Bài giảng Engineering electromagnetic - Chapter V: Energy & Potential - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Energy and Potential 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 Energy & Potential - sites.google.com/site/ncpdhbkhn 2 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 3 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: FE = 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 4 Moving a Point Charge in an Electric Field (2) • The expenditure of energy required to move Q in E: dW = – QE.dL • dW = 0 if: – Q = 0, E = 0, dL = 0, or – E is perpendicular to dL • The work needed to move the charge a finite distance: final W= − QE. d L ∫init Energy & Potential - sites.google.com/site/ncpdhbkhn 5 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 6 EL6 A The Line Integral (1) ∆L6 EL5 ∆L E EL4 5 E EL3 ∆L4 E ∆L3 EL2 E ∆L2 E ∆L1 final EL1 = − W Q∫ EL dL = + ++ E init W dW1 dW 2... dW 6 B =− ∆− ∆−− ∆ = − QE.L QEL11. L QE L 22 . L ... QE L 66 . L BA =− ∆− ∆−− ∆ (uniform E) QQQELELEL11. 22 . ... 66 . = == = EEEE1 2... 6 →WQ =−E.( ∆ L +∆ L + ... +∆ L ) 1 2 6 →WQ = − E.L ∆ +∆ + +∆ = BA LLLL1 2... 6 BA Energy & Potential - sites.google.com/site/ncpdhbkhn (uniform E) 7 EL6 A The Line Integral (2) ∆L6 E final L5 =− =− E ∆L5 E W Q EL dL QEL. BA (uniform E) L4 ∫init E EL3 ∆L4 E ∆L3 EL2 E final ∆L2 W= − QEL. d E ∫init ∆L1 EL1 E Uniform E B A = − →W = − QE. d L QE.LBA ∫B Energy & Potential - sites.google.com/site/ncpdhbkhn 8 Ex. 1 The Line Integral (3) Given E = yax + xay + zaz V/m. Find the work needed in carrying 2 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= − QE. d L ∫B = + + dL dx ax dy a y dz a z →=−A ++ ++ W2 ( yaaaxyz x z ).( dx a x dy a y dz a z ) ∫B x=0.8 y = 0.6 1 =−2ydx − 2 xdy − 2 zdz ∫x=1 ∫ y = 0 ∫ 1 x=0.8 y = 0.6 =−21 −−x2 dx 21 −− y 2 dy 0 ∫x=1 ∫ y = 0 0.8 0.6 =−xx1 −+21 sin− x  − yy 1 −+ 21 sin − y  = − 0.96 J  1  0 Energy & Potential - sites.google.com/site/ncpdhbkhn 9 Ex. 1 The Line Integral (4) Given E = yax + xay + zaz V/m. Find the work needed in carrying 2 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= − QE. d L ∫B = + + dL dx ax dy a y dz a z →=−A ++ ++ W2 ( yaaaxyz x z ).( dx a x dy a y dz a z ) ∫B x=0.8 y = 0.6 1 =−2ydx − 2 xdy − 2 zdz ∫x=1 ∫ y = 0 ∫ 1 y− y yy− =A B ( xx − ) →y =−3( x − 1) B− B xA x B x=0.8 y = 0.6 y  →=W6∫ (1)2 x −− dx ∫  1 −  dy − 0 = − 0.96 J x=1 y = 0 3  Energy & Potential - sites.google.com/site/ncpdhbkhn 10 The Line Integral (5) = + + dL dx ax dy a y dz a z (Descartes) =ρ + ρ ϕ + (Cylindrical) ddL aρ d a ϕ dz a z = +θ + θ ϕ (Spherical) dLa drr rd aθ rsin d a ϕ Energy & Potential - sites.google.com/site/ncpdhbkhn 11 z Ex. 2 The Line Integral (6) Find the work needed in carrying the charge Q about a dL circular path centered at the line charged. y ρL final x W= − Q∫ EL. d ρ init L E= a ρ πε ρ final ρ 2 0 → = − L ρ ϕ W Q∫ aρ. d a ϕ ddL=ρ aρ + ρ d ϕ a ϕ + dz a init πε ρ z 2 0 2π ρ dρ = 0 L = − Q d ϕaρ. a ϕ ∫0 πε dz = 0 2 0 o aρ. a ϕ = 1 × 1 × cos90 ρ 2π →W = − QL cos90 o d ϕ = 0 πε ∫0 2 0 Energy & Potential - sites.google.com/site/ncpdhbkhn 12 z Ex. 3 The Line Integral (7) ρL Find the work done in carrying a charge Q from ρ = a to ρ = b. a y dL final x b W= − Q∫ EL. d ρ init L E= a ρ πε ρ final ρ 2 0 L →W = − Qaρ. d ρ a ρ ddL=ρ a + ρ d ϕ a + dz a ∫init πε ρ ρ ϕ z 2 0 dϕ = 0 b ρ dρ = − Q L dz = 0 ∫a πε ρ 2 0 Qρ b = − L ln πε 2 0 a Energy & Potential - sites.google.com/site/ncpdhbkhn 13 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 14 Potential Difference & Potential (1) final W= − QEL. d ∫init • Potential difference V: work done in moving a unit positive charge from one point to another in an electric field: final Potential difference =V = − QEL. d ∫init • Potential difference between points A & B: = − A VAB E. d L ∫B • Unit: volt (V, J/C) Energy & Potential - sites.google.com/site/ncpdhbkhn 15 z Ex. Potential Difference & Potential (2) ρL Find the potential difference between ρ = a & ρ = b. Work done in carrying Q from a to b: a y x Qρ b b W = − L ln πε 2 0 a → work done in carrying Q from b to a: Qρ b W = L ln πε 2 0 a ρ → = L b Vab ln W 2πε a V = 0 ab Q Energy & Potential - sites.google.com/site/ncpdhbkhn 16 Potential Difference & Potential (3) A • Potential difference between points A & B: V= − E. d L AB ∫B • No B? • → potential (absolute potential) at A • → still need a reference point: – “ground” – Infinity • If the potential at A is VA & that at B is VB, then: VAB = VA – VB • (provided VA & VB have the same zero reference point) Energy & Potential - sites.google.com/site/ncpdhbkhn 17 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 18 The Potential Field of a Point Charge (1) A B = − A VAB E. d L ∫B rA rB Q rA Q = →V = − dr E2 a r AB ∫ 2 πε rB πε Q 4 0r 4 0r = dL dr a r Q 1 1  = −  Q 4πε r r →V = 0 A B  A πε 4 0rA → ∞ rB Q V = πε 4 0r (Potential field of a point charge) Energy & Potential - sites.google.com/site/ncpdhbkhn 19 The Potential Field of a Point Charge (2) Q V = πε 4 0r • The potential at any point distant r from a point charge Q • The zero reference is the potential at infinite radius • Q/4 πε 0r (J) must be done in carrying a 1-C charge from infinity to any point r meters from the charge Q Q Q • If = C →VC = + πε 1 πε 1 4 0rB 4 0r • The potential difference does not depend on C1 Energy & Potential - sites.google.com/site/ncpdhbkhn 20 The Potential Field of a Point Charge (3) Q V = πε 4 0r • The potential field of a point charge • A scalar field, & no unit vector • Equipotential surface : a surface composed of all those points having the same value of potential • No work is required in moving a charge around on an equipotential surface • The equipotential surfaces in the potential field of a point charge are spheres centered at the point charge Energy & Potential - sites.google.com/site/ncpdhbkhn 21 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 22 The Potential Field of a System of Charges (1) Q Q2 r – r V (r ) = 1 2 πε − r2 4 0r r 1 r – r1 Q1 QQ V (r ) =1 + 2 r πε− πε − r1 40rr 1 4 0 rr 2 Origin QQ Q n Q V (r )=1 + 2 ++ ... n = m πε− πε − πε − ∑ πε − 401rr 4 02 rr 4 0 rr n m=1 4 0 r r m =ρ ∆ Qm v v m ρ()r∆v ρ () r ∆ v ρ () r ∆ v →=V (r )v11 + v 22 ++ ... vnn πε− πε − πε − 401rr 4 02 rr 4 0 rr n Energy & Potential - sites.google.com/site/ncpdhbkhn 23 The Potential Field of a System of Charges (2) ρ()r∆v ρ () r ∆ v ρ () r ∆ v V (r )=v11 + v 22 ++ ... vnn πε− πε − πε − 401rr 4 02 rr 4 0 rr n n → ∞ ρ (r ')dv ' →V (r ) = v ∫V πε − 40 r r ' ρ (r ')dL ' V (r ) = L ∫ πε − 40 r r ' ρ (r ')dS ' V (r ) = S ∫S πε − 40 r r ' Energy & Potential - sites.google.com/site/ncpdhbkhn 24 The Potential Field of a System of Charges (3) Ex. 1 z Find the potential on the z axis. (0, 0, z) ρ (r ')dL ' V (r ) = L r r− r ' =a2 + z 2 ∫ 4πε r− r ' 0 ρ = a dL'= ad ϕ ' y = ϕ ' r’ rz a z 2 2 dL'= ad ϕ ' →−r r ' =a + z x ρL r' = a a ρ 2π ρad ϕ ' ρ a →V (r ) = ∫ L = L 0 πε 2+ 2 ε 2+ 2 4 0 a z 2 0 a z Energy & Potential - sites.google.com/site/ncpdhbkhn 25 The Potential Field of a System of Charges (4) For a zero reference at infinity, then: • The potential due to a single point charge: the work done in carrying a unit positive charge from infinity to the point at which we desire the potential, this work does not depend on the path chosen between these two points • The potential field due to a number of point charges is the sum of the individual potential fields due to each charge A • The expression for potential: V= − E. d L A ∫∞ A • The potential difference: VVV= − =− E. d L AB A B ∫B • For a static field: ∫ E.d L = 0 Energy & Potential - sites.google.com/site/ncpdhbkhn 26 The Potential Field of a System of Charges (5) Ex. 2 z Investigate the uniform line charge density ρL of finite L = ρ length 2 L centered on the z axis. z ' dQL dz ' Q R V = ρ ϕ point charge πε P( , ,) z y 4 0r ρ dQ ρ dz ' ϕ →dV = = L πε πε ρ 2+ − 2 x 4 0R 40 (z z ') −L L ρ dz ' →V = L ∫ πε ρ 2+ − 2 −L 40 (z z ') ρ zL−+ρ 2 +−( zL ) 2  = − L ln   πε ++ρ 2 +− 2  4 0 zL( zL )  Energy & Potential - sites.google.com/site/ncpdhbkhn 27 The Potential Field of a System of Charges (6) P Ex. 3 r Investigate a sphere of radius R has a uniform surface rQP charge density ρS. 2 ρ dQ = ρ = ρ θθϕ S R dQS dS S Rsin d d dQρ R2 sin θθϕ d d →dV = = S 4πεr 4 πε r 0QP 0 QP z 2= 2 + 2 − ϕ dS = rsin θdrd φaθ rQP R r2 rR cos 2 dS = r sin θdθdφar dr dS = rdrd θaφ y rd θ x rsin θdφ Energy & Potential - sites.google.com/site/ncpdhbkhn 28 The Potential Field of a System of Charges (6) P Ex. 3 r Investigate a sphere of radius R has a uniform surface rQP charge density ρS. 2 ρ dQ = ρ = ρ θθϕ S R dQS dS S Rsin d d dQρ R2 sin θθϕ d d →dV = = S πε πε 40rQP 4 0 r QP Rrsin θ d θ 2= 2 + 2 − θ → = θ θ →r = rQP R r2 rR cos 2rQP dr QP 2 rR sin d QP dr QP ρRdr d ϕ →dV = S QP πε 2 4 0r  ρ R  S ,r> R r+ R 2π ρRdr d ϕ ε r →V = S QP =  0 ∫r= r − R ∫ ϕ = 0 QP 4πε r  ρ R 0 S ,r< R  ε  0 Energy & Potential - sites.google.com/site/ncpdhbkhn 29 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 30 