Study Package in Physics for JEE Main & Advanced PDF

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 Head Office : B-32, Shivalik Main Road, Malviya Nagar, New Delhi-110017 Sales Office : B-48, Shivalik Main Road, Malviya Nagar, New Delhi-110017 Tel. : 011-26691021 / 26691713 Page Layout : Prakash Chandra Sahoo Typeset by Disha DTP Team DISHA PUBLICATION ALL RIGHTS RESERVED © Copyright Author No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried and made our best efforts to provide accurate up-to-date information in this book. For further information about the books from DISHA, Log on to www.dishapublication.com or email to [email protected]  STUDY PACKAGE IN PHYSICS FOR JEE MAIN & ADVANCED Booklet No. Title Chapter Nos. Page Nos. Ch 0. Mathematics Used in Physics Ch 1. Units and Measurements Units, Measurements & 1 Ch 2. Vectors 1-202 Motion Ch 3. Motion in a Straight Line Ch 4. Motion in a Plane Laws of Motion and Ch 5. Laws of Motion and Equilibrium 2 203-318 Circular Motion Ch 6. Circular Motion Ch 7. Work, Energy and Power Work Energy, Power & 3 Ch 8. Collisions and Centre of Mass 319-480 Gravitation Ch 9. Gravitation 4 Rotational Motion Ch 1. Rotational Mechanics 1-120 Ch 2. Properties of Matter Properties of Matter & 5 Ch 3. Fluid Mechanics 121-364 SHM Ch 4. Simple Harmonic Motion Ch 5. Thermometry, Expansion & Calorimetry 6 Heat & Thermodynamics Ch 6. Kinetic Theory of Gases 365-570 Ch 7. Laws of Thermodynamics Ch 8. Heat Transfer Ch 9. Wave – I 7 Waves 571-698 Ch 10. Wave –II Ch 0. Mathematics Used in Physics 8 Electrostatics Ch 1. Electrostatics 1-216 Ch 2. Capacitance & Capacitors Ch 3. DC and DC circuits 9 Current Electricity Ch 4. Thermal and Chemical effects of 217-338 Current" Ch 5. Magnetic Force on Moving Charges & Conductor Ch 6. Magnetic Effects of Current 10 Magnetism, EMI & AC Ch 7. Permanent Magnet & Magnetic 339-618 Properties of Substance Ch 8. Electromagnetic Induction Ch 9. AC and EM Waves Ch 1. Reflection of Light Ch 2. Refraction and Dispersion 11 Ray & Wave Optics Ch 3. Refraction at Spherical Surface, 1-244 Lenses and Photometry Ch 4. Wave optics Ch 5. Electron, Photon, Atoms, Photoelectric Effect and X-rays 12 Modern Physics 245-384 Ch 6. Nuclear Physics Ch 7. Electronics & Communication  Contents Contents Study Package Booklet 8 - Electrostatics 0. Mathematics Used in Physics I-VI 1.23 Electric dipole 64 1.24 A dipole in an electric field 67 Partial differentiation ii 1.25 Force between two short dipoles 68 Uses of differentiation and integration iii 1.26 Quantisation of charge : Millikan oil drop Area bounded by the curve iii experiment 70 Solid angle iv Review of formulae and important points 73 Exercise 1.1 - 1.6 77-106 Complex number and phasor) v Hints & solutions 107-130 1. Electrostatics 1-130 2. Capacitance & Capacitors 131-216 1.1 Basics of electrostatics 02 2.1 Capacitor or condenser : 1.2 Earthing or grounding 04 1.3 Gold leaf electroscope 04 An introduction 132 1.4 Lightning and lightning conductor 05 2.2 Capacitance or capacity 132 1.5 Coulomb’s law 08 2.3 Charging a capacitor 133 1.6 Electric field 14 2.4 Calculating capacitance 133 1.7 Electric field lines (efl) 15 2.5 Capacitance of parallel plate capacitor 135 1.8 Electric potential energy (epe) 18 2.6 Energy stored in a capacitor 137 1.9 Potential difference 19 2.7 Force between the plates of a capacitor 139 1.10 Equipotential surface 22 2.8 Capacitors in series and parallel 142 1.11 Electric flux 23 2.9 Spherical capacitor 147 1.12 Gauss’s law or gauss’s theorem 26 1.13 Applying gauss’s law : spherical symmetry 30 2.10 Cylindrical capacitor 149 1.14 Applying gauss’s law : cylindrical symmetry 34 2.11 Dielectrics 153 1.15 Line charge of finite length 35 2.12 Induced or bound charge 154 1.16 Applying gauss’s law : planar symmetry 37 2.13 Total energy of the system 159 1.17 Conductor of any shape 39 2.14 Van de graff generator 161 1.18 Mechanical force on the charged conductor 39 2.15 Kirchhoff’s laws 161 1.19 Energy density 40 Review of formulae & important points 178 1.20 Electric field due to charged ring 41 Exercise 2.1-2.6 180-200 1.21 Charged disc 44 Hints & solutions 201-216 1.22 Electric potential energy of system of charges 61  ii ELECTRICITY & MAGNETISM PARTIAL DIFFERENTIATION In physics, we often come across quantities which depend on two or more variables. For example electric potential V depends on x, y coordinates as : V = xy. For given pair of value of x and y, V has a definite value. If we differentiate quantity V w.r.t. x keeping y V constant, then it is known as partial differentiation and represented by. Similarly x V