f - Maxwell’s Equations Gauss ∇ · εE = ρ HR ε ~ E...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Maxwell’s Equations Gauss ∇ · εE = ρ HR ε ~ E · d~a = HRR ρdV ∇ · μ ~ H = 0 HR μ ~ H · d~a = 0 Faraday ∇ × ~ E =- ∂μ ~ H ∂t H ~ E · d~s =- ∂ ∂t RR μ ~ H · d~a Ampere ∇ × ~ H = ~ J + ∂ε ~ E ∂t H ~ H · d~s = RR ~ J · d~a + ∂ ∂t RR ε ~ E · d~a Boundary Conditions ~ D 2 ⊥- ~ D 1 ⊥ = σ u ~ E 2 k- ~ E 1 k = ε ( E 2- E 1 ) = σ u + σ p ~ B 2 ⊥- ~ B 1 ⊥ = ~ H 2 k- ~ H 1 k = Common Electric Fields Point Charge E r = q 4 πε r 2 Infinite Charge Plane E x = σ 2 ε Uniformly Charged Sphere E r = ρ 3 ε r,r < a E r = ( ρ 4 3 πa 3 ) 4 πε r 2 ,r > a Charge Ring E z = λ 2 πa 4 πε z 2 Line Charge E z = λL 4 πε z 2 Oppositely Charged Plates E = σ ε ↑ Charge Dipole φ ( ~ r ) = qd 4 πε r 3 (2 cos θ ˆ r + sin θ ˆ θ ) Electric Potential φ ( ~ r ) = I ∞ ~ r ~ E · d~s ~ E =-∇ φ ( ~ r ) ∇ 2 φ = ρ ε Point Charge φ ( ~ r ) = q 4 πε r ↑ Charge Dipole φ ( ~ r ) = qd 4 πε r 2 cos θ ↓↑ Charge Quadrupole φ ( ~ r ) = qd 2 4 πε r 3 sin(2 θ ) Charge Dipole ~ p = q ~ d ~ P = ~ pN = ε χ e ~ E ε = ε (1 + χ e ) φ ( ~ r ) = p cos θ 4 πε 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 = IZZ ε ~ E · ~ E 2 dV U cap = 1 2 CV 2 Vector Potential ∇ 2 ~ A =- μ ~ J ~ A ( ~ r ) = ZZZ μ ~ J ( ~ r ) 4 π | ~ r- ~ r | 2 d 3 ~ r ← two wires A z ( ~ r ) = μ I 2 π ln r- r + Biot Savart ~ H ( ~ r ) = ZZZ ~ J ( ~ r ) × ˆ n ~ r- ~ r 4 π | ~ r- ~ r | 2 d 3 ~ r Biot Savart for Line Currents ~ H ( ~ r ) = I 4 π Z d~s × ˆ n ~ s- ~ r 4 π | ~s- ~ r | 2 d ~ r d~s points in the direction of current flow ˆ n is a unit vector from the curve to observation point ~ r . Magnetic Flux λ = ZZ ~ B · d~a = μ ZZ ~ H · d~a = ZZ ( ∇ × ~ A ) · d ~ A = I ~ A · d~s Magnetic Induction L = dλ 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 ~ H ( ~ r,t ) · ~ H ( ~ r,t ) 2 | {z } m-field en. density + ε ~ E ( ~ r,t ) · ~ E ( ~ r,t ) 2 | {z } e-field en. density -∇ ~ S ( ~ r,t ) = ∂W ( ~ r,t ) ∂t + ~ J ( ~ r,t ) · ~ E ( ~ r,t )- IZ ~ S ( ~ r,t ) = ∂ ∂t IZZ W ( ~ r,t ) dV + IZZ ~ 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....
View Full Document

This note was uploaded on 04/15/2008 for the course ECE 3030 taught by Professor Rana during the Fall '06 term at Cornell.

Page1 / 6

f - Maxwell’s Equations Gauss ∇ · εE = ρ HR ε ~ E...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online