Charge-Charge, Charge-Dipole, Dipole-Charge, Dipole-Dipole Interaction

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charge-charge, charge-dipole, dipole-charge, dipole-dipole interaction Masahiro Yamamoto Department of Energy and Hydrocarbon Chemistry, Graduate School of Engineering, Kyoto University, Kyoto 606-8501, Japan (Dated: December 1, 2008) In the bulk and at the interface of polar solvents and/or ionic liquids, molecules feel the strong electric ¯eld and this strong ¯eld distort the electron cloud of molecule and induces dipole around atoms. To consider this polarization e®ect we should consider charge-charge, charge-dipole, dipole-charge, dipole-dipole interaction. I. CHARGE-CHARGE (C-C) INTERACTION The coulomb potential Á from the point charge zie at ri is given by 0 0 . 2 0 . 4 0 . 6 0 . 8 1 1 0 5 1 0 0 5 0 5 5 1 0 1 0 FIG. 1: Coulomb potential of negative charge zie 1 ÁC(jr ¡ rij) = (1) 4¼²0 jr ¡ rij The electric ¯eld is given by the gradient of the potential ¡ i ¡r j ¡ j zie r ri EC(r) = ÁC( r ri ) = 3 (2) 4¼²0 jr ¡ rij If we put the other point charge zje at at rj the electrostatic energy Vcc is given by 2 zizje 1 Vcc = zjeÁC(jrj ¡ rij) = (3) 4¼²0 jri ¡ rjj II. CHARGE-DIPOLE (C-D) INTERACTION In this interaction we consider two cases, i.e. (I) the coulomb potential interact with dipole ~¹j = ezjpj and (II) dipole ¯eld interact with point charge. In the case of (I), the C-D interaction becomes 2 2 zi(¡zj)e 1 zizje 1 Vcd = + (4) 4¼²0 jrj ¡ rij 4¼²0 jrj + pj ¡ rij ¡1=2 2 rj ¡ ri ´ rji; jrjij = rji; (1 + x) ' 1 ¡ x=2 + 3x =8::: (5) 0 1 " # z z e2 1 1 z z e2 1 V = ¡ i j @ ¡ q A = ¡ i j 1 ¡ cd ¢ 2 2 2 1=2 4¼²0 rji 2 ¢ 2 4¼²0rji (1 + 2rji pj=rji + pj =rji) rji + 2rji pj + pj 2 "#$%&' j *+,-.'()'&" ! i *+,-.' ! j "#$%&'()'&" i FIG. 2: charge-dipole(C-D) and dipole-charge(D-C) interaction " # 2 ¢ p2 '¡ zizje ¡ 1 2rji pj 1 j ¡ 3 ¢ 2 2 2 2 1 1 + 2 + 2 (2rji pj=rji + pj =rji) (6) 4¼²0rji 2 rji 2 rji 8 if we assume rji >> pj 2 ¢ 2 ¢ ¢ ¡zizje rji pj zizje rij pj zie rij ~¹j ´ Vcd = 3 = 3 = 3 ; ~¹j zjepj (7) 4¼²0 rji 4¼²0 rij 4¼²0 rij In the ordinary method the potential is given by ¡~¹ ¢ EC ¡ i ¡r j ¡ j zie r ri EC(r) = ÁC( r ri ) = 3 (8) 4¼²0 jr ¡ rij ¢ ¢ ¡ ¢ i ¡ zie rji ~¹j zie rij ~¹j Vcd = ~¹j EC(rj) = 3 = 3 (9) 4¼²0 rji 4¼²0 rij which gives the same results. In the case of (II), the D-C interaction becomes 2 2 (¡zi)zje 1 zizje 1 Vdc = + (10) 4¼²0 jrj ¡ rij 4¼²0 jrj ¡ ri ¡ pij 0 1 " # z z e2 1 1 z z e2 1 = ¡ i j @ ¡ q A = ¡ i j 1 ¡ ¡ ¢ 2 2 2 1=2 4¼²0 rji 2 ¡ ¢ 2 4¼²0rji (1 2rji pi=rji + pi =rji) rji 2rji pi + pi " # 2 ¡ ¢ 2 '¡ zizje ¡ 1 2rji pi 1 pi ¡ 3 ¡ ¢ 2 2 2 2 1 1 + 2 + 2 ( 2rji pi=rji + pi =rji) : if we assume rji >> pj 4¼²0rji 2 rji 2 rji 8 2 ¢ 2 ¢ ¢ zizje rji pi ¡zizje pi rij ¡ 1 ~¹i rijzje ´ = 3 = 3 = 3 ; ~¹i ziepi (11) 4¼²0 rji 4¼²0 rij 4¼²0 rij Please note that C-D and D-C interaction the sign is di®erent. Since the interaction energy is given by zjeÁ(jrj ¡ rij), the dipoler ¯eld is given by ¢ ¡ ¡ 1 ~¹i (r ri) ÁD(r ri) = 3 (12) 4¼²0 jr ¡ rij 3 0 . 4 0 . 2 0 0 . 2 0 . 4 1 0 5 1 0 0 5 0 5 5 1 0 1 0 FIG. 3: dipole ¯eld !"#$%&'(&%! '!"#$%&' j i FIG. 4: dipole-dipole interaction III. DIPOLE-DIPOLE(D-D) INTERACTION The dipole-dipole(D-D) interaction can be obtained in the same way. Again we assume rij >> pi; pj. (¡z )(¡z )e2 1 (¡z )z e2 1 z (¡z )e2 1 z z e2 1 V = i j + i j + i j + i j (13) dd 4¼² jr ¡ r j 4¼² jr + p ¡ r j 4¼² jr ¡ r ¡ p j 4¼² jr + p ¡ r ¡ p j 00 j i 0 j j i 0 j i i 0 j j i i 1 z z e2 1 1 1 1 = i j @ ¡ q ¡ q + q A 4¼²0 rji 2 ¢ 2 2 ¡ ¢ 2 2 ¡ ¢ ¢ 2 2 ¡ ¢ rji + 2rji pj + pj rji 2rji pi + pi rji 2rji pi + 2rji pj + pi + pj 2pi pj " z z e2 1 1 = i j 1 ¡ ¡ 2 2 2 1=2 2 2 2 1=2 4¼²0rji (1 + 2rji ¢ pj=r + p =r ) (1 ¡ 2rji ¢ pi=r + p =r ) ji j ji ji i #ji 1 + 2 2 2 2 2 2 2 1=2 (1 ¡ 2rji ¢ pi=r + 2rji ¢ pj=r + p =r + p =r ¡ 2pi ¢ pj=r ) " ji ji i ij j ji ji 2 ¢ p2 zizje ¡ 1 2rji pj 1 j ¡ 3 ¢ 2 2 2 2 = 1 1 + 2 + 2 (2rji pj=rji + pj =rji) 4¼²0rji 2 rji 2 rji 8 1 ¡2r ¢ p 1 p2 3 ¡1 + ji i + i ¡ (¡2r ¢ p =r2 + p2=r2 )2 2 r2 2 r2 8 ji i ji i ji ji ji 3 7 1 ¡2r ¢ p 1 2r ¢ p 1 p2 1 p2 1 2p ¢ p 3 ¡2r ¢ p 2r ¢ p p2 p2 2p ¢ p 7 +1 ¡ ji i ¡ ji j ¡ i ¡ j + i j + ( ji i + ji j + i + j ¡ i j )27 2 r2 2 r2 2 r2 2 r2 2 r2 8 r2 r2 r2 r2 r2 5 ji ji ji ji | {zji } | ji {z ji } ji ji ji " # survive cross term survive 2 ¢ ¢ ¢ zizje pi pj ¡ (rji pi)(rji pj) = 2 3 4 4¼²0rji r r " ji ji # ¢ ¢ 1 ¢ ¡ (~¹i rij)(rij ~¹j) Vdd = 3 ~¹i ~¹j 3 2 (14) 4¼²0rij rij 4 i If electric ¯eld ED from i-th dipole can be calculated from ÁD given above, and the interaction energy can be obtained ¡ ¢ i by ~¹j ED(rj). µ ¶ @ @ @ Ei = ¡rÁ (r ¡ r ) = ¡ i + j + k Á (r ¡ r ) D D i @x @y @z D i 1 ¹ (x ¡ x ) + ¹ (y ¡ y ) + ¹ (z ¡ z ) Á (r ¡ r ) = ix i iy i iz i D i 2 2 2 3=2 4¼²0 [(x ¡ xi) + (y ¡ yi) + (z ¡ zi) ] 1 1 ¹ i ¡rÁ (r ¡ r )j = ¡ ix D i x 3 2 