 # Sem5 em skee 2523 electromagnetic field theory lect 01-24

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Information about Sem5 em skee 2523 electromagnetic field theory lect 01-24

Published on January 18, 2016

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1. Copyright of M.H. Ibrahim, 2008 WELCOME TO THE WORLD OF ELECTROMAGNETICS Mohd Haniff Ibrahim Room Number: P03-510 Email: hanif@fke.utm.my H/P: 013-7146367 Notes are downloadable at mail.fke.utm.my/~hanif/teaching/SEE2523.htm

2. Copyright of M.H. Ibrahim, 2008 • Course Outline • Evaluation Method and Important Dates • Contents Electromagnetics Field Theory: SEE 2523 • Reference Books

3. Copyright of M.H. Ibrahim, 2008 LECTURE #1 Topics to be covered: Revision on vectors Motivation: To enable the students to refresh their knowledge on vectors and its operation Reference: Hayt & Buck, page 1-14

4. Copyright of M.H. Ibrahim, 2008 Why Vectors? AB A B

5. Copyright of M.H. Ibrahim, 2008 Vector Operations • Vector Components in Rectangular Coordinates • The Dot Product • The Cross Product

6. Copyright of M.H. Ibrahim, 2008 Vector Components in Rectangular Coordinates x y z • A (2,4,5) • B (4,5,6) =AB =BA =AB =BA =rˆ =rˆ

7. Copyright of M.H. Ibrahim, 2008 Example: D1.1 (Hayt pg.8) Given points M(-1,2,1), N(3,-3,0) and P(-2,-3,-4). Find: a) b) c) d) e) MN MPMN + MP rˆ M NM 32 −

8. Copyright of M.H. Ibrahim, 2008 The Dot Product • The first type of vector multiplication • The yield is a scalar component •Definition: ABcosBABA θ=• • In rectangular coordinates: •In general: zˆ,yˆ,xˆ zzyyxx BABABABA ++=•

9. Copyright of M.H. Ibrahim, 2008 Example: D1.3 (Hayt pg. 12) The three vertices of a triangle are located at A (6,-1,2), B(-2,3,-4) and C(-3,1,5). Find: a) b) c) The angle at vertex A AB AC BACθ

10. Copyright of M.H. Ibrahim, 2008 The Cross Product • The second type of vector multiplication • The yield is a vector component • By definition: nˆsinBABA ABθ=× • The yield vector is normal to both A and B • In rectangular coordinates: • In general: zˆ,yˆ,xˆ zyx zyx BBB AAA zˆyˆxˆ BA =×

11. Copyright of M.H. Ibrahim, 2008 Example: D1.4 (Hayt pg.14) The three vertices of a triangle are located at A(6,-1,2), B(-2,3,-4) and C(-3,1,5). Find: a) c) A unit vector perpendicular to the plane in which the triangle is located. ACAB×

12. Copyright of M.H. Ibrahim, 2008 Next Lecture Please have a preliminary observation on the following topics: a) Coordinate systems: (i) Rectangular coordinate (ii) Cylindrical coordinate (iii) Spherical coordinate Please refer to Hayt and Buck, page 14-22

13. LECTURE #2 Topics to be covered: Coordinate Systems Motivation: To familiarize the students with common coordinate systems used in solving electromagnetic problems Reference: Hayt & Buck, page 14-22

14. Why Coordinate Systems??? • Electromagnetic problems can be in any forms. e.g. Find electric/magnetic field in the wire and its surrounding. Find electric/magnetic field induced by the radio antenna Find electric/magnetic field induced by the telco tower Need proper coordinate systems to analyze the electromagnetic problems

15. Important parameters of coordinate systems • Differential line length, (scalar and vector) Application example: Obtain the electric potential (Chapter 4) dl • Differential area, (scalar and vector) Application example: Obtain the electric field using Gauss’s Law (Chapter 3) ds • Differential volume, (scalar) Application example: Obtain the electric field using Coulomb’s Law (Chapter 2) dv ∫= dl.EV enQdsD =∫ . ∫= 2 4 r dv E o v πε ρ

16. Rectangular Coordinate System (x, y, z) x y z A B From A to B: = Surface C: ds = Surface D: ds = Surface E: ds = dx dz dy surface C surface D dl surface E xˆ yˆ zˆ

17. Cylindrical Coordinate System (r, , z)φ y dr rdφ dz z x zˆ rˆ φˆ H (r, , z)φ r φ z

18. Spherical Coordinate System (r, θ, )φ y z x φˆ H (r, θ, )φ φ θ r rˆ θˆ rdθ dr φrsinθd

19. Next Lecture Please have a preliminary observation on the following topics: (i) Type of electrostatic charges (ii) Electric field intensity (iii) Coulomb’s Law Please refer to Hayt and Buck, page 26-48

20. Lecture #3 Topics to be covered (i) Type of electrostatic charges Reference: Hayt and Buck, page 26-48 Motivation: To familiarize the students with the type of sources for electrostatic field

21. Electric charges • Smallest magnitude of charge, q = 1.602 ×10-19 (Coulomb) • Charge can be positive or negative • In electromagnetic problems: electric charges can be of type: (i) Point charge (ii) Line charge (iii) Surface charge (iv) Volume charge

22. Point charge •Simplest electric charge •Normally labeled as Q. Unit of Coulomb (C) Q Line charge • Accumulated point charges along a thin line • Accumulated point charges line charge density, ρl (C/m) ρl (C/m) dldQ lρ= A B Total charge along AB ??? dl

23. Surface charge • Accumulated point charges on a surface •Accumulated point charges surface charge density, ρs (C/m2) ρs (C/m2) dsdQ sρ= ds Total charge on the surface ??? Area A

24. Volume charge • Accumulated point charges in a volume •Accumulated point charges volume charge density, ρv (C/m2) ρv (C/m2) dvdQ vρ= dv Total charge in the volume ??? Volume V

25. Example: D2.4 (b) (Hayt pg.37) Calculate the total charge for the following volume: φρπφ 6.0sin;42,0,1.00 22 zrzr v =≤≤≤≤≤≤ Example: Problem 4.5 (Sadiku pg. 155) Determine the total charge for the followings: (i) On line 0<x<5m if ρl =12x2 mC/m (ii) On the cylinder r=3, 0<z<4m if ρs=rz2 nC/m2 (iii) Within the sphere r=4m if ρv=10/rsinθ C/m3

26. Example: Problem 2.1a (Skitek pg. 51) Find the charge within a sphere of radius 0.03m when the charge density is given by: ( )3 223 / sin cos102 mC r v θ φ ρ − × =

27. Next Lecture Please have a preliminary observation on the following topics: (i) Electric field intensity (ii) Coulomb’s Law Please refer to Hayt and Buck, page 26-48

28. Lecture #4 Topics to be covered: (i)Electric Field Intensity (ii) Coulomb’s Law Reference: Hayt and Buck, page 26-48 Motivation: To understand the concept of electric field intensity and its method of calculation

29. Electric (Electrostatic) Field Intensity • Field generated by the static charges mentioned in the last lecture. • Symbol of . Unit of volt/meter, (V/m). • The first law to calculate the is Coulomb’s Law. E Electro+ static Electric field Static sources E

30. Coulomb’s Law • The law was developed by French physicist, Charles Augustin de Coulomb in 1780s. “ The magnitude of electrostatic force between two point electric charges is directly proportional to the product of the magnitudes of each charge and inversely proportional to the square of the distance between the charges” 2 21 r QQ F ∝

31. Coulomb’s Law Q1 Q2 2F1F 12 ˆr21 ˆr 122 21 2 ˆr r QQ kF =e.g. Coulomb’s experiments: Constant, o k πε4 1 = εo : Permittivity in free space/Dielectric constant of vacuum. ≈ 8.854 × 10-12 Farad/meter (F/m) (Newton)

32. Coulomb’s Law With regard to Lorentz Force Law (Chapter 9): EQF = Hence, 122 1 1 ˆ 4 r r Q E oπε = 212 2 2 ˆ 4 r r Q E oπε = Q1 12 ˆr 1E A Q2 21 ˆr 2E B Field at A; Field at B; V/m V/m r

33. Coulomb’s Law for Point Charge Electric field intensity incur by any point charge, Q r r Q E o ˆ 4 2 πε = Q rˆ E B r Unit of voltage/meter (V/m)

34. Example: Problem 4.1 (Sadiku) Point charges, Q1=5µC and Q2=-4µC are placed at (3,2,1) and (-4,0,6) respectively. Determine the force on Q1. Example: Problem D2.2 (Hayt) A charge of -0.3µC is located at A(25,-30,15) (in cm) and a second charge of 0.5µC is at B(- 10,8,12) cm. Find electric field intensity at (a) origin, (b) P(15,20,50) cm.

