f - Maxwell's Equations Gauss Faraday Ampere E = 0 H = 0 E...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Maxwell’s Equations Gauss ∇ · εE = ρ εE · da = ρdV ∇ · μ 0 H = 0 μ 0 H · da = 0 Faraday ∇ × E = - ∂μ 0 H ∂t E · ds = - ∂t μ 0 H · da Ampere ∇ × H = J + ∂εE ∂t H · ds = J · da + ∂t εE · da Boundary Conditions D 2 - D 1 = σ u E 2 - E 1 = 0 ε 0 ( E 2 - E 1 ) = σ u + σ p B 2 - B 1 = 0 H 2 - H 1 = 0 Common Electric Fields Point Charge E r = q 4 πε 0 r 2 Infinite Charge Plane E x = σ 2 ε 0 Uniformly Charged Sphere E r = ρ 3 ε 0 r, r < a E r = ( ρ 4 3 πa 3 ) 4 πε 0 r 2 , r > a Charge Ring E z = λ 2 πa 4 πε 0 z 2 Line Charge E z = λL 4 πε 0 z 2 Oppositely Charged Plates E = σ ε 0 Charge Dipole φ ( r ) = qd 4 πε 0 r 3 (2 cos θ ˆ r + sin θ ˆ θ ) Electric Potential φ ( r ) = r E · ds E = -∇ φ ( r ) 2 φ = ρ ε 0 Point Charge φ ( r ) = q 4 πε 0 r Charge Dipole φ ( r ) = qd 4 πε 0 r 2 cos θ ↓↑ Charge Quadrupole φ ( r ) = qd 2 4 πε 0 r 3 sin(2 θ ) Charge Dipole p = qd P = pN = ε 0 χ e E ε = ε 0 (1 + χ e ) φ ( r ) = p cos θ 4 πε 0 r 2 Capacitance C = dQ dV = Q V Laplacian Trial Solutions (Spherical) Constant φ ( r ) = A Spherically Symm. φ ( r ) = A r + B Uniform z-dir. φ ( r ) = Ar cos( θ ) Pt. charge dipole +z φ ( r ) = A cos( θ ) r 2 Laplacian Trial Solutions (Cylindrical) Constant φ ( r ) = A Cylindrically Symm. φ ( r ) = A ln( r ) + B Uniform x-dir. φ ( r ) = Ar cos( φ ) Pt. charge dipole +x φ ( r ) = A cos( φ ) r Forces F x = - dU dx + V dQ dx U = ε 0 E · E 2 dV U cap = 1 2 CV 2 Vector Potential 2 A = - μ 0 J A ( r ) = μ 0 J ( r ) 4 π | r - r | 2 d 3 r two wires A z ( r ) = μ 0 I 2 π ln r - r + Biot Savart H ( r ) = J ( r ) × ˆ n r - r 4 π | r - r | 2 d 3 r Biot Savart for Line Currents H ( r ) = I 4 π ds × ˆ n s - r 4 π | s - r | 2 dr ds points in the direction of current flow ˆ n is a unit vector from the curve to observation point r . Magnetic Flux λ = B · da = μ 0 H · da = ( ∇ × A ) · dA = A · ds Magnetic Induction L = dI = λ I Current-Charge Continuity Equation -∇ · J ( r, t ) = ∂ρ ( r, t ) ∂t Electromagnetic Power-Energy Continuity Equation S ( r, t ) = E ( r, t ) × H ( r, t ) W ( r, t ) = μ 0 H ( r, t ) · H ( r, t ) 2 m-field en. density + εE ( r, t ) · E ( r, t ) 2 e-field en. density -∇ S ( r, t ) = ∂W ( r, t ) ∂t + J ( r, t ) · E ( r, t ) - S ( r, t ) = ∂t W ( r, t ) dV + J ( r, t ) · E ( r, t ) dV { net power flow } = rate of increase of total energy in closed volume + rate of total energy loss thru diss.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Electromagnetic Waves f = λv, c = 1 / ε 0 μ 0 3 × 10 8 m/s, η 0 = μ 0 0 377Ω Time Harmonic Fields E ( r, t ) = ˆ nE 0 cos( ωt - k · r ) H ( r, t ) = ( k × ˆ n ) E 0 η 0 cos( ωt - k · r ) k · ˆ n = 0 , k 2 = k · k, ω = kc E ( r, t ) = E ( r ) e jωt , E ( r ) = ˆ nE 0 e - jk · r Maxwell’s Equations (Complex Phasor Notation) (Gauss, Gauss, Faraday, Ampere) ∇ · εE ( r ) = ρ ( r ) (1) ∇ · μ 0 H ( r ) = 0 (2) ∇ × E = - jωμ 0 H ( r ) (3) ∇ × H = J ( r ) + jωεE ( r ) (4) Complex Poynting Vector E ( r ) = ˆ nE 0 e - jk · r H ( r ) = ˆ k × ˆ n E 0 η 0 e - jk · r S ( r ) = E ( r ) × H * ( r ) S ( r ) = 1 2 S ( r ) = 1 2 ˆ n × ˆ k × ˆ n * E 2 0 η 0 Wave Propagation in Dielectric Medium n = ε/ε 0 k = ω μ 0 ε = ω n c = 2 π λ λ = 2 πc ωn η = μ 0 ε Wave Propagation in Conductive Medium ε eff ( ω ) = ε 1 - j σ ωε n eff ( ω ) = ε eff ( ω ) ε 0 = ε ε 0 1 - j σ ωε k = ω μ 0 ε 0 ε eff ( ω ) ε 0 = ω n eff ( ω ) c k = k - jk η eff ( ω ) = μ 0 ε eff ( ω ) 1 η eff ( ω ) = k ωμ 0 = k - jk ωμ 0 S ( r, t ) = ˆ z E 2 0 2
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern