Bài giảng Engineering electromagnetic - Chapter III: Coulomb’s Law & Electric Field Intensity - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Coulomb’s Law & Electric Field Intensity Contents I. Introduction II. Vector Analysis III. Coulomb’s Law & Electric Field Intensity IV. Electric Flux Density, Gauss’ Law & Divergence V. Energy & Potential VI. Current & Conductors VII. Dielectrics & Capacitance VIII.Poisson’s & Laplace’s Equations IX. The Steady Magnetic Field X. Magnetic Forces & Inductance XI. Time – Varying Fields & Maxwell’s Equations XII. The Uniform Plane Wave XIII.Plane Wave Reflection & Dispersion XIV.Guided Waves & Radiation Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 2 Coulomb’s Law & Electric Field Intensity 1. Coulomb’s Law 2. Electric Field Intensity 3. Field Due to a Continuous Volume Charge Distribution 4. Field of a Line Charge 5. Field of a Sheet Charge 6. Sketches of Fields Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 3 Coulomb's Law (1) • Experiment of Coulomb: QQ F= k 1 2 R2 – In free space – between 2 very small objects (compared to the separation R) – Q1 & Q2 are the positive/negative quantities of charge 1 −k = πε 4 0 −1 − – ε : permittivity of free space, ε =8.854 × 1012 = 10 9 F/m 0 0 36 π Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 4 Coulomb's Law (2) QQ1 2 a Q F F= k Q 12 R12 2 2 R2 QQ 1 → = 1 2 F 2 r2 1 4πε R r1 k = 0 πε Q & Q : alike in sign 4 0 Origin 1 2 QQ F= 1 2 a 2πε 2 12 4 0R 12 = − a F Q2 R12 r 2 r 1 12 2 Q1 R12 R R rr− r2 a =12 = 12 = 2 1 r1 12 RR r− r 12 12 2 1 Q & Q : opposite sign Origin 1 2 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 5 Ex. Coulomb's Law (3) -4 -4 Given Q1 = 4.10 C at A(3, 2, 1) & Q2 = – 3.10 C at B(1, 0, 2) in vacuum. Find the force exerted on Q2 by Q1. QQ F= 1 2 a 2πε 2 12 4 0R 12 =−=− +− +− =− − + Rrr12 2 1 (1 3) ax (0 2) a y (2 1) aaaa zxyz 2 2 = −2 +− 2 + 2 = R12 ( 2) ( 2) 1 3 −2a − 2 a + a =R12 = x y z a12 R12 3 −4− − 4 −2a − 2 a + a → =4.10 ( 3.10 ) × x y z = + − F2 80ax 80 a y 40 a z N 1 − 4π 1039 2 3 36 π Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 6 Coulomb’s Law & Electric Field Intensity 1. Coulomb’s Law 2. Electric Field Intensity 3. Field Due to a Continuous Volume Charge Distribution 4. Field of a Line Charge 5. Field of a Sheet Charge 6. Sketches of Fields Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 7 Electric Field Intensity (1) • Consider a fixed Q1 & a test Qt QQ F Q F= 1 t a →t = 1 a tπε 2 1 t πε 2 1t 4 0R 1 t QRt4 0 1 t • Electric Field Intensity : the vector force on 1C • Unit: V/m • EFI due to a single point charge Q in a vacuum: Q E= a πε 2 R 4 0 R – R: from Q to the point of E – aR: unit vector of R Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 8 Electric Field Intensity (2) Q E= a πε 2 R 4 0 R • If Q is at the center of a spherical coordinate system: Q E= a πε 2 r 4 0r • If Q is at the center of a rectangular coordinate system: Q x y z  E= aaa + +  πε 2+ 2 + 2  