Energy 2017 mk

Nguyễn Công Phương Engineering Electromagnetics Energy and Potential Contents I II III IV V VI VII VIII IX X XI XII XIII XIV XV Introduction Vector Analysis Coulomb’s Law & Electric Field Intensity Electric Flux Density, Gauss’ Law & Divergence Energy & Potential Current & Conductors Dielectrics & Capacitance Poisson’s & Laplace’s Equations The Steady Magnetic Field Magnetic Forces & Inductance Time – Varying Fields & Maxwell’s Equations Transmission Lines The Uniform Plane Wave Plane Wave Reflection & Dispersion Guided Waves & Radiation Engineering Electromagnetics - sites.google.com/site/ncpdhbkhn Energy & Potential Moving a Point Charge in an Electric Field The Line Integral Potential Difference & Potential The Potential Field of a Point Charge The Potential Field of a System of Charges Potential Gradient The Dipole Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn Moving a Point Charge in an Electric Field (1) • Moving a charge Q a distance dL in an E, the force on Q arising from the electric field: F E = QE • The component in the direction dL: FEL = F.aL = QE.aL • aL: a unit vector in the direction of dL • → the force must be applied: Feff = – QE.aL • The expenditure of energy: dW = – QE.aLdL = – QE.dL Energy & Potential - sites.google.com/site/ncpdhbkhn Moving a Point Charge in an Electric Field (2) • The expenditure of energy required to move Q in E: dW = – QE.dL • dW = if: – Q = 0, E = 0, dL = 0, or – E is perpendicular to dL • The work needed to move the charge a finite distance: W = −Q ∫ final init E.dL Energy & Potential - sites.google.com/site/ncpdhbkhn Energy & Potential Moving a Point Charge in an Electric Field The Line Integral Potential Difference & Potential The Potential Field of a Point Charge The Potential Field of a System of Charges Potential Gradient The Dipole Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn EL6 The Line Integral (1) A ΔL6 EL5 EL4 EL3 EL2 EL1 W = dW1 + dW2 + + dW6 B ΔL1 E ΔL2 E ΔL3 E ΔL4 E W = −Q ∫ = −QEL1.∆L1 − QEL ∆L2 − − QEL ∆L6 = −QE1.∆L1 − QE2 ∆L2 − − QE6 ∆L final init ΔL5 E E E LdL = −QE.L BA (uniform E) E1 = E2 = = E6 = E → W = −QE.(∆L1 + ∆L + + ∆L6 ) → W = −QE.L BA ∆L1 + ∆L2 + + ∆L = L BA Energy & Potential - sites.google.com/site/ncpdhbkhn (uniform E) EL6 The Line Integral (2) W = −Q ∫ final init W = −Q ∫ final init E L dL = −QE.L BA ΔL6 EL5 (uniform E) EL4 EL3 EL2 E.dL EL1 Uniform E A B ΔL1 ΔL2 E ΔL3 E ΔL4 E ΔL5 E E E A → W = −QE.∫ dL = −QE.L BA B Energy & Potential - sites.google.com/site/ncpdhbkhn The Line Integral (3) Ex Given E = yax + xay + zaz V/m Find the work needed in carrying C from B(1; 0; 1) to A(0.8; 0.6; 1) along: a)the shorter arc of the circle x2 + y2 = 1, z = 1; b)the straight-line path from B to A A W = −Q∫ E.dL B dL = dxa x + dya y + dza z A → W = −2 ∫ ( ya x + xa y + za z ).( dxa x + dya y + dza z ) = −2 ∫ B x =0.8 x =1 = −2 ∫ x = 0.8 x =1 ydx − ∫ y =0.6 y =0 xdy − ∫ zdz 1 − x dx − ∫ y =0.6 y =0 0.8 − y dy − 0.6 = −  x − x + sin −1 x  −  y − y + sin −1 y  = −0.96 J  1  0 Energy & Potential - sites.google.com/site/ncpdhbkhn The Line Integral (4) Ex Given E = yax + xay + zaz V/m Find the work needed in carrying C from B(1; 0; 1) to A(0.8; 0.6; 1) along: a)the shorter arc of the circle x2 + y2 = 1, z = 1; b)the straight-line path from B to A A W = −Q ∫ E.dL B dL = dxa x + dya y + dza z A → W = −2 ∫ ( ya x + xa y + za z ).