Potential Gradient (1) • 2 methods to find potential: from electric field intensity & from charge distribution • however E & ρv, S, L are often not given • → problem: finding EFI from potential • solution: potential gradient Energy & Potential - sites.google.com/site/ncpdhbkhn 31 ∆L Potential Gradient (2) θ E V= − ∫ E. d L ∆V ≐ −E. ∆ L ∆V≐ − E ∆ L cos θ dV = − E cos θ dL dV E =(cosθ = − 1) dL max Energy & Potential - sites.google.com/site/ncpdhbkhn 32 Potential Gradient (3) V = +90 +80 +70 dV +60 E = +50 aN dL max +40 P ∆L • The magnitude of E is given by the E maximum value of the rate of change of potential with distance +30 • This maximum value is obtained when +20 the direction of the distance increment +10 is opposite to E, or, in other words, the direction of E is opposite to the   = − dV direction in which the potential is E  a N increasing the most rapidly dL max  Energy & Potential - sites.google.com/site/ncpdhbkhn 33 Potential Gradient (4) V = +90 +80 dV  +70 = − +60 E  a N +50 dL max  aN dV dV dV ∆L +40 = →E = − a P N E dLmax dN dN +30 dT +20 Gradient of T = grad T = a dN N +10 E = − grad V Energy & Potential - sites.google.com/site/ncpdhbkhn 34 Potential Gradient (5) E = − grad V  ∂V E = −  x ∂ ∂V ∂ V ∂ V  x V= Vxyz(, , ) →=dV dx + dy + dz ∂ ∂ ∂ ∂ → = − V x y z Ey  ∂y dV=−E. d L =− Edx − Edy − Edz x y z  ∂V E = −  z ∂z ∂ ∂ ∂  →=−V + V + V E ax a y a z  ∂x ∂ y ∂ z  ∂V ∂ V ∂ V → grad V =a + a + a ∂xx ∂ y y ∂ z z Energy & Potential - sites.google.com/site/ncpdhbkhn 35 Potential Gradient (6) ∂V ∂ V ∂ V grad V =a + a + a ∂xx ∂ y y ∂ z z ∂ ∂ ∂ ∂T ∂ T ∂ T ∇=a + a + a →∇=T a + a + a ∂xx ∂ y y ∂ z z ∂xx ∂ y y ∂ z z → ∇T = grad T E = − grad V → E = −∇V Energy & Potential - sites.google.com/site/ncpdhbkhn 36 Potential Gradient (7) ∂V ∂ V ∂ V ∇=V a + a + a (Descartes) ∂xx ∂ y y ∂ z z ∂V1 ∂ V ∂ V ∇=V aρ + a ϕ + a (Cylindrical) ∂ρ ρ ∂ ϕ ∂ z z ∂V1 ∂ V 1 ∂ V ∇=V a + aθ + a ϕ (Spherical ) ∂rr r ∂θ r sin θ ∂ ϕ Energy & Potential - sites.google.com/site/ncpdhbkhn 37 Potential Gradient (8) ∂V ∂ V ∂ V Gradient: ∇=V a + a + a ∂xx ∂ y y ∂ z z ∂DD∂D ∂ Divergence: ∇=.D x +y + z ∂x ∂ y ∂ z Energy & Potential - sites.google.com/site/ncpdhbkhn 38 Ex. 1 Potential Gradient (9) Find the gradient of each of the following functions: =2 − 3 af)1 2 ay 5 yz =ρϕ + ρ ϕ b) f2 6 sin 4 z cos3 1 c) f= + 2 r sinθ cos ϕ 3 r Energy & Potential - sites.google.com/site/ncpdhbkhn 39 Ex. 2 Potential Gradient (10) Given a potential field V = x2 – 10 yz (V) & a point P(1, 3, 1). Find several values at P: VP, EP , the direction of EP , DP , & ρv . =2 − ××=− VP 1 10 3 1 29 V ∂ ∂ ∂  = −∇ =−V + V + V =− + + E V ax a y a z  2xax 10 z a y 10 y a z V/m ∂x ∂ y ∂ z  → =−× +× +× =− + + Ep21 a x 101 a y 103 aaaa zxyz 2 10 30V/m E −2a + 10 a + 30 a a = p = x y z =−0.063a + 0.32 a + 0.95 a E, P −2 + 2 + 2 x y z E p ( 2) 10 30 Energy & Potential - sites.google.com/site/ncpdhbkhn 40 Ex. 2 Potential Gradient (11) Given a potential field V = x2 – 10 yz (V) & a point P(1, 3, 1). Find several values