35. Example: Problem 4.2 (Sadiku) Point charges Q1 and Q2 are respectively located at (4,0,- 3) and (2,0,1). If Q2 = 4 nC, find Q1 such that: a) The electric field at (5,0,6) has no z component. b) The force on a test charge at (5,0,6) has no x component. Example: Problem 4.3 (Sadiku) Five identical 15 µC point charges are located at the center and corners of a square defined by -1<x,y<1, z=0. (a) Find the force on 10µC point charge at (0,0,2) (b)Calculate the electric field intensity at (0,0,2)

36. Next Lecture Please have a preliminary observation on the following topics: (i) Electric field intensity of continuous charge distribution. Please refer to Hayt and Buck, page 26-48

37. Lecture #5 Topics to be covered: Electric Field Intensity of Continuous Charge Distribution Reference: Hayt and Buck, page 34-48 Motivation: To calculate the electric field intensity based on continuous charge distribution

38. Recall (Coulomb’s Law) Previously, Coulomb’s Law for point charge: At A; R ao a r Q E ˆ 4 2 πε = Q A Raˆ ra What if Q is a line charge, surface charge or volume charge ??? (V/m)

39. Electric Field Intensity-Line Charge Distribution What is the field at A as incurred by the line charge? A ( )mCl /ρ l Step 1: Identify small line length, dl with element of charge, dQ dldQ lρ=dl R o l a R dl Ed ˆ 4 2 πε ρ = Step 2: Identify the electric field intensity, dE as contributed by dQ (based on Coulomb’s Law). ∫= z o total EdE Step 3: Integrate dE, over the length of line charge, z, to obtain total electric field.

40. Electric Field Intensity-Line Charge Distribution For line charge, normally cylindrical coordinate system was utilized. From previous, A 'dzdQ lρ=dz’ ( )zr ,,φ lρ r x y z φ z E field at A ??? Step 1: ??? Step 2: ??? Step 3: ??? z’ a b

41. Electric Field Intensity-Line Charge Distribution Step 1: 'dzdQ lρ= Step 2: ( )( ) ( ) ( )( )2 1 2'2 ' 2'2 ' ˆˆ 4 zzr zzzrr zzr dz Ed o l −+ −− −+ = πε ρ Step 3: Using ; ( ) ( ) mV r z r r E o l /coscos ˆ sinsin ˆ 4 1212 ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ −++= αααα πε ρ ( ) ( )2 1 2222 3 22 xcc x xc dx + = + ∫ ( ) ( )2 1 222 3 22 1 xcxc xdx + − = + ∫and

42. Electric Field Intensity-Line Charge Distribution For a ∞ and b ∞ (line charge of infinite length) ( )mVr r E o l /ˆ 2πε ρ =

43. Problem 2.19 - Hayt A uniform line charge of 2µ C/m is located on the z axis. Find electric field in Cartesian coordinates at P(1,2,3) if the charge extends from (a) z = -∞ to z= ∞; (b) z=-4 to z=4 Example: Problem 2.5.1 (Skitek) Find the electric field intensity on the axis of a circular ring of uniform charge, ρl and radius a. Let the axis of the ring be along the z axis.

44. Example: Problem 2.5.2 (Skitek) Find the electric field at P(0,0,1.5) due to two infinite and parallel line charges of uniform ρl. Let one ρl=10-6 C/m line be located at y=2 m and other ρl =-10-6 C/m line be located at y=-2m, with both lines parallel to the z axis and in the z=0 plane. Problem 2.26: Hayt A uniform line charge density of 5 nC/m is at y=0, z =2 in free space, while -5 nC/m is located at y=0, z=-2m. Find the electric field at the origin.

45. Example: Problem 2.5.2 (Skitek) Two identical uniform line charges, with ρl =75 nC/m are located in free space at x=0, y=±0.4m. What force perunit length does each line exert on the other. Example: D2.5 (Hayt page 43) Infinite uniform line charges of 5 nC/m lie along the (positive and negative) x and y axes in free space. Find electric field intensity at (a) (0,0,4) ; (b) (0,3,4)

46. Electric Field Intensity-Surface Charge Distribution ρs (C/m2) B What is the electric field at B ??? As in the case of line charge distribution; dS Step 1: Identify small surface, dS having charge element of dQ dSdQ sρ= Step 2: Identify electric field intensity, dE as contributed by dQ R o S a R dS Ed ˆ 4 2 πε ρ = Step 3: Integrate dE over the total surface, to obtain total electric field ∫= surface EdE

47. Electric Field Intensity-Surface Charge Distribution )/(ˆ 4 2 mVa R dS E R o s πε ρ = For a finite surface area, electric field incurred at any desired location: What if the surface is having infinite area ??? Consider a circular surface charge having density of ρS (C/m2) with radius a, centered at origin and lying on the z=0 plane. To solve

48. Electric Field Intensity-Surface Charge Distribution x y z a C (0,0,z) Find electric field at C ??? dS ( )φρρ rdrddSdQ sS == ( ) ( ) ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ + +− = 2 3 224 zr zˆzrˆrrdrd Ed o s πε φρ Using cylindrical coordinate; ∞→aConsider; ???E

49. Electric Field Intensity- Infinite Surface Charge Distribution For surface charge distribution of infinite area: )m/V(nˆE o s ε ρ 2 =

50. Example:D2.6 (Hayt, page 45) Three infinite uniform sheets of charge are located in free space as follows: 3 nC/m2 at z =-4, 6 nC/m2 at z=-1 and -8 nC/m2 at z=-4. Find electric field intensity at the point (a) (2,5,-5); (b) (4,2,-3); (c) (-1,-5,2) and (d) (-2,4,5) Example: (Exercise from Skitek) Find the electric field along the z axis, as incurred by the annular ring with surface charge density of ρs, having inner radius of a and outer radius of b.

51. Example:Problem 2.6-1 (Skitek) A semi-infinite sheet of charge density, ρS is described by -∞< x <0, -∞< y <∞, in the z=0 plane. Calculate the component of electric field normal to the sheet at a distance a, directly above the edge at x=0.

52. Electric Field Intensity-Volume Charge Distribution As previously noted: )/(ˆ 4 )/(ˆ 4 22 mVa R dV mVa R dQ E R o v R o πε ρ πε == However, we are not going to cover on detail….. However, electric field due to volume charge distribution will be elaborated in Chapter 3. End of Chapter 2…..

53. Next Lecture Please have a preliminary observation on the following topics: (i) Electric Flux (ii) Electric Flux Density (iii) Gauss’s Law Please refer to Hayt and Buck, page 51-67

54. Lecture #6 Topics to be covered: (i) Electric Flux/ Flux Density (ii) Gauss’s Law Reference: Hayt and Buck, page 51-67 Motivation: To understand the concept of electric flux/ flux density and to apply the Gauss’s Law in electric field calculation

55. Electric Flux Sketched field lines (imaginary lines) pointed outwards from the electric charge. Symbol of ΨE (psi). A scalar component. Flux is in the same direction as the electric field Q Electric flux about a point charge Relation: ΨE=Q Unit of electric flux: Coulomb E

56. Electric Flux -∞ ∞ Electric flux about an infinite line charge ∞ ∞ ∞ Electric flux about an infinite surface charge ∞

57. Electric Flux Density Symbol of : vector quantity Unit of Coulomb/m2 D Total electric flux, ΨSurface with area A ( )2 m/C A D Ψ =∴

58. Electric Flux Density Consider a sphere with radius r, and we locate a point charge, Q at the centre of the sphere: Q Flux density at radius r: ( )2 22 44 m/Crˆ r Q rˆ r D E ππ ψ == Electric field intensity at radius r: ( )m/Vrˆ r Q E o 2 4πε = Hence: ED oε=

59. Electric Flux Density Consider a small portion of sphere (radius of r ) surface, dS and flux (from point charge, Q, at the centre) pass through the surface. Small element of flux that pass through dS : ∫= dS,surface E dS.Ddψ Total flux that pass through the sphere: ∫ == Ssurface EE Qd , ψψ ∫ == Ssurface enclosedQdSD , .Gauss’s Law dS Flux density, D Q

60. Gauss’s Law The law was developed by Carl Friedrich Gauss, a German mathematician in the early 19th century. The law states that: “ Total electric flux passing through any closed imaginary surface, enclosing the charge Q(Coulomb), is equal to Q (Coulomb)” How to apply Gauss’s Law in calculating electric field??? Mathematically; ∫ = surface enQdSD.

