Classical Electrodynamics - Jackson Problem 4.1.pdf

PHY 5346 HW Set 6 Solutions – Kimel 3. 4.1 q lm = ∫ r lYlm∗θ,φρx⃗d3x = ∑ i q ir l lYl m∗θ i,φ i Using Yl m∗θ,φ = 2l + 1l − m! 4πl + m! Pl mxe−−mφ = N l mPl mxe−−mφ From the figure we get q lm = a lN l mPl m0q1 − −1m 1 − im  = 0, for m even, so m = 2n + 1, n = 0,1,2, . . . q lm = 2qa lN l mPl m01 − −1ni b) The figure for this system is Since the sum of the charges equals zero, l ≥ 1. q lm = qa lYl m∗x = 1,φ + Yl m∗x = −1,φ = qa lN lmPlm1 + Plm−1 From the Rodrigues formula for Pl mx, we see Pl m±1 = 0, for m ≠ 0. So q lm = qa lN l 01 + −1 l Pl1 Thus l is even, but l ≠ 0 q lm = 2qa lN l 0 c) Using the fact that N l 0 = 2l+1 4π and Yl 0 = 2l+1 4π Pl Φx⃗ = ∑ l=2 ∞ 2qa l  Plx r l+1 ≈ 2qa 2 r3 P2x = 0 on x-y plane) Φx⃗ = − qa 2 r3 Let us plot Φx⃗/−q/a, ie, 1  ra  3 = 1 x3 0 1 2 3 4 5 6 7 8 0.5 1 1.5 2 2.5 3 x The exact answer on the x-y plane is Φx⃗ = −qa 2 x − 2 x 1 + 1 x2 = −q a 1 x 3 − 3 4 1 x 5 + 5 8 1 x 7 − 35 64 1 x 9 +. . . . So let’s plot 1 x3 , 2x − 2 x 1+ 1 x2 0 0.05 0.1 0.15 0.2 0.25 0.3 1 2 3 4 5 x where the smaller is the exact answer.