Classical electrodynamics, 3rd edition 1st cd

www.elsolucionario.net 1.3 In general we take the charge density to be of the form ρ = fr⃗δ, where fr⃗ is determined by physical constraints, such as ∫ ρd x = Q a) variables: r, θ, φ d x = dφd cos θr dr ρ = fr⃗δr − R = frδr − R = fRδr − R ∫ ρd x = fR ∫ r drdΩdr − R = 4πfRR = Q → fR = ρr⃗ = Q 4πR Q δr − R 4πR b) variables: r, φ, z d x = dφdzrdr ρr⃗ = fr⃗δr − b = fbδr − b ∫ ρd x = fb ∫ dzdφrdrδr − b = 2πfbbL = λL → fb = λ 2πb λ δr − b 2πb c) variables: r, φ, z d x = dφdzrdr Choose the center of the disk at the origin, and the z-axis perpendicular to the plane of the disk ρr⃗ = fr⃗δzθR − r = fδzθR − r where θR − r is a step function ∫ ρd x = f ∫ δzθR − rdφdzrdr = 2π R2 f = Q → f = πRQ Q ρr⃗ = δzθR − r πR d) variables: r, θ, φ d x = dφd cos θr dr ρr⃗ = fr⃗δcos θθR − r = frδcos θθR − r ρr⃗ = ∫ ρd x = ∫ frδcos θθR − rdφd cos θr dr = ∫ frrrdrdφ = 2πN ∫ rdr = πR N = Q R R where I’ve used the fact that rdrdφ is an element of area and that the charge density is uniformly distributed over area Q ρr⃗ = δcos θθR − r πR r www.elsolucionario.net 1.4 Gauss’s Law: ∫ E⃗ ⋅ da⃗ = Q enclosed 0 Q a) Conducting sphere: all of the charge is on the surface σ = 4πa Q E4πr = 0, r < a E4πr =  , r ∑ a Q ̂ ⃗ = E = 0, r < a E r, r ∑ a 4π r b) Uniform charge density: ρ = Q πa 3 , r < a, ρ = 0, r ∑ a Qr Qr →E= , r