Welcome to my class I am Mr. Humayun Kabir Assistant Professor, Department of Physics,

1. Welcome to my class I am Mr. Humayun Kabir Assistant Professor, Department of Physics, Jahangirnagar University, Savar, Dhaka-1342.

2. Course No. : PHY-122 • Course Title : Physics-II (Electricity, Magnetism and Modern Physics) • Number of Credit: 3.0

3. BOOKS RECOMMENDED BOOK RECOMMENDED: • Resnick, R. and Halliday, D. and Walker, J.; Fundamental of Physics, part-2; John Wiley and Sons. • Ahmed G. U.; Physics for Engineers Part-2, Hafiz Book Center. • Beiser, A.; Concepts of Modern Physics; McGraw-Hill International.

4. Syllabus Chapter-1:Coulomb’s law and Electric field • Concept of charge, • Coulomb’s law, • Electric field, • Electric field strength, • Field due to a point charge, • Field due to a circular ring, • Field due to an electric dipole and • A dipole in an external electric field.

5. Syllabus Chapter-2:Gauss’s law and its Applications • Flux of the electric field, • Gauss’s law and • Applications of Gauss’s law. Deduction of Coulomb’s law, Field due to a line of charge, Electric field due to a charged solid sphere, Field due to a uniformly charged plane (sheet of charge)

6. Syllabus Chapter-3:Electric Potential • Electric potential, • Potential due to a point charge, • Potential due to collection of charge, • Potential due to a dipole and • Electric potential energy.

7. Syllabus Chapter-4:Capacitance and Dielectric • Capacitor, Capacitance, • Calculation of capacitance for various capacitors, • Capacitance for parallel plate, • spherical and • cylindrical capacitor. • Capacitors in series and parallel, • Energy stored in a charged capacitor, • Capacitance with dielectric.

8. Syllabus Chapter-5:Current and Resistance • Concept of electric current, • Current density, • Electron drift velocity, • Ohm’s law, • Resistivity, • Conductivity, • Resistor in series and parallel and • Electromotive force

9. Syllabus Chapter-6:D.C. Circuits • Calculation of current in a single loop, • Multi-loop circuits (Kirchhoff’s laws), • Applications of Kirchhoff’s laws, • RC circuit • Charging of a capacitor • Discharging of a capacitor.

10. Syllabus Chapter-7:The Magnetic Field • Magnetic field, • Magnetic force on a charge, • Magnetic lines of induction and • Magnetic force on a current carrying conductor.

11. Syllabus Chapter-8:Electromagnetic Induction • Laws of electromagnetic induction (Faraday’s law), • Lenz’s law, • Self induced electromotive force (self inductance), • Mutual inductance, • Energy stored in a magnetic field and • Energy density in magnetic field.

12. Syllabus Chapter-9:Magnetic Properties • Magnetization, • Magnetic susceptibility • Magnetic permeability, • Diamagnetism, • Paramagnetism and • Ferromagnetism.

13. Syllabus Chapter-10:Modern Physics • Particle and wave nature of light, • Atomic models (Thomson model, Rutherford model and Bohr model), • Radioactivity and Nuclear Phenomena (Radioactivity, mean life, half life, disintegration constant, radioactive decay law, nuclear fission and fusion), • Basis of Quantum mechanics (Failure of classical mechanics, origin of quantum mechanics, uncertainty principle) and • Relativistic mechanics (Special theory of relativity, length contraction, time dilation)

14. Chapter-1 (Coulomb’s law and Electric field) Electric Charge: Electric charge is an intrinsic characteristics of the fundamental particles whose presence generate static electricity, electric field, static electric energy in an object which can attract very tiny objects like paper and due to their motion electric current, electric and magnetic field are generated. Point Charge: When the size of the charged body is extremely small, then that charged body is called point charge. Those charged bodies are so small compared to the distances between them that they may be considered as mathematical points.

15. Chapter-1 (Coulomb’s law and Electric field) Coulomb’s law: This law states that “At a particular medium, the magnitude of the electric force (attractive or repulsive) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between the charges. The force acts along the straight line joining the two charges”. F A B r q1 q2

16. Chapter-1 (Coulomb’s law and Electric field) Explanation: If two point charges q1 and q2 are separated by distance r and the force between them is F, then according to Coulomb’s law, F ∞ q1q2, when r is constant, and F ∞ 1/r2, when q1 and q2 are constant. When q1 and q2 and r all vary, then F ∞ q1q2/ r2, or, F= Kq1q2/ r2, where K is the proportionality constant and is equal to 1/4πε in S.I. or M.K.S. unit. here, ε (epsilon) is the permittivity of the medium in which the charges are located.