at P: VP, EP , the direction of EP , DP , & ρv . = ε = ×−12 −+ + DE0 8.854 10 ( 2xax 10 z a y 10 y a z ) =− + + 2 17.71xax 88.54 z a y 88.54 y a z pC/m ρ = ∇ v .D ∂∂D ∂ =DDx +y + z ∂x ∂ y ∂ z ∂−( 17.71x ) ∂ (88.84 z ) ∂ (88.84 y ) = + + = − 17.71 pC/m3 ∂x ∂ y ∂ z Energy & Potential - sites.google.com/site/ncpdhbkhn 41 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. Energy Density in the Electrostatic Field Energy & Potential - sites.google.com/site/ncpdhbkhn 42 The Dipole (1) z P θ R1 Q1 1  Q R− R V = − = 2 1 +Q πε  πε r 401RR 2  4 012 RR R2 R1≐ R 2 d − θ y R2 R 1 ≐ d cos –Q Qd cos θ x → V = πε 2 z 4 0r R1 = −∇ θ E V +Q ∂ ∂ ∂  r =−V +1 V + 1 V R2 ar aθ a ϕ  ∂r r ∂θ r sin θ ∂ ϕ  d Qd y E=(2cosθ a + sin θ a θ ) 4πε r3 r R− R≐ d cos θ 0 x –Q 2 1 Energy & Potential - sites.google.com/site/ncpdhbkhn 43 The Dipole (2) Qd E=(2cosθ a + sin θ a ) z πε 3 r θ 0,4 4 0r Qd cos θ 0,6 V = πε 2 0,8 4 0r 1 Energy & Potential - sites.google.com/site/ncpdhbkhn 44 The Dipole (3) z P θ R1 +Q The dipole moment p= Q d r ar R2 = θ d d.a r d cos d y Qd cos θ V = –Q πε 2 x 4 0r p.a 1r− r ' →V = r = p. 4πε r 2 πε − 2 r− r ' 0 40 r r ' r : locates P r’: locates the dipole center Energy & Potential - sites.google.com/site/ncpdhbkhn 45 Energy & Potential 1. Moving a Point Charge in an Electric Field 2. The Line Integral 3. Potential Difference & Potential 4. The Potential Field of a Point Charge 5. The Potential Field of a System of Charges 6. Potential Gradient 7. The Dipole 8. 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 1 is held near the charge 2, it has a potential energy • If then the charge 1 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, 1 • V2, 1 : the potential at Q2 due to Q1 • An additional charge Q3: • (Work to position Q3) = Q3V3, 1 + Q3V3, 2 • (Work to position Q4) = Q4V4, 1 + Q4V4, 2 + Q4V4, 3 • Total positioning work = potential energy of field = = WE = Q2V2, 1 + Q3V3, 1 + Q3V3, 2 + Q4V4, 1 + Q4V4, 2 + + Q4V4, 3 + Energy & Potential - sites.google.com/site/ncpdhbkhn 48 Energy Density in the Electrostatic Field (3) WE = Q2V2, 1 + Q3V3, 1 + Q3V3, 2 + Q4V4, 1 + Q4V4, 2 + Q4V4, 3 + Q = 1 QVQ3 3,1 3 Q 4πε R →QV = Q3 = QV 0 13 3 3,1 14πε R 1 1,3 = 0 31 R13 R 31 W = Q V + Q V + Q V + Q V + Q V + Q V + + E 1 1, 2 1 1, 3 2 2, 3 1 1, 4 2 2, 4 3 3, 4 WE = Q2V2, 1 + Q3V3, 1 + Q3V3, 2 + Q4V4, 1 + Q4V4, 2 + Q4V4, 3 + = ++++ 2WE QV1 ( 1,2 V 1,3 V 1,4 ...) + ++++ QV2( 2,1 V 2,3 V 2,4 ...) + ++++ QV3( 3,1 V 3,2 V 3,4 ...) +... Energy & Potential - sites.google.com/site/ncpdhbkhn 49 Energy Density in the Electrostatic Field (4) = ++++ ++++ 2WQVVVE 1 ( 1,2 1,3 1,4 ...) QVVV 2 ( 2,1 2,3 2,4 ...) + ++++ QV2( 3,1 V 3,2 V 3,4 ...) ... + + += VVV1,2 1,3 1,4... V 1 + + += VVV2,1 2,3 2,4... V 2 + + += VVV3,1 3,2 3,4... V 3 N →=1 + + += 1 WE ( QVQVQV11 22 33 ...) ∑ QVk k 1 2 2 k =1 → = ρ WE∫ v Vdv 2 V = ρ Qk v dv Energy & Potential - sites.google.com/site/ncpdhbkhn 50 Energy Density in the Electrostatic Field (5) = 1 ρ WE∫ v Vdv 2 V st ∇ = ρ Maxwell’s 1 equation: .D v → =1 ∇ WE ∫ ( .D ) Vdv 2 V ∇.