differentiation of V w.r.t. y keeping x is constant is represented by. y V xy Thus = y x x V xy and = x y y In general if f is a function of n variables x1, x2,......xn, then partial differential coefficient of f with respect to x1, keeping all the variables except x1 as constant can be written as f. x1 dz Ex. 1 Given = a x 2 + y 2 + bz 2 , where a and b are constants. or 2 = dt Find partial differentiation of w.r.t. x, y and z. dz Sol. dt = 2 = a x2 y2 bz 2 t t x x A Thus A sin 2 t dt = sin z dz 2 = a × 2x = 2ax, 0 0 a x2 y2 bz 2 A t y = y = cos z 0 2 = a × 2y = 2ay, A t = cos 2 t 0 2 and = a x2 y2 bz 2 z z A = 1 cos 2 t Ans. = b × 2z = 2bz 2 INTEGRATION BY SUBSTITUTION T 2 Ex. 3 Evaluate sin 2 t dt , where = T t Ex. 2 Evaluate Asin2 t dt , where A and are constants. 0 0 Sol. We know that Sol. Substituting 1 cos 2 sin 2 = 2 t = z. 2 On differentiating, we have T T 1 cos 2 t sin 2 t dt = dt d 2 t dz 2 = 0 0 dt dt  MATHEMATICS USED IN PHYSICS iii 1 sin 2 t T Ex. 5 Evaluate x 2dy + 2 xydx = t 2 2 0 Sol. We know that 2 d 2 dy sin 2 T x y = x2 y 2x 1 T dx dx = T 0 0 2 2 or d x2 y = x 2dy 2 xydx 1 = T 0 2 Thus we can write T x 2dy 2 xy dx = d x2 y x2 y C Ans. = Ans. 2 Ex. 6 Evaluate sin 3 d Ex. 4 Evaluate xdy + ydx Sol. We know that Sol. sin 3 d = sin 2 sin d d(xy) = xdy+ydx = 1 cos2 d cos xdy ydx = d xy cos3 = xy + C Ans. = cos C 3 USES OF DIFFERENTIATION AND INTEGRATION Finding maxima and minima : Suppose a quantity is defined by a function of x as : y = f(x). dy For maximum and minimum value of y, put 0. On doing this the values of x can be obtained; some of them may corresponds to dx d2y maxima and some of them corresponds to minima (if x has many values). If 0 , then y is maximum for a certain value of x and if dx 2 d2y is > 0, then y will be minimum for other value of x. dx 2 Ex. 7 Divide 10 into two such parts, so that their product is For maximum value of y, dy = 0, maximum. dx Sol. Let x is one of the parts of 10, then second will be 10 – x. If y is d 10 x x2 or = 0 their product, then dx y = x (10 – x) or 10 – 2x = 0 = 10x – x2. or x = 5 Thus 5 and 5 are the two required parts. Their product is 25. Area bounded by the curve Suppose a curve which is defined by a function of x as; y = f(x). The plot of the curve is shown in figure. Take a small element of curve of width dx. The corresponding average value on y-axis is y. The area of the shaded strip (assuming rectangle) dA = y dx The area bounded between x1 and x2 is given by x2 x2 A ydx = f x dx Fig. 0.1 x1 x1  iv ELECTRICITY & MAGNETISM Ex. 8 Figure shows the parabolic curve y = 2x2. Find the area Sol. The given function decreases with increase in value of x. For y to bounded by the curve between 0 and 5. be zero, e–x = 0 Sol. The area bounded by the curve is given by which gives x = x2 At x = 0, y = e–0 = 1 A = y dx The area bounded by the curve is given by x1 x2 5 A = y dx 2 x1 = 2x dx Fig. 0.2 0 5 x3 = e x dx Fig. 0.3 = 2 3 0 0 2 3 250 x e = 5 03 unit Ans. = 3 3 1 0 Ex. 9 Find the area bounded by the curve y = e x the x-axis , 0 = e e 1 Ans. and the y-axis. Finding the average value : Suppose a quantity y is the function of x. We have to find its average value between x values from x1 to x2. The average value of y can be obtained as : x2 y dx x1 yaverage y = x2 x1 Ex. 10 Find average value of sin in one complete cycle from cos 2 0 = 0 to 2 radian. 2 Sol. The average value sin is given by cos 2 cos0 = 2 = –1+1 = 0 Ans. sin d 0 to 2 sin average = 0 2 0 SOLID ANGLE Solid angle is the analogue in three dimensions of the usual angle in two dimensions. The solid angle subtended by any surface dA at a point O at a distance r away is given by dA cos Fig. 0.4 d =. r2 The total solid angle is given by dA cos = r2 The SI unit of solid angle is steradian. The solid angle will be 4. It does not depend on Fig. 0.5 Fig. 0.6 the shape of the surface enclosed. In the following cases the total solid angle subtended is 4.  MATHEMATICS USED IN PHYSICS v Ex. 11 Find solid angle subtended at the centre O by the where h = R(1–cos ) shaded surface shown in figure. A = 2 R2(1–cos ) The solid angle subtended is given by A cos 0 = R2 Substituting value of A, we have Fig. 0.7 2 R 2 1 cos Sol. Suppose R is the radius of the spherical surface. The surface area = R2 of h height of the spherical cap is given by A = 2 Rh, = 2 1 cos Ans. Sagitta theorem In a circle of radius R and a cord of length d d d 2R h h = 2 2 2 Fig. 0.8 or 2Rh h 2 = d 4 Ex. 12 The diameter of aperture of a plano-convex lens is d In a case, t < < R, t2 can be neglected and and its maximum thickness is t. Calculate radius of curvature of d2 the lens. 2Rt = 4 