A + B & (A – B) = A – B for two comparable matrices A and B.

(iv) (A1 ± A2 ± ..... ± An) = A1 ± A2 ± ..... ± An, where Ai are comparable.

(v) Let A = [aij]m × p & B = [bij]p × n , then (AB) = BA

(vi) (A1 A2 .......An)= An. An – 1 ...........A2 . A1, provided the product is defined.

Symmetric & skew-symmetric matrix : A square matrix A is said to be symmetric if A = Ai.e. Let A = [aij]n. A is symmetric iff aij = aji i & j.

A square matrix A is said to be skew-symmetric if A = – Ai.e. Let A = [aij]n. A is skew-symmetric iff aij = – aji i & j.

e.g. A =

cfgfbhgha

is a symmetric matrix.


B =

0zyzoxyxo

is a skew-symmetric matrix.

Notes :(1) In a skew-symmetric matrix all the diagonal elements are zero.( aii = – aii aii = 0)

(2) For any square matrix A, A + A is symmetric & A – A is skew-symmetric.(3) Every square matrix can be uniqualy expressed as a sum of two square matrices of

which one is symmetric and the other is skew-symmetric.

A = B + C, where B = 21

(A + AA) & C = 21

(A – AA).

Submatrix : Let A be a given matrix. The matrix obtained by deleting some rows or columns of A is calledas submatrix of A.

eg. A =

srqpwzyxdcba

Then

rpzxca

,

sqpdba

,

rqpzyxcba

are all submatrices of A.

Determinant of a square matrix :Let A = [a]1×1 be a 1×1 matrix. Determinant A is defined as |A| = a.e.g. A = [– 3]1×1 |A| = – 3

Let A =

dcba

, then |A| is defined as ad – bc.

e.g. A =

41

35, |A| = 23

Remarks :

1. |A| = |A| for any square matrix A.

i.e. the value of a determinant remains unaltered, if the rows & columns are inter changed,

i.e. D =

321

321

321

333

222

111

cccbbbaaa

cbacbacba

= D

2. |A| = n |A|, when A = [aij]n.

3. A skew-symmetric matrix of odd order has deteminant value zero.

Singular & non singular matrix : A square matrix A is said to be singular or non-singular accordingas |A| is zero or non-zero respectively.

Cofactor matrix & adjoint matrix : Let A = [aij]n be a square matrix. The matrix obtained byreplacing each element of A by corresponding cofactor is called ascofactor matrix of A, denoted as cofactor A. The transpose of cofactormatrix of A is called as adjoint of A, denoted as adj A.

i.e. if A = [aij]nthen cofactor A = [cij]n when cij is the cofactor of aij i & j.Adj A = [dij]n where dij = cji i & j.


Properties of cofactor A and adj A:(a) A . adj A = |A| n = (adj A) A where A = [aij]n.(b) |adj A| = |A|n – 1, where n is order of A.

In particular, for 3 × 3 matrix, |adj A| = |A|2(c) If A is a symmetric matrix, then adj A are also symmetric

matrices.(d) If A is singular, then adj A is also singular.

Inverse of a matrix (reciprocal matrix) :

Let A be a non-singular matrix. Then the matrix |A|1

adj A is the

multiplicative inverse of A (we call it inverse of A) and is denoted by A–1.We have A (adj A) = |A| n = (adj A) A

A

Aadj

|A|1

= n =

Aadj

|A|1

A, for A is non-singular

A–1 = |A|1

adj A.

Remarks :

1. The necessary and sufficient condition for existence of inverse of A is that A is non-singular.

2. A–1 is always non-singular.

3. If A = dia (a11, a22, ....., ann) where aii 0 i, then A–1 = diag (a11– 1, a22

–1, ...., ann–1).

4. (A–1) = (A)–1 for any non-singular matrix A. Also adj (A) = (adj A).

5. (A–1)–1 = A if A is non-singular.

6. Let k be a non-zero scalar & A be a non-singular matrix. Then (kA)–1 = k1

AA–1.

7. |A–1| = |A|1

for |A| 0.

8. Let A be a non-singular matrix. Then AB = AC B = C & BA = CA B= C.

9. A is non-singular and symmetric A–1 is symmetric.

10. (AB)–1 = B–1 A–1 if A and B are non- singular.11. In general AB = 0 does not imply A = 0 or B = 0. But if A is non-singular and AB = 0, then B = 0.

Similarly B is non-singular and AB = 0 A = 0. Therefore, AB = 0 either both are singular or oneof them is 0.

System of linear equations & matrices : Consider the systema11 x1 + a12x2 + .......... + a1nxn = b1a21x1 + a22 x2 + ..........+ a2n xn = b2.................................................am1x1 + am2x2 + ..........+ amnxn = bn.

Let A =

mn2m1m

n22221

n11211

a..........aa.........................a..........aaa..........aa

, X =

n

2

1

x....xx

& B =

n

2

1

b......bb

.

Then the above system can be expressed in the matrix form as AX = B.The system is said to be consistent if it has atleast one solution.


System of linear equations and matrix inverse:If the above system consist of n equations in n unknowns, then we have AX = B where A is a squarematrix.

Results : (1) If A is non-singular, solution is given by X = A–1B.(2) If A is singular, (adj A) B = 0 and all the columns of A are not proportional, then the

system has infinitely many solutions.(3) If A is singular and (adj A) B 0, then the system has no solution

(we say it is inconsistent).

Homogeneous system and matrix inverse :If the above system is homogeneous, n equations in n unknowns, then in the matrix form it is AX = O.( in this case b1 = b2 = ....... bn = 0), where A is a square matrix.

Results : (1) If A is non-singular, the system has only the trivial solution (zero solution) X = 0(2) If A is singular, then the system has infinitely many solutions (including the trivial

solution) and hence it has non-trivial solutions.

Rank of a matrix :Let A = [aij]m×n. A natural number is said to be the rank of A if A has a non-singularsubmatrix of order and it has no non-singular submatrix of order more than . Rankof zero matrix is regarded to be zero.

eg. A =

050002005213

we have

2023

as a non-singular submatrix.

The square matrices of order 3 are

500200213

,

000000513

,

050020523

,

050020521

and all these are singular. Hence rank of A is 2.

Elementary row transformation of matrix :The following operations on a matrix are called as elementary row transformations.(a) Interchanging two rows.(b) Multiplications of all the elements of row by a nonzero scalar.(c) Addition of constant multiple of a row to another row.

Note : Similar to above we have elementary column transformations also.Remarks :1. Elementary transformation of a matrix does not affect its rank.2. Two matrices A & B are said to be equivalent if one is obtained from other using elementary

transformations. We write A B.

Echelon form of a matrix : A matric is said to be in Echelon form if it satisfy the followings:(a) The first non-zero element in each row is 1 & all the other elements in

the corresponding column (i.e. the column where 1 appears) arezeroes.

(b) The number of zeroes before the first non zero element in any nonzero row is not more than the number of such zeroes in succeedingrows.


Result : Rank of a matrix in Echelon form is the number of non zero rows (i.e. number of rows withatleast one non zero element.)

Remark : To find the rank of a given matrix we may reduce it to Echelon form using elementary rowtransformations and then count the number of non zero rows.

System of linear equations & rank of matrix :Let the system be AX = B where A is an m × n matrix, X is the n-column vector & B is the m-columnvector. Let [AB] denote the augmented matrix (i.e. matrix obtained by accepting elements of B asn + 1th column & first n columns are that of A). (A) denote rank of A and ([AB]) denote rank of theaugmented matrix.

Clearly (A) ([AB]).

Results : (1) If (A) < ([AB]) then the system has no solution (i.e. system is inconsistent).(2) If (A) = ([AB]) = number of unknowns, then the system has unique solution.

(and hence is consistent)(3) If (A) = ([AB]) < number of unknowns, then the systems has infinitely many solutions

(and so is consistent).

Homogeneous system & rank of matrix :

Let the homogenous system be AX = 0, m equations in 'n' unknowns. In this case B = 0 and so (A)= ([AB]). Hence if (A) = n, then the system has only the trivial solution. If (A) < n, then the systemhas infinitely many solutions.

Characteristic polynomial & characteristic equation :

Let A be a square matrix. Then the polynomial | A – x| is called as characteristic polynomial of A &the equation | A – x| = 0 is called as characteristic equation of A.

Cayley - Hamilton theorem : Every square matrix A satisfies its characteristic equationi.e. a0 xn + a, xn – 1 + ........ + an – 1x + an = 0 is the characteristic equation of A, then

a0An + a1An – 1 + ......... + an – 1 A + an = 0

Nilpotent matrix :A square matrix A is said to be nilpotent ( of order 2) if, A2 = O. A square matrix is said to be nilpotentof order p, if p is the least positive integer such that Ap = O.

Idempotent matrix :

A square matrix A is said to be idempotent if, A2 = A.

e.g.

1001

is an idempotent matrix.

Involutory matrix :

A square matrix A is said to be involutory if A2 = , being the identity matrix.

e.g. A =

1001

is an involutory matrix.

Orthogonal matrix :

A square matrix A is said to be an orthogonal matrix if,A A = = AA.

RESONANCERE SONANC E KVPY_ELECTROSTATICS - 1

ELECTROSTATICS

1. INTRODUCTION

The branch of physics which deals with electric effect of static charge is called electrostatics.

2. ELECTRIC CHARGECharge of a material body or particle is the property (acquired or natural) due to which it produces andexperiences electrical and magnetic effects. Some of naturally occurring charged particles are electrons,protons, -particles etc.

Charge is a derived physical quantity & is measured in Coulomb in S.. unit. In practice we use mC(10–3C), C (10–6C), nC(10–9C) etc.C.G.S. unit of charge = electrostatic unit = esu.1 coulomb = 3 × 109 esu of chargeDimensional formula of charge = [MºLºT11]

2.1 Properties of Charge( i ) Charge is a scalar quantity : It adds algebraically and represents excess or deficiency

of electrons.

( i i ) Charge is of two types : (i) Positive charge and (ii) Negative charge Charging abody implies transfer of charge (electrons) from one body to another. Positively charged bodymeans loss of electrons i.e. deficiency of electrons. Negatively charged body means excess ofelectrons. This also shows that mass of a negatively charged body > mass of a positivelycharged identical body.

( i i i ) Charge is conserved : In an isolated system, total charge (sum of positive and negative)remains constant whatever change takes place in that system.

( i v) Charge is quantized : Charge on any body always exists in integral multiples of afundamental unit of electric charge. This unit is equal to the magnitude of charge on electron(1e = 1.6 × 10–19 coulomb). So charge on anybody is Q = ± ne, where n is an integer and e isthe charge of the electron. Millikan's oil drop experiment proved the quantization of charge oratomicity of charge

Note : Recently, the existence of particles of charge ± 31

e and ± 32

e has been postulated. These particles are

called quarks but still this is not considered as the quantum of charge because these are unstable(They have very short span of life).(v) Like point charges repel each other while unlike point charges attract each other.(vi) Charge is always associated with mass, i.e., charge can not exist without mass though mass

can exist without charge. The particle such as photon or neutrino which have no (rest) masscan never have a charge.

(vii) Charge is relativistically invariant: This means that charge is independent of frame of referencei.e. charge on a body does not change whatever be its speed. This property is worth mentioningas in contrast to charge, the mass of a body depends on its speed and increases with increasein speed.

(viii) A charge at rest produces only electric field around itself, a charge having uniform motionproduces electric as well as magnetic field around itself while a charge having acceleratedmotion emits electromagnetic radiations.

RESONANCE KVPY_ELECTROSTATICS - 55RE SONANC E

OBJECTIVE QUESTIONS1. A charged particle q1 is at position (2, - 1, 3). The electrostatic force on another charged particle q2 at

(0, 0, 0) is :

(A) 0

21

56qq )k3ji2( (B)

0

21

1456qq

)k3ji2(

(C) 0

21

56qq )k3i2j( (D)

0

21

1456qq

)k3i2j(

2. Three charges +4q, Q and q are placed in a straight line of length at points at distance 0, /2 and respectively from one end of line. What should be the value of Q in order to make the net force on q tobe zero?(A) –q (B) –2q (C) –q/2 (D) 4q

3. Two similar very small conducting spheres having charges 40 C and –20 C are some distance apart. Nowthey are touched and kept at same distance. The ratio of the initial to the final force between them is :(A) 8 : 1 (B) 4 : 1 (C) 1 : 8 (D) 1 : 1

4. Two point charges placed at a distance r in air exert a force F on each other. The value of distance R atwhich they experience force 4F when placed in a medium of dielectric constant K = 16 is :(A) r (B) r/4 (C) r/8 (D) 2r

5. There is a uniform electric field in X-direction. If the work done by external agent in moving a charge of 0.2C through a distance of 2 metre slowly along the line making an angle of 60º with X-direction is 4 joule,then the magnitude of E is:

(A) C/N3 (B) 4 N/C (C) 5 N/C (D) 20 N/C

6. A simple pendulum has a length & mass of bob m. The bob is givena charge q coulomb. The pendulum is suspended in a uniformhorizontal electric field of strength E as shown in figure, then calculatethe time period of oscillation when the bob is slightly displaced fromits mean position.

