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This document provides full notes on Electric Charges and Fields for II PU Physics. It covers topics like types of charges, conductors and insulators and methods of charging a body. It also provides examples and explanations.
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STUDY MATERIAL Electric Charges and Fields II PU Chapter 1: Electric Charges and Fields ELECTRIC CHARGES AND FIELDS Electrostatics is a branch of physics which deals with the study of electric charge at rest...
STUDY MATERIAL Electric Charges and Fields II PU Chapter 1: Electric Charges and Fields ELECTRIC CHARGES AND FIELDS Electrostatics is a branch of physics which deals with the study of electric charge at rest Or It is a branch of physics which deals with the study of force, electric field and potential due to electric charge at rest What is an electric charge? Electric charge is a basic property of a body due to which the body attracts or repels another body. Charge is a scalar quantity.S.I. unit of charge is coulomb (C) Note: Electric charge is best understood by what it does and not by what it is Types of charges: There are two types 1) Positive charge, 2) Negative charge By convention: The charge on proton is taken as positive charge. The charge on electron is taken as negative charge. Note: Old convention 1. Charge on glass rod is called positive charge, Charge on plastic rod is called negative charge 2. The charges were named as positive and negative by an American scientist Benzamin Franckline. Neutral body or uncharged body: The body is said to be neutral if it has no charge. Ex: Atom Charged body or electrified body:The body is said to be charged if it has net charge Ex: Ion Polarity of charge: The property which differentiate the two types of charges is called polarity of charge. Note: 1. When two bodies are rubbed together, then electrons are transferred from one body to another body. The body which looses an electron becomes positively charged body and the body which gains an electrons becomes negatively charged body 2. Addition of electrons gives negatively charged body and removal of electrons gives positively charged body. Department of Physics 1 STUDY MATERIAL Electric Charges and Fields II PU 3. When the substance listed in column (1) are rubbed with the substances listed in column (2). Then the substances in column (1) acquire the positive charge and the substance in column(2) acquires negative charge. Column (1) (+ve charge) Column (2) (−ve charge) Glass rod Silk cloth Cat skin or fur Ebonite rod Woollen cloth Amber rod Woollen cloth Plastic Woollen cloth rubber 4. Thales of Miletus was the first scientist who discovered the fact that an amber rubbed with silk or wool attracts the light objects. It was around 600 B.C. 5. Mass of a body is always is positive, whereas charge can be either positive or negative. Electroscope: It is a device used to detect the charge on a body. Gold leaf electroscope: It is one type of electroscope used to detect charge on a body. Construction: It is one type of electroscope used to detect the charge on a body. It consists of a vertical metal rod kept on a box.A metal knob is present at the upper end and gold leaves are attached at bottom end of a metal rod.When a charged body touches the metal knob, the charge flows to the gold leaves and gold leaves diverge (repelled).The degree of divergence of leaves depends on charge on a body. Conductor:The substance which allows the charges to pass through it easily is called conductor Ex:All metals (Cu, Al, Ag, Fe etc…), thehumanbody, animal body and earth. Insulator: The substance which doesnot allow the charges to pass through it is called insulator. Ex:Wood, plastic, Diamond, Glass Department of Physics 2 STUDY MATERIAL Electric Charges and Fields II PU Note (1): When a conductor is charged then charges are distributed over the entire surface of the conductor Ex: Negatively charged conductor Positively charged conductor Note(2): If some charge is put on an insulator, it stay at the same place Ex: Negatively charged Insulator Positively charged Insulator That is the charge doesn’t get distributed over the surface of insulator. Electrification:The process of charging a body is called electrification. Earthing or Grounding: The process of sharing the charges with the earth is called earthing or grounding. Ex 1):When a negatively charged body is connected to earth, then the electrons flow from body to earth till the body becomes neutral. Ex 2):When a positively charged body is connected to earth, then the electron flow from earth to body till the body becomes neutral. Ex 3):When a neutral body is connected to earth, then there is no flow of charge from body to earth. Earthing is necessary for house, why? The electricity from the mains is supplied to our houses using a 3 core wiring called live wire, neutral wire and earth wire. The live and neutral wires carry the current from the power station. One end of the earth wire is connected to a thick metal plate which is buried in the earth and the other end of earth wire is connected to metallic bodies of electrical appliances (refrigerator, T.V, etc). When live wire touches the metallic body then charges flow to the earth through earth wire.Therefore, there is no damages to the electrical appliances. Department of Physics 3 STUDY MATERIAL Electric Charges and Fields II PU Note: 1. Usually the live wire is red in colour neutral wire is black in colour and the earth wire is green in colour. Methods of charging a body There are three methods to charge a body a) Charging by friction b) Charging by conduction c) Charging by induction Charging by friction: When a glass rod is rubbed with silk cloth then glass rod loses the electron and silk cloth gains electron therefore glass rod becomes positively charged body and the silkcloth becomesnegatively charged body. This process is called charging by friction Note:Normally insulator can be charged by friction method. In this method opposite charges are developed. Charging by conduction: A B A B A B (a) (b) (c) Let A and B are the two conductors A is negatively charged and B is neutral when A and B are brought in contact, free electrons flow from A to B therefore conductor B gets negatively charged.This process is called charging by conduction. Note:Charging by conduction is suitable for conductors. In this method same charges are developed. Charging by Induction: When a charged body is brought near the neutral body then opposite charges are developed on a near surface of a neutral body. This process is called charging by Induction Describe how two spheres can oppositely charged by induction method (e) The two metal spheres can be oppositely charged by following step: Step 1:Bring two metal spheres A and B in contact mounted on insulating stand as shown in fig (a). Department of Physics 4 STUDY MATERIAL Electric Charges and Fields II PU Step 2:Bring a positively charged glass rod near the sphere A. Then the free electrons in A and B are attracted towards glass rod. As a result excess negative charges and positive charges are developed on A and B as shown in fig (b). Step 3: Separate the spheres A and B by a small distance while the glass rod is held near ‘A’ as shown in fig (c). Step 4:Remove the glassrod, the negative charges on A and positive charges on ‘B’ are uniformly distributed as shown in fig (d). By induction method the two metal spheres acquired equal and opposite charges Note: Charging by induction is suitable for conductors. In this method opposite charges are developed. Note: Describe how canyou charge a single metal sphere positively without touching it (i.e. by induction) (a) (b) (c) (d) The metallic sphere can be positively charged by following steps Step 1:Consider a neutral metal sphere kept on insulating stand as shown in fig (a). Step 2:Bring a negatively charged plastic rod near the sphere. Then free electrons move away from the rod due to repulsion. As a result excess positively and negatively charges are developed (induced) on the sphere as shown in fig (b) Step 3:negatively charged part of a sphere is connected to earth withoutremoving plasticrod, then electrons flow into earth as shown in fig (c). Step 4: Remove the plasticrod and earthing, now positive charges are distributed over the surface of sphere as shown in fig (d). i.e. Metallic sphere is positively charged by induction method Note: 1. Similar steps are involved on charging a metal sphere negatively by bringing a positively charged glass rod. 2. Why charged body attracts light objects? When a charged body is brought near the light objects like piece of paper, the charged body induces opposite charges on near surface of a light objects. As a result charged body attracts light objects. Department of Physics 5 STUDY MATERIAL Electric Charges and Fields II PU What are Point charges? Charges whose sizes are very small compared to the distance between them are called point charges. Basic unit of charge or elementary charge or fundamental charge The charge on an electron or proton is called basic unit of charge or elementary charge It is denoted by the letter ‘e’ The value of elementary charge is e=1.602192 10-19C Note: 1) Charge on electron= −e = −1.602 10-19C. 2) Charge on proton = +e = +1.602 × 10−19C Magnitude of charge on electron and proton is equal 3) Charge on neutron = 0 4) The least possible amount of charge that can exist independently in the nature is called elementary charge. Basic properties of electric charge: 1. The electric charge is conserved. 2. The electric charge is quantized. 3. The charge is additive. 4. Like charges repel each other and unlike charges attract each other. 5. Charge is scalar, its SI unit is coulomb. Explain Additivity of charge: Charges can be added algebraically, i.e., if a system contains positive charges +q1 , + q 2 , + q3 and negative charges −q 4 , − q5 , − q 6. Then, total charges of a system or net charge = q1 + q 2 + q3 − q 4 − q5 − q 6. Ex: If a system contains the charges +4C, +2C, −3C and −1C Then, net charge = 4C+2C−3C−1C=+2C Explain Conservation of charge: Conservation of charge states that “The total charge of an isolated system remains always constant”, i.e. total charge of the isolated system can neither be created nor be destroyed but there is a transfer of electrons from one body to another body.Therefore charge is conserved. Ex:In unstable nucleus neutron is converted into proton and electron. i.e, neutron = proton + electron A/C to law of conservation of charge, Charge on neutron = charge on proton + charge on electron 0 = +e – e 0=0 Explain Quantization of charge: The charge on a body is always an integral multiple of charge on electron. Quantization of charge means charge on a body can be increased or decreased in steps of ‘e’ Department of Physics 6 STUDY MATERIAL Electric Charges and Fields II PU Charge on a body is given by (Expression for quantization of charge) q=n e Where, n= number of electrons added or removed from the body Where,q= charge on a body e=elementary charge Note: 1. Quantization of charge was experimentally demonstrated by R.A Millikan through his oil-drop experiment. 2. Quantization of charge was first suggested by the experimental laws of electrolysis discovered by Faraday. 3. The force of attraction or repulsion between two charges at restis called electrostatic force. State and explain coulomb’s law: It states that “the force between two point charges is directly proportional to the product of magnitudes of two chargesand inversely proportional to square of the distance between them. The force is acting along the line joining two charges.” r q1 q2 Let, q1 and q 2 = two point charges, r = distance between them, F = force between q1 and q2 1 F q1q 2 and F r2 q1q 2 F r2 qq F = K 12 2 r Where, K = proportionality constant 1 but, K = 4 0 1 q1q 2 F= 40 r 2 This is the expression for electrostatic force or columb’s law Note 1: 0 is called permittivity of free space S.I unit is C2 N-1 m-2 0 = 8.854 10-12C2 N-1 m-2 1 1 1 Note2:To find the value of , = , 4 0 4 0 4 3.14 8.854 10 −12 1 1 1 = 0.00899 1012 , = 8.99 109 , = 9 109Nm2/c2 4 0 4 0 4 0 Department of Physics 7 STUDY MATERIAL Electric Charges and Fields II PU Note3: Coulomb used a Torsion balance for measuring the force between two charged metallic spheres. Note4:Torsion balance is a sensitive device to measure a force. Note5:Cavandis also used Torsion balance to measure the very feeble gravitational force. Define one columb using coulomb’s law: According to coulomb’s law 1 q1q 2 F= if q1=q2=1C and r=1m 4 0 r 1 1 1 Then F= 40 12 1 F= 4 0 F=9×109 N Definition:One columb is the charge that when placed at a distance of 1m from another identical charge kept in vacuum experiences a force of repulsion of 9×109N. Note: Define one coulomb of charge in terms of current: One coulomb is that charge flowing through a wire in one second if the current is one ampere. Columb’s law in vector form: q1 r21 = r2 − r1 q2 1 q1q 2 F21 = rˆ21 F12 F21 4 0 r212 Where, r1 r2 F21 = Force on q2 by q1 r21 = position vector. r̂21 = unit vector which gives direction of force. Note:1) The force on q2 and q1 = force on q1 by q2 F21 = −F12 Thus,Coulomb’s law obeys Newton’s III law. Note:2) Pictorial representation of force of repulsion between two like charges (q1 q2>0) → → → F12 r21 F21 q1 q2 Department of Physics 8 STUDY MATERIAL Electric Charges and Fields II PU Note3) Pictorial representation of force of attraction between two unlike charges (q1 q2< 0). → → → F12 r21 F21 q1 q2 Principle of superposition of electrostatic forces : It states that “Force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to other charges taken one at a time. The individual forces are unaffected due to the presence of other charges. Application of Principle of superposition to find the force between system of charges: Consider a system of n stationary charges. Let force on q1by q2 is 1 qq F12 = 1 2 2 rˆ12 0 r12 Force on q1by q3 is 1 qq F13 = 12 3 rˆ13 0 r13 Similarly, force on q1 by q n is 1 qq F1n = 1 2 n rˆ1n 0 r1n According to principle of superposition, net force on q1 is F1 = F12 + F13 + F14..........F1n 1 qq 1 qq 1 qq F1 = 1 2 2 rˆ12 + 12 3 rˆ13............. + 1 2 n rˆ1n 40 r12 40 r13 40 r1n q1 q 2 q3 qn F1 = 2 rˆ12 + 2 rˆ13............. + 2 rˆ1n 40 r12 r13 r1n n q q F1 = 1 2 i rˆ1i 4 0 i = 2 r 1i Electric field or electric field intensity or strength: Electric field at a point is the force per unit positive charge. Force i.e. Electric field= unit positive charge F E= q0 S.I. unit is newton/coulomb(N/C)or volt / meter (v/m), Electric field is a vector quantity. Department of Physics 9 STUDY MATERIAL Electric Charges and Fields II PU Source charge: The charge which produces the electric field is called source charge. Test charge: The charge