Sol. If R is the radius of curvature of the lens, then d2 d d and R =. Ans. 2R t t = 8t 2 2 d2 2Rt–t2 = Fig. 0.9 4 Complex number and phasor A number (quantity), which is made of two parts; real and imaginary parts is called complex number. If j 1 is an number (operator), then a phasor A in cartesian form can be written as : A = a + jb, where a is the x-component and b is the y-component of phasor A. The magnitude of A A = a2 b2. If is the angle made by phasor A with the positive x-axis, then Fig. 0.10 b tan = a Also j 2 1; j3 j; j 4 1. In polar form the phasor A can be written as; A = A cos jA sin. vi ELECTRICITY & MAGNETISM Average value : (i) If x is the function of t, then its average value between t1 and t2 : t2 xdt t1 x av =. (t2 t1 ) (ii) Average value of sin or cos over one complete cycle : T 1 2 sin tdt = 0, T T 0 T 1 and cos tdt = 0. T 0 (iii) RMS value of i i0 sin t over the cycle : T 1 2 irms = i dt T 0 T 1 = i0 2 sin 2 tdt T 0 i0 =. 2 2 ELECTRICITY & MAGNETISM 1.1 BASICS OF ELECTROSTATICS Electric charge The word electric charge was derived from Greek word elektron, meaning amber. It was known to the ancient Greeks as long ago as 600 B.C. that amber rubbed with wool, acquires property of attracting light objects. It has now concluded that there are two kinds of electric charges; positive charge and negative charge. The experiments lead to the fundamental results that (1) like charges repel, (2) unlike charges attract. Where does charge come from? The process of acquiring charge consists of transferring something from one body to another, so that one body has an excess and the other a deficiency of that something. It was not until the end of nineteenth century that this “something” was found to consist of very small, negatively charged particles, known today as electron discovered by Sir J.J. Thomson. To give a body an excess negative charge, we may add a number of electrons. And to give excess of positive charge, we may remove the electrons from the body. Note: The “Charge” of a body refers to its excess charge only. The excess charge is always a very small fraction of the total positive or negative charge in the body. Conductors and insulators Conductors permit the passage of charge through them, while insulators do not. Metals in general are good conductors, while most of non metals are insulators. Within a metallic conductor, a few outer electrons become detached from each atom and can move freely throughout the metal in the same way that the molecule of a gas can move through the Fig. 1.1 spaces between grains of sand in sand-filled container. In fact, these free electrons are often referred to an electron gas. The positive nucleus and the remainder of the electrons remain fixed in position. On the other hand, there are no or at most very few free electrons within an insulator,. Charge given to conducting body spreads over its surface, because of repulsion between the like charges. It happens because of high mobility of charge carriers in metals. When some charge is put any where on or in the conducting body, it ultimately will speed over its surface. When charge is given to insulator, it remains at its position. Because there are no or very few charge carriers in an insulator. Fig. 1.2 Impact on body after charging 1. When body is charged positively, it means removal of electrons from the body, resulting decrease of mass of the body. 2. When body is charged negatively, it means addition of electrons on the body, resulting increase of mass of the body. 3. Body either positively charged or negatively charged, it will increase in size because of repulsion between the like charges. Fig. 1.3 Fig. 1.4  ELECTROSTATICS 3 Test of charging of body Let us consider a charged body brings near an uncharged body. It is clear from the figure, that there will be attraction between them. The repulsion can be possible when both the bodies are charged. Therefore we can conclude that repulsion is the sure test of charging. Methods of charging 1. By rubbing: Static electricity builds when two different non-metal materials rub Fig. 1.5 together. Rubbing or friction makes electrons move. This gives one material a positive charge and the other a negative charge. The charges stay, or remain static, on the surfaces of the materials, until they have a pathway along which they can flow suddenly, or discharge. The energy of rubbing or friction gives extra energy to electrons. This allows some of them to break free from their nuclei and wander off on their own. It is known as “separating charge”. Some electrons pass or transfer from one material to the other. Fig. 1.6 Fig. 1.7 When glass rod is rubbed with silk, it aquires positive charge and by convention silk will aquires some amount of negative charge. Similarly when a plastic rod rubbed with fur it acquires negative charge, and the fur itself is found to acquire a positive charge. Moreover, if the rod and fur are brought into contact with one another, the charge is found to disappear from both. Similar observations are true for the glass rod and silk cloth. 