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PHYSICS HIGHER SECONDARY SECOND YEAR Textbook

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PHYSICS HIGHER SECONDARY SECOND YEAR VOLUME - I Revised based on the recommendation of the Textbook Development Committee Untouchability is a sin Untouchability is a crime Untouchability is inhuman TAMILNADU TEXTBOOK CORPORATION COLLEGE ROAD, CHENNAI - 600 0061. Electrostatics Electrostatics is the branch of Physics, which deals with static electric charges or charges at rest. In this chapter, we shall study the basic phenomena about static electric charges. The charges in a electrostatic field are analogous to masses in a gravitational field. These charges have forces acting on them and hence possess potential energy. The ideas are widely used in many branches of electricity and in the theory of atom. 1.1 Electrostatics – frictional electricity In 600 B.C., Thales, a Greek Philosopher observed that, when a piece of amber is rubbed with fur, it acquires the property of attracting th light objects like bits of paper. In the 17 century, William Gilbert discovered that, glass, ebonite etc, also exhibit this property, when rubbed with suitable materials. The substances which acquire charges on rubbing are said to be ‘electrified’ or charged. These terms are derived from the Greek word elektron, meaning amber. The electricity produced by friction is called frictional electricity. If the charges in a body do not move, then, the frictional electricity is also known as Static Electricity. 1.1.1 Two kinds of charges (i) If a glass rod is rubbed with a silk cloth, it acquires positive charge while the silk cloth acquires an equal amount of negative charge. (ii) If an ebonite rod is rubbed with fur, it becomes negatively charged, while the fur acquires equal amount of positive charge. This classification of positive and negative charges were termed by American scientist, Benjamin Franklin. Thus, charging a rod by rubbing does not create electricity, but simply transfers or redistributes the charges in a material. 1+ + +++++ 1.1.2 Like charges repel and unlike charges attract each other – experimental verification. A charged glass rod is suspended by a silk thread, such that it swings horizontally. Now another charged glass rod is brought near the end of the suspended glass rod. It is found that the ends of the two rods repel each other (Fig 1.1). However, if a charged ebonite rod is brought near the end of the suspended rod, the two rods attract each other (Fig 1.2). The above experiment shows that like charges repel and unlike charges attract each other. Silk Silk Glass F Glass F F F Glass Ebonite Fig 1.2 Two charged rods Fig. 1.1 Two charged rods of opposite sign of same sign The property of attraction and repulsion between charged bodies have many applications such as electrostatic paint spraying, powder coating, fly−ash collection in chimneys, ink−jet printing and photostat copying (Xerox) etc. 1.1.3 Conductors and Insulators According to the electrostatic behaviour, materials are divided into two categories : conductors and insulators (dielectrics). Bodies which allow the charges to pass through are called conductors. e.g. metals, human body, Earth etc. Bodies which do not allow the charges to pass through are called insulators. e.g. glass, mica, ebonite, plastic etc. 2 - - - - - - + + + + +++ +++++++ 1.1.4 Basic properties of electric charge (i) Quantisation of electric charge The fundamental unit of electric charge (e) is the charge carried by the electron and its unit is coulomb. e has the magnitude −19 1.6 × 10 C. In nature, the electric charge of any system is always an integral multiple of the least amount of charge. It means that the quantity can take only one of the discrete set of values. The charge, q = ne where n is an integer. (ii) Conservation of electric charge Electric charges can neither be created nor destroyed. According to the law of conservation of electric charge, the total charge in an isolated system always remains constant. But the charges can be transferred from one part of the system to another, such that the total 238 charge always remains conserved. For example, Uranium ( U ) can 92 4 decay by emitting an alpha particle ( He nucleus) and transforming to 2 234 thorium ( Th ). 