Physics-II 10B11PH211 Electromagnetic Theory Thermodynamics Solid State Physics

1. Physics-II • 10B11PH211 • Electromagnetic Theory • Thermodynamics • Solid State Physics • Quantum Mechanics

2. Choice is based on symmetry of problem To understand the Electromagnetic, we must know basic vector algebra and coordinate systems. So let us start the coordinate systems. COORDINATE SYSTEMS • RECTANGULAR or Cartesian • CYLINDRICAL • SPHERICAL Examples: Sheets - RECTANGULAR Wires/Cables - CYLINDRICAL Spheres - SPHERICAL

3. Orthogonal Coordinate Systems: 1. Cartesian Coordinates z P(x,y,z) Or y Rectangular Coordinates x P (x, y, z) z z P(r, , z) 2. Cylindrical Coordinates P (r, , z) y r x Φ z 3. Spherical Coordinates P(r, θ,) θ r P (r, θ, ) y x Φ

4. Cartesian Coordinates y 6 2 3 7 x Differential quantities: Deduce the area of the lamina

5. Cartesian Coordinates Differential quantities:

6. Cylindrical coordinate system (r,φ,z) Z Z Y r φ X

7. Cylindrical Coordinates: Visualization of Volume element Differential quantities: Limits of integration of r, φ,z are 0<r<∞ , o<φ <2π ,0<z <∞

8. Spherical coordinate system (r,,φ) Radius=r 0<r<∞ • -Zenith angle • 0<θ < ( starts from +Z reaches up to –Z) ,  -Azimuthal Angle 0<φ <2 (starts from +X direction and lies in x-y plane only)

9. Spherical Coordinates P(r, θ, φ) Z dr P r cos θ r dθ θ r dθ Y dφ φ r sinθ dφ r sinθ X

10. Spherical Coordinates Differential quantities:

11. Points to remember

12. Determine a) Areas S1, S2 and S3. b) Volume covered by these surfaces.

13. Ex: Use the spherical coordinate system to find the area of the strip     on the spherical shell of radius a. What results when = 0 and  = ? Ex: Use spherical coordinates to write the differential surface areas dS1 and dS2 and then integrate to obtain the areas of the surfaces marked 1 and 2. Ans : /4,/6

14. Books: • Introduction to Electrodynamics • by D.J. Griffith • Electromagnetics • by Edminister (Schuam series) • Principles of Electromagnetics • by Matthew N. O. Sadiku • Engineering Electromagnetic • by W H Hayt & J A Buck

15. Electricity and magnetism are different aspects of electromagnetism

16. Electromagnetism A fundamental interactionbetween the magnetic field and the presence and motion of an electric charge A “Field” is any physical quantity which takes on different values at different points in space.

17. Assignment 1: Basics of fields Gradient Divergence and Curl.

18. Vector Analysis Given A = ax + ay , B=ax+2ay, and C=2ay+az, find (A x B) x C and compare it with A x (B x C). find АВ x С and compare it with A x В.С Scalar and Vector Fields voltage, current, energy, temperature velocity, momentum, acceleration and force

19. Gradient, Divergence and Curl The Del Operator  • Gradient of a scalar function is a vector quantity. • Divergenceof a vector is a scalar quantity. • Curl of a vector is a vector quantity.

20. Operator in Cartesian Coordinate System gradT: points the direction of maximum increase of the function T. Divergence: Curl: where

21. Operator in Cylindrical Coordinate System Gradient: Divergence: Curl:

22. Operator in Spherical Coordinate System Gradient : Divergence: Curl:

23. Fundamental theorem for divergence and curl • Gauss divergence theorem: • Stokes curl theorem Conversion of volume integral to surface integral and vice verse. Open S Closed L Conversion of surface integral to line integral and vice verse.

24. Coulomb’s Law Like charges repel, unlike charges attract • Exercise: A charge Q1= 1nC is located at the origin in free space and another charge Q at (2,0,0). If the X-component of the electric field at (3,1,1) is zero, calculate the value of Q. Is the Y component zero at (3,1,1)? • Calculate E due to • Dipole, • Rod (line charge), Ring (Line charge), • Circular plate (surface charge), Square sheet, • Sphere or Cylinder (Volume charge density)

25. Electric Flux The number of electric field lines through a surface A  E  E=EA, • Conclusion: • The total flux depends on • strength of the field, • the size of the surface area it passes through, • and on how the area is oriented with respect to the field.