( dxa x + dya y + dza z ) B = −2 ∫ x =0.8 x =1 ydx − ∫ y =0.6 y =0 xdy − ∫ zdz y A − yB y − yB = ( x − xB ) → y = −3( x − 1) xA − xB → W = 6∫ x = 0.8 x =1 ( x − 1)dx − ∫ y =0.6 y =0 y   −  dy − = −0.96 J   Energy & Potential - sites.google.com/site/ncpdhbkhn 10 The Dipole (2) E= Qd (2cos θ a r + sin θ aθ ) 4πε r z 0.4 Qd cos θ V = 4πε r 0.6 0.8 Energy & Potential - sites.google.com/site/ncpdhbkhn 44 The Dipole (3) z The dipole moment p = Qd +Q d.ar = d cos θ d d Qd cos θ V = 4πε r P θ ar R1 r R2 y x –Q p.a r r −r' →V = = p 2 4πε r r −r' 4πε r − r ' r : locates P r’: locates the dipole center Energy & Potential - sites.google.com/site/ncpdhbkhn 45 Energy & Potential Moving a Point Charge in an Electric Field The Line Integral Potential Difference & Potential The Potential Field of a Point Charge The Potential Field of a System of Charges Potential Gradient The Dipole Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 46 Energy Density in the Electrostatic Field (1) • Carrying a positive charge (1) from infinity into the field of another fixed positive charge (2) needs work • If the charge is held near the charge 2, it has a potential energy • If then the charge is released, it will accelerate away from the charge 2, acquiring kinetic energy • Problem: find the potential energy present in a system of charges Energy & Potential - sites.google.com/site/ncpdhbkhn 47 Energy Density in the Electrostatic Field (2) • • • • • • (Work to position Q2) = Q2V2, V2, 1: the potential at Q2 due to Q1 An additional charge Q3: (Work to position Q3) = Q3V3, + Q3V3, (Work to position Q4) = Q4V4, + Q4V4, + Q4V4, Total positioning work = potential energy of field = = WE = Q2V2, + Q3V3, + Q3V3, + Q4V4, + Q4V4, + Q4V4, + … Energy & Potential - sites.google.com/site/ncpdhbkhn + 48 Energy Density in the Electrostatic Field (3) WE = Q2V2, + Q3V3, + Q3V3, + Q4V4, 1+ Q4V4, + Q4V4, + … Q1 Q3V3,1 = Q3 Q3 4πε R13 → Q3V3,1 = Q1 = Q1V1,3 4πε R31 R13 = R31 WE = Q1V1, + Q1V1, + Q2V2, + Q1V1, 4+ Q2V2, + Q3V3, +… +W E = Q2V2, + Q3V3, + Q3V3, + Q4V4, 1+ Q4V4, + Q4V4, + … 2WE = Q1 (V1,2 + V1,3 + V1,4 + ) + + Q2 (V2,1 + V2,3 + V2,4 + ) + + Q3 (V3,1 + V3,2 + V3,4 + ) + + Energy & Potential - sites.google.com/site/ncpdhbkhn 49 Energy Density in the Electrostatic Field (4) 2WE = Q1 (V1,2 + V1,3 + V1,4 + ) + Q2 (V2,1 + V2,3 + V2,4 + ) + + Q2 (V3,1 + V3,2 + V3,4 + ) + V1,2 + V1,3 + V1,4 + = V1 V2,1 + V2,3 + V2,4 + = V2 V3,1 + V3,2 + V3,4 + = V3 1 N → WE = (Q1V1 + Q2V2 + Q3V3 + ) = ∑ QkVk 2 k =1 → WE = ∫ ρ vVdv V Qk = ρv dv Energy & Potential - sites.google.com/site/ncpdhbkhn 50 Energy Density in the Electrostatic Field (5) WE = ∫ ρ vVdv V Maxwell’s 1st equation: ∇.D = ρ v → WE = ∫ (∇.D)Vdv V ∇.(VD) ≡ V (∇.D) + D.(∇V ) → WE = ∫ [∇.(VD) − D.(∇V ) ] dv V Energy & Potential - sites.google.com/site/ncpdhbkhn 51 Energy Density in the Electrostatic Field (6) WE = ∫ [∇.