that detects the effect of source charge is called test charge. Note: UPC=Unit positive charge. In SI, UPC = 1 coulomb. Mention the expression for electric field due to a point charge: q q0 r P Let, q=source charge, q0=test charge (upc), P=any point According to coulomb’s law 1 qq F= 20 0 r F 1 q = 2 q 0 0 r F But = E = Elecric field q0 1 q E= 2 4 0 r This is the expression for electric field at a point p due to a charge. Note: 1 q 1. Electric field in vector form: E = 2 rˆ 4 0 r +q 2. + E The electric field due to positive charge is directed radially outward. -q 3. − E The electric field due to negative charge is directed radially inward. Department of Physics 10 STUDY MATERIAL Electric Charges and Fields II PU Principle of superposition of electric field: It’s state that “The net electric field at a point is the vector sum of all the electric fields at that point due to number of charges.” Application of principle of superposition to find Electric field due to system of charges: Consider a system of ‘n’ charges arranged as shown. Electric field at a point P by q1 charge is 1 q E1P = 21 rˆ1P 4 0 r1P Electric field at a point P by q2 charge is 1 q E 2P = 22 rˆ2P 4 0 r2P Similarly 1 q E 3P = 23 rˆ3P 40 r3P 1 q E nP = 2n rˆnp 40 rnP According to principle of superposition net electric field at P EP = E1P + E2P + E3P +......... + EnP 1 q 1 q 1 q EP = 21 rˆ1P + 22 rˆ2P +.......... + 2n rˆnP 40 r1P 40 r2P 40 rnP 1 q1 q2 qn EP = 2 rˆ1P + 2 rˆ2P +.......... + 2 rˆnP 40 r1P r2P rnP 1 n qi EP = rˆiP 40 i =1 riP2 Physical significance of electric field: Electric field is a vector quantity. Electric field was first introduced by Faraday. Electric field transport the energy. It is a characteristic of system of charges and it does not depend on test charge. It deals with theelectromagnetic phenomena. Electricfield lines or Electric lines of force: It is a curved path drawn such that tangent at any point on it gives the direction of net electric field at that point. Or It is an imaginary path along which unit positive charge moves. The concepts of electric field lines was developed by Faraday. Department of Physics 11 STUDY MATERIAL Electric Charges and Fields II PU Note: 1. Electric field lines due to a single positive charge are directed radially outward. 2. Electric field lines due to a single negative charge are directed radially onward. 3. The electric field lines around two opposite charges are as shown in figure. 4. The electric field lines around two same charges are as shown. Properties of electric field lines: 1. The electric field lines start from the positive charge and ends at negative charge. 2. The electric field lines donot intersect each other. 3. The electric field lines donot form a closed loop. 4. The electric field lines are normal to the surface of charged conductor. Note: If electric field lines intersect, the field at the point of intersection will not have a unique direction. That’s why two electric field lines never intersect. Electric flux: The total number of electric field lines passing through the surface area held perpendicular to the direction of field lines is called electric flux. S.I unit is NC-1 m2. Electric flux is a scalar quantity. Electric flux through an area element: Consider a surface ‘S’. Let, S = small area element of surface ‘S’ E = Electric field Department of Physics 12 STUDY MATERIAL Electric Charges and Fields II PU = Angle between ‘E’ and normal to S = Electric flux through S. The projection of area ‘ S ’ normal to E is given by S cos The electric flux through the area element S is given by = E S cos ⎯⎯→ = E S Therefore, electric flux through a surface is defined as the dot product of electric field and area element. Note 1) Electric flux through small area S is given by = ES cos ⎯⎯→ = E S Note 2) Total electric flux through entire surface is given ⎯⎯→ ⎯⎯→ = E S or = E S Note 3) We have = E S cos E = S cos ` i.e. Electric field is the electric flux per unit normal area. If = 00, cos00 = 1, then E = S Note 4) If = 00 , cos00 = 1 = EScos 00 = ES i.e. flux is maximum Note 5)If = 900 , cos900 = 0 Then, = ES 0 =0 i.e. flux is minimum. Electric dipole:A set of two equal and opposite point charges separated by a small distance is called electric dipole. +q −q +q and −q = point charges 2a = distance between +q and −q = dipole length 2a Note: Net charge of an electric dipole is zero Department of Physics 13 STUDY MATERIAL Electric Charges and Fields II PU Electric dipole moment P : () It is the product of magnitude of one of the charges and the distance between two charges of dipole. Dipolemoment = charge × distance P = q 2a S.I unit is coulomb × metre (Cm) Dipole moment is a vector quantity Note (1) Dipole moment in vector form p = q 2a P̂ Where P̂ = unit vector which gives direction of dipole moment. Note (2) The direction of dipole moment is directed from negative charge to positive charge along the dipole axis Note (3) Obtain an expression for the electric field at a point along the axis of dipole Consider a point M on the axis of electric dipole. Let O = mid point of dipole +q and −q = two point charges, q = magnitude of point charge 2a = dipole length, r = distance between O and M P = dipole moment acting from −q to +q, P̂ = unit vector of P but P = q × 2a ------------> (1) Electric field at M due to +q charge is 1 q E +q = P̂ away from +q ------------- (2) 40 (r − a )2 Electric field at M due to −q charge is Department of Physics 14 STUDY MATERIAL Electric Charges and Fields II PU 1 q E −q = − Pˆ towards −q 40 ( r + a )2 (−ve sign indicates that E −q is opposite to P ) According to the principle of superposition the total electric field at M is E = E + q + E −q 1 q 1 q E= P̂ − P̂ 40 (r − a ) 2 40 (r + a )2 q 1 1 E= − P̂ 40 (r − a ) (r + a )2 2 ( r + a) − (r − a) 2 2 q 4ar ˆ 1 1 E= P − = ( ) 40 r 2 − a 2 2 ( r − a ) (r + a ) ( r − a ) (r + a ) 2 2 2 2 If, r>>a r 2 + a 2 + 2ar − r 2 − a 2 + 2ar = ( ) ( ) 2 then r 2 − a 2 = r 2 = r 4 2 (r − a )2 (r + a )2 q 4ar 4ar E = 40 r 4 P̂ r2 − a2 2 ( ) q 4a = P̂ 40 r 3 1 2 2a q E= P̂ 40 r 3 1 2P E= P̂ 40 r 3 This is the expression for electric field at a point on the axis of dipole. Obtain an expression for the electric field at a point on the equatorial plane of electric dipole. Consider a point M on the perpendicular bisector of the axis of dipole. Let O = Midpoint of dipoles, +q and −q = two point charges q = magnitude of charge, 2a = dipole length Department of Physics 15 STUDY MATERIAL Electric Charges and Fields II PU P = dipole moment acting from −q to +q, P̂ = unit vector Magnitude of dipole moment is P = q × 2a ------ (1) From figure, AM2 = r2 + a2 and BM2 = r2 + a2 AM = BM = r + a2 2 2 2 AM = BM = (r 2 + a 2 )2 1 Electric field at M due to +q charge is 1 q E+q = pˆ 40 AM 2 1 q E+q = 2 Pˆ away from +q ( 40 r + a 2 ) Electric field at M due to −q charge is 1 q E −q = pˆ 4 0 BM 2 1 q E−q = 2 Pˆ towards −q ( 40 r + a 2 ) Hence E + q = E − q (magnitudes) Resolving E + q and E − q in to two rectangular components each. Here, perpendicular components are equal and opposite. So they cancels each other. The components parallel to dipole axis are in same direction. So they add up Total electric field at M E = − E+ q cos + E−q cos [−ve sign indicates that E is opposite to P ] E = − E + q + E −q cos 1 q 1 q = − 2 2 Pˆ + 2 2 Pˆ cos ( 40 r + a ) 40 r + a ( ) 2q E = − cos Pˆ 40 ( r + a ) 2 2 But from figure, AO a cos = = 1 (r ) AM 2 +a 2 2 ˆ E = − 2q a 1 P ( P = q 2a ) 2 ( 40 r + a 2 r2 + a2 ) ( ) 2 Department of Physics 16 STUDY MATERIAL Electric Charges and Fields II PU 1 P ˆ E=− 3 P 40 ( r2 + a2 ) 2 ( ) (r ) 3 3 If r>>a then r 2 + a 2 2 2 2 = r3 1 Pˆ E = − P 40 r 3 This is the expression for electric field at a point on the equatorial plane. Note (1)Electric field at a point on the dipole axis. 1 2P ˆ Edipoleaxies = P 40 r 3 It is acting along dipole moment Note (2)Electric field at a point on the equatorial plane 1 Pˆ Eequatorial = − P 40 r 3 −ve sign indicates that net electric field is opposite to dipole moment Note (3): From note (1) and (2). It is clear that, Edipole axis = 2 Equatorial (magnitude wise) Note (4) : Force acting on positive charge is along the electric field direction. The force acting on negative charge is opposite to the field direction. Obtain the expression for the torque acting on an electric dipole placed in a uniform electric field: Consider an electric dipole placed in a uniform electric field ‘E’. Let +q and −q = two charges of dipole AB = 2a = dipole length = angle between E and dipole moment P. The fore acting on +q is F+q=qE along E The force acting on –q is F− q = qE opposite to E i.e. F+ q and F− q are two equal and opposite force acting on the dipole. As a result, torque is acting on the dipole but, torque = force × perpendicular distance between two forces. = qE BC ----- (1) From right angled triangle BCA Department of Physics 17 STUDY MATERIAL Electric Charges and Fields II PU BC BC sin = = AB 2a BC = 2a sin equation (1) becomes = q 2a E sin = P E sin ( q 2a = P ) This is the expression for torque acting on the dipole Note-1:We have = P E sin = P E ⎯⎯→ Vector form i.e. Torque is the cross product of dipole moment and electric field, Torque is a vector. S.I unit of torque is Nm (newtonmetre) Note-2: If dipole is along the electric field Then = 00 Or = 1800 Then sin = sin00 = sin1800 = 0 = PE × sin = PE × 0 = 0 i.e no torque is acted on dipole (Torque is minimum) Note-3: If dipole is perpendicular to Electric field. Then, =900. Then sin = sin 900 = 1 = P E sin = PE Torque is maximum when the dipole is placed perpendicular to the direction of field. Linear charge density (): It is defined as the charge per unit length. charge i.e Linear charge density = length q = L S.I unit is coulomb/metre (c/m) Department of Physics 18 STUDY MATERIAL Electric Charges and Fields II PU Surface charge density (): It is defined as the charge per unit area charge i.e. surface charge density = area q = S S.I unit is C/m2 Volume charge density (): It is defined as the charge per unit volume charge i.e. volume charge density = Volume q = V S.I unit is C/m3 Mention the expression for electric field due to continuous charge distribution Electric field due to q charge present in small volume element V is 1 q E = 2 ⎯⎯→ (1) 4 0 r q But = = volume charge density V q = V - equation (1) become 1 V E = 2 40 r Electric field due to entire volume V is E = E all V 1 V E= 40 all V r 2 This is the expression for electric field due to continuous charge distribution. State Gauss law: It states that “Total electric flux flow through the closed surface is equal q to. 0 q That is, Total electric flux = 0 Where, q=total charge closed by the surface 0= permittivity of free space. Department of Physics 19 STUDY MATERIAL Electric Charges and Fields II PU Deduce Gauss law: Consider a sphere of radius ‘r’, divide the sphere into small area element as shown in figure. Let, q = charge enclosed by a sphere, r = radius of sphere. s = small area element Electric field at the surface is 1 q E= 4 0 r 2 Small electric flux flows through S is = EScos , but = 0°, cos0° = 1 = ES 1 q = S 40 r 2 Total electric flux through entire sphere. = 1 q = 2 S 40 r 1 q = 2 S 40 r But S = 4r 2 = area of sphere 1 q = 2 4r 2 40 r q = 0 This is called Gauss law. Department of Physics 20 STUDY MATERIAL Electric Charges and Fields II PU Significance of Gauss law: 1. Gauss law is true for any closed surface, no matter what its shape or size. q 2. According to Gauss law, = here q is the sum of all the charges enclosed 0 by the surface. 3. The surface that we choose for the application of gauss law is called Gaussian surface. 4. Gauss law is based on inverse square dependence on distance contained in coulombs law. Applications of Gauss law: Using Gauss law 1. Electric field due to infinite long straight uniformly charged wire can be found. 2. Electric field due to uniformly charged infinite sheet can be found. 3. Electric field due to a charged thin spherical shell can be found. Derive an expression for electric field due to infinitely long straight uniformly charged wire using Gauss law: Normal to S 1 E =900 1 E no 3 3 =0 r L nor P m to S E Normal r E mal no al no no toto r Gaussian r surface r S m m m 2 al al 0 S E al =90 no to 2 to E to Normal to Sr S S m S Consider an infinitely long straight charged wire. al Let = Linear charge density of wire, E=Electric field at atopoint ‘P’ r = Distance between wire and P. S Draw a cylindrical Gaussian surface of radius ‘r’ passing through point ‘P’ a shown in figure. The electric field is radially outward. i.e. perpendicular to wire. Electric flux through 1st circular end of a cylinder (Gaussian surface) is Department of Physics 21 STUDY MATERIAL Electric Charges and Fields II PU 1 =ES cos But =900 cos900=0 1=0 Similarly electric flux through 2nd circular end of cylinder (G.S.) is 2= 0 Electric flux through curved surface of cylinder (Gaussian surface) is 3=ES cos =00, cos00=1 3 = ES But, S = 2rL = Total surface area of cylinder, Where, L=length of cylinder. 3=E2rL Total electric flux is =1+2+3 =0+0+ E2rL =E2rL -------------- (1) Let, q= charge enclosed by cylinder (Gaussian surface) Charge But = Length q = L q=L According to Gauss law q = 0 L = --------------- (2) 0 From equation (1) and (2) L E 2rL = 0 E= 2 0 r This is the expression for electric field due to a charged long wire. Note: 1) In vector form, E = rˆ , 2 0 r r̂ = unit vector which gives directions of E and it is radially directed. Note: 2) Electric field due to a wire is radially outward if charge is +ve (or is +ve). Electric field due to a wire is radially inward if charge is negative (or is negative): Department of Physics 22 STUDY MATERIAL Electric Charges and Fields II PU Derive an expression for electric field due to a uniformly charged infinite plane sheet: P E E G.S. norm al to Plane sheet S Consider a uniformly charged infinite plane sheet. Let, = surface charge density of plane sheet E=Electric field at a point P due to plane sheet, it is directed outward. Draw a cylindrical Gaussian surface, perpendicular to plane sheet passing through a point ‘P’ as shown in figure. Let, A=cross sectional area of cylinder. Electric flux through 1st circular end of cylinder (Gaussian surface) is 1= ES cos But, S=A, =00, cos00=1 1=EA Similarly, electric flux through 2nd circular end of cylinder is 2=EA Electric flux through curved surface of cylinder is 3= ES cos =900, cos900=0 3= ES 0 3=0 Total electric flux through cylinder (Gaussian surface) is =1+2+3 =EA+EA+0 =2EA -------------- (1) Charge enclosed by cylinder = q Charge But = Area q = q= A A According to Gauss law q = 0 A = ------------- (2) 0 Department of Physics 23 STUDY MATERIAL Electric Charges and Fields II PU From equations (1) and (2) A 2EA = 0 E= 2 0 This is the expression for electric field due to plane sheet. Note: 1. In vector form E = nˆ. Where n̂ = unit vector. 