(A) g

2 (B)

mqEg

2

(C)

mqEg

2 (D) 2

2

mqEg

2

7. Charges 2Q and –Q are placed as shown in figure. The point at which electricfield intensity is zero will be: (A) Somewhere between –Q and 2Q(B) Somewhere on the left of –Q(C) Somewhere on the right of 2Q(D) Somewhere on the perpendicular bisector of line joining –Q and 2Q


8. The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R willbe :

(A) 04

1 2R33

q(B)

041 2R3

q2(C)

041 2R33

q2(D)

041 i 2R32

q3

9. A charged particle of charge q and mass m is released from rest in a uniform electric field E. Neglectingthe effect of gravity, the kinetic energy of the charged particle after time ‘t’ seconds is

(A) tEqm

(B) m2

tqE 222(C) mq

tE2 22

(D) 2

2

t2mEq

10. A flat circular fixed disc has a charge +Q uniformly distributed on the disc. A charge +q is thrown withkinetic energy K, towards the disc along its axis. The charge q :(A) may hit the disc at the centre(B) may return back along its path after touching the disc(C) may return back along its path without touching the disc(D) any of the above three situations is possible depending on the magnitude of K

11. At a certain distance from a point charge, the electric field is 500 V/m and the potential is 3000 V.What is the distance ?(A) 6 m (B) 12 m (C) 36 m (D) 144 m

12. Figure represents a square carrying charges +q, +q, –q, –q atits four corners as shown. Then the potential will be zero atpoints : (A, C, P and Q are mid points of sides)

–q –qQ

A BC

+q+qP

(A) A, B, C, P and Q (B) A, B and C

(C) A, P, C and Q (D) P, B and Q

13. Two equal positive charges are kept at points A and B. The electric potential, while moving from A to Balong straight line :(A) continuously increases (B) remains constant(C) decreases then increases (D) increases then decreases

14. A semicircular ring of radius 0.5 m is uniformly charged with a total charge of 1.5 × 10–9 coul. Theelectric potential at the centre of this ring is :(A) 27 V (B) 13.5 V (C) 54 V (D) 45.5 V

15. When a charge of 3 coul is placed in a uniform electric field, it experiences a force of 3000 newton. Thepotential difference between two points separated by a distance of 1 cm along field within this field is:(A) 10 volt (B) 90 volt (C) 1000 volt (D) 3000 volt

16. A 5 coulomb charge experiences a constant force of 2000 N when moved between two points separatedby a distance of 2 cm in a uniform electric field. The potential difference between these two points is:(A) 8 V (B) 200 V (C) 800 V (D) 20,000 V

17. The kinetic energy which an electron acquires when accelerated (from rest) through a potential differenceof 1 volt is called :(A) 1 joule (B) 1 electron volt (C) 1 erg (D) 1 watt

18. The potential difference between points A and B inthe given uniform electric field is :

a B

bE

E

C

A

(A) Ea (B) E )ba( 22

(C) Eb (D) )2/Eb(

19. An equipotential surface and an electric line of force :(A) never intersect each other (B) intersect at 45º(C) intersect at 60º (D) intersect at 90º


20. A particle of charge Q and mass m travels through a potential difference V from rest. The final momen-tum of the particle is :

(A) QmV

(B) mVQ2 (C) QVm2 (D) mQV2

21. If a uniformly charged spherical shell of radius 10 cm has a potential V at a point distant 5 cm from itscentre, then the potential at a point distant 15 cm from the centre will be :

(A) 3V

(B) 3V2

(C) V23

(D) 3V

22. A hollow uniformly charged sphere has radius r. If the potential difference between its surface anda point at distance 3r from the centre is V, then the electric field intensity at a distance 3r fromthe centre is:(A) V/6r (B) V/4r (C) V/3r (D) V/2r

23. A hollow sphere of radius 5 cm is uniformly charged such that the potential on its surface is 10 voltsthen potential at centre of sphere will be :(A) Zero(B) 10 volt(C) Same as at a point 5 cm away from the surface(D) Same as at a point 25 cm away from the centre

24. If a charge is shifted from a high potential region to low potential region, the electrical potential energy:(A) Increases (B) Decreases(C) May increase or decrease. (D) Remains constant

25. A particle of mass 2 g and charge 1C is held at rest on a frictionless horizontal surface at a distanceof 1 m from a fixed charge of 1 mC. If the particle is released it will be repelled. The speed of the particlewhen it is at distance of 10 m from the fixed charge is:(A) 100 m/s (B) 90 m/s (C) 60 m/s (D) 45 m/s

26. When the separation between two charges is decreased, the electric potential energy of the charges(A) increases (B) decreases(C) may increase or decrease (D) remains the same

27. Six charges of magnitude + q and –q are fixed at the corners of a regular hexagon of edge length a asshown in the figure. The electrostatic potential energy of the system of charged particles is :

(A) aq

0

2

415

83

(B) aq

0

2

49

23

(C) aq

0

2

215

43

(D) aq

0

2

815

23

28. You are given an arrangement of three point charges q, 2q and xq separated by equal finite distancesso that electric potential energy of the system is zero. Then the value of x is :

(A) 32

(B) 31

(C) 32

(D) 23

29. A uniformly charged sphere of radius 1 cm has potential of 8000 V at surface. The energy density nearthe surface of sphere will be:(A) 64 × 105 J/m3 (B) 8 × 103 J/m3 (C) 32 J/m3 (D) 2.83 J/m3

30. If ' n ' identical water drops (assumed spherical each) charged to a potential energy U coalesce to forma single drop, the potential energy of the single drop is(Assume that drops are uniformly charged):(A) n1/3 U (B) n2/3 U (C) n4/3 U (D) n5/3 U


31. The variation of potential with distance r from a fixed pointis shown in Figure. The electric field at r = 5 cm, is :

(A) (2.5) V/cm (B) (–2.5) V/cm

(C) (–2/5) cm (D) (2/5) V/cm

32. In the above question, the electric force acting on a point charge of 2 C placed at the origin will be :(A) 2 N (B) 500 N (C) –5 N (D) –500 N

33. The electric potential V as a function of distance x (in metre) is given byV = (5x2 + 10x – 9) volt.

The value of electric field at x = 1 m would be :(A) – 20 volt/m (B) 6 volt/m (C) 11 volt/m (D) –23 volt/m

34. A uniform electric field having a magnitude E0 and direction along positive X-axis exists. If the electricpotential V is zero at x = 0, then its value at x = +x will be :(A) Vx = xE0 (B) Vx = –xE0 (C) Vx = x2E0 (D) Vx = –x2 E0

35. Let E be the electric field and V, the electric potential at a point.(A) If E 0, V cannot be zero (B) If E = 0, V must be zero(C) If V = 0, E must be zero (D) None of these

36. The electric field in a region is directed outward and is proportional to the distance r from the origin.Taking the electric potential at the origin to be zero, the electric potential at a distance r :(A) increases as one goes away from the origin.(B) is proportional to r2

(C) is proportional to r(D) is uniform in the region

37. Due to an electric dipole shown in fig., the electric field intensity is parallel to dipole axis :

(A) at P only (B) at Q only (C) both at P and at Q (D) neither at P nor at Q

38. An electric dipole of dipole moment p is placed at the origin along the x-axis. The angle made by

electric field with x-axis at a point P, whose position vector makes an angle with x-axis, is :(where,

tan = tan21

)

(A) (B) (C) + (D) + 2

39. An electric dipole consists of two opposite charges each of magnitude 1.0 C, separated by a distance of2.0 cm. The dipole is placed in an external electric field of 1.0 × 105 N/C. The maximum torque on thedipole is :(A) 0.2 × 10–3 N-m (B) 1.0 × 10–3 N-m (C) 2.0 × 10–3 N-m (D) 4.0 × 10–3 N-m

40. A dipole of electric dipole moment P is placed in a uniform electric field of strength E. If is the anglebetween positive directions of P and E, then the potential energy of the electric dipole is largest when is:(A) zero (B) /2 (C) (D) /4

41. Two opposite and equal charges of magnitude 4 × 10–8 coulomb each when placed 2 × 10–2 cm apartform a dipole. If this dipole is placed in an external electric field of 4 × 108 N/C, the value of maximumtorque and the work required in rotating it through 180º from its initial orientation which is along electricfield will be : (Assume rotation of dipole about an axis passing through centre of the dipole):(A) 64 × 10–4 N-m and 44 × 10–4 J (B) 32 × 10–4 N-m and 32 × 10–4 J(C) 64 × 10–4 N-m and 32 × 10–4 J (D) 32 × 10–4 N-m and 64 × 10–4 J


42. At a point on the axis (but not inside the dipole and not at infinity) of an electric dipole(A) The electric field is zero(B) The electric potential is zero(C) Neither the electric field nor the electric potential is zero(D) The electric field is directed perpendicular to the axis of the dipole

43. The force between two short electric dipoles separated by a distance r is directly proportional to :(A) r2 (B) r4 (C) r–2 (D) r–4

44. If electric field is uniform, then the electric lines of forces are:(A) Divergent (B) Convergent (C) Circular (D) Parallel

45. The figure shows the electric lines of force emerging from acharged body. If the electric fields at A and B are EA and EBrespectively and if the distance between A and B is r, then

B

A(A) EA < EB (B) EA > EB

(C) r

EE BA (D) 2

BA r

EE

46. Select the correct statement :(A) The electric lines of force are always closed curves(B) Electric lines of force are parallel to equipotential surface(C) Electric lines of force are perpendicular to equipotential surface(D) Electric line of force is always the path of a positively charged particle.

47. A neutral spherical metallic object A is placed near a finite metal plate B carrying a positive charge.The electric force on the object will be :(A) away from the plate B (B) towards the plate B(C) parallel to the plate B (D) zero

48. A positive point charge q is brought near a neutral metal sphere.(A) The sphere becomes negatively charged.(B) The sphere becomes positively charged.(C) The interior remains neutral and the surface gets non-uniform charge distribution.(D) The interior becomes positively charged and the surface becomes negatively charged.

49. Three concentric conducting spherical shells carry charges as follows : + 4Q on the inner shell, - 2 Q onthe middle shell and – 5 Q on the outer shell. The charge on the inner surface of the outer shell is:(A) 0 (B) 4 Q (C) - Q (D) - 2 Q

50. A charge q is uniformly distributed over a large plastic plate. The electric field at a point P close to thecentre and just above the surface of the plate is 50 V/m. If the plastic plate is replaced by a copperplate of the same geometrical dimensions and carrying the same uniform charge q, the electric field atthe point P will become:(A) zero (B) 25 V/m (C) 50 V/m (D) 100 V/m

51. Figure shows a thick metallic sphere. If it is given a charge +Q,then electric field will be present in the region (A) r < R1 only

(B) r > R1 and R1 < r < R2

(C) r R2 only

(D) r R2 only

52. An uncharged sphere of metal is placed in a uniform electric field produced by two large conductingparallel plates having equal and opposite charges, then lines of force look like:

(A) (B) (C)

+

–

+ + + +

– – – –

(D)


53. Two small conductors A and B are given charges q1 and q2respectively. Now they are placed inside a hollow metallic con-ductor (C) carrying a charge Q. If all the three conductors A, Band C are connected by conducting wires as shown, the chargeson A, B and C will be respectively:

(A) Q,2

qq,2

qq 2121 (B) 3

qqQ,3

qqQ,3

qqQ 212131

(C) 0,2

Qqq,2

Qqq 2121 (D) 0, 0, Q + q1 + q2

54. You are travelling in a car during a thunder storm. In order to protect yourself from lightening, would youprefer to :(A) Remain in the car (B) Take shelter under a tree(C) Get out and be flat on the ground (D) Touch the nearest electrical pole

55. A positively charged body 'A' has been brought near aneutral brass sphere B mounted on a glass stand asshown in the figure. The potential of B will be:

(A) Zero (B) Negative

(C) Positive (D) Infinite

56. The amount of work done in joules in carrying a charge +q along theclosed path PQRSP between the oppositely charged metal plates is:(where, E is electric field between the plates)

(A) zero (B) q

(C) qE (PQ + QR + SR + SP) (D) 0/q

57. Five balls, numbered 1 to 5, are suspended using separate threads. Pairs (1, 2), (2, 4), (4, 1) showelectrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion. Therefore ball 1 :(A) Must be positively charged (B) Must be negatively charged(C) May be neutral (D) Must be made of metal

58. Two point charges of same magnitude and opposite sign are fixed at points A and B. A third small pointcharge is to be balanced at point P by the electrostatic force due to these two charges. The point P:(A) lies on the perpendicular bisector of line AB(B) is at the mid point of line AB(C) lies to the left of A(D) none of these.

59. A particle A has charge +q and particle B has charge + 4q with each of them having the same mass m.When allowed to fall from rest through same electrical potential difference, the ratio of their speed vA :vB will be :(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

60. A charge ' q ' is placed at the centre of a conducting spherical shell ofradius R, which is given a charge Q. An external charge Q is alsopresent at distance R (R > R) from ' q '. Then the resultant field will bebest represented for region r < R by: [where r is the distance of thepoint from q ]

(A) (B) (C) (D)


61. In the above question, if Q' is removed then which option is correct :

(A) (B) (C) (D)

62. The volume charge density as a function of distance X from one faceinside a unit cube is varying as shown in the figure. Then the total flux(in S.I. units) through the cube if (0 = 8.85 1012 C/m3) is:

(A) 1/4 (B) 1/2

(C) 3/4 (D) 1

63. A positive point charge Q is kept (as shown in the figure) inside aneutral conducting shell whose centre is at C. An external uniformelectric field E is applied. Then : (A) Force on Q due to E is zero(B) Net force on Q is zero(C) Net force acting on Q and conducting shell considered as a system is zero(D) Net force acting on the shell due to E is zero.

64. A point charge q is brought from infinity (slowly so that heat developedin the shell is negligible) and is placed at the centre of a conductingneutral spherical shell of inner radius a and outer radius b, then workdone by external agent is:

(A) 0 (B) k q

b

2

2

(C) k q

b

2

2 – k q

a

2

2 (D) k q

a

2

2 - k q

b

2

2

65. The electric field above a uniformly charged nonconducting sheet is E. If the nonconducting sheet isnow replaced by a conducting sheet, with the charge same as before, the new electric field at thesame point is :

(A) 2E (B) E (C) 2E

(D) None of these

66. The linear charge density on upper half of a segment of ring is and at

lower half, it is – . The direction of electric field at centre O of ring is:

(A) along OA (B) along OB(C) along OC (D) along OD

67. A charged particle ‘q’ is shot from a large distance with speed v towards a fixed charged particle Q. Itapproaches Q upto a closest distance r and then returns. If q were given a speed ‘2v’, the closestdistance of approach would be :

(A) r (B) 2r (C) 2r

(D) 4r


68. A total charge of 20 C is divided into two parts and placed at some distance apart. If the chargesexperience maximum coulombian repulsion, the charges should be :

(A) 5 C , 15 C (B) 10 C , 10 C (C) 12 C , 8 C (D) C3

20,C340

69. The magnitude of electric force on 2 c charge placed at the centre O oftwo equilateral triangles each of side 10 cm, as shown in figure is P. Ifcharge A, B, C, D, E & F are 2 c, 2 c, 2 c, -2 c, - 2 c, - 2 c respectively,then P is: O

A

B

C

D

E

F

(A) 21.6 N (B) 64.8 N(C) 0 (D) 43.2 N

70. Two point charges a & b, whose magnitudes are same are positionedat a certain distance from each other with a at origin. Graph is drawnbetween electric field strength at points between a & b and distance xfrom a. E is taken positive if it is along the line joining from a to b. Fromthe graph, it can be decided that (A) a is positive, b is negative (B) a and b both are positive(C) a and b both are negative (D) a is negative, b is positive

71. The net charge given to an isolated conducting solid sphere:(A) must be distributed uniformly on the surface (B) may be distributed uniformly on the surface(C) must be distributed uniformly in the volume (D) may be distributed uniformly in the volume.