222x 222 y 222 z  40 (x y z )  xyz++ xyz ++ xyz ++  Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 9 Q Electric Field Intensity (3) E= a πε 2 R 4 0 R 3.5 3 2.5 2 1.5 1 0.5 0 4 3 2 4 3 1 2 0 1 -1 0 -2 -1 -2 -3 -3 -4 -4 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 10 Electric Field Intensity (4) R = r – r’ E • If Q is not at the origin: Q Q E( r ) = a x’, y’, z’ r P(x, y, z) πε 2 R r’ 4 0 R R =r − r ' R= r − r ' r− r ' Origin a = R r− r ' Q r− r ' Q(r− r ') →E( r ) = . = πε − 2 r− r ' πε − 3 40 r r ' 40 r r ' Qxx[(− ')a +− ( yy ') a +− ( zz ') a ] = x y z πε −2 +− 2 +− 23/2 40 [(xx ') ( yy ') ( zz ') ] Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 11 Electric Field Intensity (5) QQ Er( ) =1 a + 2 a πε−21 πε − 2 2 401rr 4 02 rr z Q2 r2 r – r2 Q1 r – r1 P a1 E n r 1 y Q r1 = k a E( r ) ∑ 2 a k 2 = πε − k 1 4 0 r r k E2 E(r) x Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 12 Ex. Electric Field Intensity (6) -9 -9 -9 Given Q1 = 4.10 C at P1(3, – 2, 1), Q2 = – 3.10 C at P2(1, 0, – 2), Q3 = 2.10 C -9 at P3(0, 2, 2), Q4 = – 10 C at P4(– 1, 0, 2). Find E at P(1, 1, 1). n Q = k E∑ 2 a k = πε − k 1 4 0 r r k QQQQ E=1 a + 2 a +3 a + 4 a πε−21 πε − 2 2 πε − 2 3 πε − 2 4 401rr 4 02 rr 4 03 rr 4 04 rr − = − +− +− r r 1 (xx1 )(ax yy 1 )( a y zz 1 ) a z =− +−− +− (1 3)ax (1 ( 2)) a y (1 1) a z = − + 2ax 3 a y Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 13 Ex. Electric Field Intensity (7) -9 -9 -9 Given Q1 = 4.10 C at P1(3, – 2, 1), Q2 = – 3.10 C at P2(1, 0, – 2), Q3 = 2.10 C -9 at P3(0, 2, 2), Q4 = – 10 C at P4(– 1, 0, 2). Find E at P(1, 1, 1). QQQQ E=1 a + 2 a +3 a + 4 a πε−21 πε − 2 2 πε − 2 3 πε − 2 4 401rr 4 02 rr 4 03 rr 4 04 rr r−= r ( − 2)2 + 3 2 = 3.32 − =− + 1 rr1 2 ax 3 a y r− r − 2 3 a=1 = aaaa + =−0.60 + 0.91 1 − x y x y r r 1 3.32 3.32 − = = + r r 2 3.16 a2 0.32 ay 0.95 a z − = = − − r r 3 1.73 a3 0.58 ax 0.58 a y 0.58 a z − = = + − r r 4 2.45 a4 0.82 ax 0.41 a y 0.41 a z Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 14 Ex. Electric Field Intensity (8) -9 -9 -9 Given Q1 = 4.10 C at P1(3, – 2, 1), Q2 = – 3.10 C at P2(1, 0, – 2), Q3 = 2.10 C -9 at P3(0, 2, 2), Q4 = – 10 C at P4(– 1, 0, 2). Find E at P(1, 1, 1). 4× 10 −4 E=( − 0.60 a + 0.91 a ) πε × 2 x y 40 3.32 −3 × 10 −4 +(0.32a + 0.95 a ) + πε × 2 y z 40 3.16 2× 10 −4 +(0.58a −−+ 0.58 a 0.58 a ) πε × 2 x y z 40 1.73 −10 −4 +(0.82a + 0.41 a − 0.41 a ) πε × 2 x y z 40 2.45 = + − 24.66ax 9.99 a y 32.40 a z V/m Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 15 Coulomb’s Law & Electric Field Intensity 1. Coulomb’s Law 2. Electric Field Intensity 3. Field Due to a Continuous Volume Charge Distribution 4. Field of a Line Charge 5. Field of a Sheet Charge 6. Sketches of Fields Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 16 Volume Charge (1) • Volume charge density (unit C/m3): ∆ ρ = Q v lim ∆v →0 ∆v Q= ρ dv ∫V v Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 17 z Ex. Volume Charge (2) z = 4 cm Find the total