2. By conduction (by touch without rubbing): Because of having excess free electrons in metals they can be charged by conduction. When two conductors, one charged and other uncharged bring in contact, the same type of charge will appear on both the conductors. Charge on any conductor is proportional to its size. 4 ELECTRICITY & MAGNETISM Fig. 1.8 3. By induction (without touch): If a charged body is brought near an uncharged body, then neutral body becomes oppositely charged. By induction method, we can charge any type of material body. From the figure one can understand the steps follow in the charging of the body by induction. The magnitude of induced charge depends on the material of the body being charged. If q is the inducing charge and q is the induced charge, then 1 | q'| = q 1 k For metals k = q = q, and for non-metals k < , q < q 1.2 EARTHING OR GROUNDING We know that sea is the very large reservoir of water; from which if few buckets of water is taken out, the level of water in sea will remain almost same, practically same. On the same way earth can be regarded as a large reservoir of oppositely charged particles; Fig. 1.10 electrons and protons, on a whole, is electrically neutral. By any means, if we take out or add some electrons to the earth, its charge will remain almost zero, practically zero. Thus the earth is regarded as a large capacitor with zero potential under all practical situations. Any body connected to earth becomes its part and hence of zero potential. (i) When a positively charged body is connected to the earth, the electrons start flowing from the earth till the positive charge of the body becomes zero, and hence potential of the body V becomes zero. (ii) When a negatively charged body is connected to earth, the electrons from the body start flowing to the earth, till the body becomes neutral and hence its Fig. 1.11 potential becomes zero. Note: The charge of the connected body to earth becomes zero only if there is no other inducing charge nearby the body. 1.3 GOLD LEAF ELECTROSCOPE It is used to detect the charge on the body. It consists of a gold leaf attached to the brass strip which is suspended with the help of a brass rod in a glass jar (see figure). Uses of gold leaf electroscope 1. To detect the charge : Bring the body near the brass cap of the electroscope. If the gold leaf diverges outward, the body is electrically charged. However if no Fig. 1.12 divergence takes place, then body has no charge.  ELECTROSTATICS 5 2. Identification of charge : (a) By conduction : First of all, charge the electroscope positively. Then touch the body under consideration to the brass cap of the electroscope. If the divergence of the gold leaf increases then the body is positively charged. However if divergence of gold leaf decreases then body will be negatively charged. Fig. 1.13 When positively charged body touches the brass cap of the electroscope, the positive charge on the leaf will increase and hence divergence of it increases. But when negatively charged body touches the brass cap of the electroscope, the positive charge on the leaf will decrease and hence divergence of it decreases. (b) By induction : First charge the electroscope positively. Then bring the body under consideration near the brass cap of the electroscope. If the divergence of the gold leaf increases, then the object is positively charged. However if divergence of the gold leaf decreases then the object is negatively charged. Fig. 1.14 When positively charged body brings near the brass cap of the electroscope, the negative charge induces on the brass cap and positive charge on the gold leaf and hence divergence of the leaf will increase. However when negatively charged body brings near the brass cap, the positive charge induces on the brass cap and negative charge on gold leaf and hence divergence of the leaf will decrease. 