72. The net charge given to a solid insulating sphere:(A) must be distributed uniformly in its volume(B) may be distributed uniformly in its volume(C) must be distributed uniformly on its surface(D) the distribution will depend upon whether other charges are present or not.

73. A charge Q is kept at the centre of a conducting sphere of inner radius R1 and outer radius R2. A pointcharge q is kept at a distance r (> R2) from the centre. If q experiences an electrostatic force 10 N thenassuming that no other charges are present, electrostatic force experienced by Q will be:(A) – 10 N (B) 0 (C) 20 N (D) none of these

74. Two short electric dipoles are placed as shown (r is the distance betweentheir centres). The energy of electric interaction between these dipoleswill be:

C

(C is centre of dipole of moment P2)

(A) 321

rcosPPk2

(B) 321

rcosPPk2

(C) 321

rsinPPk2

(D) 321

rcosPPk4

75. The given figure gives electric lines of force due to two charges q1 and q2. What are the signs of the twocharges?

(A) Both are negative (B) Both are positive(C) q1 is positive but q2 is negative (D) q1 is negative but q2 is positive


76. A positively charged pendulum is oscillating in a uniform electric fieldas shown in Figure. Its time period of SHM as compared to that whenit was uncharged. (mg > qE) (A) Will increase(B) Will decrease(C) Will not change(D) Will first increase then decrease

77. A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shellhaving charge –Q. The radius of the sphere is a and that of the spherical shell is b(>a). What is theelectric field at a distance r(a < r < b) from the centre?

(A) 04

1

rQ

(B) 04

1

rQ3

(C) 04

1 2r

Q3(D)

041 2r

Q

78. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conductinghollow spherical shell. Let the potential difference between the surface of the solid sphere and that ofthe outer surface of the hollow shell be V. If the shell is now given a charge –3Q, the new potentialdifference between the same two surfaces is :(A) V (B) 2V (C) 4V (D) –2V

79. For an infinite line of charge having charge density lying along x-axis,the work required in moving charge q from C to A along arc CA is :

(A) 0

q

loge 2 (B) 04

q

loge 2

(C) 04

q

loge 2 (D) 02

q

loge 21

80. In an electron gun, electrons are accelerated through a potential difference of V volt. Taking electroniccharge and mass to be respectively e and m, the maximum velocity attained by them is :

(A) meV2

(B) meV2

(C) 2 m/eV (D) (V2 /2em)

KVPY PROBLEMS (PREVIOUS YEARS)1. Figure shows two graphs (X) and (Y) related to a charged conducting sphere of radius a and charge Q

[KVPY_2007_] [1 Marks]

+Q

a

X

Y

(A) X represents potential versus distance (from the centre) graph while Y represents electric field versusdistance graph.(B) Y represents potential versus distance graph while X represents electric field versus distance graph(C) Both graphs show that potential and electric field are continuous throughout.(D) Both graphs show that potential and electric field have continuous first order derivatives.


2. Figure below show a charge disribution of three charge q, –2q and q located along the y-axis with the charge–2q at centre and the other charges symetrically placed about it : [KVPY_2007] [1 Marks]

q

yx

–2q

q

The electric field along the x axis. for large x compared to the size of the distribution, varies as :

(A) x1

(B) 2x1

(C) 3x1

(D) 4x1

3. In a certain region of space, electric field is along the positive z direction throughout. The field is, however,non-uniform ; its magnitude increases uniformly along the positive z-direction at the rate of 105NC–1m1. Theforce and torque experienced by a system having a total dipole moment of 10–7 C × m in the negative z-direction are - [KVPY_2007] [2 Marks](A) 10–2 N in the negative z-direction ; torque = 0(B) 10–2 N in the positive z-direction torque = 0(C) Force = 0 ; torque = 10–2 Nm so as to decreases potential energy.(D) Force = 10–2 N in the negative Z-direction, torque = 10-2 Nm so as to decrease the potential energy

4. In one model of an atom, a positively charged point nucleus of charge Ze is surrounded by a uniform densityof negative charge up to a radius R. The atom as a whole is neutral. In this model the electric field at adistance r from the nuelus is given by - [KVPY_2007] [2 Marks]

(A) E(r) = 30 R

r4Ze r < R = 2

0r4Ze r > R (B) E(r) = 3

0R4Zer r < R = 0 r > R

(C) E(r) = 04

Ze

32 Rr–

r1

r < R = 0 r > R (D) E(r) = 20r4

Ze

Rr–1 , for all r

5. Consider two cases : [KVPY_2008] [1 Marks](i) A point charge q at the origin(ii) A uniformly charged solid sphere of radius R (wth its centre at the origin) and total charge q.Given below are graphs of a property X versus distance r from the origin for the two cases. The graphscoincide for r R.

(A) X is electric potential due to point charge/charged sphere.(B) X is magnitude of electric field due to point charge/charged sphere(C) X is electrostatic potential energy of point charge/charged sphere(D) X is charge density in space.


6. A charge Q is spread non uniformly on the surface of a hollow sphere of radius R, such that the chargedensity is given by = 0 (1 – sin), where is the usual polar angle. The potential at the centre of the sphereis [KVPY_2009] [1 Marks]

(A) R2Q

0 (B) RQ

0 (C) R8Q

0 (D) R4Q

0

7. Two identical conducting spheres carry identical charges. If the spheres are set at a certain distance apart,they repel each other with a force F. A third conducting sphere, identical to the other two, but initiallyuncharged, is then touched to one sphere, and then to the other before being removed. The force-betweenthe original two spheres is now [KVPY_2009] [1 Marks]

(A) 2F

(B) 4F

(C) 4F3

(D) 8F3

8. A point electric dipole placed at the origin has a potential given by V(r, ) = 20r4

cosp

where is the angle

made by the position vector with the direction of the dipole. Then [KVPY_2009] [1 Marks]

(A) since the potential vanishes at = 2

, the electric field is zero everywhere on the = 2

plane

(B) the electric field everywhere on the = 2

plane is normal to the plane.

(C) the electric field everywhere on the = 2

plane is along the plane

(D) the electric field vanishes on the = 0

9. Consider a uniform spherical volume charge distribution of radius R. Which of the following graphs correctlyrepresents the magnitude of the electric field E at a distance r from the center of the sphere ?

[KVPY_2010] [1 Marks]

(A) (B) (C) (D)

10. A charge +q is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius R1

and outer radius R2. A charge +Q is placed at a distance r > R2 from the center of the shell. Then the electricfield in the hollow cavity. [KVPY_2010] [1 Marks](A) depends on both +q and +Q (B) is zero(C) is only that due to +Q (D) is only that due to +q

11. Three equal charges +q are placed at the three vertices of an equilateral triangle centred at the origin. Theyare held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is aconstant. What is the distance of the three charges from the origin ? [KVPY_2010] [2 Marks]

(A)

3/12

0 kq

61

(B)

3/12

0 kq

123

(C)

3/22

0 kq

61

(D)

3/22

0 kq

43

12. Two identical particles of mass m and charge q are shot at each other from a very great distance with aninitial speed v. The distance of closest approach of these charges is : [KVPY_2010] [2 Marks]

(A) 20

2

mv8q

(B) 20

2

mv4q

(C) 20

2

mv2q

(D) 0


13. An isolated sphere of radius R contains uniform volume distribution of positive charge. Which of the curvesshown below correctly illustrates the dependence of the magnitude of the electric field of the sphere as afunction of the distance r from its centre ? [KVPY_2011] [2 Marks]

(A) I (B) II (C) III (D) IV

14. The surface of a planet is found to be uniformly charged. When a particle of mass m and no charge is thrownat an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontalrange L. A particle of mass m and charge q, with the same initial conditions has a range L/2. The range ofparticle of mass m and charge 2q with the same initial conditions is : [KVPY_2011] [2 Marks](A) L (B) L/2 (C) L/3 (D) L/4

15. At a distance form a uniformly charged long wire , a charged particle is thrown radially outward with avelocity u in the direction perpendicular to the wire . When the particle reaches a distance 2 from the wire

its speed is found to be u2 . The magnitude of the velocity , when it is a distance 4l away from the wire ,is (ignore gravity) [KVPY_2011] [2 Marks]

(A) u3 (B) 2 u (C) u22 (D) 4 u

16. Consider three concentric metallic spheres A, B and C of radii a, b, c respectively where a<b<c. A and B areconnected whereas C is grounded. The potential of the middle sphere B is raised to V then the charge on thesphere C is [KVPY_2012] [2 Marks]

(A) – 40V bcbc

(B) + 40V bcbc

(C) – 40V acac

(D) zero

Exercise # 11. (D) 2. (A) 3. (A) 4. (C) 5. (D) 6. (D) 7. (B)8. (C) 9. (B) 10. (D) 11. (A) 12. (B) 13. (C) 14. (A)15. (A) 16. (A) 17. (B) 18. (C) 19. (D) 20. (C) 21. (B)22. (A) 23. (B) 24. (C) 25. (B) 26. (C) 27. (D) 28. (A)29. (D) 30. (D) 31. (A) 32. (D) 33. (A) 34. (B) 35. (D)36. (C) 37. (C) 38. (C) 39. (C) 40. (C) 41. (D) 42. (C)43. (D) 44. (D) 45. (B) 46. (C) 47. (B) 48. (C) 49. (D)50. (C) 51. (C) 52. (C) 53. (D) 54. (A) 55. (C) 56. (A)57. (C) 58. (D) 59. (B) 60. (A) 61. (A) 62. (C) 63. (D)64. (C) 65. (B) 66. (C) 67. (D) 68. (B) 69. (D) 70. (A)71. (A) 72. (B) 73. (B) 74. (B) 75. (A) 76. (A) 77. (C)78. (A) 79. (A) 80. (B)

Exercise # 21. (B) 2. (B) 3. (A) 4. (C) 5. (B) 6. (D) 7. (D)8. (B) 9. (A) 10. (D) 11. (B) 12. (B) 13. (B) 14. (C)15. (A) 16. (A)


2.2 Charging of a body

A body can be charged by means of (a) friction, (b) conduction, (c) induction, (d) thermionic ionizationor thermionic emission (e) photoelectric effect and (f) field emission.

(a) Charging by Friction :When a neutral body is rubbed against other neutral body then some electrons are transferred from onebody to other. The body which can hold electrons tightly, draws some electrons and the body which cannot hold electrons tightly, loses some electrons. The body which draws electrons becomes negativelycharged and the body which loses electrons becomes positively charged.

For example : Suppose a glass rod is rubbed with a silk cloth. As the silk can hold electrons more tightlyand a glass rod can hold electrons less tightly (due to their chemical properties), some electrons willleave the glass rod and get transferred to the silk. So, in the glass rod their will be deficiency ofelectrons, therefore it will become positively charged. And in the silk, there will be some extra electrons,so it will become negatively charged

(b) Charging by conduction (flow): There are three types of materials in nature(i) Conductor : Conductors are the material in which the outer most electrons are very loosely bound,so they are free to move (flow). So in a conductor, there are large number of free electrons.Ex. Metals like Cu, Ag, Fe, Al.............

(ii) Insulator or Dielectric or Nonconductor : Non-conductors are the materials in which outer mostelectrons are very tightly bound, so that they cannot move (flow). Hence in a non-conductor there are nofree electrons. Ex. plastic, rubber, wood etc.(iii) Semi conductor : Semiconductors are the materials which have free electrons but very less in num-ber.

Now lets see how the charging is done by conduction. In this method, we take a charged conductor 'A'and an uncharged conductor 'B'. When both are connected, some charge will flow from the charged bodyto the uncharged body. If both the conductors are identical & kept at large distance and connected toeach other, then charge will be divided equally in both the conductors otherwise they will flow till theirelectric potential becomes same. Its detailed study will be done in last section of this chapter.


(c) Charging by Induction : To understand this, let’s have introduction to induction.

We have studied that there are lot of free electrons in the conductors. When a charged particle +Q isbrought near a neutral conductor, due to attraction of +Q charge, many electrons (–ve charges) comecloser and accumulate on the closer surface.On the other hand, a positive charge (deficiency of electrons) appears on the other surface. The flow ofcharge continues till the resultant force on free electrons of the conductor becomes zero. This phenom-ena is called induction and charges produced are called induced charges.

(d) Thermionic emission : When the metal is heated at ahigh temperature then some electrons of metals are ejectedand the metal becomes positively charged.

(e) Photoelectric effect : When light of sufficiently high fre-quency is incident on metal surface then some electrons gainenergy from light and come out of the metal surface and re-maining metal becomes positively charged.

(f) Field emission : When electric field of large magnitudeis applied near the metal surface then some electrons comeout from the metal surface and hence the metal gets posi-tively charged.


Example 1.If a charged body is placed near a neutral conductor, will it attract the conductor or repel it?

Solution :

If a charged body (+ve) is placed left side near a neutral conductor, (–ve) charge will induce at left surface and(+ve) charge will induce at right surface. Due to positively charged body –ve induced charge will feel attrac-tion and the +ve induced charge will feel repulsion. But as the –ve induced charge is nearer, so the attractiveforce will be greater than the repulsive force. So the net force on the conductor due to positively charged bodywill be attractive. Similarly, we can prove for negatively charged body also.

From the above example we can conclude that. "A charged body can attract a neutral body."

If there is attraction between two bodies then one of them may be neutral. But if there is repulsion be-tween two bodies, both must be charged (similarly charged).So "repulsion is the sure test of electrification".

Example 2.A positively charged body 'A' attracts a body 'B' then charge on body 'B' may be:(A) positive (B) negative (C) zero (D) can't say

Ans. B, C

Example 3.Five styrofoam balls A, B, C, D and E are used in an experiment. Several experiments are performedon the balls and the following observations are made :(i) Ball A repels C and attracts B.(ii) Ball D attracts B and has no effect on E.(iii) A negatively charged rod attracts both A and E.For your information , an electrically neutral styrofoam ball is very sensitive to charge induction andgets attracted considerably, if placed nearby a charged body. What are the charges, if any, on eachball ?

A B C D E(A) + – + 0 +(B) + – + + 0(C) + – + 0 0(D) – + – 0 0

Ans. CSol. From (i), as A repels C, so both A and C must be charged similarly. Either both are +ve or both are

–ve. As A also attract B, so charge on B should be opposite of A or B may be uncharged conductor.From (ii) as D has no effect on E, so both D and E should be uncharged and as B attracts unchargedD, so B must be charged and D must be an uncharged conductor.From (iii), a –vely charged rod attracts the charged ball A, so A must be +ve and from exp. (i) C mustalso be +ve and B must be –ve.