charge in the cylinder, given 5 ρ= − 5e−10ρ z µ C/m. 3 v z = 1 cm = ρ Q∫ v dV V x y ρ = 1 cm 5 →Q = − 5 e−10 ρz dV ∫V dV = ρdρdφdz 0.04 2π 0.01 5 →=Q −×5 10 −6 edddz − 10 ρz ρ ρ ϕ ∫0.01 ∫ 0 ∫ 0 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 18 z Ex. Volume Charge (3) z = 4 cm Find the total charge in the cylinder, given 5 ρ= − 5e−10ρ z µ C/m. 3 v z = 1 cm 5 Q= − 5 e−10 ρ z dV ∫V x y ρ = 1 cm 0.04 2π 0.01 5 = −5 × 10 −6e − 10 ρz ρ dddz ρ ϕ ∫0.01 ∫ 0 ∫ 0 2π π dϕ= ϕ2 = 2 π ∫0 0 0.04 0.01 5 →Q = − 10 −5π e − 10 ρz ρ ddz ρ ∫0.01 ∫ 0 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 19 z Ex. Volume Charge (4) z = 4 cm Find the total charge in the cylinder, given 5 ρ= − 5e−10ρ z µ C/m. 3 v z = 1 cm 5 Q=∫ − 5 e−10 ρ z dV V y 0.04 0.01 5 x =∫ ∫ − 10 −5πe − 10 ρz ρ ddz ρ ρ = 1 cm 0.01 0 0.04 0.04 5 π 5 −10 −5πe − 10 ρ z dz = 10 −10e − 10 ρ z ∫0.01 ρ π 0.01 =10(−10e − 4000ρ − e − 1000 ρ ) ρ 0.01 π →Q =10−10 ( eed − 4000ρ − − 1000 ρ ) ρ ρ ∫0 ρ Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 20 z Ex. Volume Charge (5) z = 4 cm Find the total charge in the cylinder, given 5 ρ= − 5e−10ρ z µ C/m. 3 v z = 1 cm 5 Q=∫ − 5 e−10 ρ z dV V y 0.01 −π −ρ − ρ x =1010 (e 4000 − e 1000 ) ρ d ρ ρ = 1 cm ∫0 ρ 0.01 =10(−10πe − 4000ρ − e − 1000 ρ ) d ρ ∫0 0.01 −4000ρ − 1000 ρ π − e e  3 −12 =10 10 π  −  = − 10 = 0.236 pC −4000 − 1000 40   0 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 21 Volume Charge (6) • EFI at r due to ∆Q at r’ : Q r− r ' ∆Q r − r ' E( r )= . → ∆E( r ) = . πε − 2 r− r ' πε − 2 r− r ' 40 r r ' 40 r r ' ∆ =ρ ∆ Qv v ρ ∆v r− r ' → ∆E( r ) = v . πε − 2 r− r ' 40 r r ' • → EFI at r due to a volume charge: ρ (r ')dv ' − = v r r ' E( r )∫ 2 . V πε − r− r ' 40 r r ' Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 22 Volume Charge (7) ρ (r ')dv ' − = v r r ' E( r )∫ 2 . V πε − r− r ' 40 r r ' • r : the vector locates E • r’: the vector locates the volume charge ρ(r’) dv ’ Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 23 Coulomb’s Law & Electric Field Intensity 1. Coulomb’s Law 2. Electric Field Intensity 3. Field Due to a Continuous Volume Charge Distribution 4. Field of a Line Charge 5. Field of a Sheet Charge 6. Sketches of Fields Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 24 Line Charge (1) • Line charge density ρL (unit: C/m) • Cylindrical coordinate system • EFI of an infinite uniform line charge has only an Eρ component & it varies only with ρ Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 25 Q E= a Line Charge (2) = ρ πε 2 R EEρ( ) a ρ 4 0 R EFI of an infinite uniform line charge has only an Eρ component & it varies only with ρ z z z φ = const ρL ρL ρL z = const z = var z = const φ = var φ = const ρ = const y y y x x ρ = const x ρ = var ρ = const  ρ = const  ρ = var     ϕ =var →E = const ϕ =const →E = const ϕ =const →E = var    z = const  z = var  z = const  Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 26 z Line Charge (3) z ' dQ Q(r− r ') dQ(r− r ') r ' E( r ) = →dE = πε − 3 πε − 3 P (0, y, 0) 40 r r ' 40 r r ' r dQ x y ρ = →dQ = ρ dz ' ρ L dz ' L L ρ dz '(r− r ') →dE = L 4πε r− r ' 3 0 ρdz'( ρ a− z ' a ) →dE = Lρ z = = ρ rr−' =ρ a − z ' a 2 23/2 ry ay a ρ  ρ z 4πε ( ρ + z ') →  0 = − =ρ 2 + 2 r'z ' a z  r r 'z ' Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 27 z Line Charge (4) z ' dQ ρdz'( ρ aρ − z ' a ) dE = L z r ' πε ρ 2+ 23/2 40 (z ') P (0, y, 0) E is not a function of z x r y ρ ρ ρ dz ' L →dE = L ρ πε ρ 2+ 23/2 40 (z ') ∞ ρ ρ dz ' ρ →E = L = L ρ ∫−∞ πε ρ 2+ 23/2 πε ρ 40 (z ') 2 0 ρ → E= L a πε ρ ρ 2 0 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 28 ρ E= L a Line Charge (5) πε ρ ρ 2 0 8 6 4 2 0 -2 -4 -6 -8 2.5 2 1.5 2.5 1 2 0.5 1.5 0 1 0.5 -0.5 0 -1 -0.5 -1.5 -1 -1.5 -2 -2 -2.5 -2.5 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 29 Ex. 1 Line Charge (6) Infinite uniform line charge of 10 nC/m lie along the x & y axes in free space. Find E at (0, 0, 3). Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 30 Ex. 2 Line Charge (7) The x & y axes are charged with uniform line charge of 10 nC/m. A point charge of 20nC is located at (3, 3, 0). The whole system is in free space. Find E at (0, 0, 3). Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 31 z Ex. 3 Line Charge (8) dE p Given a circular hoop of radius a with uniform line charge ρL centered about the origin in the z = 0 plane. Find EFI at P? θ dE1 dE2 = dQ 1 P (0, 0, z) dE 1 2 πε r2 4 0r 1 ρ ϕ → = Lad dE 1 dQ dQ=ρ dL = ρ ad ϕ πε 2+ 2 r1 1 1 L L 40 (a z ) 2 2 ϕ a y r= a + z dE=2 dE = 2 dE cos θ dQ 1 Pz1 z 1 2 ρ L z x cos θ = a2+ z 2 π ρazd ϕ ρazd ϕ ρ az →dE = L →E = L = L Pz πε 2+ 23/2 Pz ∫ πε 2+ 23/2 ε 2+ 23/2 20 (a z ) ϕ =0 20 (a z ) 20 (a z ) ρ az →E = L a Pε 2+ 23/2 z 20 (a z ) Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 32 Coulomb’s Law & Electric Field Intensity 1. Coulomb’s Law 2. Electric Field Intensity 3. Field Due to a Continuous Volume Charge Distribution 4. Field of a Line Charge 5. Field of a Sheet Charge 6. Sketches of Fields Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 33 Sheet Charge (1) • Charge is distributed on the surface of a plate (e.g. of a parallel-plate capacitor) 2 • Sheet/surface charge density ρS (unit: C/m ) ρ = dQ S dS Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 34 Sheet Charge (2) z ρ ρ dy ' = L → = L E a R dE a ρ 2πε ρ πε 2+ 2 R S y ' 0 20 x y ' dQ=ρ dS = ρ Ldy ' S S P( x ,0,0) θ y L → ∞ dE R= x2 + y ' 2 dQ ρ Ldy ' dEx →ρ = =S = ρ dy ' x L L L S ρ dy 'a →dE = S R ρ θ πε 2+ 2 → = S dy 'cos 20 x y ' dE x πε 2+ 2 20 x y ' dE x = dE cos θ Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 35 Sheet Charge (3) z dy ' ρdy 'cos θ = S ρ dE x S y ' πε 2+ 2 20 x y ' x cos θ = P( x ,0,0) θ y x2+ y ' 2 dE 2 2 ρ xdy ' dE R= x + y ' →dE = S . x x πε 2+ 2 x 20 x y ' ρ ∞ xdy ' ρ ρ →E = S = S → E= S a x πε ∫−∞ 2+ 2 ε ε N 20 x y ' 2 0 2 0 (aN: vector perpendicular to the sheet) Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 36 ρ E= S a Sheet Charge (4) ε N 2 0 2.5 2 1.5 1 0.5 0 10 8 6 10 4 8 2 6 4 0 2 -2 0 -4 -2 -6 -4 -6 -8 -8 -10 -10 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 37 Sheet Charge (5) ρ E= S a 2ε N – ρ 0 S ρS z x> a x < 0 ρ ρ E= S a 0 <x < a E= − S a + ε x a 0 + ε x 2 0 2 ρ x y ρ 0 = − S ρ E= S a E- a x = S - x 2ε E+ a x 2ε 0 2ε 0 ρ 0 →= + = = S EEE+ − 0 E a →=EEE + = 0 - 2ε x + − 0 ρ →=EEE + = S a + − ε x 0 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 38 Ex. 1 Sheet Charge (6) Given 3 infinite uniform sheets (all parallel to x0y) at z = – 3, z = 2 & z = 3. Their surface charge density are 4 nC/m2, 6 nC/m2 & –9 nC/m2 respectively. Find E at P(5, 5, 5). Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 39 z dy ' Ex. 2 Sheet Charge (7) A uniformly volume charge density charge ρv of infinite extent in the x & z directions & of width − a 2a is centered about the y axis. Find EFI? a 0 y ρ ρ= ρ v S v dy ' ρ ρ dy ' a ρdy' a ρ y≤− a: dE =−S =− v →E =−v =− v y ε ε y ∫ ε ε 20 2 0 −a 2 0 0 ρ ρ dy ' a ρdy' a ρ y≥ a: dE =S = v →E =v = v y ε ε y ∫ ε ε 20 2 0 −a 2 0 0 y ρdy'a ρ dy ' ρ − ≤ ≤ →=Ev − v = v y a y a : y ∫ε ∫ ε ε −a20 y 2 0 0 Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 40 Ex. 3 Sheet Charge (8) Given a circular disk of radius a with uniform surface charge z ρ centered about the origin in the z = 0 plane. Find EFI at P? S P (0, 0, z) ρ ρ ρ= ρd ρ S L S y ρ ρ ρ z d a dE = L Pz ε ρ 2+ 23/2 20 (z ) x (ρd ρ ) ρ z a (ρd ρ ) ρ z ρ z z  →dE = S →E = S = −S  −  Pz ε ρ 2+ 23/2 Pz ∫ ε ρ 2+ 23/2 ε 2 2 20 (z ) 0 20 (z ) 2 0 a+ z z  Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 41 Ex. 4 Sheet Charge (9) z A hollow cylinder of radius a & length 2 L with uniform P (0, 0, z) surface charge ρS on its outer surface. Find EFI at P? L dz' z' y ρ= ρ L S dz ' –L ρ az dE = L x a Pz ε 2+ 23/2 ρS 20 (a z ) (ρ dz ') az (− z ') L (ρ dz ') az (− z ') →dE = S →E = S Pz ε 2+ − 23/2 Pz ∫ ε 2+ − 23/2 20 [a ( z z ') ] −L 2[0 a ( z z ')] ρ a 1 1  =S  −  ε 2+− 22 ++ 2 2 0 a() zL a () zL  Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 42 Ex. 5 Sheet Charge (10) z A cylinder (of radius a & length 2 L) is uniformly charged P (0, 0, z) throughout the volume with volume charge density ρv. Find L EFI at P? y ρv –L x a Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 43 Basic Charge Configurations Point Charge Line Charge − ρ Q r r ' L E = . E= a ρ πε − 2 r− r ' 2πε ρ 40 r r ' 0 Sheet Charge Volume Charge ρ ρ (r ')dV ' r− r ' = S = v E a N E ∫ 2 . 2ε V πε − r− r ' 0 40 r r ' Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 44 Coulomb’s Law & Electric Field Intensity 1. Coulomb’s Law 2. Electric Field Intensity 3. Field Due to a Continuous Volume Charge Distribution 4. Field of a Line Charge 5. Field of a Sheet Charge 6. Sketches of Fields Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 45 Sketches of Fields • To “picture” a field • A set of vectors of a field Coulomb’s Law & Electric Field Intensity - sites.google.com/site/ncpdhbkhn 46