90 238 234 4 U −−−−→ Th + He 92 90 2 Total charge before decay = +92e, total charge after decay = 90e + 2e. Hence, the total charge is conserved. i.e. it remains constant. (iii) Additive nature of charge The total electric charge of a system is equal to the algebraic sum of electric charges located in the system. For example, if two charged bodies of charges +2q, −5q are brought in contact, the total charge of the system is –3q. 1.1.5 Coulomb’s law The force between two charged bodies was studied by Coulomb in 1785. Coulomb’s law states that the force of attraction or repulsion between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between 3q q 2 1 them. The direction of forces is along F F the line joining the two point charges. r Let q and q be two point charges 1 2 Fig 1.3a Coulomb forces placed in air or vacuum at a distance r apart (Fig. 1.3a). Then, according to Coulomb’s law, qq qq 12 12 F α or F = k 2 2 r r where k is a constant of proportionality. In air or vacuum, 1 k = , where ε is the permittivity of free space (i.e., vacuum) and o 4πε o −12 2 −1 −2 the value of ε is 8.854 × 10 C N m . o 1 qq 12 F = …(1) 2 4πε r o 1 9 2 −2 and = 9 × 10 N m C 4πε o In the above equation, if q = q = 1C and r = 1m then, 1 2 11 × 9 9 F = (9 × 10 ) = 9 × 10 N 2 1 One Coulomb is defined as the quantity of charge, which when placed at a distance of 1 metre in air or vacuum from an equal and 9 similar charge, experiences a repulsive force of 9 × 10 N. If the charges are situated in a medium of permittivity ε, then the magnitude of the force between them will be, qq 1 12 F = …(2) 2 m 4πε r Dividing equation (1) by (2) F ε == ε r F ε m ο 4ε The ratio = ε , is called the relative permittivity or dielectric r ε ο constant of the medium. The value of ε for air or vacuum is 1. r ∴ ε = ε ε o r F Since F = , the force between two point charges depends on m ε r the nature of the medium in which the two charges are situated. Coulomb’s law – vector form q q 2 1 r → 12 If F is the force exerted on charge 21 ++ r F F 12 21 q by charge q (Fig.1.3b), 2 1 → qq 12 F r q = k q 2 21 12 1 2 r 12 r + 12 F F 12 21 where r is the unit vector 12 r from q to q . 1 2 Fig 1.3b Coulomb’s law in → If F is the force exerted on 12 vector form q due to q , 1 2 qq 12 → k F = r 2 12 21 r 21 where r is the unit vector from q to q . 21 2 1 Both r and r have the same magnitude, and are oppositely 21 12 directed qq 12 → = k ∴ F 2 (– r ) 12 12 r 12 qq 12 → =− k or F 2 r 12 12 r 12 → → or F = – F 12 21 So, the forces exerted by charges on each other are equal in magnitude and opposite in direction. 51.1.6 Principle of Superposition The principle of superposition is to calculate the electric force experienced by a charge q due to other charges q , q ……. q . 1 2 3 n The total force on a given charge is the vector sum of the forces exerted on it due to all other charges. The force on q due to q 1 2 1 qq 12 → = F r 2 12 4πε 21 r ο 21 Similarly, force on q due to q 1 3 1 qq 13 → = F r 2 13 4πε 31 r ο 31 The total force F on the charge q by all other charges is, 1 1 → → → → → F = F + F + F F 1 12 13 14 ......... + 1n Therefore, qq qq q q 1⎡⎤ 12 13 1 n → ˆˆ ˆ rr++ ....... r = 21 31 n1 F⎢⎥ 22 2 1 rr r 4πε ο⎣⎦ 21 31 n1 1.2 Electric Field Electric field due to a charge is the space around the test charge in which it experiences a force. The presence of an electric field around a charge cannot be detected unless another charge is brought towards it. When a test charge q is placed near a charge q, which is the o source of electric field, an electrostatic force F will act on the test charge. Electric Field Intensity (E) Electric field at a point is measured in terms of electric field intensity. Electric field intensity at a point, in an electric field is defined as the force experienced by a unit positive charge kept at that point. 6F = E It is a vector quantity. . The unit of electric field intensity q o −1 is N C . The electric field intensity is also referred as electric field strength or simply electric field. So, the force exerted by an electric field on a charge is F = q E. o 1.2.1 Electric field due to a point charge Let q be the point charge +q +q 0 placed at O in air (Fig.1.4). A test E r O P charge q is placed at P at a o distance r from q. According to Fig 1.4 Electric field due to a Coulomb’s law, the force acting on point charge q due to q is o 1 qq o F = 2 4πε r o The electric field at a point P is, by definition, the force per unit test charge. Fq 1 = E = 2 q 4πε r oo The direction of E is along the line joining O and P, pointing away from q, if q is positive and towards q, if q is negative. → 1 q = In vector notation E r, where r is a unit vector pointing 2 4πε r o away from q. 1.2.2 Electric field due to system of charges If there are a number of stationary charges, the net electric field (intensity) at a point is the vector sum