1.4 LIGHTNING AND LIGHTNING CONDUCTOR The occurance of lightning depends on the atmospheric conditions. It needs a weather in which moisture-laden warm air near the earth rises rapidly. As the moist air rises, it collides with the other drops of moisture and air molecules and so lose or gain electric charges by friction. Usually, the moisture drops become positive and the air molecules negative. When moist air becomes cold enough, it condenses to form a cloud. The lower part of the cloud usually becomes negative and upper most part of the cloud becomes positive. The sudden movements in the cloud cause this charge to discharge in a flash of lightening to another cloud or the earth. When such a charged cloud moves over a tall building, the positive charge will induce over it. Since this induce charge is of very high order, so the building experiences a bursting force and hence cracks will develope in it. The building can be saved from destruction by using lightning conductor. A lightning conductor is made of copper and has sharp pointed edges. It is connected to the earth by the metal conductor running down the building and terminating in a buried metal plate. In the presence of lightning conductor, most of the charge will induce on it, which soonly move to the earth without harming the building. Fig. 1.15  6 ELECTRICITY & MAGNETISM Ex. 1 Fig. 1.16 shows an uncharged, insulated conductor AB, which is in contact with gold leaf electroscope C. A positively charged rod is brought near the end A. Answer the following questions. Fig. 1.16 (i) What charge is developed at A and why? (ii) What charge is present at B, C and D, and why? (iii) How is gold leaf affected? Fig. 1.18 (iv) If positively charged rod is taken away, give your observations. Ex. 3 A metal sphere, a cylindrical conductor and a conical conductor are mounted on insulated stands such that they are in Sol. electrical contact. A highly charged ebonite rod is brought near (i) The electrons of the atoms of the rod are attracted and so end A of sphere X. Answer the following questions : the rod becomes negative and end B thus have deficiency of electrons and so becomes positive if the cap C was not connected. As cap C is connected to the end B of the rod, so deficiency of electrons is created at C which will carry on to D. So end D becomes positively charged. (ii) The end B and C will not have any charge, where as end D has positive charge. Fig. 1.19 (iii) The gold leaf will diverge, because positive charge on gold leaf (i) What will be the charges on X, Y and Z? repels positive charge on the plate P. (ii) If conductor Z is earthed for a moment and the ebonite rod is (iv) When positively charged rod is taken away, the leaves collapse removed, state the charges on X, Y, Z. because the electron from end A flow back to the end D and hence (iii) In the presence of ebonite rod, conductors Y and Z are removed. no deficiency of electrons remain at the end. What is the charge on each of them ? (iv) What will you observe if a lighted candle is placed near the Ex. 2 A glass rod rubbed with silk is brought close to two pointed end of Z ? uncharged metallic spheres in contact with each other, inducing (v) The ebonite rod is made to touch X. What charges are on X, Y charges on them as shown in fig. 1.17. Describe what happens when and Z. Sol. (i) The negatively charged ebonite rod repels the electrons and so electrons from X move to the Z. Therefore X becomes positively charged and Z becomes negatively charged, while Y will have no charge. (ii) When conductor Z is earthed, its negative charge flows to the earth. After removal of the ebonite rod, all the three become Fig. 1.17 positively charged (the charge of X now spread to all the three). (i) the spheres are slightly separated, and (iii) When Z is removed in the presence of ebonite rod, it will have (ii) the glass rod is subsequently removed, and finally negative charge. As soon as Z is removed more electrons from X (iii) the sphere are separated far apart. will flow into Y. Thus now Y becomes negatively charged. So when Y is now removed, it will have negative charge. Sol. (iv) Because of the excessive charge density at the pointed end of (i) Little change in the distribution of charges. conductor Z, it ionise the surrounding air and so charges leak from (ii) The charges on the sphere are redistributed; positive and negative it and set up an electric wind. Thus the candle flame will bend charges will face each other. away from Z. (iii) Charge on each sphere will be uniformly distributed. (v) When ebonite rod is touched with X, the negative charge will conduct in all the three.  