17. Chapter-1 (Coulomb’s law and Electric field) Therefore, Coulomb’s law becomes, F= q1q2/4πε r2. In air or vacuum, Coulomb’s law is F0 = q1q2/4πε0 r2. In S.I. unit F and F0 is expressed in Newton (N), charge in Coulomb (C) and distance in meter (m), so the experimental value of εo is 8.85X10-12 C2N-1m-2 and 1/4πε0 =1/(4π×8.854×10-12) = 9 × 109 Nm2C-2. Thus, F0 = 9 × 109 (q1q2/ r2)

18. Chapter-1 (Coulomb’s law and Electric field) Example-1: Two charge having magnitude 3 × 10-6 C and 8 × 10-6 C are separated at a distance of 2 × 10-3 m. Calculate the force exerted by one another. Solution: Given, q1=3X10-6 C, q2= 8X10-6 C, r= 2 X10-3 m, εo =8.85X10-12 C2/(Nm2). We know, F= q1q2/4πεo r2, or, F=(3X10-6 X 8X10-6 /(4X3.14X8.85X10-12X2 X10-3) = 107.95 N

19. Chapter-1 (Coulomb’s law and Electric field) Example-2: Determine the force between two free electrons spaced 1 A0 (10-10 m) apart. Solution: Given, q1= q2= -1.6 × 10-19, r= 1 A0 = 1 ×10-10 m, εo =8.85X10-12 C2/(Nm2). We know, F= 9 × 109 (q1q2/ r2) = 9 × 109× (-1.6 × 10-19)2/(1 ×10-10)2 = 2.3 × 10-8 N. Self Assesent-1: The distance between the electron and the proton in the hydrogen atom is about 5.3 × 10-11 meter. What is the magnitude of the electric force? [ Ans: 8.1 × 10-8 N].

20. Chapter-1 (Coulomb’s law and Electric field) Electric field: The space around a charged body where its influence is experienced is called the electric field of that charged body. Electric field strength / Electric field intensity/ Electric intensity: It is defined as the amount of electric force acting on a unit charge at a point in the electric field. It is denoted by E. It is a vector quantity. Mathematically, it can be written as, E = F/q0 or, F = q0E. Its S.I. unit is N/C. Example-3: What is the magnitude of the electric field strength E such that an electron placed in the field, would experience an electrical force equal to its weight? Solution: We know, E = F/q0 = mg/ q0 = (9.1 × 10-31 × 9.8) / 1.6 × 10-19 = 5.6 × 10-11 N/C.

21. Chapter-1 (Coulomb’s law and Electric field) Example-4: A pith ball of mass 0.002 kg is charged with 10-4 C. What is the magnitude of electric field needed to keep the ball at rest in gravitational field? Solution: When the weight of the ball is equal to the electric force, the ball will remain at rest. Now, We know, W = mg and F = q0E so, q0E = mg or, E = mg/ q0 = 0.002 × 9.8 / 10-4 = 196 N/C. Self Assesent-2: A plastic ball of charge 3.23 × 10-19 C is kept hanging in an uniform electric field of 2.6 × 104 V/m. If the acceleration due to gravity at that place is 9.8 ms-2, find the mass of the ball. [ Ans: 8.4 × 10-16 kg]

22. Chapter-1 (Coulomb’s law and Electric field) Calculation of electric field strength (E):a) For a point charge: Let a test charge (a charge of exceedingly small magnitude whose presence does not affect other charges) q0 be placed at a distance r from a point charge q. The magnitude of the force acting on q0 is given by Coulomb’s law is The electric field strength at the site of the test charge is given by, The direction of E is on a radial line from q, pointing outward if q is positive and inward if q is negative.

23. Chapter-1 (Coulomb’s law and Electric field) b) For a group of point charge: If there are more than one charge in the field then (i) the electric field strength at the given point should be calculated for each charge as if it were the only charge present and (ii) these separately calculated fields should be added vectorially to find the resultant field strength E at the point. In equation form E = E1 + E2 +E3 + ……= where n = 1, 2, 3, …. The sum is a vector sum, taken over all the charges.

24. Chapter-1 (Coulomb’s law and Electric field) If the charge distribution is not discrete but continuous, then the charge must be divided into infinitesimal elements of charge dq. The field dE due to each element of charge at the point in question is then calculated. The magnitude of dE is given by, where r is the distance of the given point from the charge element dq. The resultant field at the point is then obtained by adding (that is by integrating) the field contributions due to all charge element. Thus, In carrying out the integration,the only suitable method is to resolve the field contributions from the charge elements into components, add the components by integration giving say Ex, Eyand Ezfrom which E can be obtained.

25. Chapter-1 (Coulomb’s law and Electric field) c) For at a point on the axis of a charged circular ring: Let us consider a circular ring of radius ‘a’ carrying a charge ‘q’. We would like to calculate E at a point P on the axis of the ring at a distance x from its center as shown in fig-1. Fig-1: Field at appoint P due to a charged circular ring.