61. Gauss’s Law-Point Charge Q Step 1: Draw a close imaginary surface with radius of r, enclosing the point charge Q. Spherical surface can be used. Imaginary surface en surface QdSD =∫ . The aim is to find an electric field at a distance r from the point charge. r Step 2: Apply the Gauss’s Law, and obtain the flux density, D Step 3: Use the relation of and obtain . The obtained electric field is the electric field at a distance r from the point charge. ED oε= E

62. Gauss’s Law- Line Charge Strictly for special case of line charge – Infinite length of line charge. The aim is to find an electric field at a distance r from the line charge. -∞ ∞ r Step 1: Draw a close imaginary surface with radius of r, enclosing the line charge ρl. Cylindrical surface can be used. Step 2: Apply the Gauss’s Law, and obtain the flux density, D Step 3: Use the relation of and obtain . The obtained electric field is the electric field at a distance r from the line charge. ED oε= E L ( )mCl /ρ

63. Gauss’s Law- Surface Charge Strictly for special case of surface charge – Infinite area of surface charge. The aim is to find an electric field above and below the surface charge. a Step 1: Draw a close imaginary surface enclosing the surface charge ρS. Cylindrical surface with radius r can be used. ρS (C/m2) Step 2: Apply the Gauss’s Law, and obtain the flux density, D Step 3: Use the relation of and obtain . The obtained electric field is the electric field above the surface charge. ED oε= E r

64. Exercise #1: Skitek Through the use of Gauss’s Law, find the expression for , inside and outside of a sphere of ro radius and of uniform ρV (C/m3) distribution. D E Exercise #2: Skitek Through the use of Gauss’s Law, find the expression for , inside and outside an infinite cylinder of ro radius and of uniform ρV (C/m3) distribution. D E

65. Example 1 Let (nC/m2) in free space. (a) Find at r = 0.2 m. (b)Find the total charge within the sphere, r = 0.2 m. (c) Find the total electric flux leaving the sphere, r = 0.3 m. r r D ˆ 3 = E Example 2 The cylindrical surfaces, r = 1, 2 and 3 cm carry uniform surface charge densities of 20, -8 and 5 nC/m2, respectively. (a) How much electric flux passes through the closed surface, r=5 cm, 0<z<1m (b)Find at P(1cm, 2 cm, 3 cm)D

66. Example 3 The spherical surfaces r = 1, 2 and 3 cm carry surface charge density of 20, -9 and 2 nC/m2, respectively. (a) How much electric flux leaves the spherical surface r =5 cm. (b)Find at P (1cm, -1 cm, 2 cm).D Example 4: (Hayt) Spherical surfaces at r = 2, 4 and 6 m carry uniform surface charge densities of 20 nC/m2, -4 nC/m2 and ρso, respectively. (a) Find at r = 1, 3 and 5 m. (b)Determine ρso such that = 0 at r = 7m. D D

67. Example 5: D 3.5 (Hayt) A point charge of 0.25 µC is located at r =0, and uniform surface charge densities are located as follows: 2 mC/m2 at r =1 cm and -0.6 mC/m2 at r = 1.8 cm. Calculate at:D (a) r = 0.5 cm (b)r = 1.5 cm (c) r = 2.5 cm (d)What uniform surface charge density should be established at r = 3 cm to cause = 0 at r = 3.5 cm.D

68. Example 6: (Hayt) A uniform volume charge density of 80 µC/m3 is present throughout the region 8 mm < r < 10 mm. Let ρV = 0 for 0 < r < 8 mm. (a) Find the total charge inside the spherical surface, r = 10mm. (b)Find Dr at r = 10 mm. (c) If there is no charge for r > 10 mm, find Dr at r = 20 mm.

69. Next Lecture Please have a preliminary observation on the following topics: (i) Divergence (ii) Divergence Theorem Please refer to Hayt and Buck, page 67-75

70. Lecture #7 Topics to be covered: (i) Divergence (ii) Divergence Theorem Reference: Hayt and Buck, page 67-75 Motivation: To understand the divergence concept and the application of divergence theorem

71. Divergence Consider the Gauss’s Law; en surface QdSD =•∫ Graphically; ∫ • surface dSD Qen What if the closed surface volume shrink to zero ? Let the charge be located at (xo, yo, zo) (xo, yo, zo)

72. Divergence Consider the closed surface volume shrink to zero. Take ∆v as the closed surface volume. Divided both sides of Gauss’s Law with ∆v and take the limit as ∆v 0. ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∆ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ∆ • →∆→∆ ∫ v Q v dSD en v surface v 00 limlim Mathematically; Charge density, ρV at point (xo, yo, zo)Divergence of orD D•∇

73. Divergence Using a divergence concept, the Gauss’s Law can be stated as: VD ρ=•∇ Physically; 1st Maxwell’s equation for static field “ It relates the rate of change of the component to the charge density, ρV at a point” D Positive divergence “flux source” Negative divergence “sink of flux” +ρV D -ρV D VD ρ=•∇ VD ρ−=•∇

74. Divergence Divergence operator, is a vector quantity:( )•∇ Heavily depending upon coordinate system. Rectangular: Cylindrical: Spherical: z D y D x D D zyx ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ ( ) z DD r rD rr D z r ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ φ φ11 ( ) ( ) φθ θ θθ φ θ ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ D r D r Dr rr D r sin 1 sin sin 11 2 2

75. Example 1 Find the volume charge density that is associated with each of the following fields: ( )222 /ˆˆˆ mCzzyyxxxyD ++=(a) (b) ( )222222 /ˆsinˆcossinˆsin mCzzrrzrrzD φφφφφ ++= Example 2 If , find:( )23 /ˆ2ˆ2ˆ4 mCzyyzxxD −−= (a) div D (b) ρV at P(x, y, z) (c) the total charge lying within the region -1<x,y,z<1 (d)the total charge lying within that region without finding ρV first.

76. Example 3 Evaluate above an infinite sheet of uniform charge distribution, ρS (C/m2) located at the z = 0 plane. D•∇ Example 4 Evaluate at a distance r from an infinite length of line charge of uniform charge distribution, ρl (C/m). The line charge is located at z-axis. D•∇

77. x y z b Example 5 a ρV (C/m3) (a) Find the expression of at r<a, a<r<b and r>b (b)Evaluate at r<a, a<r<b and r>b D D•∇

78. Divergence Theorem A theorem used to relate the surface integral to a volume integral involving the same vector. ρV ∫ • surface dSD ∫= volume Ven dVQ ρ Consider a volume charge and applying the Gauss Law. D•∇ Divergence Theorem∫∫ •∇=• volumesurface dVDdSD⇒ ∫∫ ==•⇒ volume Ven surface dVQdSD ρ Gauss’s Law:

79. Summary of Chapter 3 Students are expected to be well-versed in the followings: (d) Extension of Gauss’s Law that lead to the concept of divergence (1st Maxwell’s equation for static field) (a) Concept of electric flux and flux density (b) Relation between flux density and electric field (c) Application of Gauss’s Law in calculating electric field induced by special cases of charge distribution The end of chapter 3

80. Next Lecture Please have a preliminary observation on the following topics: (i) Potential Difference (ii) Absolute Potential Please refer to Hayt and Buck, page 87-105

81. Lecture #8 Topics to be covered: (i) Potential difference (ii) Absolute Potential Reference: Hayt and Buck, page 80-105 Motivation: To enable the students to understand the concept of potential difference and absolute potential induced by point charge and continuous charge system

82. Work in Electric Field E Q Point B Point A ( )JouledlFdW •−= Small work done in moving Q along the length dl: dl Work done in moving Q from point B to point A:∴ ( )JouledlEQW A B ∫ •−=

83. Work in Electric Field Vector of length; (Discussed in our first and second lecture)dl Depending on the coordinate used: ( ) ( ) ( )Sphericaldrrdrdrdl lCylindricazdzrdrdrdl Cartesianzdzydyxdxdl φφθθθ φφ ˆsinˆˆ ˆˆˆ ˆˆˆ ++= ++= ++=

84. Example (2,0,0) x y z (0,4,0) (0,0,4) Path 1 Path 2 Obtain the work done in moving 2 C charge in existence of (i) From (2,0,0) to (0,4,0) along path 1 (ii) From (2,0,0) to (0,4,0) along path 2 (iii) Round trip movement at (2,0,0) along path 1 and path 2. zzyyxxE ˆˆˆ ++=

85. Work in Electric Field From previous example: (i) = (ii) The work done does not depend on the path taken. From (iii); 0=•−= ∫ dlE Q W 2nd Maxwell’s equation for static field

86. Potential Difference/Potential Work done in moving a unit of charge from point B to point A: = Q W ∫ •− A B dlE (volt) Absolute potential (potential): (if point B is in infinity) From above: ∫∞ ∞ •−=−= A A dlEVV Q W Potential difference between point A and B BA A B AB VVdlEV −=•−= ∫