ELECTROSTATICS 7 Ex. 4 The diagram shows a gold coated pith-ball suspended Sol. between two plates and closer to the plate A. (i) The opposite charges are induced on both sides of the ball. On being close to plate A, the ball experiences a net force towards this plate, and so string becomes slent from the vertical. (ii) When plate B is earthed, it becomes neutral. The opposite charges are induced on the both sides of the ball. Now plate A exerts net attractive force on the ball. The ball will touch the plate A and becomes positively charged due to conduction. Now plate A repels Fig. 1.20 the ball. The ball moves to mean position, but due to inertia, it (i) Plates A and B are connected to the positive and negative crosses the mean position and then touches the plate B. There it poles of a high tension battery. What happens to the pith- becomes neutral. The ball swing back and repeats its motion till ball ? Give reason. (ii) What would happen if B is earthed ? Give reasons. entire charge of plate A flows into the earth. Properties of charge 1. Charge always be associated with mass. 2. It can be transferred from one body to another. 3. Charge is conserved. It can neither be created nor be destroyed. 4. Charge is quantised. That is, any body can have charge in integral multiple of smallest charge called electronic charge or q ne , where n = 1, 2, 3,....... ,e = 1.6 × 10–19 C. 5. Charge is relativistically invariant. That is, qrest = qmotion. 6. Moving charge produces magnetic field in addition to electric field. 7. Accelerated charge radiates energy. Types of charge densities 1. Point charge : When linear size of charged body is much smaller than the distance under consideration, the size may be ignored and the charge body is called point charge. 2. Linear charge density: Charge per unit length of the object is called linear charge density. It is defined as; dq Fig. 1.21 d q = ( simply) SI unit of is C/m. 3. Surface charge density: The charge per unit surface of the body is called surface charge density. It is defined as ; dq dA q = (simply) A SI unit of is C/m2. Fig. 1.22 8 ELECTRICITY & MAGNETISM 4. Volume charge density: The charge per unit volume of the body is called volume charge density. It is defined as ; dq = dV q = V (simply) Fig. 1.23 SI unit of is C/m3. Charge versus mass Charge Mass 1. Electric charge can be positive 1. Mass of body is always positive. or negative. 2. Charge is conserved. 2. Mass is not conserved as it can be converted into energy and vice versa by E = mc2. 3. Charge is quantized. 3. Quantization of mass is yet to be established. 4. Charge is relativistically invariant, 4. Mass is relativistically variant. Mass m0 i.e., qrest = qmotion. in motion m , where m0 1 v2 / c2 is rest mass object. 5. Force between charges may be 5. Force between masses always be attractive or repulsive. attractive. 1.5 COULOMB’S LAW The force between two point charges at rest is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Thus, if two point charges q1, q2 are separated by a distance r in free space, the magnitude of the force (F) between them is given by q1q2 F = k , r2 where k is a constant. For historical reason and so simplifies many other formulas in electricity 1 and magnetism, the constant k is usually written. Thus Coulomb's law can be written as 4 0 Fig. 1.24 1 | q1 || q2 | F = … (1) 4 r2 0 If the particles repel each other, the force on each particle is directed away from the other particle. If the particles attract each other, the force on each particle is directed toward the other particle. Coulomb's law has proved its validity by experimental tests; no exceptions to it have ever been found. It holds even within the atom, correctly describing the force between the positively charged nucleus and each of the negatively charged electrons. This law also correctly accounts for the forces that bind atoms together to form molecules, and for the forces that bind atoms and molecules together to form solids and liquids. Note: 1. In using Coulomb's law in scalar form, one should not put charges with signs; only place their numerical values. 2. Signs of charges are put in the vector form of Coulomb's law.  ELECTROSTATICS 9 More about Coulomb’s law 1. The force between any two charges is independent of the presence of other charges. 2. Coulomb’s law holds for point charges (whose dimensions) are small compared to their separation. 3. Coulomb’s law holds for charges at rest or nearly at rest. 4. Coulomb’s law holds over wide range from atomic distances (10–15 m) to many kilometers. 