2 0 2. E is directed outward if is +ve (i.e. q is +ve) and E is directed inward if is negative (i.e. q is negative). 3. E is independent of distance from plane sheet. Derive an expression for electric field due to a charged thin spherical shell using Gauss law: Case 1: Electric field outside the shell: Consider a charged thin spherical shell. Let, R= radius of spherical shell, q=charge on spherical shell E= electric field at a point ‘P’, =surface charge density Area of spherical shell = 4R2 charge = area q = 4R 2 q=4R2----------- (1) Consider a point P out side the shell. Imagine that a spherical Gaussian surface around the shell passing through point ‘P’ with the centre ‘O’ as shown in figure. r=radius of Gaussian surface Let S=small area element around point ‘P’ The electric flux through S is Department of Physics 24 STUDY MATERIAL Electric Charges and Fields II PU =ES cos But =0, cos0=1 =ES Total electric flux through entire Gaussian surface is = = ES = E S (Electric field E is same at all point on Gaussian surface) But S = r 2 = area of Gaussian surface. = E 4r 2 -------------- (2) According to Gauss law q = [ from equation (1), q=4R2] 0 4R 2 = ----------- (3) 0 From equations (1) and (2), 4R 2 E 4r 2 = 0 R 2 E= 0 r 2 This is the expression for electric field at a point ‘P’ Case 2:Electric field inside the shell: Consider a point P inside the spherical shell. Consider a Gaussian surface passing through point ‘P’ as shown in figure. Electric flux through the Gaussian surface is = E 4r 2 (Refer equation 2) q According to Gauss law, = 0 q E 4r 2 = 0 Charge enclosed by Gaussian surface. i.e. q=0. (No charge is present inside the shell) E=0 i.e. Electric field due to a uniformly charged thin spherical shell is zero at all point inside the shell. Department of Physics 25 STUDY MATERIAL Electric Charges and Fields II PU R 2 Note: E= 0 r 2 q But = 4R 2 q R2 E = 4R 2 2 0 r q E= 40 r 2 1 q E= 4 0 r 2 This is the expression for electric field at a point out side the charged spherical shell. One mark questions 1. What is an electric charge? 2. What are point charges? 3. Define polarity of charge. 4. What is gold leaf electro scope? 5. What is meant by earthing? 6. What is significance of earthing? 7. Define elementary charge. 8. What is meant by additivity of charge? 9. The net charge of a system of point charges -4, +3, -1 & +4 (S.I.units) =? 10. What is meant by conservation of charge? 11. What is quantisation of charge? 12. Mention the S.I. unit of charge. (March-2014) 13. State coulomb’s law (March-2017) 14. Define one coulomb of charge. (March-2015) 15. State the principal of super position of electro statics. 16. Which principle is employed in finding the force between multiple charges? 17. Define electric field. 18. Is electric field a scalar/vector? 19. Mention the S.I. unit of electric field. 20. What is the direction of electric field due to a point positive charge 21. What is the direction of electric field due to a point negative charge? 22. What is a source charge? 23. What is a test charge? 24. How do you pictorially map the electric field around a configuration of charges? 25. What is an electric field line? 26. What is electric flux? 27. Mention the S.I.unit of electric flux. 28. What is an electric dipole? 29. What is the net charge of an electric dipole? Define dipole moment. 30. Is dipole moment a vector / scalar? Department of Physics 26 STUDY MATERIAL Electric Charges and Fields II PU 31. What is the direction of dipole moment? 32. What is the net force on an electric dipole placed in a uniform electric field? 33. When is the torque acting on an electric dipole placed in a uniform electric field maximum? 34. When is the torque acting on an electric dipole placed in a uniform electric field minimum? 35. State Gauss’s law. 36. What is a Gaussian surface? 37. What happens to the force between two point charges if the distance between them is doubled? 38. If two charges kept in ‘air’ at a certain separation, are now kept at the same separation in ‘water’ of dielectric constant 80, then what happens to the force between them? 39. On a macroscopic scale is charge discrete or continuous? Two mark questions 1. Explain construction of gold leaf electroscope. 2. Earthing is necessary for house, Why? 3. Write the expression for quantisation of charge and explain the terms in it. 4. State and explain Coulomb’s law of electrostatics. (March-2014, July-2015, March-2017) 5. Write Coulomb’s law in vector notation and explain the terms. (March-2015) 6. Write the pictorial representations of the force of repulsion and attraction, between two point charges. 7. Explain the principle of superposition to calculate the force between multiple charges. 8. Mention the expression for the electric field due to a point charge placed in vacuum. 9. Write the expression for the electric field due to a system of charges and explain it. 10. Draw electric field lines in case of a positive point charge. 11. Sketch electric field lines in case of a negative point charge. 12. Sketch the electric field lines in case of an electric dipole. 13. Sketch the electric field lines in case of two equal positive point charges. 14. Mention any two properties of electric field lines. (July-2015, March-2015, March-2017) 15. Write the expression for the torque acting on an electric dipole placed in a uniform electric field and explain the terms in it. 16. Define linear density of charge and mention its SI unit. 17. Define surface density of charge and mention its SI unit. 18. Define volume density of charge and mention its SI unit. 19. What is the effect of a non-uniform electric field on an electric dipole? Three mark questions 1. Mention three properties of electric charge. (July-2014) 2. Draw a diagram to show the resultant force on a charge in a system of three charges. 3. Why is the electric field inside a uniformly charged spherical shell, zero? Explain. 4. Obtain the expression for the torque acting on an electric dipole placed in a uniform electric field. Department of Physics 27 STUDY MATERIAL Electric Charges and Fields II PU Five mark questions 1. Obtain an expression for the electric field at a point along the axis of an electric dipole.(Mar-16) 2. Obtain an expression for the electric field at a point on the equatorial plane of an electric dipole.(March-2015) 3. State Gauss’s law. Obtain an expression for the electric field due to an infinitely long straight uniformly charged conductor. (July-2015) 4. State Gauss’s law.Obtain an expression for the electric field due to a uniformly charged infinite plane sheet. 