of the individual electric fields due to each charge. → → → → → E = E + E + E ...... E 1 2 3 n q q q 1⎡⎤ 1 2 3 rr ++ r +......... = 12 3 ⎢⎥ 22 2 rr r 4πε o⎣⎦ 12 3 7 1.2.3 Electric lines of force The concept of field lines was introduced by Michael Faraday as an aid in visualizing electric and magnetic fields. Electric line of force is an imaginary straight or curved path along which a unit positive charge tends to move in an electric field. The electric field due to simple arrangements of point charges are shown in Fig 1.5. +q +q -q +q +q (a) (b) (c) Isolated charge Unlike charges Like charges Fig1.5 Lines of Forces Properties of lines of forces: (i) Lines of force start from positive charge and terminate at negative charge. (ii) Lines of force never intersect. (iii) The tangent to a line of force at any point gives the direction of the electric field (E) at that point. (iv) The number of lines per unit area, through a plane at right angles to the lines, is proportional to the magnitude of E. This means that, where the lines of force are close together, E is large and where they are far apart, E is small. 1 (v) Each unit positive charge gives rise to lines of force in free ε o space. Hence number of lines of force originating from a point q charge q is N = in free space. ε o 81.2.4 Electric dipole and electric dipole moment Two equal and opposite charges separated by a very small distance p -q +q constitute an electric dipole. 2d Water, ammonia, carbon−dioxide and Fig 1.6 Electric dipole chloroform molecules are some examples of permanent electric dipoles. These molecules behave like electric dipole, because the centres of positive and negative charge do not coincide and are separated by a small distance. Two point charges +q and –q are kept at a distance 2d apart (Fig.1.6). The magnitude of the dipole moment is given by the product of the magnitude of the one of the charges and the distance between them. ∴ Electric dipole moment, p = q2d or 2qd. It is a vector quantity and acts from –q to +q. The unit of dipole moment is Cm. 1.2.5 Electric field due to an electric dipole at a point on its axial line. AB is an electric dipole of two point charges –q and +q separated by a small distance 2d (Fig 1.7). P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole. E E 2 1 AB O P -q x axis +q 2d r Fig 1.7 Electric field at a point on the axial line The electric field at the point P due to +q placed at B is, 1 q E = (along BP) 2 1 4πε () rd − o 9The electric field at the point P due to –q placed at A is, 1 q E = (along PA) 2 2 4πε () rd + o E and E act in opposite directions. 1 2 Therefore, the magnitude of resultant electric field (E) acts in the direction of the vector with a greater magnitude. The resultant electric field at P is, E = E + (−E ) 1 2 11qq ⎡⎤ − E = ⎢⎥ along BP. 22 44 πε πε () rd−+(r d) oo ⎣⎦ 11 ⎡⎤ q − E = ⎢⎥ along BP 22 4πε () rd−+(r d) o⎣⎦ ⎡⎤ 4rd q E = ⎢⎥along BP. 222 4πε() rd − o⎣⎦ If the point P is far away from the dipole, then d r qr44 d q d = ∴ E = 43 44 πε πε rr oo 12p E = along BP. 3 4πε r o Electric dipole moment p = q x 2d ∵ E acts in the direction of dipole moment. 101.2.6 Electric field due to an electric dipole at a point on the equatorial line. Consider an electric dipole AB. Let 2d be the dipole distance and p be the dipole moment. P is a point on the equatorial line at a distance r from the midpoint O of the dipole (Fig 1.8a). M E 1 E 1 Esin 1 E P R Ecos 1 E 2 P N R Ecos r 2 B A Esin 2 -q +q E O 2 dd (a) Electric field at a point on (b) The components of the equatorial line electric field Fig 1.8 Electric field at a point P due to the charge +q of the dipole, 1 q E = along BP. 2 1 4πε BP o 1 q 2 2 2 = along BP ( BP = OP + OB ) 22∵ 4πε () rd + o Electric field (E ) at a point P due to the charge –q of the dipole 2 1 q E = along PA 2 2 4πε o AP 1 q E = along PA 22 2 4πε () rd + o The magnitudes of E and E are equal. Resolving E and E into 1 2 1 2 their horizontal and vertical components (Fig 1.8b), the vertical components E sin θ and E sin θ are equal and opposite, therefore 1 2 they cancel each other. 11The horizontal components E cos θ and E cos θ will get added 1 2 along PR. Resultant electric field at the point P due to the dipole is E = E cos θ + E cos θ (along PR) 1 2 = 2 E cos θ ( E = E ) ∵ 1 1 2 1 q E= × 2 cos θ 22 4πε () rd + o d But cos θ = 22 rd + 12 qd 12 qd × E= = 22 2 21/2 223/2 4πε 4πε () rd++() rd () rd + o o 1 p = ( p = q2d) ∵ 223/2 4πε () rd + o For a dipole, d is very small when compared to r 1 p ∴ E = 3 4πε r o The direction of E is along PR, parallel to the axis of the dipole and directed opposite to the direction of dipole moment. 