87. Potential Difference (Point Charge) Q rA rB A B Let us move a unit of positive charge, 1C from point B to point A: The potential difference between point B and point A ∫ •−= A B AB dlEV ( ) rˆdrdl;m/Vrˆ r Q E o == 2 4πε From: ???VVV BAAB =−=∴

88. Q rA ∞ A ∞ ( ) ∫∞ ∞ •−= A A dlEV Absolute Potential/Potential (Point Charge) What if B=∞ ???VVA ==∴

89. Potential (Charge System) Q4 What is total absolute potential/potential at A ??? A Q1 Q2 Q3 R1 R2 R3 R4 1 1 4 R Q oπε 2 2 4 R Q oπε 3 3 4 R Q oπε 4 4 4 R Q oπε ∑= =+++=∴ 4 14 4 3 3 2 2 1 1 44444 i io i oooo A R Q R Q R Q R Q R Q V πεπεπεπεπε

90. Potential (Charge System) What if we have a system of charges; i.e. line charge, surface charge, volume charge ??? A Surface charge; ρS (C/m2) From previous slide: ∑= = 4 1 4i io i A R Q V πε ∫=∴ surface o A R dQ V πε4 What is the potential at A due to the surface charge ?? Continuous charge system

91. Potential (Charge System) In general; potential at a point (let point A) due to continuous system of charges: Line Charge: Surface Charge: Volume Charge: ∫= surface o S A R dS V πε ρ 4 ∫= line o l A R dl V πε ρ 4 ∫= volume o v A R dv V πε ρ 4

92. Example 1: Skitek Obtain the expression for the absolute potential along the z-axis of a ring of uniform ρl (C/m) and whose radius is ro. The ring is located at z = 0. ( ) zˆ zr zr E oo ol 2 3 22 2 + = ε ρ Given; electric field at any point on z axis

93. Example 2: Skitek Obtain the absolute potential inside and outside a thin shell, centered at the origin, of uniform ρS (C/m2) and of radius ro. Obtain the potential at: ( ) ( ) ( ) o o o rriii rrii r ri 2 2 = = =

94. Next Lecture Please have a preliminary observation on the following topics: (i) Potential Gradient (ii) Energy Density in the Electrostatic Field Please refer to Hayt and Buck, page 95-110

95. Lecture #9 Topics to be covered: (i) Potential Gradient (ii) Energy Density in Electrostatic Field Reference: Hayt and Buck, page 95-110 Motivation: To enable the students to calculate the electric field using the potential value and to understand the concept of energy density in electrostatic field

96. Recall … Charge Distribution ( )vSlQ ρρρ ,,, Coulomb’s Law Electric Field Intensity E Gauss’s Law Electric Potential V ∫= R dQ V oπε4 Potential Gradient

97. Potential Gradient Consider… R Q V o A πε4 = Q Point A R Alternatively, electric field at point A can be obtained using gradient of potential at point A AA VE ∇−= gradient or grad Gradient operator: z z V y y V x x V V ˆˆˆ ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ z z VV r r r V V ˆˆ1 ˆ ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ φ φ φ φθ θ θ ˆ sin 1ˆ1 ˆ ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ V r V r r r V V (Cartesian) (Cylindrical) (Spherical)

98. Example 1 Please refer to Example 1 in Lecture #8. Given the potential at any point on z-axis is: ( )2 1 22 2 zr r V oo ol + = ε ρ Obtain the electric field term at any point on z-axis and prove that the answer is similar to the given term in Example 1 (Lecture #8).

99. Example 2 Using Example 2 in Lecture #8, obtain the electric field terms at the following location (radius) based on the previously obtained potential terms: ( ) ( ) ( ) o o o rriii rrii r ri 2 2 = = = Verify your results using the common law.

100. Energy Density in Electrostatic Field ρl (C/m) ρS (C/m2) ρv (C/m3) What is the total energy conserved by the charge distributions?

101. Energy Density in Electrostatic Field For simplicity, consider… We have a system of four point charges. The work done by an external source in positioning (locating) the charges What is the energy (potential energy) present in such system of charges ??? Q4 Q3 Q2 Q1

102. Energy Density in Electrostatic Field How to calculate the work done by external source in positioning the charges ??? Step 1: Assume that the system is empty (no charge exist) Step 2: Bring point charge, Q1 from infinity into the system. No work required !!!! Q1 01 == WW

103. Energy Density in Electrostatic Field Step 3: Bring point charge, Q2 from infinity into the system. Requires work !!!! Q1 Q2 R21 21 1 2 212 21 4 0 0 R Q Q VQ WWW oπε += += += Step 4: Bring point charge, Q3 from infinity into the system. Requires work !!!! Q1 Q2 R21 Q3 R31 R32

104. ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +++= +++= ++= 32 2 3 31 1 3 21 1 2 323313212 321 444 0 0 R Q Q R Q Q R Q Q VQVQVQ WWWW ooo πεπεπε Energy Density in Electrostatic Field Step 4: Bring point charge, Q4 from infinity into the system. Requires work !!!! Q1 Q2 R21 Q3 R31 R32Q4 R41 R43 R42

105. Energy Density in Electrostatic Field ( ) ( ) ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ++ +⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +++= ++++++= +++= 43 3 4 42 2 4 41 1 4 32 2 3 31 1 3 21 1 2 434424414323313212 4321 444 444 0 0 R Q Q R Q Q R Q Q R Q Q R Q Q R Q Q VQVQVQVQVQVQ WWWWW ooo ooo πεπεπε πεπεπε (1)

106. Energy Density in Electrostatic Field Notice that… 121 12 2 1 21 1 2212 44 VQ R Q Q R Q QVQ oo === πεπε By interchange the subscripts in (1), the following energy term can be obtained: 343242141232131121 VQVQVQVQVQVQW +++++= (2) Hence, the average energy: ( ) ( ) 2 21 + ( ) ( ) ( ) ( ) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ +++++ ++++++ = 44342413343231 22423211141312 2 1 QVVVQVVV QVVVQVVV W

107. Energy Density in Electrostatic Field It can be written that the total energy required: ( ) ∑= = +++= 4 1 44332211 2 1 2 1 k kkVQ VQVQVQVQW In terms of continuous charge distribution: ldVW line l∫= ρ 2 1 SdVW surface S∫= ρ 2 1 vdVW volume v∫= ρ 2 1 V is the potential at the charge distribution !!!

108. Energy Density in Electrostatic Field Alternatively; the previously obtained equations of energy can be written as: (Using vector identity) ( )dvEDW volume •= ∫ 2 1 The equation state that energy in the system of charges can be obtained if the electric field distribution (which is incurred by the charge itself) is known

109. Example A spherical charge distribution of radius ra and uniform ρv is centered at the origin. If the total charge is Q: Find the energy needed to build up the charge system.

110. Summary of Chapter 4 The students should be able to understand the followings: (i) The concept of work and energy in electrostatic field (ii) The concept of potential difference and absolute potential due to point charges and continuous charge system. (iii) Able to apply the potential gradient concept in calculating the electric field at a specific point (iv) Ability to calculate the stored energy in a electrical charge system

111. Next Lecture Please have a preliminary observation on the following topics: (i) The concept of electric current (ii) Conductors and conductivity Please refer to Hayt and Buck, page 114-123

112. Lecture #10 Topics to be covered: (i) Electric Current (ii) Conductors and conductivity Reference: Hayt and Buck, page 114-123 Motivation: To refresh the students understanding regarding the concept of electric current, conductors and conductivity

113. Electric Current Current: Moving electric charges. : Source of magnetostatic field. (Chapter 8) Definition: Rate of movement of charge passing a given reference point/plane per second. dt dQ I = (Ampere) Let us introduce the current density, . Unit of (A/m2)J Q Q Q dS dI ( )2 / mA dS dI JJ == ∫=∴ surface dsJI .