Example 4.Charge conservation is always valid. Is it also true for mass?

Sol. No, mass conservation is not always. In some nuclear reactions, some mass is lost and it is convertedinto energy.

Example 5.What are the differences between charging by induction and charging by conduction ?

Sol. Major differences between two methods of charging are as follows :

(i) In induction, two bodies are close to each other but do not touch each other while in conductionthey touch each other. (Or they are connected by a metallic wire)

(ii) In induction, total charge of a body remains unchanged while in conduction it changes.

(iii) In induction, induced charge is always opposite in nature to that of source charge while in conductioncharge on two bodies finally is of same nature.

Example 6.If a glass rod is rubbed with silk, it acquires a positive charge because :(A) protons are added to it (B) protons are removed from it(C) electrons are added to it (D) electrons are removed from it.

Ans. D

3 . COULOMB’S LAW (INVERSE SQUARE LAW)

On the basis of experiments Coulomb established the following law known as Coulomb's law :The magnitude of electrostatic force between two point charges is directly proportional to the productof charges and inversely proportional to the square of the distance between them.

i.e. F q1q2 and F 2r1

F 221

rqq

F = 221

rqKq

Important points regarding Coulomb's law :(i) It is applicable only for point charges.

(ii) The constant of proportionality K in SI units in vacuum is expressed as 04

1

and in any other

medium expressed as 41 . If charges are dipped in a medium then electrostatic force on one

charge is r04

1

221

rqq

; where 0 and are called permittivity of vacuum and absolute

permittivity of the medium respectively. The ratio 0/ = r is called relative permittivity of themedium, which is a dimensionless quantity.

(iii) The value of relative permittivity r is constant for a medium and can have values between 1 to. For vacuum, by definition it is equal to 1. For air it is nearly equal to 1 and may be taken tobe equal to 1 for calculations. For metals, the value of r is and for water is 81. The materialin which more charge can induce r will be higher.

(iv) The value of 04

1 = 9 × 109 Nm2 C–2 & 0 = 8.855 × 10–12 C2/Nm2.

Dimensional formula of is [M–1 L–3 T4 A2 ]

(v) The force acting on one point charge due to the other point charge is always along the linejoining these two charges. It is equal in magnitude and opposite in direction on two charges,irrespective of the medium in which they lie.


(vi) The force is conservative in nature i.e., work done by electrostatic force in moving a pointcharge along a closed loop of any shape is zero.

(vii) Since the force is a central force, in the absence of any other external force, angular momentumof one particle w.r.t. the other particle (in two particle system) is conserved.

(viii) In vector form formula can be given as below.

F =

r041

r|r|qq

321

= r04

1

r|r|

qq221

; (q1 & q2 are to be substituted with sign.)

Here, r

is position vector of the test charge (on which force is to be calculated) with respect tothe source charge (due to which force is to be calculated).

Example 7. Find out the electrostatic force between two point charges placed in air (each of +1 C) if theyare separated by 1m .

Sol. Fe = 221

rqkq

= 2

9

111109

= 9×109 N

From the above result, we can say that 1 C charge is too large to realize. In nature, charge is

usually of the order of CExample 8.

A particle of mass m carrying charge q1 is revolving around a fixed charge –q2 in a circular path ofradius r. Calculate the period of revolution and its speed also.

Sol.04

1 2

21

rqq

= mr2 = 2

2

Tmr4

'

T2 = 21

220

qq)mr4(r)4(

or T = 4r 21

0

qqmr

and also we can say that

20

21

r4qq

=

rmv2

V = mr4qq

0

21

Example 9.A point charge qA = + 100 µc is placed at point A (1, 0, 2) m and another point charge qB = +200µc isplaced at point B (4, 4, 2) m. Find :(i) Magnitude of electrostatic interaction force acting between them

(ii) Find AF

(force on A due to B) and BF

(force on B due to A) in vector form

Sol. (i)

Value of F : 2BA

rqkqF = 2

222

669

)22()04()14(

)10200()10100()109(

= 7.2 N


(ii) Force on B, BF

= r|r|qkq3BA

= k)22(j)04(i)14()22()04()14(

)10200()10100()109(3

222

669

= 7.2

j

54i

53

N

Similarly AF

= 7.2

j

54i

53

N

Action( AF

) and Reaction ( BF

) are equal but in opposite direction.

4 . PRINCIPLE OF SUPERPOSITION

The electrostatic force is a two body interaction i.e. electrical force

between two point charges is independent of presence or absence of other charges and so the principleof superposition is valid i.e. force on charged particle due to number of point charges is the resultant offorces due to individual point charges. Therefore, force on a point test charge due to many charges is

given by ...........FFFF 321

.

Example 10Three equal point charges of charge +q each are moving alonga circle of radius R and a point charge –2q is also placed atthe centre of circle (as shown in figure). If charges are revolvingwith constant and same speed in the circle then calculate speedof charges

Sol.

F2 – 2F1 cos 30º = R

mv2

2R)q2()q(K

– 2

2

)R3()Kq(2

cos 30 = R

mv2

312

Rmkqv

2


Example 11Two equally charged identical small metallic spheres A and B repel each other with a force 2 × 10–5Nwhen placed in air (neglect gravitational attraction). Another identical uncharged sphere C is touched toB and then placed at the mid point of line joining A and B. What is the net electrostatic force on C?

Sol. Let, initially the charge on each sphere be q and separation between their centres be r. Then accordingto given problem:

F = 04

1 2r

qq = 2 × 10–5 N

When sphere C touches B, the charge of B i.e. q will distribute equally on B and C as sphere areidentical how charges on spheres;

qB = qC = (q/2)So sphere C will experience a force

FCA = 04

1 2)2/r(

)2/q(q = 2F along AB due to charge on A.A.

and, FCB = 04

1 2)2/r(

)2/q)(2/q( = F, along BA due to charge on B :

So the net force FC on C due to charges on A and B,

FC = FCA – FCB = 2F – F = 2 × 10–5 N along AB .

Example 12Five point charges, each of value q are placed on five vertices of a regular hexagon of side L. What isthe magnitude of the force on a point charge of value – q coulomb placed at the centre of the hexagon?

Sol. Method : IIf there had been a sixth charge +q at the remaining vertex of hexagon, force due to all the six chargeson –q at O would have been zero (as the forces due to individual charges will balance each other), i.e.,

0FR

Now if f

is the force due to sixth charge and F

due to remaining five charges.

From F

+ f

= 0 i.e. F

= – f

or, |F| = |f| = 04

1 2L

qq = 2

2

0 Lq

41

NetF

= ODF

= 2

2

Lq

41

along OD

qq C

qqDE

O

-q

qBA

F

L

Method : II

In the diagram, we can see that force due tocharge A and D are opposite to each other

oFF

+ OCF

= 0 ....(i)

Similarly OBF

+ OEF

= 0 ....(ii)

So oFF

+ OBF

+ OCF

+ ODF

+ OEF

= NetF

Using (i) and (ii) NetF

= ODF

= 2

2

Lq

41

along OD.

qq C

qqDE

O

-q

qBA

F

L


Example 13A thin straight rod of length carrying a uniformly distributed charge q islocated in vacuum. Find the magnitude of the electric force on a pointcharge 'Q' kept as shown in the figure.

Sol. As the charge on the rod is not point charge, therefore, first we have to find force on charge Q due to charge over a very small part on thelength of the rod. This part, called element of length dy can be consideredas point charge.

Charge on element, dq = dy = dyq

Electric force on 'Q' due to element = 2yQ.dq.K

= .ydy.q.Q.K

2

All forces are along the same direction,

F = dF . This sum can be calculated using integration,

therefore, F =

a

ay2y

KQqdy =

KqQ

a

ay1

=

a

1a1q.KQ

= )a(aKQq

Note : (1) The total charge of the rod cannot be considered to be placed at the centre of the rodas we do in mechanics for mass in many problems.

Note : (2) If a >> then, F = 2aKQq

i.e. Behaviour of the rod is just like a point charge.

5 . ELECTROSTATIC EQUILIBRIUMThe point where the resultant force on a charged particle becomes zero is called equilibrium position.

5.1 Stable Equilibrium : A charge is initially in equilibrium position and is displaced by a smalldistance. If the charge tries to return back to the same equilibrium position then this equilibrium iscalled position of stable equilibrium.

5.2 Unstable Equilibrium : If charge is displaced by a small distance from its equilibrium positionand the charge has no tendency to return to the same equilibrium position. Instead it goes away fromthe equilibrium position.

5.3 Neutral Equilibrium : If charge is displaced by a small distance and it is still in equilibriumcondition then it is called neutral equilibrium.

Example 14Two equal positive point charges 'Q' are fixed at points B(a, 0) and A(–a, 0). Another test charge q0 isalso placed at O(0, 0). Show that the equilibrium at 'O' is(i) Stable for displacement along X-axis.(ii) Unstable for displacement along Y-axis.

Sol. (i)O

Q Qq0

y

xFAOFBOA B(– a, 0) (a, 0)


Initially AOF

+ BOF

= 0 |F| AO

= |F| BO

=

20

aKQq

When charge is slightly shifted towards + x axisby a small distance x, then.

|F| AO

< |F| BO

Therefore, the particle will move towards origin (its original position). Hence, the equilibrium is stable.(ii) When charge is shifted along y axis:

After resolving components, net force will be along y axis So, the particle will not return to itsoriginal position & it is unstable equilibrium. Finally, the charge will move to infinity.

Example 15.Two point charges of charge q1 and q2 (both of same sign) and each of mass m are placed such thatgravitational attraction between them balances the electrostatic repulsion. Are they in stable equilibrium?If not then what is the nature of equilibrium?

Sol. In given example :

221

rqqK

= 2

2

rGm

We can see that irrespective of distance between them charges will remain in equilibrium. If nowdistance is increased or decreased then there is no effect in their equilibrium. Therefore it is a neutralequilibrium.

Example 16.A particle of mass m and charge q is located midway between two fixed charged particles each havinga charge q and a distance 2 apart. Prove that the motion of the particle will be SHM if it is displacedslightly along the line connecting them and released. Also find its time period.

Sol. Let the charge q at the mid-point is displaced slightly to the left.The force on the displaced charge q due to charge q at A,

F1 = 04

1

2

2

)x(q

The force on the displaced charge q due to charge at B,

F2 = 04

1 2

2

)x(q

Net restoring force on the displaced charge q.

F = F2 – F1 or F = 04

1

2

2

)x(q

– 04

1

2

2

)x(q

or F = 0

2

4q

22 )x(1

)x(1

=

0

2

4q 222 )x(

x4

Since >> x, F = 40

2 xq

or F = 30

2xq


Hence we see that F x and it is opposite to the direction of displacement. Therefore, the motion isSHM.

T = km2 , (here k = 3

0

2q ) T = 2

30

qm

2

Example 17.Find out mass of the charge Q, so that it remains

in equilibrium for the given configuration.

qq

h qq

Q

Sol. 4 Fcos = mg

4× 2/32

2h

2

KQq

h = mg m = 2/3

22

h2

g

KQqh4

Example 18.Two identical charged spheres are suspended by strings of equal length. Each string makes an angle with the vertical. When suspended in a liquid of density = 0.8 gm/cc, the angle remains the same.What is the dielectric constant of the liquid? (Density of the material of sphere is = 1.6 gm/cc.)

Sol. Initially as the forces acting on each ball are tension T,weight mg and electric force F, for its equilibrium along vertical

T cos = mg ...(1)and along horizontal

T sin = F ...(2)Dividing Eqn. (2) by (1), we have

tan = mgF

... (3)

When the balls are suspended in a liquid of density and dielectric constant K, the electric force willbecome (1/K) times, i.e., F' = (F/K) while weight

mg' = mg – FB = mg – Vg [as FB = Vg, where is density of material of sphere]

i.e. mg' = mg

1

mVas

So, for equilibrium of ball,

tan ' = 'mg'F

= )]/(1[KmgF

... (4)

According to given information ' = ; so from equations (4) and (3), we have :

K = )(

= )8.06.1()6.1(

= 2 Ans.

6 . ELECTRIC FIELD

Electric field is the region around charged particle or charged body in which if another charge is placed,it experiences electrostatic force.

6.1 Electric field intensity E

: Electric field intensity at a point is equal to the electrostatic forceexperienced by a unit positive point charge both in magnitude and direction.


If a test charge q0 is placed at a point in an electric field and experiences a force F

due to somecharges (called source charges), the electric field intensity at that point due to source charges is given

by 0q

FE

If the E

is to be determined practically then the test charge q0 should be small otherwise it will affectthe charge distribution on the source which is producing the electric field and hence modify the quantitywhich is measured.

Example 19.A positively charged ball hangs from a long silk thread. We wish to measure E at a point P in the samehorizontal plane as that of the hanging charge. To do so, we put a positive test charge q0 at the pointand measure F/q0. Will F/q0 be less than, equal to, or greater than E at the point in question?

Sol. When we try to measure the electric field at point P then after placing the testcharge at P, it repels the source charge (suspended charge) and the measured

value of electric field Emeasured = 0q

F will be less than the actual value Eact ,that

we wanted to measure.

6.2 Properties of electric field intensity E

:(i) It is a vector quantity. Its direction is the same as the force experienced by positive charge.

(ii) Direction of electric field due to positive charge is always away from it while due to negativecharge, always towards it.

(iii) Its S.. unit is Newton/Coulomb.

(iv) Its dimensional formula is [MLT–3A–1]

(v) Electric force on a charge q placed in a region of electric field at a point where the electric field

intensity is E

is given by EqF

.

Electric force on point charge is in the same direction of electric field on positive charge and inopposite direction on a negative charge.

(vi) It obeys the superposition principle, that is, the field intensity at a point due to a system ofcharges is vector sum of the field intensities due to individual point charges.

i.e. 321 EEEE

+ .....

(vii) It is produced by source charges. The electric field will be a fixed value at a point unless wechange the distribution of source charges.

Example 20.Electrostatic force experienced by –3C charge placed at point 'P' due

to a system 'S' of fixed point charges as shown in figure is )j9i21(F

µN.

(i) Find out electric field intensity at point P due to S.(ii) If now, 2C charge is placed and –3 C is removed at point P

then force experienced by it will be.


Sol. (i) EqF

µN)j9i21( = -3µC )E(

E= – 7 i – 3 j

CN

(ii) Since the source charges are not disturbed the electric field intensity at 'P' will remain same.