1.2.7 Electric dipole in a uniform electric field Consider a dipole AB of +q B dipole moment p placed at an F=qE angle θ in an uniform electric field E (Fig.1.9). The charge +q 2d θ E experiences a force qE in the p direction of the field. The charge –q experiences an equal force in A the opposite direction. Thus the F=-qE -q C net force on the dipole is zero. Fig 1.9 Dipole in a uniform field The two equal and unlike 12parallel forces are not passing through the same point, resulting in a torque on the dipole, which tends to set the dipole in the direction of the electric field. The magnitude of torque is, τ = One of the forces x perpendicular distance between the forces = F x 2d sin θ = qE x 2d sin θ = pE sin θ ( q × 2d = P) ∵ → → → In vector notation, τ = p× E Note : If the dipole is placed in a non−uniform electric field at an angle θ, in addition to a torque, it also experiences a force. 1.2.8 Electric potential energy of an electric dipole in an electric field. Electric potential energy E of an electric dipole in an F=qE B electrostatic field is the work +q 2d done in rotating the dipole to the desired position in the p A field. -q When an electric dipole F=-qE of dipole moment p is at an Fig 1.10 Electric potential angle θ with the electric field energy of dipole E, the torque on the dipole is τ = pE sin θ Work done in rotating the dipole through dθ, dw = τ.dθ = pE sinθ.dθ The total work done in rotating the dipole through an angle θ is W = ∫dw W = pE ∫sinθ.dθ = –pE cos θ This work done is the potential energy (U) of the dipole. ∴ U = – pE cos θ 13o When the dipole is aligned parallel to the field, θ = 0 ∴ U = –pE This shows that the dipole has a minimum potential energy when it is aligned with the field. A dipole in the electric field experiences a → → → torque ( τ = p × E) which tends to align the dipole in the field direction, dissipating its potential energy in the form of heat to the surroundings. Microwave oven It is used to cook the food in a short time. When the oven is operated, the microwaves are generated, which in turn produce a non− uniform oscillating electric field. The water molecules in the food which are the electric dipoles are excited by an oscillating torque. Hence few bonds in the water molecules are broken, and heat energy is produced. This is used to cook food. 1.3 Electric potential +q Let a charge +q be placed at a E O B A point O (Fig 1.11). A and B are two x dx points, in the electric field. When a unit Fig1.11 Electric potential positive charge is moved from A to B against the electric force, work is done. This work is the potential difference between these two points. i.e., dV = W . A → B The potential difference between two points in an electric field is defined as the amount of work done in moving a unit positive charge from one point to the other against the electric force. The unit of potential difference is volt. The potential difference between two points is 1 volt if 1 joule of work is done in moving 1 Coulomb of charge from one point to another against the electric force. The electric potential in an electric field at a point is defined as the amount of work done in moving a unit positive charge from infinity to that point against the electric forces. Relation between electric field and potential Let the small distance between A and B be dx. Work done in moving a unit positive charge from A to B is dV = E.dx. 14The work has to be done against the force of repulsion in moving a unit positive charge towards the charge +q. Hence, dV = −E.dx −dV E = dx The change of potential with distance is known as potential gradient, hence the electric field is equal to the negative gradient of potential. The negative sign indicates that the potential decreases in the direction of electric field. The unit of electric intensity can also be −1 expressed as Vm . 1.3.1 Electric potential at a point due to a point charge Let +q be an isolated dx +q p E point charge situated in air at O B A r O. P is a point at a distance r from +q. Consider two points Fig 1.12 Electric potential due to a point charge A and B at distances x and x + dx from the point O (Fig.1.12). The potential difference between A and B is, dV = −Edx The force experienced by a unit positive charge placed at A is 1 q . E = 2 4πε x o 1 q ∴ dV = − .dx 2 4πε x o The negative sign indicates that the work is done against the electric force. The electric potential at the point P due to the charge +q is the total work done in moving a unit positive charge from infinity to that point. r q q V = − .dx = 2 4π ε r ∫ o 4πε x o ∞ 151.3.2 Electric potential at a point due to an electric dipole Two charges –q at A and +q at B separated by a small P distance 2d constitute an electric dipole and its dipole r 2 moment is p (Fig 1.13). r r 1 Let P be the point at a distance r from the midpoint 180- A of the dipole O and θ be the p B angle between PO and the -q +q O dd axis of the dipole OB. Let r 1 and r be the distances of the Fig 1.13 Potential due to a dipole 2 point P from +q and –q charges respectively. 