114. x y z Electric Current ρv dvdQ vρ= yUU y ˆ= From: dt dv dt dQ dI vρ == dS dy dt dydSvρ = yU dSUyvρ= yJ ( )2 /mAUJ vρ= In general: Charge in motion constitutes a current E

115. Continuity of Current Let us consider: ρV ∫= volume Ven dVQ ρ Close surface Rate of charge flow outwardly from the close surface current {dt dQ I en −= Q Reduction of charge Due to close surface, current flow outwardly: ∫ •= surface dSJI

116. Continuity of Current It can be written: dv dt d dv dt d dt dQ dSJ volume v volume v en surface ∫∫∫ −= ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −=−=• ρ ρ ( )∫ ∫−=•∇ volume volume v dv dt d dvJ ρ Divergence Theorem dt d J vρ −=•∇⇒ Continuity current equation Current per unit volume, emanating from a point equals the time rate decrease of volume charge density at the same point J Will be used in the study of conductor

117. Conductors & Conductivity Consider the atomic structure of conductor: E EqF e−= The electron will moved with constant velocity, called drift velocity, dU The electron drift velocity and the applied electric field are linearly related by the electron mobility, µe EU ed µ−=

118. Conductors & Conductivity From relation between current density and charge velocity: UJ vρ= In conductor: EUJ eede µρρ −== Conductivity; Unit: Siemens per meter σEJ σ= Relation between induced current flow and applied electric field in conductors

119. Conductor Properties in Electrostatic Field E conductor What will happen to the free charge, ρv in conductor due to applied electrostatic field ? vρ Consider the continuity current equation: dt d J vρ −=•∇ Due to applied electric field, current density will exist; EJ σ= dt d E vρ σ −=•∇∴ (1)

120. Conductor Properties in Electrostatic Field From: vD ρ=•∇ (Divergence) 0 1 =+ dt d v o v ρ σε ρ (1) Can be further written as: 0=+ o v o v dt d ε ρ ε σρ PDE !!! Solution will give: ( ) t vov o et ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − = ε σ ρρ Initial charge density at t= 0 ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = o rt ε σ 1 Relaxation time constant

121. Conductor Properties in Electrostatic Field For bad conductor (  ↓ tr ↑ charge (ρv) decay slowly) For good conductor (  ↑ tr ↓ charge (ρv) decay rapidly) E conductor vρ E Sρ−Sρ+ 0=vρ 0 0 =∴ =•∇ E D (inside)

122. Next Lecture Please have a preliminary observation on the following topics: (i) Conductor – Free Space (Boundary condition) (ii) Resistance in Conductor Please refer to Hayt and Buck, page 121-128

123. Lecture #11 Topics to be covered: (i) Conductor – Free Space (Boundary condition) (ii) Resistance in Conductor Reference: Hayt and Buck, page 121-128 Motivation: To enable the students to understand the concept of boundary condition and to calculate the resistance in conductor

124. Boundary Conditions Consider heat flow (thermodynamics…) conductor heater air at 27oC To calculate the temperature at conductor-air boundary, need to use the boundary conditions. Boundary conditions: Essential in finding physical quantities (temperature, electric field, magnetic field, etc..) if two different materials are in contact.

125. Boundary Conditions (Conductor – Free Space) Let us consider conductor in free space: conductor Free space E What is the resultant electric field at the conductor-free space boundaries if we apply an external electric field ??? Consider the upper boundary: #2 (conductor) #1 (free space) 2E 1E tE2 tE1 nE2 nE1

126. Boundary Conditions (Conductor – Free Space) (1) Consider the normal component: #1 #2 nE1 nE2 Relation between & ?nE1 nE2 Use Gauss’s Law en surface QdSD =•∫ ρS o S nE ε ρ =1 (1) Consider the tangential component: tE1 tE2 #1 #2 Relation between & ?tE1 tE2 Use conservation of electric field ∫ =• loop dlE 0 a b cd 01 =tE

127. Boundary Conditions (Conductor – Free Space) From previous slide: ( )mVn EEE o S tn /ˆ 111 ε ρ = += In general: conductor free space E

128. Example 1: Skitek At a point on a conductor-free space boundary, Find: (a) ρS at the point on the boundary (b) at the boundary. ( )mVyxE /ˆ3ˆ2 += D Example 2: D5.5 (Hayt, pg 128) Given the potential field in free space, V = 100 sinh 5x sin 5y V and a point P(0.1,0.2,0.3). Find at P: (a) V (b) (c) ρS if it is known that P lies on a conductor surface. E

129. Resistance in Conductor R How to calculate the resistance for the conductor ??? a a b I ( )Ω= I V R ab l J E y ( )ASE dSE dSJI y surface surface σ σ = •= •= ∫ ∫ lEdlEV y a b ab ∫ =•−= S l R σ =∴ Conductor with uniform cross section

130. Resistance in Conductor For conductor with non-uniform cross section: ∫ ∫ ∫ ∫ • •− = • •− == surface a b surface a bab dSE dlE dSJ dlE I V R σ I a b J E

131. Example 1: Skitek Find the resistance between the =0 and = π/2 surfaces of the truncated wedge section shown below. Given,  = 4x10-7 S/m and V/m φ φ ( )φˆ1 −= r E x y z 0.9 m0.1 m 1 m

132. Example 2: Skitek For the truncated wedge of example 1, find the resistance between faces at z = 0 and z = c , when the inner radius is ra and the outer radius is rb. Example 3: Skitek For the truncated wedge of example 1, find the resistance between the curved faces at radius of ra and that at a radius of rb, where rb > ra and the length in the z direction is c meter.

133. Next Lecture Please have a preliminary observation on the following topics: (i) Dielectrics (ii) Polarization in Dielectrics Please refer to Hayt and Buck, page 136-143

134. Lecture #12 Topics to be covered: (i) Dielectrics (ii) Polarization in Dielectrics Reference: Hayt and Buck, page 136-143

135. Objectives 1) To understand the concept of dielectric material 2) To investigate the effect of applied electric field to the dielectric material Polarization 3) To understand the concept of bound charge densities in dielectric due to polarization 4) To investigate the effect of polarization on the propagating electric fields in dielectric material

136. Dielectric Previously, we learn about conductors free charges to produce conduction current, J=E Dielectrics or insulators differ from conductors no free charges and charges are confined/bounded to molecular structure. Hence very small value of . For example, glass, quartz, polymer, etc… +ve Bound charges Atomic structure of dielectric

137. Polarization in Dielectric Material Dielectric Macrosopic view Microscopic view0=aE +ve No effect on atomic structure

138. Polarization in Dielectric Material Dielectric Macrosopic view 0≠aE⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - +ve ⇔ ⊕ - Microsopic view Q ( )CmdQp = dipole moment 0≠aE 0≠aE -Q d pnP = Electric polarization vector

139. Bound Charge Densities 0≠aE Consider the dielectric being polarized by the application of external field Dipole moment will appear and thus the polarization vector ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕ - ⊕⊕ ⊕⊕ ⊕ ⊕ ⊕ ⊕ ⊕ Two type of bound charges appear: P Bound surface charge density, ρsb sbρ+ sbρ− ( )2 m/CnˆPsb •=ρ vbρ Bound volume charge density, ρvb ( )3 m/CPvb •−∇=ρ

140. Effect of Polarization on Fields From previous, we have identify the existence of bound surface charge density, ρsb and bound volume charge density, ρvb due to polarization. Let us apply the Maxwell’s equation in polarized dielectric to find the electric field. From: vo ED ρε =•∇=•∇ Due to existence of ρvb: ( )vbvo E ρρε +=•∇ Free charge density From definition of ρvb: ( )PE vo •∇−=•∇ ρε Further manipulation: vo PE ρε =•∇+•∇ Hence: PED o += ε

141. Effect of Polarization on Fields Previously, in free space: PED o += ε ED oε= Presently, in polarized dielectric: Alternatively, it can be written as: EED ro εεε == Polarization vector is eliminated by introducing εr Relative permittivity of dielectric Depending on type of dielectric. For example: εr (air)=1.0006 εr (glass) =6.0

142. Example: Skitek An ungrounded spherical configuration of concentric spherical dielectric shells, enclosed by a conductor shell, is shown below. Regions 1 and 3 are free space, region 2 is a dielectric whose relative permittivity is εr1 and region 4 is a conductor. If a charge Q is placed at the center, find (a) D in all region (b) E in all region (c) P in all region (d) ρS on the conductor surface (e) ρsb on the dielectric surface (f) ρvb within the dielectric a b c d #4 #3 #2 #1

143. Next Lecture Please have a preliminary observation on the following topics: (i) Boundary Conditions (Dielectric-Dielectric) (ii) Capacitance Please refer to Hayt and Buck, page 143-159

144. Lecture #13 Topics to be covered: (i) Boundary Conditions (dielectric-dielectric) (ii) Capacitance Reference: Hayt and Buck, page 143-159

145. Objectives 1) To understand the concept of boundary conditions between dielectrics 2) To investigate the fundamental aspects of capacitance and to calculate the capacitance for a given structure

146. Boundary Conditions (Dielectric-Dielectric) Let us consider two bounded dielectrics : What is the resultant electric field at the dielectrics boundary if we apply an external electric field ??? Consider the upper boundary: Assume surface charge, ρS at the boundary 2E 1E tE2 tE1 nE2 nE1 #2 (Dielectric 2), ε2 #1 (Dielectric 1), ε1 ρS Dielectric 2 E Dielectric 1