(VVVDDD ) ≡ (.) ∇ + .( ∇ ) 1 → =[] ∇ − ∇ WE ∫ .( VDD ) .( Vdv ) 2 V Energy & Potential - sites.google.com/site/ncpdhbkhn 51 Energy Density in the Electrostatic Field (6) 1 =[] ∇ − ∇ WE ∫ .( VDD ) .( V ) dv 2 V 1 1 =∫ ∇.()VDD dv − ∫ .( ∇ V ) dv 2V 2 V 1 ∫ ∇.(VD ) dv 2 V Div. theorem: DSD.d= ∇ . dv ∫S ∫ V 1 1 →∫ ∇.(VdvD ) = ∫ ( Vd D ). S 2V 2 S → =1 − 1 ∇ WE ∫( VDSD ). d ∫ .( V ) dv 2S 2 V Energy & Potential - sites.google.com/site/ncpdhbkhn 52 Energy Density in the Electrostatic Field (7) =1 − 1 ∇ WE ∫( VDSD ). d ∫ .( V ) dv 2S 2 V Q V = :
0 with 1/ r 4πε r 0 →1 = Q 2 (VD ). d S 0 = :
0 with 1/ r ∫S D2 a r 2 4π r dS : increases with r2 → = −1 ∇ WE D.( Vdv ) 1 1 ∫V → = = ε 2 2 WE ∫DE. dv ∫ 0 E dv 2V 2 V E = −∇V (pot. grad) Energy & Potential - sites.google.com/site/ncpdhbkhn 53 Ex. 1 Energy Density in the Electrostatic Field (8) Given a coaxial cable, the surface charge density of the outer ρ = a ρ = b surface of the inner cylinder is ρS . Find its potential energy? 1 Method 1: = ε 2 WE ∫ 0 Edv 2 V aρ aρ D=S ( a <ρ < b ) →E = S ρ ρ ε ρ 0 2 z= Lϕ =2 π ρ = b aρ  → = 1 ε S WE 0   dv ∫z=0 ∫ϕ = 0 ∫ ρ = a ε ρ 2 0  dv= ρ d ρ d ϕ dz 2 2 2 2 1 z= Lϕ =2 π ρ = b a ρ πLa ρ b →W = εS ρρϕ d d dz = S ln E ∫z=0 ∫ϕ = 0 ∫ ρ = a 0 ε2 ρ 2 ε 2 0 0 a Energy & Potential - sites.google.com/site/ncpdhbkhn 54 Ex. 1 Energy Density in the Electrostatic Field (9) Given a coaxial cable, the surface charge density of the outer ρ = a ρ = b surface of the inner cylinder is ρS . Find its potential energy? 1 Method 2: = ρ WE∫ v Vdv 2 V final = − a VAB ∫ EL. d → = − ρ init Va Eρ d ∫b V = 0 aρ b E = S ρ ε ρ 0 a aρ aρ b → = − S ρ = S ln Va ∫ d aρ b ε ρ ε a → = 1 ρ S b 0 0 WE∫ v ln dv 1 V ε = ρ 2 0 a WE∫ v Vdv 2 V Energy & Potential - sites.google.com/site/ncpdhbkhn 55 Ex. 1 Energy Density in the Electrostatic Field (10) Given a coaxial cable, the surface charge density of the outer ρ = a ρ = b surface of the inner cylinder is ρS . Find its potential energy? 1 Method 2: = ρ WE∫ v Vdv 2 V aρ = 1 ρ S b ∫ v ln dv 2 v ε a ρ t0 t ρ=S ,a −≤≤+ ρ a , ta≪ v t 2 2 1 zL=ϕ =2 π ρ =+ at / 2 ρ ρ b →W = S a S ln ρ dddz ρ ϕ E ∫z=0 ∫ϕ = 0 ∫ ρ =− a t /2 ε 2 t0 a πLa 2 ρ 2 b = S ln ε 0 a Energy & Potential - sites.google.com/site/ncpdhbkhn 56 Ex. 2 Energy Density in the Electrostatic Field (11) A metallic sphere of radius 10cm has a surface charge density of 10nC/m 2. Calculate the electric energy stored in the system. 1 Method 1: = ε 2 WE ∫ 0 Edv 2 V ρ R2 ρ R2 D.d S = Q →D(4π r2 ) = ρ (4 π R 2 ) →D = S →E = S ∫S total S 2 ε 2 r 0r 2 1 ρ R2  → = ε S WE 0   dv ∫V ε 2 2 0r  2− 18 1∞ π2 π (0.1)× 10 = r2 sin θ drd θ d ϕ ∫r=0.1 ∫θ = 0 ∫ ϕ = 0 ε 4 2 0r = 71.06nJ Energy & Potential - sites.google.com/site/ncpdhbkhn 57 Ex. 2 Energy Density in the Electrostatic Field (12) A metallic sphere of radius 10cm has a surface charge density of 10nC/m 2. Calculate the electric energy stored in the system. Method 2: ? Energy & Potential - sites.google.com/site/ncpdhbkhn 58