hapter 10 Summary of time independent electrodynamics 10.1 Electrostatics Physical law oulomb s law charges as origin of electric field Superposition principle ector of the electric field E(x) in vacuum due to charge distribution in volume given by the volume density of free charges ρ(x) E(x) = 1 ε 0 ρ(x ) x x x x 3 d3 x d 3 x volume element (= dx dy dz in artesian coordinates) harge density of a discrete set of charged point particles at positions x i n ρ(x) = q i δ(x x i ) Medium as phenomenological model i=1 Basic differential and integral equations P polarization vector, D electric displacement vector, n outward normal unit vector D(x) = ρ(x), D n = ρ(x)d 3 x = Q encl (Gauss law) S E(x) = 0, E dl = 0 D = ε 0 E+P, D = εe Bounded charges or polarization charges ρ b = P, σ b = P n 153
Boundary conditions at interface between two media n 21 normal unit vector from medium 1 to medium 2 σ free surface charge density Electrostatic scalar potential Φ(x) (D 2 D 1 ) n 21 = σ, n 21 (E 2 E 1 ) = 0 E(x) = 0 E(x) = Φ(x) Poissonequationforisotropicandlinearmedia(εpositionanddirectionindependent) (ρ = 0: Laplace equation) 2 Φ Φ = ρ ε Solution without bounding surfaces Φ(x) = 1 ε ρ(x ) x x d3 x Solution of the Poisson equation in the presence of bounding surfaces Use Green s theorems and Green functions Green functions solution with δ-function as inhomogeneous part interpretation: potential due to a unit point charge (in units ε) G(x,x ) = 2 G(x,x ) = δ(x x ) 1 x x +F(x,x ), 2 F(x,x ) = 0 x Use freedom in the definition of G via F to satisfy appropriate bounday conditions Dirichlet: specification of potential on closed surface S G D (x,x ) = 0, x S Potential for x using Dirichlet boundary conditions Φ(x) = 1 ρ(x )G D (x,x )d 3 x 1 Φ(x ) G D(x,x ) da, ε n S G D n G D n Neumann: specification of electric field (normal derivative of potential) on S Solution of Laplace equation with boundary conditions for problems with symmetries Method of image charges: mimic boundary conditions by placing image charges of appropriate magnitude at positions in a region external to region of interest and take into account their potentials 154
Method of separation of variables: construct an ansatz for the solution Example: Boundary-value problem with azimuthal symmetry using Legendre polynomials (unknown constants A l, B l determined from boundary conditions) Φ(r,θ) = 0 Φ(r,θ) = ( Al r l +B l r l 1) P l (cosθ) l=0 Electrostatic energy for linear media W e = 1 E Dd 3 x = 1 2 2 ρφd 3 x, w e = 1 2 E D Multipole expansion of a localized charge distribution in artesian coordinates ( 1 Q Φ(x) = ε 0 x + p x + 1 ) x 3 2 Q x i x j ij +, x = x x ( 5 ) 1 Qx E(x) = ε 0 x + 3(p x)x x2 p + 3 x 5 Q total charge, p electric dipole moment, Q ij traceless quadrupole moment tensor Q = ρ(x)d 3 x, p = xρ(x)d 3 x, Q ij = (3x i x j x 2 δ ij )ρ(x)d 3 x 10.2 Magnetostatics Physical laws Biot and Savart law absence of magnetic monopoles (individual magnetic charges) B vector of magnetic induction or magnetic flux density due to some free current distribution with density J (SI-unit A/m 2 ) B(x) = µ 0 J(x ) (x x ) x x 3 d 3 x Ampere s law free currents are origins of magnetic fields, H magnetic field vector H dl = I encl, in vacuum H = 1 B µ 0 Basic equations (current divergenceless: J = 0, M magnetization) B(x) = 0, B nda = 0 S H(x) = J(x), H dl = J nda = I encl H = 1 µ 0 B M S 155
Diamagnetic and paramagnetic material: H = 1 µ B Ferromagnetic material: Nonlinear relation between H and B Effective volume and surface currrent densities (SI-units A/m 2 and A/m) J e = M, λ e = M n Boundary conditions at interface between two media (usually λ 0) ector potential A(x) (B 2 B 1 ) n 21 = 0, n 21 (H 2 H 1 ) = λ A defined up to the gradient of an arbitrary scalar function ψ (gauge freedom) B(x) = 0 B(x) = A(x) For isotropic and linear media (µ constant) and choosing oulomb gauge A = 0 derive from the partial differential equation for A H = 1 µ B = J 2 A = A = µj Solution in unbounded space (oulomb gauge) A(x) = µ J(x ) x x d3 x, ψ = const Method of scalar magnetic potential Use in regions with J = 0 (µ constant) H = 0 H = Φ m Laplace equation for Φ m B = 0 2 Φ m = Φ m = 0 arious quantities for a localized current distribution J(x) Force on J in an external B F = J(x) B(x)d 3 x 156
Total torque N = x (J B)d 3 x Magnetic moment m = 1 2 x J(x)d 3 x ector potential and magnetic field at large distances (in vacuum) A = µ 0 m x x 3, B = µ 0 3(m x)x x 2 m x 5 Magnetostatic energy for linear diamagnetic and paramagnetic media in derivation Faraday s law of induction required W m = 1 H Bd 3 x = 1 J Ad 3 x, w m = 1 2 2 2 H B 157