Classical Electrodynamics - Jackson Problem 1.8.pdf

PHY 5346 HW Set 2 Solutions – Kimel 2. 1.8 We will be using Gauss’s law ∫ E⃗ ⋅ da⃗ = Qenc 0 a) 1) Parallel plate capacitor From Gauss’s law E = σ 0 = Q A0 = φ12 d → Q = A0φ12 d W = 0 2 ∫ E 2d3x = 0E 2Ad 2 = 0 Q A0 2 Ad 2 = 1 20 Q2 A d = 1 20 A0φ12 d 2 A d = 1 2 0A φ12 2 d 2) Spherical capacitor From Gauss’s law, E = 1 4π0 Q r2 , a < r < b. φ12 = ∫ a b Edr = Q 4π 0 ∫ a b r−2dr = 1 4 Q π 0 b − a ba → Q = 4π 0baφ12 b − a W = 0 2 ∫ E 2d3x = 0 2 Q 4π 0 2 ∫ a b 4π r 2dr r4 = 0 2 Q 4π 0 2 −4π −b + a ba = 1 80 Q2 π b − a ba W = 1 80 4π0baφ12 b−a 2 π b − a ba = 2π 0ba φ12 2 b − a 3) Cylindrical conductor From Gauss’ s law, E2πrL = λL 0 = Q 0 φ12 = ∫ a b Edr = Q 2π 0L ∫ a b dr r = Q 2π 0L ln ba W = 0 2 ∫ E 2d3x = 0 2 Q 2π 0L 2 2πL ∫ a b rdr r2 = 1 4π 0 Q2 L ln ba W = 1 4π 0 2π0Lφ12 ln b a 2 L ln ba = π 0L φ12 2 ln ba b) w = 02 E 2 1) wr = 02 Q A0 2 = 1 20 Q2 A2 0 < r < d, = 0 otherwise. 2) wr = 02 1 4π0 Q r2 2 = 1 320π2 Q2 r4 , a < r < b, = 0, otherwise 3) wr = 02 Q 2π0Lr 2 = 1 80 Q2 π2L2r2 , = 0 otherwise.