147. Boundary Conditions (Dielectric-Dielectric) (1) Consider the normal component: #1 #2 nE1 nE2 Relation between & ?nE1 nE2 Use Gauss’s Law ρS en surface QdSD =•∫ Snn EE ρεε =− 2211 (1) Consider the tangential component: tE1 tE2 #1 #2 tE1 tE2Relation between & ? Use conservation of electric field a b cd ∫ =• loop dlE 0 tt EE 21 =

148. Example 1: Skitek In the region y>0, we find εr1=4, and in the region y<0, we find εr2=3. If at the boundary, find: (a) (b) (c) (d) (e) (f) ρsb Assume that ρS =0 at the boundary ( )m/VzˆE 101 = 1D 2E 2D 1P 2P

149. Example 2: Skitek A boundary between two dielectrics is found at x=0 plane. If material 1 exists for x>0 with εr1=4 and material 2 exists for x<0 with εr2=7, and when (at boundary), find: (a) (b) (c) (d) (e) (f) ρsb Assume that ρS =0 at the boundary ( )m/VzˆyˆxˆE 6321 −+= 1D 1P 2E 2D 2P

150. Example 3: Skitek Region 4x+3y >10 consists of a dielectric material with permittivity ε1 and region 4x+3y<10 is consists pf dielectric material with permittivity ε2. At boundary, the following electric field is recorded: ( )mVzyxE /ˆ5ˆ7ˆ21 +−= Obtain and ρsb at the boundary. Assume ρS =0 at the boundary. 2211 ,,, PDPD

151. Capacitance + + + + + + + + + ++ + ++ - - - - - - - - -- - -- - - Conductor a Conductor b Dielectric, ε E Capacitance of a two conductor capacitor is defined as the ratio of magnitude of charge on one of the conductors to the magnitude of potential difference between the conductors. ( )Farad V Q C ab = Vab ∫=+ dSQ Sρ ∫=− dSQ Sρ Sρ+ Sρ−

152. Capacitance From: abV Q C = By definition: ∫ ∫ •− = a b S dlE dS C ρ Or: ∫ ∫ •− • = a b dlE dSE C ε (1) (2) (1) or (2) can be used to calculate the capacitance for any given capacitor structure.

153. Example 1: Skitek Calculate the capacitance of two concentric metallic spheres separated by a dielectric with permittivity ε. Let rb be the inner radius of the outer sphere and ra be the outer radius of the inner sphere. Example 2: Skitek Find the capacitance of an L length coaxial capacitor shown below. Conductor a Conductor b ε rb ra

154. Example 3: Skitek Find the expression for the capacitance permeter length of a cylindrical capacitor whose cross section is given below. a b c d #4 #3 #2 #1 #1: conductor #2: dielectric, ε2=6εo #3: free space #4: conductor

155. End of Chapter 5 Thank you Please recap the objectives stated in each lecture to ensure that you fully understood the topics.

156. Next Lecture Please have a preliminary observation on the following topics: (i) Uniqueness Theorem (ii) Laplace’s and Poisson Equations Please refer to Hayt and Buck, page 172-188

157. Lecture #14 Topics to be covered: (i) Uniqueness Theorem (ii) Laplace’s and Poisson’s Equations Reference: Hayt and Buck, page 172-188

158. Objectives 1. To understand the concept of uniqueness theorem 2. To calculate the electric potential/electric field at a given point using the solution of Poisson’s and Laplace’s equations

159. Methods to find Electric Field Electric Field Charge Distribution Coulomb’s Law Gauss’s Law Electric Potential Potential Gradient ∫= r dQ V πε4 Uniqueness Theorem Boundary Condition Area Properties

160. Uniqueness Theorem Solution for electrostatic problem (electric potential) can be unique for a specified region if uniqueness theorem is obeyed. Potential, V Boundary Conditions Area properties 1. Poisson’s Equation 2. Laplace’s Equation

161. Poisson’s & Laplace’s Equation From Maxwell’s equation: v v v E E D ρε ρε ρ =•∇ =•∇ =•∇ Consider homogeneous material ( ) vV ρε =∇−•∇ Laplacian operator (coordinate dependence) Poisson’s When ρv=0: 02 =∇ V Laplace’s ε ρv V −=∇2

162. Solution of Poisson’s and Laplace’s Equation Electric potential, V can be obtained by solving the Laplace’s and Poisson’s equations. For example, for Laplace’s equation: 0 0 2 2 2 2 2 2 2 = ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ z V y V x V V 3-dimensions But, we only interested in solving a 1-dimensional problem easier and less complicated. 000 2 2 2 2 2 2 = ∂ ∂ = ∂ ∂ = ∂ ∂ z V @ y V @ x V

163. Solution of Poisson’s and Laplace’s Equation For instance; V only varied in x direction: So, we choose only: 02 2 = ∂ ∂ x V A x V = ∂ ∂ ⇒1st integration 2nd integration BAxV +=⇒ Variables (A and B) can be solved by using the given boundary conditions

164. Example: Skitek A parallel plate capacitor has a separation of 0.5mm between plates and a potential difference between plates of 50V. If ρv =0 and ε=4εo between plates, find: (a) V(y) for 0≤y≤0.5mm if the potential at y=0 is 25 V and at at y=0.5 mm is 75 V. (b)Electric field between plates (c) Capacitance per square meter

165. Example: Skitek Two infinite length, concentric and conducting cylinders of radius ra=0.02 m and rb=0.05 m are located with axes on the z-axis. If ε=4εo, ρv =0 between the cylinders, V=50 V at ra, V=100 V at rb, find: (a) V in the range 0.02≤r≤0.05 (b)Electric field intensity (c) Electric flux density (d) ρs at rb (e) Capacitance per meter length

166. End of Chapter 6 & End of Electrostatics Topics

167. Next Lecture Please have a preliminary observation on the following topics: (i) Magnetostatics Field and Sources (ii) Biot-Savart’s Law Please refer to Hayt and Buck, page 210-218

168. Lecture #15 Topics to be covered: (i) Magnetostatic Field Intensity and Field Sources (ii) Biot-Savart’s Law Reference: Hayt and Buck, page 210-218

169. Objectives i) To understand the concept of magnetostatic field intensity ii) To learn about the source of magnetostatic field intensity iii) To calculate the magnetostatic field intensity using the Biot Savart’s law

170. Magnetic (Magnetostatic) Field Intensity Magneto + static Magnetic field intensity source Symbol of . Unit of Ampere/meter (A/m). Field sources: Steady current. Type of field sources: (i) Filament current (ii) Surface current (iii) Volume current H

171. Filament Current I dl Filamentary current element: dlI Surface Current sJ dS Surface current element: dSJS Surface current density : unit (A/m) (A.m) (A.m)

172. Volume Current J Volume current element: dvJ dv Those elements (filament, surface and volume) will create magnetic field. But, how ??? (A.m) volume current density : unit (A/m2)

173. Recall Previously; in electric field calculation Charge density; Electric fieldCoulomb’s Law vSlQ ρρρ ,,, Current density; JJI S ,, Similarly, in magnetic field calculation Magnetic field Biot-Savart’s Law

174. Biot-Savart’s Law The law was co-originated by Felix Savart, a professor at College de France and Jean Baptiste Biot, a French physicist in 1820. I R Observations: rradius zI H ˆ ˆ ˆ → → → φ 2 1 R H IH ∝ ∝

175. Biot-Savart’s Law Hence, Biot-Savart’s Law can be stated as: )/( 4 ˆ , 2 mA R adlI H llength R ∫ × = π (filament current) )/( 4 ˆ , 2 mA R dSaJ H Ssurface RS ∫ × = π )/( 4 ˆ , 2 mA R dVaJ H Vvolume R ∫ × = π (surface current) (volume current)

176. Biot-Savart’s Law I B At any point, let B: dl R Raˆ )/( 4 ˆ , 2 mA R adlI H llength R ∫ × = π Consider a filamentary current;

177. Biot-Savart’s Law-Filamentary Current Consider a finite length of filamentary current: )',,( zr φ y x z φ z’ r Step 3: Integrate over filament current length to get total H Rdl Step 1: Select current element (0,0,z) a b Step 2: Identify dH Hd I

178. Biot-Savart’s Law-Filamentary Current Step 1: Current element dlI Step 2: Identify dH 2 4 ˆ R adlI Hd R π × = zdzdl ˆ= ( )( ) rrzzzR ˆˆ' +−−= Hence, ( )( )2 3 22 '4 ˆ zzr Irdz Hd −+ = π φ Using: ( ) ( ) ∫ + = + 2 1 2222 3 22 xcc x xc dx

179. Biot-Savart’s Law-Filamentary Current Step 3: ( )( ) ( )( ) φ π ˆ ' ' ' ' 4 2 1 222 1 22 ⎪⎭ ⎪ ⎬ ⎫ ⎪⎩ ⎪ ⎨ ⎧ −+ − − −+ − = zar za zbr zb r I H Alternatively: ( ) )/(ˆsinsin 4 21 mA r I H φαα π += 1α 2αFor infinite length; ( )mA r I H /ˆ 2 φ π = r

180. Example 1 Consider AB in figure below as part of an electric circuit. Find H at the origin due to AB. x y 1 1 A B 6 A 0 Example 2 A square conducting loop of side 2a lies in the z=0 plane and carries a current I in the counterclockwise direction. Find a magnetic field intensity at center of the loop.

