Classical Electrodynamics - Jackson Problem 2.13.pdf

PHY 5346 HW Set 5 Solutions – Kimel 1. 2.13 The system is pictured in the following figure: a) Notice from the figure, Φρ,−φ = Φρ,φ; thus from Eq. (2.71) in the text, Φρ,φ = a0 +∑ n=1 ∞ anρn cosnφ ∫ −π/2 3π/2 Φb,φ = 2πa0 = πV1 + πV2 → a0 = V1 + V2 2 Using ∫ −π/2 3π/2 cosmφ cosnφdφ = δnmπ Applying this to Φ, only odd terms m contribute in the sum and am = 2V1 − V2  πmbm −1 m−1 2 Thus Φρ,φ = V1 + V2 2 + 2V1 − V2  π Im ∑ m odd imρme imφ mbm Using 2 ∑ m odd xm m = ln 1 + x 1 − x and Im lnA + iB = tan−1B/A we get Φρ,φ = V1 + V2 2 + V1 − V2  π tan−1 2 ρ b cosφ 1 − ρ 2 b2 as desired. b) σ = −0 ∂∂ρ Φρ,φ|ρ=b σ = −0 V1 − V2  π ∂ ∂ρ tan−1 2 ρ b cosφ 1 − ρ 2 b2 ρ=b σ = −20 V1 − V2 π bcosφ b2 + ρ2 b4 − 2b2ρ2 + ρ4 + 4ρ2b2 cos2φ |ρ=b = −0 V1 − V2 πbcosφ :