26. Gauss's Law • The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity (eo.). • eo = the permittivity of free space 8.854x10-12 C2/(N m2) da E +q Integral Form Differential Form where

27. Electric lines of flux and Derivation of Gauss’ Law using Coulombs law • Consider a sphere drawn around a positive point charge. Evaluate the net flux through the closed surface. Net Flux For a Point charge dA Gauss’ Law

28. +q C A da E da D B Where dΩ is solid angle Asmnt 2: Proof of the Gauss’s law for the charge inside da E +q and outside the Gaussian surface

29. -ve flux +ve Flux

31. Differential form of Gauss Law: Proof: Gauss Law Gauss divergence theorem: or Note: Gauss law is also known as Maxwell’s first equation.

32. Quiz: • (a) 1.12 x 105 V/m, (b) 1.86 x 104 V/m, • -10-4 V/m

33. Remember • Electric Flux (φ=EA=EA cosθ= q/ε0) • Flux is independent of the distance of a point from position of charge. • Electric Flux is the number of electric field lines crossing per unit area. • For max; EA θ=0o. • For min; EA=0 θ=90o • Gauss Law is Maxwell’s first equation. • Conductors in electric Field; • E inside=0 as net charge is distributed over the surface of a conductor. • At the surface of conductor; perpendicular E only, no parallel component of E.

34. Applications of Gauss law -Spherical and Cylindrical symmetries

35. Applications of Gauss law(Spherical distribution systems) • Conducting Sphere of charge ‘q’ and radius ‘R’: • E at an external point: Eo • E at the surface: Es • E at an internal point: Ei • Nonconducting Sphere • E at an external point: Eo • E at the surface: Es • E at an internal point: Ei

36. (Spherical systems: Conducting Sphere) Gaussian surface • Conducting Sphere of charge ‘q’ and radius ‘R’: • E at an external point: Eo r>R • E at the surface: Es r=R • E at an internal point: Ei r<R Case-I: E at an external point; Net electric fux through ‘P’: R P r S1 The Electric field strength at any point outside a spherical charge distribution is the same as through the whole charge were concentrated at the centre.

37. Gaussian surface r=R Gaussian surface R r (Spherical systems: Conducting Sphere) Case-II: E at the Surface; Case-III: E at an internal point;

38. Gaussian surface R P r S1 (Spherical systems: Nonconducting Sphere) • Nonconducting Sphere of charge ‘q’ and radius ‘R’: • E at an external point: Eo r>R • E at the surface: Es r=R • E at an internal point: Ei r<R Case-I: E at an external point; Net electric flux through ‘P’:

39. Gaussian surface r=R Gaussian surface R r (Spherical systems: Nonconducting Sphere) Case-II: E at the Surface; Case-II: E at an internal point;

40. R R P P r r E E r=0 r=R r=R r r (Spherical systems: Conducting Sphere) (Spherical systems: Nonconducting Sphere) r=0

41. Numerical R Ei Eo Es Solid sphere

42. Problems: Spherical Symmetry 2. Non conducting spherical shell of inner radius r1, outer radius r2 and charge density ρ= k/r2 , where k is a constant. Also determine Max E at any value of r . 1. Non conducting solid sphere of radius R and charge density ρ=k/r2, Where k is a constant. Determine Electric field everywhere by using Gauss Law for the following; E3 R E1 E5 Ei Eo E2 E4 Es Spherical shell Solid sphere

43. Applications of Gauss law(Cylindrical distribution systems) • Conducting long Cylinder of charge ‘q’ and radius ‘R’: • E at an external point: Eo • E at the surface: Es • E at an internal point: Ei • Nonconducting long Cylinder • E at an external point: Eo • E at the surface: Es • E at an internal point: Ei

44. Cylindrical distribution systems: Conducting Cylinder • Conducting long Cylinder of charge ‘q’ and radius ‘R’ : • E at an external point: Eo r>R • E at the surface: Es r=R • E at an internal point: Ei r<R Gaussian surface Case-I: E at an external point; Net electric flux through ‘P’: E l R O P r

45. Case-II: E at the Surface; Case-III: E at an internal point; E l R O P E Es Eo Ei=0 r=0 r=R r

46. E l R O P r Cylindrical distribution systems: Nonconducting Cylinder • Nonconducting Cylinder of radius ‘R’, height ‘l’ and charge density ‘ρ’: • E at an external point: Eo r>R • E at the surface: Es r=R • E at an internal point: Ei r<R Gaussian surface Case-I: E at an external point; Net electric flux through ‘P’:

47. Case-II: E at the Surface; Case-III: E at an internal point; E l R O P E Es Eo Ei r=0 r=R r

48. For infinite long line charge density ‘λ’

49. Numerical: Non conducting Cylindrical shell (r1, r2 and height h) having volume charge density ρ=k/r. Determine E everywhere. Case-I: E at an external point r0; E0 Gaussian surface E l O P r1 r2 r0

50. Applications of Gauss law(Infinitely long sheet of Charge) The plane is infinitely large, any point can be treated as the center point of the plane, so E at that point must be normal to the surface and must have the same magnitude at all points equidistant from the plane.