Download Midterm 2 | Electricity and Magnetism I | PHY 481 and more Exams Physics in PDF only on Docsity! Midterm II - October 31st 2008 Maxwell’s equations in integral and differential form are given by, ∮ ~E · d ~A = q 0 ; ~∇ · ~E = ρ 0 (Gauss′s law) (1) ∮ ~B · d ~A = 0; ~∇ · ~B = 0 (no official name) (2) ∮ ~E · d~l = −∂φB ∂t ; ~∇∧ ~E = −∂ ~B ∂t (Faraday′s law) (3) ∮ ~B · d~l = µ0i+ µ00 ∂φE ∂t ; ~∇∧ ~B = µ0~j + µ00 ∂ ~E ∂t (Ampere−Maxwell law) (4) In dielectric materials we introduce the polarization and the displacement field, with 0 ~E = ~D− ~P , and we break the charge appearing in Maxwell’s equations into two parts ρ = ρf +ρb and q = qf + qb. The bound charge is related to the polarization through σb = n̂ · ~P , ρb = −~∇ · ~P . The definition of the displacement field is,∫ ~D · d ~A = qf ; ~∇ · ~D = ρf (5) Finally in linear isotropic dielectrics we have the constitutive law ~D = ~E = κ0 ~E. An integral that might be useful, ∫ 2π 0 sin(nφ)sin(n′φ)dφ = πδnn′ (6) 1 Problem 1 a) Derive the differential form of Eq. (3) from the integral form of Eq. (3). Show how this, along with Eq. (1), leads to Poisson’s Equation in electrostatics. Solution. Using Stokes Law and the definition of magnetic flux in the integral form of (3), i.e., ∮ ~E · d~l = ∫ ~∇∧ ~E · d ~A; φB = ∫ ~B · d ~A (7) leads to the differential form of Faraday’s law. The relation ~∇ · ~E = 0 implies that ~E = −~∇V (actually the minus sign is by convention). Subsitution of this into ~∇· ~E = ρ/0, yields Poisson’s equation. b) A metal sphere of radius R is placed in a constant electric field ~E0 = E0k̂. (i) Find the electric field outside the sphere ~Eext(r, θ, φ). (ii) Find the charge density at the surface of the sphere. Solution. Assuming that the sphere is grounded, the solutions for the electrostatic potential are given by, V int = 0; V ext = −E0rcos(θ) + AR3 r2 cos(θ) (8) The tangential electric field must be zero at the surface of the sphere, so that, Eextt (r = R) = −1 R ∂V ext ∂θ |R = −E0sin(θ) + Asin(θ) = 0 so A = E0 (9) The electric field outside the sphere is then, Er = − ∂V ext ∂r = E0cosθ(1 + 2 R3 r3 ); Eθ = −1 r ∂V ext dθ = E0sinθ(−1 + R3 r3 ) (10) The charge density at the surface of the conducting sphere is given by, σ = 0En(r = R) = 0 −∂V ext ∂r |R = 0(E0cos(θ) + 2E0cos(θ)) = 30E0cos(θ) (11) Problem 2 a) From the integral forms of Eq. (1) and Eq. (3), derive the boundary conditions on the normal and tangential electric fields at the surface of a dielectric material. Using Eq. (5) 2