181. Example 3 An infinitely long conductor is bent into an L shape as shown in figure below. If a direct current of 5 A flows in the conductor, find the magnetic field intensity at (a) (2,2,0), (b) (0,0,2) x y 5 A 0 5 A Example 4 The y and z axes, respectively, carry filamentary currents 10 A along and 20 A along . Find H at (-3,4,5). yˆ zˆ−

182. Example 5 Find the expression for the H field along the axis of the circular current loop (radius of a), carrying a current I.

183. Next Lecture Please have a preliminary observation on the following topics: (i) Ampere’s Circuital Law Please refer to Hayt and Buck, page 218-225

184. Lecture #16 Topics to be covered: (i) Ampere’s Circuital Law Reference: Hayt and Buck, page 218-225 Motivation: To understand the concept of Ampere’s Circuital Law and its application in calculating magnetic field

185. Ampere’s Circuital Law (ACL) Discovered by Andre-Marie Ampere, a French physicist in 1800s. Analogous to Gauss’s Law. Gauss’s Law Electric fieldSpecial case of charge distribution Magnetic field Special case of current distribution Ampere’s Circuital Law

186. Ampere’s Circuital Law (ACL) Amperian loopGaussian surface Infinite length of filament current Infinite surface current Infinite length of line charge Infinite surface charge Ampere’s Circuital LawGauss’s Law ∫ = surface Gaussian enQdsD. ∫ = loop Amperian enIdlH .

187. Ampere’s Circuital Law- Filament current I z Step 1: Create an amperian loop (closed loop that surround the filament current) Step 2: Identify the loop element, dl Step 3: Use the ACL to find the magnetic field ∫ = loop Amperian enIdlH . dl ∞+ ∞−

188. Ampere’s Circuital Law- Filament current From previous slide; φφ ˆHH = φφ ˆrddl = IIen = Hence: )/(ˆ 2 2 ˆ.ˆ . mA r I H r I H IrdH IdlH en φ π π φφφ φ φ = = = = ∫ ∫ As proved from B-S Law

189. Ampere’s Circuital Law- Surface Current y x z yJJ SS ˆ= Step 1: Create an amperian loop Step 2: Identify the loop element Step 3: Use the ACL ( )xdx ˆ− ( )xdx ˆ ( )zdz ˆ ( )zdz ˆ− 1 2 3 4

190. Ampere’s Circuital Law- Surface Current From previous slide; ( ) ( ) ( ) ( ) )/(ˆ 2 1 2 2 ˆ.ˆˆ.ˆ ˆ.ˆˆ.ˆ . 1 4 4 3 3 2 2 1 mAnJH J H JH LJzdzxHxdxxH zdzxHxdxxH IdlH S s x sx sxx xx en ×=∴=⇒ =⇒ =−+−− +−+⇒ =⇒ ∫∫ ∫∫ ∫

191. Example 1 Plane x=10 carries current 100mA/m along while line x=1, y=-2 carries filamentary current 20π mA along . Determine at (4,3,2). zˆ zˆ H Example 2 An infinitely long solid conductor of radius a is placed along the z-axis. If the conductor carries current I in the direction, show that: φ π ˆ a Ir H 2 2 = zˆ

192. Example 3 Through the use of Ampere’s Circuital Law, find the field in all regions of an infinite length coaxial cable carrying a uniform and equal current I in opposite directions in the inner and outer conductors. Assume the inner conductor to have a radius of a (m) and the outer conductor to have an inner radius of b (m) and an outer radius of c (m). Assume that the cable’s axis is along the z axis. H Example 4 Through the use of Ampere’s Circuital Law, find the field inside and outside an infinite-length hollow conducting tube whose radius is 0.002 m that carries a current I = 10-7 A, directed in the direction. Consider the thickness of the tube very small. H zˆ+

193. Example 5 Consider two wire transmission line whose cross section is illustrated in figure below. Each wire is of radius 2 cm and the wires are separated by 10 cm. The wire centered at (0,0) carries current 5 A in the direction of , while the other centered at (10cm,0) carries the return current. Find at: (5cm, 0) and (10cm, 5 cm) zˆ+ H x y 10 cm

194. Next Lecture Please have a preliminary observation on the following topics: (i) Stokes’ Theorem (ii) Magnetic Flux Density (iii) Maxwell’s equation for static field (iv) Vector Magnetic Potential Please refer to Hayt and Buck, page 225-246

195. Lecture #17 Topics to be covered: (i) Stokes’ Theorem (ii) Magnetic Flux Density (iii) Maxwell’s equation for static field (iv) Vector Magnetic Potential Reference: Hayt and Buck, page 225-246 Motivation: To understand the concept of magnetic flux density and magnetic potential

196. Recall… In electrostatic field analysis: Divergence Theorem ∫ ∫∇= surface volume dvD.dS.D Gauss’s Law ∫ = surface enQdS.D Stokes’ Theorem Ampere’s Circuital Law en loop Idl.H =∫ ( )∫ ∫ ×∇= loop surface dS.Hdl.H In magnetostatic field analysis: Another form of GL Another form of ACL Details on the curl operator: Hayt and Buck, page 225-232

197. Magnetic Flux and Flux Density Electric Flux Density, Electric Flux, ψE Unit: Coulomb Electrostatic field D ∫= surface E dS.Dψ en surface E QdS.D == ∫ψ Magnetic Flux Density, Magnetic Flux, ψM Unit: Weber Magnetostatic field B ∫= surface M dS.Bψ 0== ∫surface M dS.Bψ ED oε= HB oµ= µo : magnetic permeability (free space). 4π × 10-7 (H/m)

198. Example Determine the magnetic flux through a rectangular loop (a × b) due to an infinitely long conductor carrying current I as shown below. The loop and the straight conductors are separated by a distance d. z d a bI

199. Example: D8.7 (Hayt) A solid conductor of circular cross section is made of a homogeneous nonmagnetic material. If the radius a = 1mm, the conductor axis lies on the z axis, and the total current in the direction is 20A, find: (a) H at r=0.5 mm (b) B at r=0.8 mm (c) The total magnetic flux perunit length inside the conductor (d) The total flux for r < 0.5 mm (e) The total magnetic flux outside the conductor. zˆ

200. Maxwell’s Equations for Static Field Electrostatic Field Magnetostatic Field ∫ ∫∫ ===∇ surface volume ven volume dvQdS.DdvD. ρ vD. ρ=∇ Point form ( ) 0==∇ ∫∫ loopsurface dl.EdS.E. 0=×∇ E Point form ( ) ∫ ∫∫ ===×∇ loop surface en surface dS.JIdl.HdS.H JH =×∇ Point form ( ) 0==∇ ∫∫ surfacevolume dS.BdvB. 0=∇ B. Point form The equations describe the relationship between electric field, magnetic field, static charges and steady current. For time- varying field, the equations distinct in certain aspects.

201. Example: D 8.5 (Hayt) Calculate the value of current density: (a) In rectangular coordinates at (2,3,4) if (b) In cylindrical coordinates at (1.5, 90o, 0.5) if (c) In spherical coordinates at (2, 30o, 20o) if zˆxyyˆzxH 22 −= ( )rˆ.cos r H φ20 2 = θ θ ˆ sin H 1 =

202. In electrostatic field analysis: Vector Magnetic Potential VE −∇=Charge Distribution Electric Potential, V Similarly, in magnetostatic field analysis: Magnetic Potential, A AB ×∇= From: (Maxwell’s) and (identity)0. =∇ B ( ) 0. =×∇∇ A Current Distribution How ???