26. Chapter-1 (Coulomb’s law and Electric field) We consider a differential element of the ring of length dl at the top of the ring. Then the charge contained in this element is given by dq= (q/2πa) × dl, where 2πa is the circumference of the ring. This element produces a field dE at the point P. The resultant field E at P is found by integrating the effects of all the elements that make up the ring. Thus, The component of dE perpendicular to the axis is cancelled out by an equal opposite component established by the charge element on the opposite side of the ring. Thus only the component of dE parallel to the axis of the ring contributes to the resultant field. Thus,

27. Chapter-1 (Coulomb’s law and Electric field)

28. Chapter-1 (Coulomb’s law and Electric field)

29. Chapter-1 (Coulomb’s law and Electric field) d) For at an electric dipole: An electric dipole consists of two equal but opposite charges separated by a small distance. Fig-2 shows an electric dipole. The charges are +q and –q and they are separated by a distance 2a. We would like to calculate E at a point P due to the dipole. P is located at a distance x along the perpendicular bisector of the line joining the charges. We assume that x>> a. Let the distance of the point P be y from both the charges and E1 and E2 are the electric field at the point P due to the charge +q and –q respectively. The total field at P due to the dipole is obtained by vector addition of E1 and E2, that is, E = E1 + E2.

30. Chapter-1 (Coulomb’s law and Electric field) Since the charges have the same magnitude and the distance of P from the charges is also same, the magnitude of E1 and E2 are equal, i.e., The direction of E1 and E2 are indicated by arrows in fig-2. The horizontal components of E1 and E2 cancel each other. Hence the vector sum of E1 and E2 points vertically downwards and has the magnitude, E = 2E1cosθ………..(1) From fig-2,

31. Chapter-1 (Coulomb’s law and Electric field) Since x >> a, a2 can be neglected in the denominator. The equation above then reduces to Putting the values of E1 and cosθ in equ. (1), we get,

32. Chapter-1 (Coulomb’s law and Electric field) The product 2aq is called the electric dipole moment and is denoted by ‘p’. It is a vector having the direction along the axis of the dipole from negative to the positive charge. Thus, equ (2) becomes,

33. Chapter-1 (Coulomb’s law and Electric field) The product 2aq is called the electric dipole moment and is denoted by ‘p’. It is a vector having the direction along the axis of the dipole from negative to the positive charge. Thus, equ (2) becomes,

34. Chapter-1 (Coulomb’s law and Electric field) A dipole in an external electric field: Let an electric dipole is placed in a uniform external electric field E; its dipole moment p making an angle θ with this field as shown in fig-3. The two forces (F and -F) acting on the charges are equal and opposite where F = qE. The net force on the dipole is clearly zero. But since the forces do not act along the same line, there is a net torque on the dipole about an axis passing through the center of the dipoles and is given by, τ = magnitude of the force × perpendicular distance between the forces. = F × d sinθ = qE × d sinθ = pE sinθ = p × E. [As, p = d sinθ]

35. Chapter-1 (Coulomb’s law and Electric field)

36. Chapter-1 (Coulomb’s law and Electric field) This torque tends to align the dipole with the electric field. Hence, work (positive or negative) must be done by an external agent to change the orientation of the dipole. This work is stored as potential energy U in the system. The work done to change the orientation of the dipole through a small angle dθ is given by, dW = τdq where is the torque exerted by the agent and does the work. Then the work done to turn the dipole from an initial orientation θ0 to a final orientation θ is given by, where θ0 and θ is the initial and final values of the angle between the dipole axis and external field.

37. Chapter-1 (Coulomb’s law and Electric field) This work is stored as potential energy U. Thus, Since we are interested only the change in potential energy, the reference orientation θ0 can be chosen to have any convenient value. This gives, U = -pE cosθ = - p . E .

38. Chapter-1 (Coulomb’s law and Electric field) Example-5: A molecule of water vapour (H2O) has an electric dipole moment of magnitude 6.2 ×10-30 C.m. The dipole moment arises because the effective center of positive charge does not collide with the effective center of negative charge. (i) How far apart are the effective centers of positive and negative charge in a molecule of H2O? (ii) What is the maximum torque on a molecule of H2O in an electric field of magnitude 1.5 × 104 N/C? (iii) Suppose the dipole moment of a molecule of H2O is initially pointing in a direction opposite to the field. How much work is done by the electric field in rotating the molecule into alignment with the field?

39. Chapter-1 (Coulomb’s law and Electric field) Solution: (i)There are 10 electrons and correspondingly, 10 positive charges in this molecule. Thus, we have, p = iqd = 10e ×d or, d = p/10e = 6.2 ×10-30/(10 × 1.6 × 10-19) = 3.9 × 10-12m = 3.9 pm. (ii) Maximum torque, τ = pE sin 900 = 6.2 ×10-30 × 1.5 × 104 × 1 = 9.3 10-26 Nm. (iii) Work done,

40. Chapter-1 (Coulomb’s law and Electric field) Self Assesent-3: An electric dipole consists of two opposite charges of magnitude 2.0 × 10-6 C separated by a distane 1.0 cm. It is placed in an external electric field of 2.0 × 105 N/C. What maximum torque does the field exert on the dipole? [4 × 10-3 Nm] How much work must an external agent do to turn the dipole from its initial alignment 00 to 900? [Ans: 4 × 10-3 Joules]

41. Chapter-1 (Mechanics) Thank you all