F

2C = +2( E = 2(–7 i – 3 j )

= (–14 i – 6 j ) N

Example 21.Calculate the electric field intensity which would be just sufficient to balance the weight of a particle ofcharge –10 c and mass 10 mg. (take g = 10 ms2)

Sol. As force on a charge q in an electric field E

is

F

q = qE

So, according to given problem:

q E

W

A

Fe

[W = weight of particle]

|W| |F| q i.e., |q|E = mg

i.e., E = |q|mg

= 10 N/C., in downward direction.


List of formula for Electric Field Intensity due to various types of charge distribution :


Example 22.Find out electric field intensity at point A (0, 1m, 2m) due to a point charge –20C situated at point

B( 2 m, 0, 1m).

Sol. E = r|r|

KQ3

= r

|r|KQ

2 r

= P.V. of A – P.V. of B (P.V. = Position vector)

= (- 2 i + j + k ) | r

| = 222 )1()1()2( = 2

E = 8

)1020(109 69 (– 2 i + j + k ) = – 22.5 × 103 (– 2 i + j + k ) N/C.

Example 23.Two point charges 2c and – 2c are placed at points A and Bas shown in figure. Find out electric field intensity at points Cand D. [All the distances are measured in meter].

Sol. Electric field at point C(EA, EB are magnitudes only and arrows represent directions) Electric field due to positive charge is away from it while dueto negative charge, it is towards the charge. It is clear that EB > EA . ENet = (EB – EA) towards negative X-axis

= 22 )23()c2(K

)2()c2(K

towards negative X-axis = 8000 (– i ) N/C

Electric field at point D :Since magnitude of charges are same and also AD = BDSo, EA = EB

Vertical components of AE

and BE

cancel each other while horizontalal

components are in the same direction.

So, Enet = 2EA cos = 22)c2(K.2

cos450

= 2

10K 6 = i

29000

N/C.

Example 24.Six equal point charges are placed at the corners of a regular hexagonof side ‘a’. Calculate electric field intensity at the centre of hexagon?

Ans Zero (By symmetry)


Similarly electric field due to a uniformly charged ring at the centre of ring :

Note : (i) Net charge on a conductor remains only on the outer surface of a conductor. This propertywill be discussed in the article of the conductor. (article no.17)(ii) On the surface of isolated spherical conductor charge is uniformly distributed.

6.3 Electric field due to a uniformly charged ring and arc.

Example 25.Find out electric field intensity at the centre of a uniformly charged semicircular ring of radius R andlinear charge density .

Sol. = linear charge density.The arc is the collection of large no. of point charges.Consider a part of ring as an element of length Rd which subtendsan angle d at centre of ring and it lies between and + d

dE = dEx i + dEy j ; Ex = 0dEx (due to symmetry)

& Ey = ydE =

0

sindE = RK

0

d.sin = RK2

Example 26.Find out electric field intensity at the centre of uniformly charged quarter ring of radius R and linearcharge density .

Sol. Refer to the previous question dE = dEx i + dEy j on solving Enet = )ji(R

K

,

By use of symmetry and from the formula of electric field due to half ring.

Above answer can be justified.

(ii) Derivation of electric field intensity at a point on the axis at a distance x from centre of uni-formly charged ring of radius R and total charge Q.


Consider an element of charge dq. Due to this element, the electric field at the point on axis, which isat a distance x from the centre of the ring is dE.There are two components of this electric field

dE

dEX dEY

The y-component of electric field due to all the elements will be cancelled out to each other. So netelectric field intensity at the point will be only due to X-component of each element.

Enet = xdE = dECos = 22

Q

O22

XR

xxR)dq(K

=

Q

O2/322 dq

)xR(xk

Enet = 2/322 ]xR[KQx

Graph for variation of E with r.

E will be max when dxdE

= 0, that is at x= 2R

and Emax = 2R 33KQ2

Case (i) : if x>>R, E = 2xKQ

Hence the ring will act like a point charge

Case (ii) : if x<<R, E = 3Rx KQ

Example 27.Positive charge Q is distributed uniformly over a circular ring of radius a. A point particle having a mass m anda negative charge –q, is placed on its axis at a distance y from the centre. Find the force on the particle.Assuming y << a, find the time period of oscillation of the particle if it is released from there. (Neglect gravity)

Sol. When the negative charge is shifted at a distance x from the centre of the ring along its axis then force actingon the point charge due to the ring:FE = qE (towards centre)

= q

2/322 )ya(KQy

If a >> y thena2 + y2 ~ a2

FE = 04

1

3aQqy

(Towards centre)

Since, restoring force FE y, therefore motion of charge the particle will be S.H.M.Time period of SHM:

T = 2km

= 2

3

0a4Qqm

πε =

2/130

3

Qqma16

επ


6.4 Electric field due to uniformly charged wire:

(i) Line charge of finite length : Derivation of expression for intensity of electric field at a pointdue to line charge of finite size of uniform linear charge density . The perpendicular distanceof the point from the line charge is r and lines joining ends of line charge distribution makeangle 1 and 2 with the perpendicular line.

r P1

2

r

P

dEY

dEX

dE

x

dx

A

Consider a small element dx on line charge distribution at distance x from point A (see fig.).The charge of this element will be dq = dx. Due to this charge (dq), the intensity of electricfield at the point P is dE.

Then dE = 22 xr)dq(K

= 22 xr

)dx(K

There will be two components of this field :

dE

dEX dEY

Ex =

cos.

xrdxKcosdEdE 22x

Assuming, x = r tan dx = r sec2. d

so Ex =

1

2

222

2

tanrrdcossecrK

= r

K

1

2

d.cos = r

K [sin1+ sin2] ...........(1)

Similarly y-component.

Ey = r

K

1

2

d.sinθ

θ

θθ =r

K[cos2 - cos1]

Net electric field at the point:

Enet = 2y

2x EE

(ii) We can derive a result for infinitely long line charge:In above eq. (1) & (2), if we put 1 = 2 = 90º, we can get required result.

Enet = Ex = rK2

Er

E

r

r1E


(iii) For Semi- infinite wire :1 = 90º and 2 = 0º, so,

Ex = r

K , Ey =

rK

Ey

r

Ex

Example 28.A point charge q is placed at a distance r from a very long charged thread of uniform linear chargedensity . Find out total electric force experienced by the line charge due to the point charge. (Neglectgravity).

Sol. Force on charge q due to the thread,

F =

rK2 .q

By Newton's law, every action has equal and

opposite reaction So, force on the thread = q.rK2

(away from point charge)

Example 29.Figure shows a long wire having uniform charge density as shown in figure. Calculate electric fieldintensity at point P.

Sol. 1 = 90º and 2 = 360º – 37º So

Ex = r

K [sin1+ sin2]

Ey = rK

[cos2 - cos1]


6.5 Electric field due to uniformly charged spherical shell

E = 2rKQ r R For the outside points & point on the surface the uniformly

charged spherical shell behaves as a point charge placed atthe centre

E = 0 r < R

Electric field due to spherical shell outside it is always along the radial direction.

Example 30.Figure shows a uniformly charged sphere of radius R and total chargeQ. A point charge q is situated outside the sphere at a distance r fromcentre of sphere. Find out the following :(i) Force acting on the point charge q due to the sphere.(ii) Force acting on the sphere due to the point charge.

Sol. (i) Electric field at the position of point charge

rrKQE 2

So, rr

KqQF 2

; 2rKqQ |F|

(ii) Since we know that every action has equal and opposite reaction so

F

sphere = rr

KqQ2

| F

sphere| = 2rKqQ

.

Example 31.Figure shows a uniformly charged thin sphere of total charge Q andradius R. A point charge q is also situated at the centre of the sphere.Find out the following :(i) Force on charge q(ii) Electric field intensity at A.(iii) Electric field intensity at B.

Sol. (i) Electric field at the centre of the uniformly charged hollow sphere = 0 So force on charge q = 0(ii) Electric field at A

AE

= qSphere EE

= 0 + 2rKq

; r = CA

E due to sphere = 0 , because point lies inside the charged hollow sphere.


(iii) Electric field BE

at point B = qSphere EE

= r.rKqr.

rKQ

22 = r.r

)qQ(K2

; r = CB

Note : Here, we can also assume that the total charge of sphere is concentrated at the centre, for calculationof electric field at B.

Example 32.Two concentric uniformly charged spherical shells of radius R1 and R2 (R2 > R1)have total charges Q1 and Q2 respectively. Derive an expression of electric field asa function of r for following positions.(i) r < R1 (ii) R1 r < R2 (iii) r R2

Sol. (i) For r < R1,therefore, point lies inside both the spheres

Enet = EInner + Eouter = 0 + 0(ii) For R1 r < R2,

point lies outside inner sphere but inside outer sphere: Enet = Einner + Eouter

= 21

rKQ r + 0 = r

rKQ

21

(iii) For r R2

point lies outside inner as well as outer sphere.Therefore, ENet = Einner + Eouter

= rr

KQ2

1 rr

KQ2

2 = rr

)QQ(K2

21

Example 33.A spherical shell having charge +Q (uniformlydistributed) and a point charge + q0 are placed asshown. Find the force between shell and the pointcharge(r>>R).

(i) Force on the point charge + q0 due to the

shell = q0 E

shell = (q0) rrKQ

2

= rr

KQq2

0

where r , is unit vector along OP..From action - reaction principle, force on the shell due to the point charge will

also be : Fshell = )r(r

KQq2

0

Conclusion :- To find the force on a hollow sphere due to outside charges , we can replace the sphere by

a point charge kept at centre.

Example 34.Find force acting between two shells of radius R1 and R2 which have uniformly distributed charges Q1 and Q2

respectively and distance between their centres is r.

+ Q , R1 1

r

+ Q , R2 2


Sol. The shells can be replaced by point charges kept at centre,so force between them

F = 221

rQKQ

;

6.6 Electric field due to uniformly charged solid sphere:

Derive an expression for electric field due to solid sphere of radiusR and total charge Q which is uniformly distributed in the volume,at a point which is at a distance r from centre for given two cases.

dq

(i) r R (ii) r RAssume an elementary concentric shell of charge dq. Due to this shell,the electric field at the point (r > R) will be:

dE = 2rKdq

[from above result of hollow sphere]

Enet = dE = 2rKQ

For r < R, there will be no electric field due to shell of radius greater than r, so electric field at the pointwill be present only due to shells having radius less than r.

E´net = 2r'KQ

Here, Q' = 3

3r

34

R34

Q

= 3

3

RQr

E´net = 32 RKQr

r´KQ ; away from the centre.

Note : The electric field inside and outside the sphere is always in radial direction.

Example 35.A Uniformly charged solid non-conducting sphere of uniform volume charge density and radius R is having a concentric spherical cavity of radius r. Find out electricfield intensity at following points, as shown in the figure :(i) Point A (ii) Point B(iii) Point C (iv) Centre of the sphere

Sol. Method I :(i) For point A :

We can consider the solid part of sphere to be made of large number of spherical shells which haveuniformly distributed charge on its surface. Now, since point A lies inside all spherical shells soelectric field intensity due to all shells will be zero.

AE = 0


(ii) For point B :All the spherical shells for which point B lies inside will make electric field zero at point B. Soelectric field will be due to charge present from radius r to OB.

So, BE = 3

33

OB

)rOB(34K

OB = 03

3

33

OB]rOB[

OB

(iii) For point C, similarly we can say that for all the shell points C lies outside the shell

So, CE = 3

3334

]OC[

)]rR([K OC =

03

3

33

]OC[rR

OC

Method : IIWe can consider that the spherical cavity is filled with charge density and also –, thereby making netcharge density zero after combining. We can consider two concentric solid spheres: One of radius R andcharge density and other of radius r and charge density –. Applying superposition principle :

+

(i) AE = E + E = 03

)OA(

+

03)]OA([

= 0

(ii) BE = E + E = 03

)OB(

+ OB

)OB(

)(r34K

3

3

= OB)OB(3

r3 3

0

3

0

= OBOBr1

3 3

3

0

(iii) CE = E + E = 3

3

OC

R34K

OC + 3

3

OC

)(r34K

OC = 30 )OC(3

OC]rR[ 33

(iv) OE = E + E = 0 + 0 = 0

Example 36.In above question, if cavity is not concentric and centered at point P then repeat all the steps.

Sol. Again assume and – in the cavity, (similar to the previous example) :

(i) AE = E + E = 03

]OA[

+

03PA)(

= 03

[ OA – PA ] =

03 ]OP[

Note : Here, we can see that the electric field intensity at point A is independent of position of point A inside thecavity. Also the electric field is along the line joining the centres of the sphere and the spherical cavity.

(ii) BE = E + E = 03

)OB(

+ 3

334

]PB[)](r[K

PB

(iii) CE = E + E = 3

334

]OC[]R[K

OC + 3

334

]PC[)](r[K

PC

(iv) OE = E + E = 0 + 3

334

]PO[)](r[K

PO


7. ELECTRIC POTENTIAL :

In electrostatic field, the electric potential (due to some source charges) at a point P is defined as thework done by external agent in taking a unit point positive charge from a reference point (generallytaken at infinity) to that point P without changing its kinetic energy..

7.1 Mathematical representation :If (W P)ext is the work required in moving a point charge q from infinity to a point P, the electricpotential of the point P is

q)W(

q)W

V pelc

0K

extpp

Note : (i) (W P)ext can also be called as the work done by external agent against the electric force on aunit positive charge due to the source charge.

(ii) Write both W and q with proper sign.

7.2 Properties :(i) Potential is a scalar quantity, its value may be positive, negative or zero.

(ii) S.. Unit of potential is volt = coulmbjoule

and its dimensional formula is [M1L2T–3–1].

(iii) Electric potential at a point is also equal to the negative of the work done by the electric fieldin taking the point charge from reference point (i.e. infinity) to that point.

(iv) Electric potential due to a positive charge is always positive and due to negative charge it isalways negative except at infinity. (taking V = 0).

(v) Potential decreases in the direction of electric field.(vi) V = V1 + V2 + V3 + .......

7.3 Use of potential :If we know the potential at some point ( in terms of numerical value or in terms of formula) then we canfind out the work done by electric force when charge moves from point 'P' to by the formula

Wep )p = qVp

Example 37A charge 2C is taken from infinity to a point in an electric field, without changing its velocity. If workdone against electrostatic forces is –40J, then find the potential at that point.

Sol. V = qWext = C2

J40

= –20 V

Example 38When charge 10 C is shifted from infinity to a point in an electric field, it is found that work done byelectrostatic forces is –10 J. If the charge is doubled and taken again from infinity to the same pointwithout accelerating it, then find the amount of work done by electric field and against electric field.