1 q Potential at P due to charge (+q) = 4πε r o 1 1 ⎛⎞ q − Potential at P due to charge (−q) = ⎜⎟ r 4πε ⎝⎠ 2 o 11qq − Total potential at P due to dipole is, V = 44 πεrr πε oo12 11 q⎛⎞ − V = ⎜⎟ ...(1) rr 4πε ⎝⎠ 12 o Applying cosine law, 2 2 2 r = r + d – 2rd cos θ 1 2 ⎛⎞ cosθ d 2 2 ⎜⎟ 12−+ d r = r 1 2 r ⎝⎠ r 2 d Since d is very much smaller than r, can be neglected. 2 r 1 ⎛⎞ 2d 2 ∴ r = r 1c − osθ ⎜⎟ 1 ⎝⎠ r 16−1/2 11 2d ⎛⎞ = 1c − os θ or ⎜⎟ rr⎝⎠ r 1 Using the Binomial theorem and neglecting higher powers, 11 ⎛⎞ d = 1c + osθ ⎜⎟ ∴ …(2) rr ⎝⎠ r 1 Similarly, 2 2 2 r = r + d – 2rd cos (180 – θ) 2 2 2 2 or r = r + d + 2rd cos θ. 2 1/2 2 d 2d ⎛⎞ r = r 1c + osθ ( is negligible) ⎜⎟ 2 2 ⎝⎠ r r −1/2 11 ⎛⎞ 2d = 1c + osθ or ⎜⎟ rr ⎝⎠ r 2 Using the Binomial theorem and neglecting higher powers, 11 d ⎛⎞ = 1c − osθ ⎜⎟ ...(3) rr⎝⎠ r 2 Substituting equation (2) and (3) in equation (1) and simplifying qd 1 d ⎛⎞ 1c+− osθ1+ cosθ V= ⎜⎟ 4πεrr r ⎝⎠ o qd 2 cosθ 1 p . cosθ = ∴ V= …(4) 22 4πε 4πε . rr o o Special cases : 1. When the point P lies on the axial line of the dipole on the side of +q, then θ = 0 p ∴ V = 2 4πε r o 2. When the point P lies on the axial line of the dipole on the side of –q, then θ = 180 p − ∴ V = 2 4πε r o 3. When the point P lies on the equatorial line of the dipole, then, o θ = 90 , ∴ V = 0 17 ∴1.3.3 Electric potential energy The electric potential energy of two q q 1 2 point charges is equal to the work done to AB assemble the charges or workdone in r bringing each charge or work done in Fig 1.14a Electric bringing a charge from infinite distance. potential energy Let us consider a point charge q , 1 placed at A (Fig 1.14a. The potential at a point B at a distance r from the charge q is 1 q 1 V = 4πε r o Another point charge q is brought from infinity to the point B. 2 Now the work done on the charge q is stored as electrostatic potential 2 energy (U) in the system of charges q and q . 1 2 ∴ work done, w = Vq 2 qq 12 Potential energy (U) = 4π ε r o r 23 Keeping q at B, if the charge q is 2 1 q q 3 2 imagined to be brought from infinity to the point A, the same amount of work is done. r r 13 12 Also, if both the charges q and q are 1 2 brought from infinity, to points A and B q 1 respectively, separated by a distance r, then Fig 1.14b Potential potential energy of the system is the same as the energy of system of previous cases. charges For a system containing more than two charges (Fig 1.14b), the potential energy (U) is given by qq q q ⎡⎤ qq 1 12 13 2 3 ++ U = ⎢⎥ rr r 4πε o⎣⎦ 12 13 23 1.3.4 Equipotential Surface If all the points of a surface are at the same electric potential, then the surface is called an equipotential surface. (i) In case of an isolated point charge, all points equidistant from the charge are at same potential. Thus, equipotential surfaces in this 18B E A E +q (a) Equipotential surface (b) For a uniform field (spherical) Fig 1.15 (plane) case will be a series of concentric spheres with the point charge as their centre (Fig 1.15a). The potential, will however be different for different spheres. If the charge is to be moved between any two points on an equipotential surface through any path, the work done is zero. This is because the potential difference between two points A and B is defined W AB as V – V = . If V = V then W = 0. Hence the electric field B A A B AB q lines must be normal to an equipotential surface. (ii) In case of uniform field, equipotential surfaces are the parallel planes with their surfaces perpendicular to the lines of force as shown in Fig 1.15b. 1.4 Gauss’s law and its applications S E Electric flux Consider a closed surface S in a ds non−uniform electric field (Fig 1.16). ds normal Consider a very small area on this ds surface. The direction of ds is drawn normal to the surface outward. The Fig1.16 Electric flux electric field over ds is supposed to be a → → constant E. E and make an angle θ with each other. ds The electric flux is defined as the total number of electric lines of force, crossing through the given area. The electric flux dφ through the 19