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Today in Physics 217: electric dipoles and their interactions Origin of coordinates for multipole moments Force, torque, and potential energy for dipoles in uniform electric fields Felectron Force on dipoles in nonuniform electric fields Example: dipole vs. dipole Permanent and induced dipoles in atoms and molecules 23 October 2002 Physics 217, Fall 2002 Fproton Induced dipole moment E 1 Choice of coordinate origin matters a lot in multipole expansion Changing the origin doesn’t change the physics, but it can radically change the terms in the series. Consider a point charge at the origin: q V= Monopole only, of course r and one not at the origin, say at ( r ,θ , φ ) = ( a ,θ 0 , 0 ) : 2 3 q a a a V = 1 + cosθ 0 + P2 ( cosθ 0 ) + P3 ( cosθ 0 ) + … r r r r How did a point charge get all these nonzero multipole moments? It didn’t, really; the situation still amounts to one point charge, but the accounting is more complicated, when 1 r ≠ 1 r. 23 October 2002 Physics 217, Fall 2002 2 Choice of coordinate origin matters a lot in multipole expansion (continued) So when someone gives you a charge distribution and asks what all the moments are, they also have to tell you what coordinate system to use. One useful exception: if the total charge is zero, then the monopole moment is zero, and the dipole moment is independent of the choice of coordinate origin. Consider such a situation, and suppose the dipole moment were p originally and pa in a coordinate system with its origin displaced by some vector a: pa = ra′ ρ ( r ′ ) dτ ′ = ∫ ∫ ( r ′ − a ) ρ ( r ′) dτ ′ 0 = ∫ r ′ρ ( r ′ ) dτ ′ − a ∫ ρ ( r ′ ) dτ ′ = p − qa = p 23 October 2002 Physics 217, Fall 2002 . 3 Force and torque on dipole in uniform E field If the dipole moment is constant, the net force is zero, because the charges get pulled equally and oppositely. There is a torque, though, that tends to align the dipole moment vector with the applied field: x N = r+ × F+ + r− × F− +q d d = × qE + − × ( − qE ) 2 2 = qd × E = p × E = − pEyˆ , in this case. 23 October 2002 d -qE Physics 217, Fall 2002 qE z E -q 4 Potential energy of dipole in uniform E field (Griffiths problem 4.7) Consider a dipole initially perpendicular to the field (0). The field tends to pull it into alignment (toward 1); we have to push to make it move toward 2. Thus the work we do is θ W= ∫ p × E dθ ′ x π 2 θ = pE ∫ 2 sin θ ′dθ ′ π 2 = − pE cos θ = − p ⋅ E 0 θ 1 z E U= . (Work we do = potential energy.) 23 October 2002 Physics 217, Fall 2002 5 Force on dipole in nonuniform E field If E changes with position, the forces on each charge in a simple dipole will no longer in general be equal in magnitude, leading to a net force F on the dipole. Suppose the dipole is very short; infinitesimal, in fact: F = q ( E+ − E− ) = q∆E x +q Definition of gradient: ∆Ez = ( —Ez )x ⇒ ∆Ez = d ⋅ —Ez d cos α ∆Ex = d ⋅ —Ex (if we had chosen to ∆Ey = d ⋅ —Ey give E a y or an x component) -qE α d qE z E -q ∴ F = q (d ⋅ —) E = ( p ⋅ —) E 23 October 2002 Physics 217, Fall 2002 6 Torque on dipole in nonuniform E field x The torque on a small dipole, about its own center, is still θ N = p× E since the same arguments apply as before. But since there’s a net force F on the dipole, there’s an extra torque of r×F about any other point: N = p× E + r × F F ƒ N F ƒ z E x r N E z 23 October 2002 Physics 217, Fall 2002 7 Dipole vs. dipole: force and torque Griffiths problems 4.5 and 4.29: Two perfect (infinitesimal) dipoles p1 and p2 are perpendicular and lie a distance r apart. What is the torque on p1 (about its center) due to p2? What is the torque on p2 (about its center) due to p1? What are the forces on each, due to each? Why are the torques not equal and opposite? z p2 p1 r y x 23 October 2002 Physics 217, Fall 2002 8 Dipole vs. dipole: force and torque (continued) The field of each, at each other’s position: 2 p1 cosθ p1 sin θ ˆ p1 ˆ p1 p1 E1 ( 2 ) = rˆ + θ = 3 θ = − 3 zˆ = − 3 zˆ 3 3 r r r y a 2 p2 cosθ ′ 2 p2 2 p2 2 p2 p2 sin θ ′ ˆ ˆ ˆ ˆ ˆ ′+ ′=− ′= E2 ( 1 ) = r θ r y y = r ′3 r ′3 r ′3 y3 a3 The torque on each: 2p p ˆy = − 1 2 xˆ a3 a3 p1 p2 p1 ˆ ˆ N 2 = p2 × E1 ( 2 ) = p2 y × − 3 z = − 3 xˆ a a N 1 = p1 × E2 ( 1 ) = p1 zˆ × 23 October 2002 2 p2 Physics 217, Fall 2002 9 Dipole vs. dipole: force and torque (continued) Force on 2 due to field of 1: ∂ F2 = ( p2 ⋅ — ) E1 ( 2 ) = p2 E1 ( 2 ) ∂y y=a ∂ p1 = p2 − 3 ∂y y 3 p1 p2 = zˆ zˆ . 4 a y=a By Newton’s third law, the force on 2 by 1 is F1 = ( p1 ⋅ — ) E2 ( 1 ) = − 23 October 2002 3 p1 p2 Physics 217, Fall 2002 a 4 zˆ . 10 Dipole vs. dipole: force and torque (continued) The forces the dipoles exert on each other are equal and opposite. Why aren’t the torques? Because we calculated the torques about different centers. If we refer both torques to the coordinate origin (i.e. the position of dipole 1), then N 1 ( 0 ) = p1 × E2 ( 1 ) = − 2 p1 p2 a 3 xˆ (still) N 2 ( 0 ) = p2 × E1 ( 2 ) + r × F2 = − p1 p2 = − 3 xˆ + a 3 p1 p2 a3 xˆ = p1 p2 3 a 2 p1 p2 a3 xˆ + ayˆ × 3 p1 p2 a 4 zˆ xˆ = − N 1 ( 0 ) , as you always thought it should be. 23 October 2002 Physics 217, Fall 2002 11 Permanent and induced dipoles in atoms and molecules Some naturally-occurring charge distributions, such as many simple molecules, have permanent, built-in dipole moments: H + H + O -- p H + H + N -- H + p and the electrical properties of materials made with these components are decisively influenced by the behaviour of these dipoles in each others’ fields. Some materials are of course made up of neutral, non-polar components, but even these can have dipole moments induced by external fields. Consider even the lowly hydrogen atom: 23 October 2002 Physics 217, Fall 2002 12 Permanent and induced dipoles in atoms and molecules (continued) Electron Fproton Proton Felectron a Induced dipole moment E x 23 October 2002 Physics 217, Fall 2002 13 Permanent and induced dipoles in atoms and molecules (continued) The electron and proton move in opposite directions until the force on the proton by the displaced electron balances the force on the proton by the external field. Suppose (crudely) that the electron has uniform charge density and stays spherical through this process. If the equilibrium position of the proton is a distance d from the center of the electron, then 4 3 πd qd pinduced 1 3 ˆ ˆ ˆ E− q ( q ) = − x 2 q = −x 3 = −x = − Eexternal , or 3 4 d a a π a3 3 pinduced = α Eexternal , α = a3 ( ) = 4πε 0 a3 in MKS . α, called the polarizability, is thus proportional to atomic volume. 23 October 2002 Physics 217, Fall 2002 14 Atomic polarizabilities 23 October 2002 ( Element α 10 −24 cm -3 H Li Na K 0.66 12.0 27 34 Physics 217, Fall 2002 ) 15