5. State Gauss’s law.Obtain an expression for the electric field at an outside point due to a uniformly charged thin spherical shell. (March-2014, July- 2014). Department of Physics 28 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Chapter 2: ELECTROSTATIC POTENTIAL AND CAPACITANCE Electrostatic potential energy: Consider a source charge ‘q’, Let E=electric field due to a charge q (q>0 i.e., q is the +ve charge) q 0 =test charge ( q 0 > 0 i.e., q 0 is +ve charge) When we bring a test charge q 0 from a point R to P, then q charge produces repulsive force on ‘ q 0 ’ An external force is applied on charge ‘ q 0 ’ to bring it from R to P. i.e., external force = - electric force i.e., Fext =- Fele Net force on test charge q 0 =0, As a result the test charge moves with constant speed and with zero acceleration. The work done by the external force in moving a charge ‘ q 0 ’ from R to P is P→ → WRP = Fext.dr Let dr = small displacement of q 0 R → Where dr =displacement dw= small workdone P→ → WRP = − Fele.dr dw=Fext.dr, R → → Total workdone= W= dw = Fext.dr -ve sign indicates that work done is against electrostatic repulsive force. This work done is stored as a potential energy. At every point in the electrostatic field the charge ‘q’ possesses certain potential energy. P If R is at , then WP = − Fele.dr. But, workdone= potential energy WP = UP P U P = − Fele.dr Potential energy of a charge ( q 0 ) at a point is the work done by the external force to bring that charge from infinity to that point against the electric field. Note: 1. Electrostatic potential energy is a scalar. S.I unit is joule. 2. Electrostatic potential energy is zero at infinity. DEPARTMENT OF PHYSICS 1 STUDY MATERIAL Electrostatic Potential and Capacitance II PU 3. Coulomb force between two charges is a conservative force. Because the work done by an electrostatic force in moving a charge from one point to another depends only on the initial and final points and is independent of the path taken to go from one point to the other. Electrostatic potential: Electric potential at a point is the work done by the external force to bring a unit positive charge from infinity to that point against the electric field. work done i.e., Electric potential = charge W V= q0 S.I unit is J/C or volt, potential is a scalar. Note: 1Volt = 1 Joul / Coulomb Potential difference: Potential difference between the two points is the work done by an external force to bring the unit positive charge from one point to another point against the electric field. S.I. unit is J/C or volt. Note: Potential goes on decreases along the direction of electric field. Derive an expression for potential due to a point charge. q q0 r dx x Let q = point charge at origin ‘O’ E = electric field V= Electric potential at point ‘P’ r = distance between ‘O’ and ‘P’, A and B = any two points, x=distance between O & A dx= distance between A and B q 0 = unit positive charge (UPC) =+1C According to coulomb’s law 1 qq F= 20 4 0 x 1 q F = but q 0 =1 unit, 4 0 x2 Small work done in bringing a unit positive charge ‘ q 0 ’ from A to B is dw = - F dx DEPARTMENT OF PHYSICS 2 STUDY MATERIAL Electrostatic Potential and Capacitance II PU -ve sign indicates that force is against field 1 q dw= − 2 dx 4 0 x To get total work done, integrating from x= to x=r. r r r W = dw 1 q W = dw = − dx 4 0 x2 r r 1 q q 1 = − 4 0 x2 dx W=− 4 0 x 2 dx x n +1 r 1 q q 1 W= =− − x dx = n + 1 n x 4 0 r 4 0 x But, V = W but q 0 = 1 = upc q 1 1 = − q0 4 0 r V = W q 1 = 1 q 4 0 r i.e., V= 4 0 r 1 q W= x 4 0 r This is the expression for electric potential. Compare the variation of electric potential and electric field with distance: The graph shows the variation of potential V with r and field E with r for 1 1 a point charge q. They vary as v and E 2. r r DEPARTMENT OF PHYSICS 3 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Note 1: If q>0( i e q is + ve) then V >0 ( i e V is + ve) if q (2) E = − d dL dV V From equation (1) & (2) | E |= dL = d q V = 0 A d q 0 A q Or = But =C=Capacitance of capacitor V d V 0A C= d This is the expression for capacitance of a parallel plate capacitor Note: In case of a parallel plate capacitor, near the outer boundaries of the plates, the field lines bend outward at the edges, this effect is called Fringing of the field. Derive the expression for Capacitance of Parallel capacitor when dielectric medium is present: Case 1: Consider a Parallel plate Capacitor. The space between the plate is Vacuum. Let q = Charge on capacitor V0 = Potential Difference E0 = Electric Field between plates = Surface charge density A = Area of each plate d = distance between two plates E0 = 0 DEPARTMENT OF PHYSICS 18 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Case 2: When a dielectric medium is filled between two plates of a capacitor, then dipoles are induced. Let, D = Surface charge density due to polarization of dielectric ED = electric field due to polarisation But, ED = D 0 Net electric field E = E0− ED ED D = − 0 0 − D E= 0 It is found that − D − D = K Where, K = dielectric constant (K>1) E= ----------- ① 0K V We know that, E= ----------- ② d V = 0K d V 0 K q = But = d A q V 0 K = A d q 0 KA = V d 0 KA q C= = C d V This is the expression for capacitance of a parallel plate capacitor when dielectric is present. KA Note 1: we have C = 0. d Where 0 K = = permittivity of the dielectric medium. K = 0 Where K= dielectric constant, ‘K’ has no unit. DEPARTMENT OF PHYSICS 19 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Note 2 : For vacuum, K = 1, K>1 for any dielectric medium. A Note 3 :We have C0 = 0 -----① for vacuum d 0 KA And C= For dielectric medium d A Or C= 0 XK d C = C0K from eqn ① C Or K= C0 Define dielectric constant: Dielectric constant is the ratio of capacitance of capacitor when space between the plates is filled with dielectric to the capacitance of same capacitor when space between the plates is vacuum. Equivalent capacitor and equivalent capacitance (Effective capacitance): A single capacitor which produces same effect as that of set of capacitor is called equivalent capacitor and its capacitance is called equivalent capacitance. Combination of capacitor: There are two combination. i) Series combination, ii) Parallel combination Series combination: A set of capacitors is said to be in series if they are connected end to end such that charge on each capacitor remain same. Derive an expression for effective capacitance of a series combination: Consider two capacitor of capacitance C1 and C2 connected in series as shown in diagram. Let +q and –q be the charge on two plates of a capacitors. In series combination, charge on each capacitor is same (say ‘q’ ). Let V1 = potential difference across C1 V2 = potential difference across C2 V = potential difference across combination DEPARTMENT OF PHYSICS 20 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Cs= effective capacitance of series In series, V = V1 + V2 ------- ① q q q q=CV But V1 = , V2 = and V = q C1 C2 Cs V = C equ ① becomes q q q = + Cs C1 C 2 q 1 1 = q + Cs C1 C2 1 1 1 = + Cs C1 C2 This is the Expression for effective capacitance of two Capacitors in series. For n –Capacitors in Series V = V1 +V2 + V3 + --------- + Vn q q q q = + + +−−−−−−+ C1 C2 C3 Cn 1 1 1 1 1 = + + +−−−−−−+ CS C1 C 2 C3 Cn This is the Expression for effective capacitance of ‘n’ capacitors in series Derive an expression for effective capacitance of a parallel combination Consider two capacitor of capacitance C1 and C2 connected in parallel as shown in diagram. Potential difference across each capacitor is same in parallel. Let q1 = charge on capacitor C1 q2 = charge on capacitor C2 q = total charge of combination DEPARTMENT OF PHYSICS 21 STUDY MATERIAL Electrostatic Potential and Capacitance II PU V = potential difference Cp = effective capacitance of parallel In parallel, q = q1 + q2 -------① But q1 = C1V, q2 = C2V and q = CpV eqn ① becomes q = CV CpV = C1V + C2V V is same CpV = V (C1 + C2) Cp = C1 + C2 This is the expression for effective capacitance of two capacitors in parallel. For ‘n’ capacitors q = q1 + q2 + q3+ ----------- + qn CpV = C1V + C2V + C3V + ------- + CnV Cp = C1 + C2 + C3 + ------- + Cn This is the expression for effective capacitance of n capacitors in parallel. Note: 1. In series combination, charge on each capacitor remains constant. 