5. In the CGS system the constant k = 1, without unit. The unit of charge is stat coulomb or the esu (electro static unit). The conversion factor 1C = 3 × 10 9 esu. Thus q1q2 F =. (in CGS system of units) r2 1 6. In SI system of unit: In SI system k , where 0 is called permittivity of 4 0 1 free space. The introduction of will simplify some formulas to be 4 0 encountered later. Thus coulomb’s law is usually written as 1 q1q2 F = 4 0 r2 1 with 4 = 9 109 N.m 2 / C2 , 0 12 C2 and 0 = 8.854 10. N-m 2 Note: In electrostatics problems, charges as large as one coulomb are unusual; the microcoulomb is often used. 7. When a dielectric medium is completely filled in between the charges, the force between the same charges decreases by a factor k (dielectric constant). Thus 1 q1q2 Fmed = 4 0 k r2 Ffree space = k As Ffree space Fair, Fig. 1.25 Fair or Fmed k Above equation can also be written as 1 q1q2 F = 4 r2 where, k r , is called permittivity of the medium, and r is the relative permittivity of the medium. 10 ELECTRICITY & MAGNETISM Dielectric constant k: Medium k Vacuum (free space) 1 Air (one atmosphere) 1.00546 1 Water 80 Metals 8. If dielectric is inserted in some part of the space between the charges, then force between them can be calculated as follows: When dielectric is filled completely between the charges, the force between them for a separation r is given by 1 q1q2 F = 4 k r2. 0 1 This force is decreased by a factor in comparison to the force between them in k free space. This effect of decrease of force can be achieved, if we remove the dielectric and increases the separation between the charges. Let required separation is r. Thus we can write 1 q1q2 1 q1q2 2 = 4 0 k r 4 0 r '2 which gives r = kr Now move to the situation in which dielectric k is occupied a length t. In this case we can remove the dielectric and effectively increase the distant by k r. Thus The force between the charges now 1 q1q2 Fig. 1.26 F = 4 0 [(r t ) kt ]2 1 9. 1 coulomb of charge = 1.6 10 19 = 6.25 × 1018 electrons Vector form of Coulomb’s law (a) In terms of unit vector : If r̂ is the unit vector of r from 1 towards 2, and charges are placed at separation r, then force on charge q2 due to charge q1 1 q1q2 F 21 = rˆ … (2) r2 4 0 (b) In terms of position vector : Consider a system of two charges q1 and q2 placed at Fig. 1.27 position r1 and r 2 with respect to the origin of the coordinate system. The force on one charge by the other can be obtained as r1 r21 = r2 or r21 = r2 r1 1 q1q2 r 12 F 21 4 0 r123 Fig. 1.28 1 q1q2 r2 r1 or F 21 …(3) 4 3 0 r2 r1  ELECTROSTATICS 11 If more than two charges are present, then the force exerted on any one say q1, by all others q2, q3,.... etc., can be obtained by adding vectorially all the forces. i.e., F1 = F 12 F 13........ 1 q1q2 q1q3 = 2 rˆ12 2 rˆ13... 4 0 r21 r31 Let us consider situations where the charge is spread over a region instead of being concentrated at particular point, then 1 q1dq F1 = rˆ …(4) 4 0 r2 Two shell theorems Theorem 1 : A shell of uniform charge attracts or repels a charged particle that is outside the shell as if all the shell’s charge were concentrated at its centre. Thus Theorem 2 : A shell of uniform charge exerts no electrostatic force on a charged particle that is located inside it. Note: 1. When two identical spheres with charges q1 and q2 touch together and separated, q q Fig. 1.29 the charge on each sphere will be 1 2. If q2 is negative, then each one has a 2 (q1 – q2 ) charge. 2 2. If one sphere has charge q and other identical sphere has no charge, then each sphere will have a charge q/2. F 0 3. If sphere 1 with a charge q touches to the uncharged sphere 2, of half the radius of 1, then charge on them are :q1 + q2 = q …(i) q 1 = 2q2 …(ii) Fig. 130 2q q On solving both the equations, we get q1 , q2. 3 3 Ex. 5 A charge Q is to be divided into two such parts, so that force between them when placed at some separation is maximum. Q Q Therefore the charges are and. Ans. Find their values. 2 2 Sol. Ex. 6 Consider a fixed charge Q and another charge q is placed Let one of the part is q, the other will be (Q – q). The force between them at a distance x0 from Q on a smooth plane surface. Find the velocity for separation r is of charge q as a function of x. 1 q Q q Sol. 1 F = Qq q 2 Let at any instant the separation between them is x. The force between 4 0 r2 4 0 them is dF For, F to be maximum, = 0 dq Fig. 1.31 d Qq q 2 or = 0 1 Qq dq F = 4 0 x2 or Q – 2q = 0 Q 1 Qq or q = and acceleration a = 2 4 0 m x2  12 ELECTRICITY & MAGNETISM or v dv = 1 Qqx 2 1 q2 dx 4 m F = 0 4 0 r2 When half the space is filled with a dielectric k, the force between them v x 1 becomes vdv Qq x 2 dx or = 4 m 0 0 x0 1 q2 F = 4 2 0 r r 2 1 1 x k v Qq 2 2 or = 2 4 0 m x x0 1 q2 = 1 4 2 Qq 1 1 2 0 r r or v =. Ans. 4 2 x m 0 x 2 2 0 Ex. 7 Two equally charged identical metal spheres A and B repel 4 1 q2 = each other with a force 2.0 × 10–5N. Another identical uncharged 94 0 r2 sphere C is touched to A and then placed at the mid point between 4 A and B. What is the net electric force on C? = F. Ans. 9 Sol. Ex. 9 Let us consider two, small balls of masses m1 and m2 and Suppose charge of each sphere is q, the force between spheres A and B for having charges q1 and q2 respectively are suspended with the help separation r, of two strings with the same point. Discuss about the equilibrium of the balls. 