Sol. Wext) p = –wel) p = wel)p = 10 Jbecause KE = 0

Vp = C20)W( pext

= C10J10

= 1V

So, if now the charge is doubled and taken from infinity then

1 = C20)w pext

or Wext) P = 20 J Wel ) P = –20 J

Example 39A charge 3C is released from rest from a point P where electric potential is 20 V then its kineticenergy when it reaches infinity is :

Sol. Wel = K = Kf – 0 Wel)P = qVP = 60 J So, Kf = 60 J


Electric Potential due to various charge distributions are given in table.

Name / Type Formula Note Graph

Point charge

Ring (uniform/nonuniformcharge distribution)

* q is source charge.* r is the distance of the point

from the point charge.

* Q is source chage.* x is the distance of the point on the axis from centre f ring

rKq

at centre:R

KQ

22 xR

KQ

Uniformly charged hollow conducting/nonconducting /solid conducting sphere

Uniformly charged so lid nonconducting sphere .

rkQV

RkQV

rkQV

3

22

R2)rR3(KQ

06

23

34

Infinite line charge

Infinite nonconducting thin sheet

Not defined

Not defined

Infinite charged conducting thin sheet

Not defined

AB0

AB rrVV

at the axis:

AB0

AB rrVV


7.4 Potential due to a point charge :

Derivation of expression for potential due to point charge Q, at a point which is at a distance r from thepoint charge.

r PQFrom definition of potential

V = o

r

0

o

)p(ext

q

rd)Eq(

qW

=

r

drE

V = – º180cos)dr(rKQ

r

2

= r

KQ

Example 40Four point charges are placed at the corners of a square of side Calculatepotential at the centre of square.

Sol. V = 0 at 'C'. [Use V =r

Kq]

Example 41Two point charges 2C and – 4C are situated at points (–2m, 0m) and(2 m, 0 m) respectively. Find out potential at point C (4 m, 0 m) and D

(0 m, 5 m).

Sol. Potential at point C

VC = 21 qq VV = 6)C2(K

+ 2

)C4(K =

6102109 69

– 2

104109 69 = –15000 V..

Similarly, VD = 21 qq VV = 22 2)5(

)C2(K

+ 22 2)5(

)C4(K

= 3

)C2(K + 3

)C4(K = – 6000 V..

Finding potential due to continuous charges

If formula of E is tough, then we take If formula of E is easy then, we use

a small element and integrate V =

rr

rrd·E–

V = dv (i.e. for sphere, plate, infinite wire etc.)


Example 42A rod of length is uniformly charged with charge q Calculate potential at point P.

Sol. Take a small element of length dx, at a distance x from left end. Potential due to this small element

dV = x

)dq(K Total potential V =

x

0xxdqk

Where dq =

qdx V =

rx

rx x

dxqK = elogKq

rr

7.5 Potential due to a ring :

(i) Potential at the centre of uniformly charged ring :Potential due to the small element dq

dV = R

Kdq

Net potential: V = RdqK

V = RK

dq = RKq

(ii) For non-uniformly charged ring potential at the center is

V = R

Kqtotal

(iii) Potential due to half ring at center is :

RKqV


(iv) Potential at the axis of a ring:Calculation of potential at a point on the axis which is a distance x from centre of uniformlycharged (total charge Q) ring of radius R.

R

dq

x P

Consider an element of charge dq on the ring.Potential at point P due to charge dq will be

dV = 22 xR

)dq(K

Net potential at point P due to all such element will be:

V =

22 xR

KQdv

Example 43

Figure shows two rings having charges Q and – 5 Q. Find Potential difference between A and B i.e. (VA -

VB).

Sol. VA = R

KQ+

22 RR2

Q 5K

; VB =

R 2

Q 5K +

22 RR

Q K

From above, we can easily find VA – VB.


Example 44A point charge q0 is placed at the centre of uniformly charged ring of total charge Q and radius R. If thepoint charge is slightly displaced with negligible force along axis of the ring then find out its speedwhen it reaches a large distance.

Sol. Only electric force is acting on q0

Wel = K = 21

mv2 – 0 Now Wel)c= q0 Vc = q0 . RKQ

R

QKq0 =21

mv2 v = mR

QKq2 0

7.6 Potential Due To Uniformly Charged Spherical shell :

Derivation of expression for potential due to uniformly charged hollow sphere of radius R andtotal charge Q, at a point which is at a distance r from centre for the following situation(i) r > R (ii) r < R

Assume a ring of width Rd at angle from X axis (as shown in figure).Potential due to the ring at the point P will be

d

dq

P

r – R cos

dV = 22 )sinR()cosRr(

)dq(K

Where dq = 2R sin (Rd)where Q, = 4R2then, net potential

V = 2

KQdV

022 )sinR()cosRr(

d.sin

Solving this eq.we find

V = r

KQ (for r > R)

& V = R

KQ for (r < R)

Alternate Method :

As the formula of E is easy , we use V =

rr

r

rd·E–

(i) At outside point (r > R):

Vout = dr rQ K–

rr

r2

Vout = centre from cetanDis

KQ r

KQ

For outside point, the hollow sphere acts like a point charge.

(ii) Potential at the centre of the sphere (r=0) :As all the charges are at a distance R from the centre ,

So, Vcentre = sphere the of RadiusKQ

RKQ


(iii) Potential at inside point ( r<R ) :Suppose we want to find potential at point P , inside the sphere.

+Q, R

r

P

O

Potential difference between Point P and O :

VP - VO = P

Oin rd·E–

Where, Ein = 0

So VP - VO = 0

VP = VO = R

KQ

Vin = sphere the of RadiusKQ

RKQ

7.7 Potential Due To Uniformly Charged Solid Sphere :Derivation of expression for potential due to uniformly charged solid sphere of radius R and total chargeQ (distributed in volume), at a point which is at a distance r from centre for the following situations.(i) r R (ii) r RConsider an elementary shell of radius x and width dx(i) For r R :

R

0

2

rdxx4KV =

rKQ

(ii) for r Rdx

x

R

r

2r

0

2

xdxx4K

rdxx4KV

= 3R2KQ

(3R2 – r2)

3R34

Q

From definition of potential(i) For r R :

V = rKQdrr

rKQ

r

2

(ii) For r R :

V = –

R

2 dr.rKQ

– r

R3 dr

RKQr

V = R

KQ – 3R2

KQ[r2 – R2] = 3R2

KQ [2R2 – r2 + R2] = 3R2

KQ (3R2 – r2)

Example 45Two concentric spherical shells of radius R1 and R2 (R2 > R1) are having uniformly distributed charges Q1 andQ2 respectively. Find out potential(i) at point A(ii) at surface of smaller shell (i.e. at point B)(iii) at surface of larger shell (i.e. at point C)(iv) at r R1 (v) at R1 r R2 (vi) at r R2


Sol. Using the results of hollow sphere as given in the table 7.4.

(i) VA = 1

1

RKQ

+ 2

2

RKQ

(ii) VB = 1

1

RKQ

+ 2

2

RKQ

(iii) VC = 2

1

RKQ

+ 2

2

RKQ

(iv) for r R1 , V = 1

1

RKQ

+ 2

2

RKQ

(v) for R1 r R2 , V = r

KQ1 + 2

2

RKQ

(vi) for r R2 , V = r

KQ1 + r

KQ2

Example 46Two hollow concentric non-conducting spheres of radius the a and b (a > b) contain charges Qa and Qbrespectively. Prove that potential difference between the two spheres is independent of charge on outersphere. If outer sphere is given an extra charge, is there any change in potential difference?

Sol. Vinner sphere = bKQb + a

KQa

Vouter sphere = aKQb + a

KQa

Vinner sphere – Vouter sphere = bKQb – a

KQb

V = KQb

a1

b1

Which is independent of charge on outer sphere.If outer sphere in given any extra charge, then there will be no change in potential difference.

8 . POTENTIAL DIFFERENCE

The potential difference between two points A and B is work done by external agent against electricfield in taking a unit positive charge from A to B with no change in kinetic energy between initial andfinal points ie. K = 0 or Ki = Kf

( a ) Mathematical representation :If (WA B)ext = Work done by external agent against electric field in taking the unit charge fromA to B

Then, VB – VA = 0K

extBA

q)W(

= q

)W( electricBA = q

UU AB = q

rd.FB

Ae

= B

A

rd.E

Note : Take W and q both with sign

( b ) Properties :(i) The difference of potential between two points is called potential difference. It is also called

voltage.

(ii) Potential difference is a scalar quantity. Its S.I. unit is also volt.

(iii) If VA and VB be the potential of two points A and B, then work done by an external agent intaking the charge q from A to B is

(Wext)AB= q (VB – VA) or (Wel) AB = q (VA – VB) .(iv) Potential difference between two points is independent of reference point.


8.1 Potential difference in a uniform electric field :

VB – VA = – ABE

VB – VA = – |E| |AB| cos = – |E| d= – Ed

where d = effective distance between A and B along electric field.

or we can also say that E = dV

Special Cases :

Case 1. Line AB is parallel to electric field.

VA – VB = EdCase 2. Line AB is perpendicular to electric field.

VA – VB = 0 VA = VB

Note : In the direction of electric field potential always decreases.

Example 471C charge is shifted from A to B and it is found that work done by an external force is 40J in doing soagainst electrostatic forces, then find potential difference VA – VB

Sol. (WAB)ext = q(VB – VA) 40 J = 1C (VB – VA) VA – VB = – 40 V

Example 48A uniform electric field is present in the positive x-direction. f the intensity of the field is 5N/C then find thepotential difference (VB –VA) between two points A (0m, 2 m) and B (5 m ,3 m)

Sol. VB – VA = – E

.

AB = – (5 i ) . (5 i + j ) = –25V..

The electric field intensity in uniform electric field, E = dV

Where V = potential difference between two points.d = effective distance between the two points.

(projection of the displacement along the direction of electric field.)

Example 49Find out following(i) VA – VB(ii) VB – VC(iii) VC – VA(iv) VD – VC(v) VA – VD(vi) Arrange the order of potential for points A, B, C and D.

Sol. (i) EdVAB = 20 × 2 × 10–2 = 0.4so, VA – VB = 0.4 Vbecause In the direction of electric field potential always decreases.

(ii) EdVBC = 20 × 2 × 10–2 = 0.4 so, VB – VC = 0.4 V

(iii) EdVCA = 20 × 4 × 10–2 = 0.8 so, VC – VA = – 0.8 Vbecause In the direction of electric field potential always decreases.


(iv) EdVDC = 20 × 0 = 0 so, VD – VC = 0

because the effective distance between D and C is zero.

(v) EdVAD = 20 × 4 × 10–2 = 0.8 so, VA – VD = 0.8 Vbecause In the direction of electric field potential always decreases.

(vi) The order of potential is :VA > VB > VC = VD .

8.2 Potential difference due to infinitely long wire :

Derivation of expression for potential difference between two points,which have perpendicular distance rA and rB from infinitely long linecharge of uniform linear charge density :

From definition of potential difference :

VAB = VB – VA = B

A

r

r

drE

drrrK2B

A

r

r

VAB = –2K n

A

B

rr

8.3 Potential difference due to infinitely long thin sheet:

Derivation of expression for potential difference between twopoints, having separation d in the direction perpendicularly toa very large uniformly charged thin sheet of uniform surfacecharge density .

Let the points A and B have perpendicular distance rA and rB

respectively then from definition of potential difference.

VAB = VB – VA = B

A

r

r

drE

drr2

B

A

r

r 0

VAB = –o2

(rB – rA) = –o2d

9. EQUIPOTENTIAL SURFACE : 9.1 Definition : If potential of a surface (imaginary or physically existing) is same throughout, then such

surface is known as an equipotential surface.

9.2 Properties of equipotential surfaces :(i) When a charge is shifted from one point to another point on an equipotential surface, then work done

against electrostatic forces is zero.

(ii) Electric field is always perpendicular to equipotential surfaces.

(iii) Two equipotential surfaces do not cross each other.

9.3 Examples of equipotential surfaces :(i) Point charge :

Equipotential surfaces are concentric and spherical as shown infigure. In figure, we can see that sphere of radius R1 has potentialV1 throughout its surface and similarly for other concentric spherepotential is same.


(ii) Line charge :Equipotential surfaces have curved surfaces as that ofcoaxial cylinders of different radii.

(iii) Uniformly charged large conducting / non conducting sheets:Equipotential surfaces are parallel planes.

Note : In uniform electric field equipotential surfaces are always parallel planes.

Example 50Some equipotential surfaces are shown in figure. What can you say about the magnitude and the directionof the electric field ?

Sol. Here, we can say that the electric will be perpendicular to equipotential surfaces.

Also, |E|

= dV

Where, V = potential difference between two equipotential surfaces.d = perpendicular distance between two equipotential surfaces.

So |E|

= 210)º37sin10(60

= 1000 V/m

Now there are two perpendicular directions: either direction 1 or direction 2 as shown in figure, but since weknow that in the direction of electric field, electric potential decreases, so the correct direction is direction 2.Hence E = 1000 V/m, making an angle 127° with the x-axis


Example 51Figure shows some equipotential surfaces produce by some charges. At which point, the value of electricfield is greatest?

(50

V)

(40

V)

(30

V)

(20

V)

A

B

C

Sol. E is larger where equipotential surfaces are closer. ELOF are to equipotential surfaces. In the figure, wecan see that for point B, they are closer so E at point B is maximum

10 . ELECTROSTATIC POTENTIAL ENERGY 10.1 Electrostatic potential energy of a point charge due to many charges :

The electrostatic potential energy of a point charge at a pointin electric field is the work done in taking the charge fromreference point (generally at infinity) to that point without changein kinetic energyIts Mathematical formula is

U = WP)ext|K = 0 = qV = – WP)ele

Here, q is the charge whose potential energy is being calculated and V is the potential at its positiondue to the source charges.

Note : Always put q and V with sign.

10.2 Properties :(i) Electric potential energy is a scalar quantity but may be positive, negative or zero.

(ii) Its unit is same as unit of work or energy i.e joule (in S.. system).Some times energy is also given in electron-volts.Where, 1eV = 1.6 × 10–19 J

(iii) Electric potential energy depends on reference point. (Generally Potential Energy at r= istaken zero)

Example 52The four identical charges (q each) are placed at the corners of a square of side a.Find the potential energy of one of the charges due to the remaining charges.

Sol. The electric potential of point A due to the charges placed at B, C and D is

V = 04

1 a

q +

041 a2

q +

041 a

q =

041

212 a

q

Potential energy of the charge at A is = qV = 04

1

212

aq2

.