2. In parallel combination, potential difference across each capacitor remains constant. 3. If C1 and C2 connected in series. 1 1 1 C 2 + C1 C1C2 then, = + = , Cs =. Cs C1 C 2 C1C 2 C1 + C2 For three capacitors 1 1 1 1 C2C3 + C3C1 + C1C 2 C1C2C3 = + + = , Cs = Cs C1 C 2 C3 C1C 2C3 C1C2 + C2C3 + C3C1 4. For ‘n’ capacitors in series 1 1 1 1 1 = + + +−−−−−−+ CS C1 C 2 C3 Cn For ‘n’ identical capacitors, C1=C2=C3=…….=Cn=C, 1 1 1 1 1 1 1 C = + + +−−−−−−+ , = n , CS = CS C C C C Cs C n 5. If ‘n’ identical capacitor are connected in parallel CP = C1 + C2 + C3 +...... + Cn For ‘n’ identical capacitor, CP=C+C+C+…….+C CP=nC DEPARTMENT OF PHYSICS 22 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Derive an expression for energy stored in a capacitor Consider a capacitor of capacitance C. The capacitor consists of two conductors, initially these are uncharged. When a positive charge is transferred from conductor ② to ① bit by bit, finally conductor ① gets charge ‘+q’ and conductor ② has –q. In transferring positive charge from conductor ② to ①, work will be done externally. At intermediate situation, Let, +Q and -Q' be the charge on conductor ① and ② V'=Potential difference, Potential difference, V' = Q C At this stage, dQ = a small charge is transferred from conductor ② to ①. dW=Small work done to transfer dQ but, dW = V' dQ dW Q V' = dW = V'dQ dW = dQ dQ C total work done to transfer a charge from zero to q is q q W = dW 0 W= 1 C 0 Q dQ q2 q 1 Q1+1 = W = C 1 + 1 2C x n +1 x dx = 0 n q Work done is stored as a potential energy 1 Q2 n +1 = i.e. work done = potential energy C 2 0 W = U = 1 q 2 − 0 2C q2 q2 U = W= 2C 2C This is the expression for energy stored. DEPARTMENT OF PHYSICS 23 STUDY MATERIAL Electrostatic Potential and Capacitance II PU Note: q2 1. W.K.T. U = --------------- (1) 2C but, q=CV ( CV ) 2 U= 2C CV 2 U= 2 1 U = CV 2 --------------- (2) 2 q but, C= V 1 q U = V2 2 V 1 U = qV --------------- (3) 2 q2 1 1 U= = CV 2 = qV 2C 2 2 Energy density: The energy stored per unit volume is called Energy density. Energy Energy density= Volume S.I. Unit is Joule/meter3. Derive an expression for energy density: 02 E 2 A 2 U= q2 A 2 0 We have, U= …………..(1) d 2C E Ad 2 q U= 0 But, = , 2 A U 0 E 2 q=A ----------------(2) = Ad 2 But, E= 0 Ad = Area X distance = 0 E Ad = volume of region between plates Equation (2) becomes U q = 0 EA --------------(3) = energy/ volume Ad Substituting Equation (3) in (1) U = energy density Ad DEPARTMENT OF PHYSICS 24 STUDY MATERIAL Electrostatic Potential and Capacitance II PU ( 0EA ) 2 U= 2C E2A2 2 U= 0 --------------(4) 2C A But, C = 0 d Above equation (4) becomes 0 E 2 Energy density = 2 This is the expression for energy density. 1 Q q VB = + 40 R r ******** One mark questions: 1. What do you mean by the conservative nature of electric field? 2. Define Electrostatic Potential. 3. What is the SI unit of Electric Potential? 4. What are the equipotential surfaces of a point charge? 5. Draw the Equipotential surfaces for a point charge. 6. Give the condition for equipotential surface in terms of the direction of the electric field. 7. Define Electrostatic Potential energy of a system of charges. 8. Write the expression for potential energy of two point charges in the absence of external electric field. 9. Write the expression for potential energy of two point charges in the presence of external electric field. 10. Write the expression for Electric field near the surface of a charge conductor. 11. What happens when Dielectrics are placed in an electric field? 12. What is electric polarisation? 13. What is Parallel plate capacitor? 14. Mention the expression for capacitance of a Parallel plate capacitor without any dielectric medium between the plates. 15. Mention the expression for capacitance of a Parallel plate capacitor with a dielectric medium between the plates. 16. Define Dielectric constant of a substance. 17. What is an electric dipole (March-2016)? 18. What is the electric field strength inside a charged spherical conductor? 19. What is a capacitor? (July-2014) Two mark questions: 20. How does the electric field and electric potential vary with distance from a point charge? DEPARTMENT OF PHYSICS 25 STUDY MATERIAL Electrostatic Potential and Capacitance II PU 21. How does the electric potential at a point due to an electric dipole vary with distance measured from its centre? Compare the same for a point charge. 22. Using superposition principle, write the expression for electric potential at a point due to a system of charges. 23. What is an equipotential surface? Give an example. 24. Explain why the equipotential surface is normal to the direction of the electric field at that point. 25. Explain why Electric field inside a conductor is always zero. 26. Explain why Electrostatic field is always normal to the surface of charged conductor. 27. Explain why Electric charges always reside on the surface of a charge conductor. 28. Explain why Electrostatic potential is constant throughout the volume. 29. What is Electrostatic shielding? Mention one use of it. 30. What are non-polar Dielectrics? Give examples. 31. What are polar Dielectrics? Give examples. 32. Define capacitance of a capacitor. What is SI unit? 33. On what factors does the capacitance of a parallel plate capacitors depends? (March-2017) Three marks questions: 34. Write the expression for electric potential at a point due to an electric dipole and hence obtain the expression for the same at any point on its axis and any point on its equatorial plane. 35. Obtain the relation between the electric field and potential. (July-2015, July-2014) OR Show that electric field is in the direction in which the potential decreases steepest. 36. Derive the expression for potential energy of two point charges in the absence of external electric field. (March-2016) 37. Mention the expression for potential energy of an electric dipole placed in an uniform electric field. Discuss its maximum and minimum values. 38. Derive the expression for effective capacitance of two capacitors connected in series. 39. Derive the expression for effective capacitance of two capacitors connected in parallel. 40. Derive the expression for energy stored in a capacitor. (March-2016, March- 2017) Five marks questions: 41. Derive the expression for electric potential at a point due to a point charge. 42. What are Dielectrics? Mention the types of Dielectrics. 43. Derive the expression for capacitance of a Parallel plate capacitor without any dielectric medium between the plates. (or Parallel plate air capacitor ). 