By Zangwill A.

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DO NOT DISTRIBUTE Electric Multipoles + − + + + − + + + − + + − − + + + − + − + − Chapter 4 + + − + + Source: L. Mederos, E. Chac´ on, and P. Tarazona, Physical Review B 42, 8571 (1990). 14 Practice with Cartesian Multipole Moments (a) The total charge Q = 0. There is no dipole moment because the charge is distributed symmetrically about the origin. The components of the quadrupole moment tensor are Qij = d 3 r ri rj ρ(r). 1 2 Since ρ(x, y) = qδ(z) {δ(x − a)δ(y − a) + δ(x + a)δ(y + a)} − qδ(z) {δ(x − a)δ(y + a) + δ(x − a)δ(y + a)} , all four terms contribute equally to both Qxy and Qy x .

Finally, Qm j = Qj m so the total torque can be written as N = 2(Q · ∇) × E + r × F, where (Q · ∇)i = Qij ∇j . 17 d 3 r ϕ(r)ρ(r) = Qij d 3 r δ(r − r0 )∂i ∂j ϕ(r) = −Qij ∂i Ej (r0 ). Interaction Energy of Nitrogen Molecules The leading contribution to the interaction energy may be calculated by treating each molecule as a point quadrupole. The potential produced by molecule A is ϕA (r) = 1 A 1 Q ∇i ∇j . 2 ij |r − rA | The charge density associated with molecule B is ρB (r) = QB k m ∇k ∇m δ(r − rB ).

Thus ∞ ϕ(r) = =0 m = − R r +1 Y m (Ω) ϕ(R, Ω )Y m (Ω )dΩ r > R. (b) By examining ﬁgure (b), it is clear that we need the potential to change signs every time φ is an integer multiple of π/2. Thus, m = ±2, which in turn implies that ≥ 2. For the asymptotic form of the potential we need only keep the lowest value of necessary. Examining ﬁgure (a), we can see that the potential must change signs every time θ is 61 FOR ENDORSEMENT PURPOSES ONLY. DO NOT DISTRIBUTE Chapter 4 Electric Multipoles an integer multiple of π/2 as well.

### Modern electrodynamics: Solutions manual by Zangwill A.

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