(VD) − D.(∇V ) ] dv V 1 = ∫ ∇.(VD)dv − ∫ D.(∇V ) dv V V ∇.(VD) dv ∫ V Div theorem: ∫ D.dS = ∫ ∇.Ddv S V 1 → ∫ ∇.(VD)dv = ∫ (VD).dS V S 1 → WE = ∫ (VD).dS − ∫ D.(∇V )dv S V Energy & Potential - sites.google.com/site/ncpdhbkhn 52 Energy Density in the Electrostatic Field (7) WE = V= Q : 4πε r Q D= a : r 4π r dS ∫ S (VD).dS − ∫V D.(∇V )dv with 1/r with 1/r2 → ∫ S (VD).dS = : increases with r2 → WE = − ∫ D.(∇V ) dv 1 V → WE = ∫ D.Edv = ∫ ε E dv V V 2 E = −∇V (pot grad) Energy & Potential - sites.google.com/site/ncpdhbkhn 53 Ex Energy Density in the Electrostatic Field (8) Given a coaxial cable, the surface charge density of the outer surface of the inner cylinder is ρS Find its potential energy? ρ=a ρ=b Method 1: WE = ∫ ε E dv V Dρ = aρ S ρ (a < ρ < b) → E = aρS ε0ρ z = L ϕ = 2π ρ =b  a ρ S  → WE = ∫ ∫ ε0   dv ∫ ϕ ρ z = = = a  ε0ρ  dv = ρ d ρ dϕ dz z = L ϕ = 2π ρ = b a ρ S2 π La ρ S2 b → WE = ∫ ∫ ε 2 ρ d ρ dϕ dz = ln ∫ z = = = a ϕ ρ ε0 ρ ε0 a Energy & Potential - sites.google.com/site/ncpdhbkhn 54 Ex Energy Density in the Electrostatic Field (9) Given a coaxial cable, the surface charge density of the outer surface of the inner cylinder is ρS Find its potential energy? ρ=b WE = ∫ ρvVdv V Method 2: V AB = − ∫ ρ=a final E.dL a → Va = − ∫ Eρ d ρ b a ρS Vb = Eρ = ε0ρ a aρ aρ S b S ln → Va = − ∫ dρ = a ρS b b ε ρ ε a → WE = ∫ ρ v ln dv V a ε0 WE = ∫ ρ vVdv V init Energy & Potential - sites.google.com/site/ncpdhbkhn 55 Ex Energy Density in the Electrostatic Field (10) Given a coaxial cable, the surface charge density of the outer surface of the inner cylinder is ρS Find its potential energy? ρ=a ρ=b Method 2: WE = ∫ ρvVdv V a ρS b = ∫ ρv ln dv v a ε0 ρS t t ρv = , a − ≤ ρ ≤ a + , t ≪ a t 2 z = L ϕ = 2π ρ =a +t / ρS ρ S b → WE = ∫ ∫ a ln ρ d ρ d ϕ dz ∫ z =0 ϕ =0 ρ = a −t / t ε a π La ρ S2 b = ln ε0 a Energy & Potential - sites.google.com/site/ncpdhbkhn 56 Ex Energy Density in the Electrostatic Field (11) A metallic sphere of radius 10cm has a surface charge density of 10nC/m2 Calculate the electric energy stored in the system Method 1: WE = ∫ ε E dv V 2 ρ R ρ R 2 S S D d S = Q → D (4 π r ) = ρ (4 π R ) → D = → E = total ∫S S ε 0r r2  ρS R  → WE = ∫ ε  dv  2 V  ε 0r  ∞ π 2π (0.1)2 × 10−18 = ∫ ∫ ∫ r sin θ drdθ dϕ r =0.1 θ = ϕ = ε 0r = 71.06 nJ Energy & Potential - sites.google.com/site/ncpdhbkhn 57 Ex Energy Density in the Electrostatic Field (12) A metallic sphere of radius 10cm has a surface charge density of 10nC/m2 Calculate the electric energy stored in the system Method 2: ? Energy & Potential - sites.google.com/site/ncpdhbkhn 58 ... potential energy • If then the charge is released, it will accelerate away from the charge 2, acquiring kinetic energy • Problem: find the potential energy present in a system of charges Energy. .. System of Charges Potential Gradient The Dipole Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 46 Energy Density in the Electrostatic Field (1) •... The expenditure of energy: dW = – QE.aLdL = – QE.dL Energy & Potential - sites.google.com/site/ncpdhbkhn Moving a Point Charge in an Electric Field (2) • The expenditure of energy required to