Example 53A particle of mass 40 mg and carrying a charge 5 × 10–9 C is moving directly towards a fixed positivepoint charge of magnitude 10–8 C. When it is at a distance of 10 cm from the fixed point charge it hasspeed of 50 cm/s. At what distance from the fixed point charge will the particle come momentarily torest? Is the acceleration constant during the motion?

Sol. If the particle comes to rest momentarily at a distance r from the fixed charge, then from conservationof energy, we have ?

2mu21

+ 04

1 a

Qq =

041

rQq

Substituting the given data, we get:

21

× 40 × 10–6 × 21

× 21

= 9 × 109 × 5 × 10–8 ×10–9

10r1

or,r1

–10 = 9100

1059105

8

6

r1

= 9190

r = 1909

m

or, i.e., r = 4.7 × 10–2 m

As here, F = 04

1 2r

QqSo, acc. =

mF

2r1

i.e., Acceleration is not constant during the motion.

Example 54A proton moves from a large distance with a speed u m/s directly towards a free proton originally at rest.Find the distance of closest approach for the two protons in terms of mass of proton m and its charge e.

Sol. As here the particle at rest is free to move, when one particle approaches the other, due to electrostaticrepulsion other will also start moving and so the velocity of first particle will decrease while of other willincrease and at closest approach, both will move with same velocity. So, if v is the common velocity ofeach particle at closest approach, then by 'conservation of momentum' of the two proton system.

mu = mv + mv i.e., v = 21

u

And by conservation of energy,

21

mu2 = 21

mv2 + 21

mv2 + 04

1

re2

21

mu2 – m 2

2u

= 04

1

re2

[as v = 2u

] 41

mu2 = r4e

0

2

r = 20

2

ume

11. ELECTROSTATIC POTENTIAL ENERGY OF A SYSTEM OF CHARGES

(This concept is useful when more than one charges move.)It is the work done by an external agent against the internal electric field required to make a system ofcharges in a particular configuration from infinite separation without change in their kinetic energies.


11.1Types of system of charges :(i) Point charge system(ii) Continuous charge system.

11.2Derivation for a system of point charges:

(i) Keep all the charges at infinity. Now bring the charges one by one to its correspondingposition and find work required. PE of the system is algebraic sum of all the works.

Let W1 = Work done in bringing first charge.W2 = Work done in bringing second charge against force due to 1st charge.W3 = Work done in bringing third charge against force due to 1st and 2nd charge.

PE = W1 + W2 + W3 + ...... . (This will contain 2

)1n(n = nC2 terms)

(ii) Method of calculation (to be used in problems) :U = sum of the interaction energies of the charges.= (U12 + U13 + ........ + U1n) + (U23 + U24 + ........ + U2n) + (U34 + U35 + ........ + U3n) .... .

(iii) Method of calculation useful for symmetrical point charge systems.Find PE of each charge due to rest of the charges.If U1 = PE of first charge due to all other charges.= (U12 + U13 + ........ + U1n) U2 = PE of second charge due to all other charges.

= (U21 + U23 + ........ + U2n) then U = PE of the system = 2

U....UU n21

Example 55Find out potential energy of the two point charge system having charges q1 and q2 separated bydistance r.

Sol. Let both the charges be placed at a very large separation initially.Let, W1 = work done in bringing charge q1 in absence of q2 = q1(Vf – Vi) = 0

W2 = work done in bringing charge q2 in presence of q1 = q2(Vf – Vi) = q2(Kq1/r – 0) PE = W1 + W2 = 0 + Kq1q2 / r = Kq1q2 / r

Example 56Figure shows an arrangement of three point charges. The total potential

energy of this arrangement is zero. Calculate the ratio Qq

.

-Q +q+q

2rr

Sol. Usys = 041

r)q(Q

r2)q)(q(

rqQ

= 0

–Q + 2q

– Q = 0 or 2Q = 2q

or Qq

= 14

.

Example 57Two point charges, each of mass m and charge q are released when they are at a distance r from eachother. What is the speed of each charged particle when they are at a distance 2r?

Sol. According to momentum conservation, both the charge particles willmove with same speed. Now applying energy conservation:

VV

2r0 + 0 +

rKq2

= 221

mv2 + r2

Kq2 v =

rm2Kq2

Example 58Two charged particles each having equal charges 2 × 10–5 C are brought from infinity to within aseparation of 10 cm. Calculate the increase in potential energy during the process and the work requiredfor this purpose.


Sol. U = Uf – Ui = Uf – 0 = UfWe have to simply calculate the electrostatic potential energy of the given system of charges

U = Uf = 04

1

rqq 21 =

10100102102109 559

J = 36 J

Work required = 36 J = equal to change in potential energy of system

Example 59Three equal charges q each are placed at the corners of an equilateral triangle of side a.(i) Find out potential energy of charge system.

A

CBq q

q

a

(ii) Calculate work required to decrease the side of triangle to a/2.(iii) If the charges are released from the shown position and each of them

has same mass m then find the speed of each particle when they lieon triangle of side 2a.

Sol. (i) Method I (Derivation)Assume all the charges are at infinity initially.Work done in putting charge q at corner A W1 = q (vf – v i) = q (0 – 0)

Since potential at A is zero in absence of charges, work done in putting q at corner B in presence of chargeat A :

W2 =

0

aKq

q = a

Kq2

Similarly work done in putting charge q at corner C in presence of charge at A and B.

W3 = q(v f – v i) = q

0

aKq

aKq

= a

Kq2 2

So, net potential energy PE = W1 + W2 + W3

= 0 + a

Kq2

+ a

Kq2 2

= a

Kq3 2

Method II (using direct formula):

U = U12 + U13 + U23 = a

Kq2

+ a

Kq2

+ a

Kq2

= a

Kq3 2

(ii) Work required to decrease the sides

W = Uf – Ui = 2/a

Kq3 2

– a

Kq3 2

= a

Kq3 2

Joules

(iii) Work done by electrostatic forces = Change is kinetic energy of particles.

Ui – Uf = Kf – Ki a

Kq3 2

– a2

Kq3 2

= 3(21

mv2) – 0 v = amKq2

Example 60Four identical point charges q each are placed at four corners of a square of side a.Find out potential energy of the charge system

4 3

21q q

qq

Sol. Method 1 (using direct formula) :U = U12 + U13 + U14 + U23 + U24 + U34

= a

Kq2

+ 2a

Kq2

+ a

Kq2

+ a

Kq2

+ 2a

Kq2

+ a

Kq2

=

2aKq2

aKq4 22

=

212

aKq2 2


Method 2 [Using, U = 21

(U1 + U2 + ......)] :

U1 = total P.E. of charge at corner 1 due to all other charges.U2 = total P.E. of charge at corner 2 due to all other charges.U3 = total P.E. of charge at corner 3 due to all other charges.U4 = total P.E. of charge at corner 4 due to all other charges.Since, due to symmetry, U1 = U2 = U3 = U4

UNet = 2

UUUU 4321 = 2U1 = 2

a2Kq

aKq

aKq 222

= a

Kq2 2

212

Example 61Six equal point charges q each are placed at six corners of a hexagon of side a.Find out potential energy of charge system

Sol. UNet = 2

UUUUUU 654321

Due to symmetry U1 = U2 = U3 = U4 = U5 = U6 So Unet = 3U1 =

21

322

aKq3 2

11.3 Energy density :

Def: Energy density is defined as energy stored in unit volume in any electric field. Its mathematical formulais given as following :

Energy density = 21E2

where E = electric field intensity at that point. = 0r electric permittivity of medium

Example 62Find out energy stored in an imaginary cubical volume of side a in front of a infinitely large non-conductingsheet of uniform charge density .

Sol. Energy stored :

U = dVE21 2

0 ; where dV is small volume

U = 20E

21 dV

E is constant .

U = 20

2

0 421

. a3 .= 0

32

8a

Example 63Find out energy stored in the electric field of uniformly charged thin spherical shell of total charge Q andradius R.

Sol. We know that electric field inside the shell is zero, so the energy is stored only outside the shell, which canbe calculated by using energy density formula.

Uself =

x

Rx2

0 dVE21


Consider an elementary shell of thickness dx and radius x (x > R).Volume of the shell = (4x2dx) = dV

U =

R

2

20 xKQ

21

. 4x2 dx = 021 4QK 22

R 2x

1 dx

= 2

4 0 2

0

2

)4(Q .

R1

= R8Q

0

2

= R2KQ2

.

Example 64Find out energy stored inside a solid non-conducting sphere of total charge Q and radius R. [Assume chargeis uniformly distributed in its volume.]

Sol. We can consider solid sphere to be made of large number of concentric spherical shells. Also electric fieldintensity at the location of any particular shell is constant.

Uinside = R

0

20 dVE

21

Consider an elementary shell of thickness dx and radius x.Volume of the shell = (4x2dx)

Uinside =

R

0

2

30 RKQx

21

. 4x2 dx = 021 6

22

R4QK

R

04dxx

= 60

R24

20

2

)4(Q .

5R5

= R40Q

0

2

= R10KQ2

.

12 . RELATION BETWEEN ELECTRIC FIELD INTENSIT Y AND ELECTRICPOTENTIAL :

12.1 For uniform electric field :

(i) Potential difference between two points A and B

VB – VA = – E .

AB

12.2 Non uniform electric field

(i) Ex = – xV

, Ey = – yV

, Ez = – zV

E

= Ex i + Eyj + Ez k

= –

V

zkV

yjV

xi = – V

zk

yj

xi

= – V = –grad V

Where, xV

= derivative of V with respect to x (keeping y and z constant)


yV

= derivative of V with respect to y (keeping z and x constant)

zV

= derivative of V with respect to z (keeping x and y constant)

12.3 If electric potential and electric field depends only on one coordinate, say r :

(i) E

= – rrV

where, r is a unit vector along increasing r..

(ii) dV = – drE

VB – VA = –

B

A

r

rdr.E

dr is along the increasing direction of r..

(iii) The potential of a point

V = –

r

dr.E

Example 65A uniform electric field is along x – axis . The potential difference VA– VB = 10 V is between two points A (2m, 3m) and B (4m, 8m). Find the electric field intensity.

Sol. E = dV

= 2

10= 5 V / m. (It is along + ve x-axis)

Example 66

V = x2 + y , Find E

.

Sol.xV

= 2x, 1yV

and 0zV

E = –

zVk

yVj

xVi = –(2x i + j ) Electric field is non-uniform.

Example 67 For given E

= jy3ix2 , find the potential at (x, y) if V at origin is 5 volts.

Sol. drEdVv

5

= –

x

0xdxE –

y

0ydyE

V – 5 = – 2y3

2x2 22

V = – 2y3

2x2 22

+ 5.


13 . ELECTRIC DIPOLE 13.1 Electric Dipole

If two point charges, equal in magnitude ‘q’ and opposite in sign separated by a distance ‘a’ such thatthe distance of field point r>>a, the system is called a dipole. The electric dipole moment is defined asa vector quantity having magnitude p = (q × a) and direction from negative charge to positive charge.

Note: [In chemistry, the direction of dipole moment is assumed to be from positive to negative charge.] TheC.G.S unit of electric dipole moment is debye which is defined as the dipole moment of two equal andopposite point charges each having charge 10–10 Franklin and separation of 1 Å, i.e.,

1 debye (D) = 10–10 × 10–8 = 10–18 Fr × cm

or 1 D = 10–18 × 9103C

× 10–2 m = 3.3 × 10–30 C × m.

S.I. Unit is coulomb × metre = C . m

Example 68A system has two charges qA = 2.5 × 10–7 C and qB = – 2.5 × 10–7 C located at points A :(0, 0, – 0.15 m) and B ; (0, 0, + 0.15 m) respectively. What is the net charge and electric dipolemoment of the system ?

Sol. Net charge = 2.5 × 10–7 – 2.5 × 10–7 = 0Electric dipole moment,

P = (Magnitude of charge) × (Separation between charges) = 2.5 × 10–7

[0.15 + 0.15] C m = 7.5 × 10–8 C mThe direction of dipole moment is from B to A.

13.2 Electric Field Intensity Due to Dipole :(i) At the axial point :-

)P the along( 2

2ar

Kq2

2ar

KqE

= P2

42a2r

)ra2(Kq

If r >> a then,

E

= P4rra2Kq

= 3rPK2

,

As the direction of electric field at axial position is along the dipole moment ( P )

So, axialE

= 3rPK2


(ii) Electric field at perpendicular Bisector (Equatorial Position) Enet = 2 E cos (along – P )

netE

= 2 )P(

2ar

2a

2ar

Kq2

22

22

= )P(

2ar

Kqa2/32

2

If r >> a then

netE

= )P(rKP

3

As the direction of E at equatorial position is opposite of P so we can write in vector form:

eqtE

= – 3rPK

(iii) Electric field at general point (r, ) :

For this, let’s resolve the dipole moment P into components.

One component is along radial line (=P cos) and other component is r to the radial line (=Psin)

From the given figure Enet = 2

3

2

32

t2

r rsinKP

rcosKP2EE

= 2

3 cos31rKP

tan =

3

3

r

t

rcosKP2

rsinKP

EE

= 2

tan Enet = 2

3 cos31rKP

; tan = 2

tan


Example 69The electric field due to a short dipole at a distance r, on the axial line, from its mid point is the same

as that of electric field at a distance r', on the equatorial line, from its mid-point. Determine the ratio ´rr

.

Sol.04

1 3r

p2 = 04

1

3'rp or 3r

2 = 3´r

1or 3

3

´rr

= 2 or,´rr

= 21/3

Example 70Two charges, each of 5 C but opposite in sign, are placed 4 cm apart. Calculate the electric fieldintensity of a point that is at a distance 4 cm from the mid point on the axial line of the dipole.

Sol. We cannot use formula of short dipole here because distance of the point is comparable to the distancebetween the two point charges.

q = 5 × 10–6 C, a = 4 ×10–2 m, r = 4 × 10–2 m

Eres = E+ + E– = 2cm2)C5(K

– 2cm6)C5(K

= 810144144

NC–1 = 108 N C–1

Example 71Two charges ± 10 C are placed 5 × 10–3 m apart as shown in figure. Determine the electric field at apoint Q which is 0.15 m away from O, on the equatorial line.