1 q2 Sol. Consider the equilibrium of first ball, we have F = 2.0 × 10–5 = …(i) 4 0 r2 T1 sin = Fe....(i) 1 and T1 cos 1 = m1 g...(ii) Fig. 1.32 When uncharged sphere C is touched to the sphere A, the charge on each one becomes q/2. Thus force between A and C 1 (q / 2)(q / 2) FAC = F, Fig. 1.33 4 0 (r / 2) 2 Fe and force between B and C, From (i) and (ii), tan 1 =...(iii) m1 g q Fe q Similarly tan 2 =...(iv) 1 2 m2 g FBC = 2F. 4 2 0 r (i) If m1 = m2 , then 1 2. 2 (ii) If m1 < m2 , 1 2. Thus the net force on the sphere C (iii) If m1 > m2 then 1 2. = 2F – F = F Ex. 10 Two small equally charged spheres, each of mass m, = 2.0 × 10–5 N from C to A. Ans. are suspended from the same point by silk threads of length. The Ex. 8 The force between two point charges placed at a separation dq r is F. Find the force between the same charges, if half the space distance between the spheres x > R, R 2 x2 x2 1 q E =. 4 0 x2 This is the field as the ring "looks like" a point charge. It is interesting to note that if we place an electron of charge e close to the centre of the ring, it would experience a force Frest = – Ee 1 qx = 2 (e) 4 0 (R x 2 )3/ 2 For x > R, we can write by Binomial 1/ 2 R2 V = [x 1 x] 2 0 x2 R2 = x x 2 0 2x R2 ( R2 ) = = 4 0 x 4 0x 1 q =. 4 0 x Here q ( R 2 ) is the total charge of the disc. 46 ELECTRICITY & MAGNETISM Potential at the edge of the disc : To calculate the potential at point P, we can divide the disc in large number of rings with P as the centre. Figure shows an element of radial width dr at radial position r. The arc length AB = 2r The charge on the element, dq = (2r dr) The potential due to this element 1 dq dV = 4 0 r 1 (2r dr ) = 4 0 r 1 = dr …(i) 2 0 Fig. 1.123 Potential due to the whole disc 2R V = dr 2 0 0 From the figure r = 2R cos and dr = – 2R sin d Substituting the value of dr in equation (i), we have 2 0 V = 2 R sin d 2 0 1 /2 R or V =. 0 Charged soap bubble We know that the excess pressure in a soap bubble of radius r due to surface tension T is given by 4T PO = r If now bubble is given a charge q, then the outward mechanical pressure 2 2 1 q q2 = 2 0 2 0 4 r2 32 2 0 r4 2 PO 4T For the equilibrium of the bubble Pi = 2 0 r 2 4T 4T q2 Pi PO P … (1) r 2 0 r 32 2 0 r 4 Thus decrease in pressure due to charge on the bubble Fig. 1.124 q2 –dP = 2 … (2) 32 0 r4 We see that effective value of pressure decreases therefore bubble will expand. By Boyle's law P V = constant  ELECTROSTATICS 47 4 3 or P r = constant 3 or Pr3 = constant … (3) Differentiating equation (3), we get dP 3dr = P r r dP or dr = … (4) 3 P r q2 1 = 3 2 4 P 32 0 r q2 or dr = 2. … (5) 96 0 r3P Special case : when net excess pressure is to be zero, from equation (1) we have 2 4T 0 = r 2 0 8T 0 which gives =. r Ex. 36 Two tiny conducting balls of identical mass m and Dividing equation (i) by (ii), we get identical charge q hang from non conducting threads of length L. Assume that is so small that tan can be replaced by its q2 tan = … (iii) approximate equal, sin. 4 0 x 2 mg (a) Show that for equilibrium, It is given that tan = sin x = [q 2 L /(2 0 mg )]1/ 3 where x is the separation between the balls. = ( x / 2) / L (b) What happens to the balls if one of them is discharged, and Substituting this value in (iii) and rearranging for x, we get find the new equilibrium separation x. 1/ 3 q2L x =. (2 0 mg ) (b) When one ball is discharged, it will touch the other ball which has charge q on it. Due to conduction the charge divided equally on both the balls (because both the balls are identical). Therefore each ball will have the charge q/2. They repel each other, and let they Fig. 1.125 are now have separation x0 between them. Sol. (a) Consider the equilibrium of the either ball. Electric force between 1 (q / 2) 2 Electric force between them is = the balls 4 0 ( x0 ) 2 1 q2 q2 Fe = = 4 x2 tan 0 16 0 x02mg From the free body diagram, we have q2 1 q2 or ( x0 / 2) / L = T sin = … (i) 16 0 x02mg 4 0 x2 and T cos = mg … (ii)  48 ELECTRICITY & MAGNETISM Solving above equation for x0, we get (b) The rod exerts no vertical force on the bearing, if Fy 0, 1/ 3 q2L 1 Qq 1 (2Qq) x0 =. or 2 =W 8 0 mg 4 0 h 4 0 h2 Ex. 37 Fig. 1.126 shows a long nonconducting, massless rod of or