203. Vector Magnetic Potential Caution: Mathematical derivation starts here From Maxwell’s; JH =×∇ ( ) JA JB o o µ µ =×∇×∇⇒ =×∇⇒ From the vector identity: ( ) ( )AAA ×∇×∇−∇∇=∇ .2 By simplifying: 0. =∇ A JA oµ−=∇⇒ 2 (Vector Poisson’s Equation) Recall: (PE) ∫= volume v o R dv V ρ πε4 1 o v V ε ρ −=∇ 2 Hence: ∫= volume o R dvJ A π µ 4 (Wb/m)

204. Vector Magnetic Potential In general; Volume current Surface current Filament current ∫= volume o R dvJ A π µ 4 (Wb/m) ∫= surface So R dSJ A π µ 4 (Wb/m) ∫= line o R dlI A π µ 4 (Wb/m)

205. Example 1: Skitek A conductor of radius a carries a uniform current with . Show that the magnetic vector potential for r < a is: zJJ o ˆ= Example 2: Skitek ( )mWbzrJA oo /ˆ 4 1 2 µ−= Find for an infinite line of current and derive the field from that current. A H

206. Example 3: Skitek For the filamentary circular current loop carrying current I, with radius a at z=0, find the (a) the vector magnetic potential, (b) through the use of ; along the axis of the current loop. A B A Example 4: D 8.9 (Hayt) The magnetic vector potential of a current distribution in free space is given by: Find H at (3, π/4, -10). Calculate the flux through r =5, 0 ≤Φ ≤ π/2, 0 ≤ z ≤10. ( )mWbzeA r /ˆsin15 φ− =

207. Next Lecture Please have a preliminary observation on the following topics: (i) Magnetic Forces (ii) Magnetic Material Please refer to Hayt and Buck, page 259-281

208. Lecture #18 Topics to be covered: (i) Magnetic Forces (ii) Lorentz Force Equation Reference: Hayt and Buck, page 259-273 Motivation: To demonstrate the force exerted by the magnetic field and to understand the nature and application of Lorentz force equation

209. Recall… In electrostatic field; E Q Exerted force; EQF = (N) Now, in magnetostatic field; B I Exerted force; dl BdlIFd ×=

210. Magnetic Forces-Current Elements Let two infinite and parallel filament current-carrying conductors: x y z d I2 I1 Find force per unit length between the two conductors ??? dl2 Force experienced by dl2: 1222 BdlIFd ×= ( ) ( )xˆ d I zˆdzIFd o −×−= π µ 2 1 22 ( )m/Nyˆ d II dz Fd o π µ 2 21 =

211. Magnetic Forces-Current Elements Similarly; from the previous equation: BdSJFd s ×= Force on a surface currentBdVJFd ×= Force on a volume current BdlIFd ×= Force on a filament current

212. Example 1 A rectangular loop carrying current I2 is placed parallel to an infinitely long filamentary wire carrying current, I1 as shown. Find the force experienced by the loop. I1 I2 ro a b x z Example 2 A current element of length 2 cm is located at the origin in free space and carries current 12 mA along . A filamentary current of 15 is located along x =3, y =4. Find the force on the current element. xˆ zˆ

213. Example 3 A three phase transmission line consists of three conductors that are supported at points A, B and C to form an equilateral triangle as shown below. At one instant, conductors A and B both carry a current of 75 A while conductor C carries a return current of 150 A. Find the force per meter on conductor C at that instant. A B C x y 2 m

214. Force on Moving Point Charge Consider a volume current distribution, J Current – Moving point charges Relation between volume current distribution and charge velocity, (chapter 5)U UJ Vρ= (1) Previously, magnetic force on volume current distribution: BdvJFd ×= (2) From (1) and (2): ( ) BUdQ BUdvFd v ×= ×= ρ (3) Magnetic force on moving charges Extension from volume current to moving point charge

215. Force on Moving Point Charge Equation (3) is a fundamental equation for working principle of electric motors and electric generators. B U ( )BUqF e ×= B B conductor B a b aF U BU × ( )BUqe ×

216. Lorentz Force Equation When the charge is immersed in combined and fields, the combined force becomes E B BUdQEdQFd ×+= (N)

217. Example: D 9.1 (Hayt) The point charge, Q=18 nC has a velocity 0f 5×106 m/s in the direction . Calculate the magnitude of the force exerted on the charge by the field: zˆ.yˆ.xˆ. 300750600 ++ mTzˆyˆxˆB 643 ++−=(a) (b) m/kVzˆyˆxˆE 643 ++−= (c) and acting together.B E

218. Next Lecture Please have a preliminary observation on the following topics: (i) Magnetic Material (ii) Magnetic Boundary Conditions Please refer to Hayt and Buck, page 273-290

219. Lecture #19 Topics to be covered: (i) Magnetic Material (ii) Magnetic Boundary Condition Reference: Hayt and Buck, page 273-290

220. Objectives 1. To understand the concept of magnetic polarization; i.e: the effect of atomic orientation due to applied magnetic field 2. To investigate on the bound magnetization current in polarized material 3. To investigate the effect of magnetic polarization to the resultant magnetic field in the material 4. To understand the concept of boundary condition between two magnetic materials

221. Magnetic Polarization Magnetic material 0=B Macroscopic view 0== ∑mM Magnetization vector (A/m) (i) Dipole moment randomly oriented (ii) Total dipole moment From macroscopic view; :0=B Microscopic view Magnetic dipole moment Bound current due to electron movement m I

222. Magnetic Polarization Characteristics of dipole moment will determine type of magnetic materials: (i) Diamagnetic (ii) Paramagnetic (iii)Ferromagnetic (iv)Antiferromagnetic (v) Ferrimagnetic (vi)Superparamagnetic *Interested readers: please refer to page 274-276

223. Magnetic Polarization Magnetic material 0≠B Macroscopic view (i) Orientation of dipole moment according to the magnetic field direction (polarized) (ii) Total dipole moment From macroscopic view; : 0≠= ∑mM 0≠B 0≠= ∑mM

224. Bound Magnetization Current 0≠B M I m Bound magnetization surface current density (on surface) ( )mAnMJ sm /ˆ×= Bound magnetization current density (within the material) ( )2 / mAMJ m ×∇=

225. Example 1: Skitek A long cylinder of magnetic material of radius a (m) is found along the z axis. When the magnetization, within the cylinder, find (a) (b) at r = a on the magnetic material surface. )/(ˆ10 mAzM = mJ smJ Example 2: Skitek A long bar of magnetic material is found parallel to the y axis with its cross section defined by 0 ≤ x ≤ 0.1 and 0 ≤ z ≤ 0.2. If , find (a) within the material (b) on the four surfaces (c) total bound current that flows through a cross section of the bar at z = 0.1 plane for a length of 1 meter in the y direction. )/(ˆ3 mAyxM = mJ smJ

226. Effect of Magnetization on Magnetic Fields (Recall) From free space condition: J B o =×∇ µ mJIn magnetic material, need to include (due to magnetization) ( )m o JJ B +=×∇∴ µ From ;MJm ×∇= ( )MJ B o ×∇+=×∇ µ

227. Effect of Magnetization on Magnetic Fields Further manipulation: JM B o =⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −×∇ µ M B H o −=⇒ µ ( ) ( )2 m/WbHH MHB ro o µµµ µ == +=⇒ rµ : magnetic permeability of the material

228. Example In certain material for which µ=6.5µo, ( )m/AzˆyˆxˆH 402510 −+= Find: The magnetic flux density and magnetization

229. Boundary Conditions (Two magnetic materials) Let us consider two magnetic materials: H Magnetic material 1, µ1 Magnetic material 2, µ2 What is the resultant magnetic field at the boundary if we apply an external magnetic field ??? 2H 1H tH 2 nH 2 tH1 nH1 Consider the boundary: Assume surface current in inwards direction (current due to free charges) sJ Magnetic material 2 Magnetic material 1 sJ ⊗⊗⊗⊗⊗ ⊗

230. Boundary Conditions (Two Magnetic Materials) (1) Consider the normal component: #1, µ1 #2, µ2 nH1 nH 2 Relation between & ?nH1 nH2 Use Gauss’s Law for magnetic field 0=•∫surface dSB nn BB 21 = (1) Consider the tangential component: tH1 tH 2 #1 #2 Relation between & ?tH1 tH2 Use Ampere’s Circuital Law a b cd ∫ =• loop enIdlH stt JHH =− 21 ⊗⊗ ⊗⊗⊗⊗⊗⊗ ⊗ sJ l∆

231. Boundary Conditions (Two Magnetic Materials) HB µ=From: The boundary conditions can be further written as: nn HH 2211 µµ = Normal component s tt J BB =− 2 2 1 1 µµ Tangential component

232. Example The interface 2x+y=8 between two media carries no current. If medium 1 (2x+y≥8) is nonmagnetic with: ( )m/AzˆyˆxˆH −+−= 341 Find: (a) The magnetic energy density in medium #1 (b) and in medium 2, 2x+y≤8, with µ2=10µo (c) The angles and make with the normal to the interface 2M 2B 1H 2H

233. Next Lecture Please have a preliminary observation on the following topics: (i) Time Varying Fields (ii) Faraday’s Law (iii) Displacement Current Please refer to Hayt and Buck, page 306-313

234. Lecture #20 Topics to be covered: (i) Time Va

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