Sol. In the given problem, r >> a

E = 04

1

3rp

= 04

1 3r

)a(q

or E = 9 × 109 ×15.015.015.0

1051010 36

NC–1

= 1.33 ×105 NC–1

13.3 Electric Potential due to a small dipole :

(i) Potential at axial position :

V =

2ar

)q(K

2ar

Kq

V =

22

2ar

Kqa

If r >> a then

V = 2rKqa

; where, qa = p

Vaxial = 2rKP


(ii) Potential at equatorial position :

V = 22

22

2ar

)q(K

2ar

Kq

= 0

Veqt = 0

(iii) Potential at general point (r,) :

Lets resolve the dipole moment P into

components: Pcos along radialline and Psin r to the radial line

For the Pcos component, the point A is an axial point,

So, potential at A due to Pcos =

2rcosPK

And for Psin component, the point A is an equatorial point,So potential at A due to Psin = 0

Vnet =

2rcosPK

V = 3r

r.P K

Example 72(i) Find potential at point A and B due to the small charge - system fixed near origin.(Distance between thecharges is negligible).(ii) Find work done to bring a test charge q0 from point A to point B, slowly. All parameters are in S.I. units.

Sol. (i) Dipole moment of the system is

P = (qa) i + (qa) jPotential at point A due to the dipole

VA = K 3r)r·P(

= 35)j3i4(·]j)qa(i)qa[(K

= 7125

)qa(k

VB =

3)5(j4–i3·j)qa(i)qa(K

= 125)qa(K–

(ii) WA B = UB – UA = q0 (VB – VA) = q0

125)7()qa(K–

125)qa(K– WA B = )8(

125aqqK– 0


13.4 Dipole in uniform electric field

(i) Dipole is placed along electric field :

In this case, Fnet = 0, net = 0, so it is an equilibrium state. And it is a stable equilibrium position.

(ii) If the dipole is placed at angle from E : -

In this case Fnet = 0 but

Net torque = (qEsin) (a)

Here qa = P = PE sin

in vector form : EP

Example 73A dipole is formed by two point charge –q and +q, each of mass m, and both the point charges are connectedby a rod of length and mass m. This dipole is placed in uniform electric field E

. If the dipole is disturbed bya small angle from stable equilibrium position, prove that its motion will be almost SHM. Also find its timeperiod.

Sol.If the dipole is disturbed by angle,

net = –PE sin (Here – ve sign indicates that directionof torque is opposite to )

If is very small, sin

net = –(PE)

net (–) so motion will be almost SHM & C = PE (where, P = q)

T = 2 C

T = 2m–

E.P2

m212m 2

= 2 Eq2

m12m 2

= 2 Eq12m7 2

= 2 qE12

m7

T = qE3m7


(iii) Potential energy of a dipole placed in uniform electric field :

UB – UA = – B

A

F

. dr

(for translational motion)

Here, UB – UA = – d.

B

A

(for rotational motion)

In the case of dipole, at = 90° , P.E. is assumed to be zero.

U – U90° = –

90

)d()sinPE( (As the direction of torque is opposite of )

U – 0 = – PE cos

= 90° is chosen as reference,so that the lower limit comes out to be zero.

U

= 0 = 180°

= 90°

U = – EP

From the potential energy curve, we can conclude :(i) At = 0, there is minimum of P.E. so it is a stable equilibrium position.(ii) At = 180° , there is maxima of P.E. so it is a position of unstable equilibrium.

Example 74Two point masses of mass m and equal and opposite charge ofmagnitude q are attached on the corners of a non-conducting uniformrod of mass m and the system is released from rest in uniform electricfield E as shown in figure from = 53° (i) Find angular acceleration of the rod just after releasing(ii) What will be angular velocity of the rod when it passes

through stable equilibrium.(iii) Find work required to rotate the system by 180°.

Sol. (i) net = PE sin53° = I

=

m 35qE48

2m

2m

12m

54E)q(

222

(ii) From energy conservation :

Ki + Ui = Kf + Uf

0 + (– PE cos 53°) = 212 + (–PE cos 0°)

where I = 12

m 2+ m

2

2

+ m2

2

= 12m7 2

21

2 = PE (1– 3/5) = 52

PE

21

× 12m7 2

×2 = 52

qE or = m 35

qE48


(iii) Wext = Uf – Ui

Wext = (–PE cos(180° + 53°)) – (–PEcos 53°)

or Wext = (q)E

53

+ (q)E

53

Wext =

56

qE

13.5 Dipole in non-uniform electric field :

If the dipole is placed along E , (shown in figure)

Then, Net force on the dipole :Fnet = q E(x + dx) – q E(x)

Fnet = q dx)x(E–)dxx(E

(dx) ; here (q (dx) = P)

Fnet = P

dxdE

Example 75

A short dipole is placed on the axis of a uniformly charged ring (total charge –Q, radius R) at a distance 2R

from centre of ring as shown in figure. Find the Force on the dipole due to the ring

+qP

+Q,R

x = R

2

Sol. F = P

dxdE

F = P dxd

2/322 )xR(KQx

; (at x = 2R

)

Solving we get, F = 0

14 . ELECTRIC LINES OF FORCE (ELOF)The line of force in an electric field is an imaginary line, the tangent to which at any point on it repre-sents the direction of electric field at the given point.


14.1 Properties :(i) Line of force originates out from a positive charge and terminates on a negative charge. If there

is only one positive charge, then lines starts from positive charge and terminates at . If thereis only one negative charge, then lines starts from and terminates at negative charge.

(ii) Two lines of force never intersect each other because there cannot be two directions of E at a single

Point

impossilble

(iii) Electric lines of force produced by static charges do not form closed loop.If lines of force make a closed loop, then work done to move a +q charge along the loop will be non-zero. So it will not be conservative field. So these type of lines of force are not possible in electrostat-ics.

(iv) The Number of lines per unit area (line density) representsthe magnitude of electric field.If lines are dense E will be moreIf Lines are rare E will be lessand if E = O, no line of force will be found there

(v) Number of lines originating (terminating) at a charge is proportionalto the magnitude of charge


Example 76If number of electric lines of force from charge q are 10, then find out number of electric lines of forcefrom 2q charge.

Sol. No. of ELOF charge10 q

20 2qSo, number of ELOF will be 20.

(vi) Electric lines of force end or start perpendicularly on the surface of a conductor.(vii) Electric lines of force never enter into conductors.

Example 77Some electric lines of force are shown in figure. For points A and B

A

B(A) EA > EB (B) EB > EA

(C) VA > VB (D) VB > VA

Sol.: Lines are more dense at A, so EA > EB In the direction of Electric field, potential decreases so VA > VB

Example 78If a charge is released in electric field, will it follow lines of force?

Sol. Case I :If lines of force are parallel (in uniform electric field) :-

In this type of field, if a charge is released, force on it will be qoE and its direction will be along E .So the

charge will move in a straight line , along the lines of force.Case II : -If lines of force are curved (in non-uniform electric field) :-

The charge will not follow lines of force

Example 79A charge + Q is fixed at a distance d in front of an infinite metal plate. Draw the lines of force indicating thedirections clearly.

Sol. There will be induced charge on two surfaces of conducting plate, so ELOFwill start from +Q charge and terminate at conductor and then will againstart from other surface of conductor.


15. CONDUCTOR AND IT'S PROPERTIES [FOR ELECTROSTATIC CONDITION](i) Conductors are materials which contain large number of free electrons which can move freely

inside the conductor.(ii) n electrostatics conductors are always equipotential surfaces.(iii) Charge always resides on outer surface of conductor.(iv) f there is a cavity inside the conductor having no charge then charge will always reside only

on outer surface of conductor.(v) Electric field is always perpendicular to conducting surface.(vi) Electric lines of force never enter into conductors.(vii) Electric field intensity near the conducting surface is given by formula

E =0

n

nE0

AA

; nE

0

BB

and nE

0

CC

(viii) When a conductor is grounded its potential becomes zero.

(ix) When an isolated conductor is grounded then its charge becomes zero.(x) When two conductors are connected there will be charge flow till their potentials become

equal.(xi) Electric pressure : Electric pressure at the surface of a conductor is given by formula

P = 0

2

2

, where is the local surface charge density..

1 5.1 Some other important results for a closed conductor:

(i) f a charge q is kept in the cavity then –q will be induced on theinner surface and +q will be induced on the outer surface of theconductor (it can be proved using Gauss theorem)

(ii) If a charge q is kept inside the cavity of a conductor and conductoris given a charge Q then –q charge will be induced on inner sur-face and total charge on the outer surface will be q + Q. (it can beproved using Gauss theorem)

(iii) Resultant field, due to q (which is inside the cavity) and inducedcharge on S1, at any point outside S1 (like B,C) is zero. Resultantfield due to q + Q on S2 and any other charge outside S2 , atany point inside of surface S2 (like A, B) is zero


(iv) Resultant field in a charge free cavity in a closed conductor iszero. There can be charges outside the conductor and on thesurface also. Then also, this result is true. No charge will be in-duced on the inner most surface of the conductor.

(v) Charge distribution for different types of cavities in conductors

(A) (B)

(C) (D)

(E) (F)

(G) (H)

Using the result that resE

in the conducting material should be zero and using result (iii) we can show that

Case A B C D E F G HS1 Uniform Nonuniform Nonuniform Nonuniform Uniform Nonuniform Nonuniform NonuniformS2 Uniform Uniform Uniform Uniform Nonuniform Nonuniform Nonuniform NonUniform

Note : In all cases, charge on inner surface S1 = –q and on outer surface S2 = q. The distribution of charge on ‘S1’will not change even if some charges are kept outside the conductor (i.e. outside the surface S2). But thecharge distribution on ‘S2’ may change if some charges(s) is/are kept outside the conductor.


Example 80An uncharged conductor of inner radius R1 and outer radius R2 containsa point charge q at the centre as shown in figure

(i) Find E

and V at points A,B and C(ii) If a point charge Q is kept outside the sphere at a distance ‘r’

(>>R2) from centre, then find out resultant force on charge Q andcharge q.

Sol. At point A :

VA = OAKq +

2RKq +

1R)q(K

, AE

= OAOAKq

3

Note : Electric field at ‘A’ due to –q of S1 and +q of S2 is zero individually because they are uniformly distributed

At point B : VB = OBKq

+ OB)q(K

+ 2R

Kq =

2RKq

, EB = 0

At point C : VC = OCKq

, CE

= OCOCKq

3

(ii) Force on point charge Q :(Note : Here, force on ‘Q’ will be only due to ‘q’ of S2 (see result (iii) )

QF

= rr

KqQ2 (r = distance of ‘Q’ from centre ‘O’)

Force on point charge q:

qF

= 0 (using result (iii) & charge on S1 uniform)

Example 81An uncharged conductor of inner radius R1 and outer radius R2 containsa point charge q placed at point P (not at the centre) as shown in figureFind out the following :(i) VC (ii) VA (iii) VB (iv) EA (v) EB(vi) Force on charge Q, if it is placed at B.

Sol. (i) VC = CPKq

+ 1R

)q(K +

2RKq

Note : –q on S1 is non-uniformly distributed. Still it produces potential 1R

)q(K at ‘C’ because ‘C’ is at distance ‘R1’

from each point of ‘S1’.

(ii) VA = 2R

Kq(iii) VB = CB

Kq(iv) EA = O (point is inside metallic conductor)

(v) EB = 2CBKq ^

CB (vi) FQ = ^

2 CBCBKQq

(vi) Sharing of charges :

Two conducting hollow spherical shells of radii R1 and R2 having charges Q1 and Q2 respectivelyand separated by large distance & are joined by a conducting wire

Let final charges on spheres are q1 and q2 respectively.Potential on both spherical shell becomes equal after joining. Therefore,

1

1

RKq

= 2

2

RKq

2

1

qq

= 2

1

RR

......(i)


and, q1 + q2 = Q1 + Q2 ......(ii)

from (i) and (ii) : q1 = 21

121

RRR)QQ(

q2 = 21

221

RRR)QQ(

ratio of charges :2

1

qq

= 2

1

RR

222

211

R4R4

=

2

1

RR

ratio of surface charge densities : 2

1

= 1

2

RR

Ratio of final charges :2

1

qq

= 2

1

RR

Ratio of final surface charge densities :2

1

= 1

2

RR

Example 82The two conducting spherical shells are joined by a conducting wire which is cut aftersome time when charge stops flowing.Find out the charge on each sphere after that.

Sol. After cutting the wire, the potential of both the shells is equal

Thus, potential of inner shell, Vin = RKx

+ R2

)xQ2(K =

R2

Q2xk

and potential of outer shell, Vout = R2

Kx +

R2)xQ2(K

= RKQ

As, Vout = Vin

RKQ– =

R2

Q2–xK –2Q = x – 2Q x = 0

So, charge on inner spherical shell = 0and outer spherical shell = – 2Q.


OBJECTIVE QUESTIONS1. A charged particle q1 is at position (2, - 1, 3). The electrostatic force on another charged particle q2 at

(0, 0, 0) is :

(A) 0

21

56qq )k3ji2( (B)

0

21

1456qq

)k3ji2(

(C) 0

21

56qq )k3i2j( (D)

0

21

1456qq

)k3i2j(

2. Three charges +4q, Q and q are placed in a straight line of length at points at distance 0, /2 and respectively from one end of line. What should be the value of Q in order to make the net force on q tobe zero?(A) –q (B) –2q (C) –q/2 (D) 4q

3. Two similar very small conducting spheres having charges 40 C and –20 C are some distance apart. Nowthey are touched and kept at same distance. The ratio of the initial to the final force between them is :(A) 8 : 1 (B) 4 : 1 (C) 1 : 8 (D) 1 : 1

4. Two point charges placed at a distance r in air exert a force F on each other. The value of distance R atwhich they experience force 4F when placed in a medium of dielectric constant K = 16 is :(A) r (B) r/4 (C) r/8 (D) 2r

5. There is a uniform electric field in X-direction. If the work done by external agent in moving a charge of 0.2C through a distance of 2 metre slowly along the line making an angle of 60º with X-direction is 4 joule,then the magnitude of E is:

(A) C/N3 (B) 4 N/C (C) 5 N/C (D) 20 N/C

6. A simple pendulum has a length & mass of bob m. The bob is givena charge q coulomb. The pendulum is suspended in a uniformhorizontal electric field of strength E as shown in figure, then calculatethe time period of oscillation when the bob is slightly displaced fromits mean position.

(A) g

2 (B)

mqEg

2

(C)

mqEg

2 (D) 2

2

mqEg

2

7. Charges 2Q and –Q are placed as shown in figure. The point at which electricfield intensity is zero will be: (A) Somewhere between –Q and 2Q(B) Somewhere on the left of –Q(C) Somewhere on the right of 2Q(D) Somewhere on the perpendicular bisector of line joining –Q and 2Q