Today in Physics 217: Electric Dipoles and Their Interactions

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 Fproton fields Force on dipoles in Felectron nonuniform electric fields Induced dipole Example: dipole vs. dipole moment Permanent and induced dipoles in atoms and E molecules 23 October 2002 Physics 217, Fall 2002 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 ()ra,,θφ= () , θ0 ,0: 23 q aa a VPP=+1 cosθθ02030 +() cos + () cos θ +… rr 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 11.r ≠ 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: prra ==−a′′′ρτ()dd ( rar ′ )() ρτ ′′ ∫∫ 0 =−=−=∫∫rr′′′ρτ()ddq a ρτ () r ′′ pap. 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 Nr=×++−− F +× r F +q qE dd =×qqEE +−×−() 22 E =×=×qdE pE d z -qE =−pEyˆ , in this case. -q 23 October 2002 Physics 217, Fall 2002 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 Wd=×pEθ ′ ∫ x π 2 0 θ 2 1 = pE∫ sinθθ′′ d θ π 2 E z =−pEcosθ =−pE ⋅ 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: FEE=−=∆qq()+− E x +q Definition of gradient: qE ∆E α z =⇒∆=⋅()——EEEd d cosα zzzx d E (if we had chosen to z ∆=⋅EExxd — give E a y or an x -qE ∆=⋅EEyyd — component) -q ∴=FdEpEq () ⋅—— =() ⋅ 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 NpE=× F ƒ E N z 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: x NpErF=×+× F ƒ E N r 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 p p1 2 y r 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: 2cosppppp11θθ sinˆˆ 111 Er1 ()2 =+==−=−ˆˆˆθθ zz rrrya33333 2cospp22θθ′′ sinˆ 2 ppp 222 2 2 Er2 ()1 =+=−==ˆˆˆˆ′′′θ ryy rrrya′′′33333 The torque on each: 22ppp212 NpE112=×()1 =p 1 zˆˆ × y =− x ˆ aa33 ppp112 Np221=×pE()2 = 2 yˆˆ ×− z =− x ˆ aa33 23 October 2002 Physics 217, Fall 2002 9 Dipole vs. dipole: force and torque (continued) Force on 2 due to field of 1: ∂ Fp=⋅()— E()22 =p E() 22 1 21∂y ya= ∂ ppp1123 =−p2 zzˆˆ = . ∂y ya34 ya= By Newton’s third law, the force on 2 by 1 is 3pp12 Fp11=⋅()— E 2()1. =− zˆ a4 23 October 2002 Physics 217, Fall 2002 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 2pp12 NpE112()0=× () 1 =− xˆ (still) a3 pp123 pp 12 Na2212()02=×pE () +×=− rF xyˆˆ +× z ˆ aa34 pp1232 pp 12 pp 12 =−xxˆˆ + = xN ˆ =− 1 ()0, aa33 a 3 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 HH H + H + + ++ p p O N - - 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 F Proton electron 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 π d3 1 3 qd pinduced Ex()qq=−ˆˆˆ =− xx =− =− E, or −q 2334 external daaπ a3 3 33 pEinduced===ααπε external ,4aa()0 in MKS. α, called the polarizability, is thus proportional to atomic volume. 23 October 2002 Physics 217, Fall 2002 14 Atomic polarizabilities α −24 -3 Element ()10 cm H 0.66 Li 12.0 Na 27 K34 23 October 2002 Physics 217, Fall 2002 15.

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