44. What is Van De Graff generator? Write its labelled diagram. What is the principle of its working? Mention its use. DEPARTMENT OF PHYSICS 26 STUDY MATERIAL Current Electricity II PU Chapter: 3 CURRENT ELECTRICITY Current electricity: It is a branch of physics which deals with the study of electric charge in motion. Electric Current: The net electric charge passing per unit time is called electric current net charge i.e electric Current = time taken q I= t S.I unit is coulomb/sec or ampere (A) Note: 1. Electric current is a scalar quantity q ne 2. We have I = , but q =ne I= , where n = number of free t t electrons q 3. I = , this Formula is for steady current (i.e for constant current). t 4. If current is variable (not steady), then, we take instantaneous current. Let ∆q = small charge flows ∆t = small time interval And ∆t→ 0 Then instantaneous current is defined as lim q I= t → 0 t Electric Current in Conductor: The solid conductors have large number of free electrons. In the absence of applied electric field, The free electron move in a random direction due to thermal energy. Therefore, number of free electrons travelling in any direction will be equal to the number of free electrons travelling in the opposite direction. That is net flow of free electron in a specific direction is zero. Therefore, in the absence of electric field net current flows through the conductor is zero. DEPARTMENT OF PHYSICS Page 1 STUDY MATERIAL Current Electricity II PU Take positive and negatively charged circular dielectric plates. Now, attach them on the two ends A and B of a conductor. then end A is positively charged (+q) and end B is negatively charged (− q). Therefore end A is at higher potential (+ve) and end B is at lower potential (−ve). As a result electric field is acted from A to B. Therefore, force is acted on free electron against the electric field. So all the free electrons move in a direction opposite to electric field (ie from lower potential to higher potential). As a result net current flows through the conductor. During the motion, the free electrons collide with fixed ions and atoms due to thermal excitation, and they lost kinetic energy in the form of heat. Drift Velocity ( v d ) : The average velocity with which the free electrons move in a conductor under the influence of electric field is called drift velocity. The SI Unit of drift velocity is ms-1. Mobility (µ): The ratio of magnitude of drift velocity to the applied electric field is called mobility. vd µ= E where µ = Mobility, vd = drift velocity, E = applied electric field. S.I Unit is m2/Vs Note: Mobility is positive. Relaxation time (): The time interval between the two successive collisions of free electrons in a conductor is called relaxation time. State and explain ohm’s law: It states that “The electric current flowing through a conductor is directly proportional to potential difference across its ends if temperature and dimensions of conductor kept constant.” i.e. current potential difference or, potential difference current. VI V = RI Where R = resistance of the conductor. Note: Ohm’s law was discovered by G S Ohm in 1828. DEPARTMENT OF PHYSICS Page 2 STUDY MATERIAL Current Electricity II PU Electrical resistance: V According to ohm’s law, V = RI R= I Resistance of a conductor is the ratio of potential difference across its ends to the current flows through it. S.I unit is Volt/ampere (or) ohm () Note: one ohm=1volt/ampere. Explain how resistance depends on length and cross-sectional area of conductor (dimensions of conductor): Consider a conductor of length L and cross-sectional area ‘A’. Let I = current, V = potential difference. R = resistance of a conductor V According to ohm’s law, R =. I Case ①: Consider two identical conductors in contact side by side as shown below Now, same current I flows through each conductor and potential difference across each conductor is V. Total potential difference across the combination is V| =V+V=2V Current through the combination is, I| = I Resistance of the combination is V| R = | | I 2V = I V R | = 2R = R I i.e. R L ① Resistance is proportional to length of conductor. DEPARTMENT OF PHYSICS Page 3 STUDY MATERIAL Current Electricity II PU Case ②: Now divide the conductor into two parts by cutting lengthwise as shown Length of each slab =L, cross sectional area of each slab = A/2 Current through each slab = I| = I/2, Potential Difference = V| = V. Resistance of each slab = R1 R| = V| / I| V = I/2 2V = I R = 2R (∵V/I = R) | 1 R ② A L from eq ① and ② R A L R= , Where, = resistivity of a conductor A Note-1 : a) Resistance of a conductor is directly proportional to Length of a conductor (RL) b) Resistance of a conductor is inversely proportional to Cross - sectional area 1 of a conductor R . A c) Nature of material of a conductor. Note-2: Resistance of a conductor is directly proportional to temperature of a conductor. (RT) Resistivity of a conductor L We have R = A If L =1m, and A = 1m2, then R = or = R Resistivity of a conductor is the resistance of conductor of length 1m and cross- sectional area 1m2 S.I Unit of resistivity is m (ohm-metre). DEPARTMENT OF PHYSICS Page 4 STUDY MATERIAL Current Electricity II PU Note-1: Resistivity does not depend on dimension (length and area) of a conductor but it depends on nature of material of conductor. Note-2: Resistivity of a conductor is directly proportional to temperature. (T) Conductivity: The reciprocal of resistivity is called conductivity. 1 i.e Conductivity = resistivity 1 = S.I Unit is ohm−1 m−1 (−1 m−1) or siemen m−1 Current density (J): The current per unit area taken normal to the current is called current density. current Current density = Area I J= A S.I Unit is ampere/metre2 (A/m2). Current density is a vector. Ohm’s law in terms of electric field and current density. L we have V=RI but R = A L L I I V= I = , but =J A A A V V= LJ but V=EL (∵E= ) L EL = LJ E = J ① 1 1 or J= E but = = Conductivity J=E ② eq: ① and ② represent Ohm’s law Note: in vector form, E = J and J = E. DEPARTMENT OF PHYSICS Page 5 STUDY MATERIAL Current Electricity II PU Drift of Electron and the origin of resistivity of conductor (Deduce ohm’s law): (Obtain the expression for conductivity): Consider a conductor of cross sectional area `A’. In the absence of electric Field, The free electrons move in random direction. They will suffer collisions with fixed ions. After collision, their average velocity will be zero, since their directions are random. 1 n i.e. vi = 0 , N i =1 Where, N=number of electrons, vi=velocity of ith electron When an electric field is applied, all the free electrons are drifted towards positive potential, force on electron is, F = −eE ① Where, −e = charge on electron, E=electric field. but F = ma ----------- ② F Where, a =acceleration, m = mass of electron, E = , F = qE, q form eq: ① & ②, ma = − eE but q = −e for electron − eE a= ------- ③ m W.k.t, v = v0 + at ----------- ④ Where, v0 = velocity of ith electron immediately after last collision at time `t’ v = final velocity Collisions of electrons do not occur at regular intervals, therefore the time ti is not constant. So consider average time and average velocity. i.e (t)average= = relaxation time (v)average = vd = drift velocity and (v0)average =0 eq. (4) becomes v d = 0 +a − eE v d = a but a = (from eqn ③) m DEPARTMENT OF PHYSICS Page 6 STUDY MATERIAL Current Electricity II PU − eE vd = --------------- ⑤ m eE vd = m This is the expression for drift velocity. Let, x = small length of conductor, t = small time interval. x Then, vd = t Velocity = dis tan ce time x = vdt d V= Volume of conductor = x A t (Volume=lengthArea) Volume of conductor = vdt A Let, n = number of free electrons per unit volume. Total free electrons transported = n volume Total free electrons transported = n vdt A Total Charge transported = number of electrons Charge on electron total charge transported = n vdt A (-e) q = −neAvdt (This is the charge transported opposite to E ) The total charge transported along E is q = − ( −neAvd t ) q = neAvd t q But current, I= t neAv d t I= t I = neAvd ⑥ This is the expression for drift current. eE But |vd| = vd = m eq