2 2 3=2 4¼²0 jr ¡ rij [(x ¡ xi) + (y ¡ yi) + (z ¡ zi) ] 1 ~¹ ¢ (r ¡ r ) ¡3 ¡ i i 2(x ¡ x )i 4¼² ¡ 2 ¡ 2 ¡ 2 5=2 2 i 0 [(x xi)· + (y yi) + (z zi) ] ¸ 1 ~¹ ¢ (r ¡ r )(r ¡ r ) Ei (r) = ¡ ~¹ ¡ 3 i i i (15) D 4¼² jr ¡ r j3 i jr ¡ r j2 0 i " i # ¢ ¡ ¡ ¢ ¡ ¢ i 1 ¢ ¡ ~¹i (ri rj)(ri rj) ~¹j Vdd = ~¹j Ed(rj) = 3 ~¹i ~¹j 3 2 (16) 4¼²0rij rij This equation is the same as the equation given above. IV. UNIFIED DEFINITION OF C-C, C-D, D-C, AND D-D INTERACTIONS The total electrostatic potential is given by [check sign!OK] 0 1 B C X B X X X C 4¼² V = Bq T q ¡q T ®¹ + ¹ T ®q ¡ ¹ T ®¯¹ C (17) 0 tot B|i {zij }j i ij j;® i;® ij j i;® ij j;¯C i>j @ ® ® ®;¯ A C¡C | {z } | {z } | {z } C¡D D¡C D¡D qi = ezi; ~¹i = (¹ix; ¹iy; ¹iz) = ezipi = ezi(pix; piy; piz) (18) Here the interaction tensors are given by [check this!OK] 1 Tij = (19) rij ® r ¡ ¡3 Tij = ®Tij = rij;®rij (20) ®¯ r r ¡ 2 ¡5 Tij = ® ¯Tij = (3rij;®rij;¯ rij±®¯)rij (21) ®¯γ r r r ¡ ¡ 2 ¡7 Tij = ® ¯ γ Tij = [15rij;®rij;¯rij,γ 3rij(rij;®±¯γ + rij;¯±γ® + rij,γ ±®¯)]rij (22) @ ®¯γ Here r® means for ® = x; y; z. T will be used in the force formulation. @rij;® ij V. POLARIZATION When a strong electric ¯eld is applied to an atom i, the electrons around atom i starts to deform and a dipole moment may be induced. j The dipole moment of atom i in the ® direction may be written as the superposition of the electric ¯eld EC generated j by the charge of atom j and that ED by the dipole of atom j stat ind ' ind ¹i;® = ¹i;® + ¹i;® ¹i;® (23) static Usually we take ¹i;® = 0. The electric ¯eld at atom i can be written as X j j E(ri) = [EC(ri) + ED(ri)] (24) j(=6 i) 5 + + + + + + + + d i p o l e i n d u c e d FIG. 5: What's polarization? ( " #) X ind 1 r ¡ r 1 ~¹ ¢ (ri ¡ rj)(ri ¡ rj) = z e i j ¡ ~¹ind ¡ 3 j (25) 4¼² j jr ¡ r j3 jr ¡ r j3 j jr ¡ r j2 0 6 i j i j i j j(=i) 0 1 X X 1 @¡ ® ®¯ indA E®(ri) = Tij qj + Tij ¹j;¯ (26) 4¼²0 j(=6 i) ¯ ind The induced dipole moment ¹i;® is given by ind i ¹i;® = ®®;¯E¯(ri) (27) i Here ®®;¯ polarizability tensor of atom i. If we assume 0 1 ®i 0 0 [®i] = @ 0 ®i 0 A (28) 0 0 ®i then, we have ~¹ind = ®iE(r ); ¹ind = ®iE (r ) (29) i i 0 i;® ® i 1 i X X ind ® @¡ ® ®¯ ind A ¹i;® (out) = Tij qj + Tij ¹j;¯ (input) (30) 4¼²0 6 ¯ j(=i) 0 1 i X X ind ® @¡ x x¯ ind A ¹i;x (out) = Tijqj + Tij ¹j;¯ (input) (31) 4¼²0 6 ¯ j(=i) 0 1 i X X ® @xij 3xijrij;¯ ¡ ±x¯ ind A = 3 qj + [ 5 3 ]¹j;¯ (input) (32) 4¼²0 r r r 6 ij ¯ ij ij j(=i) 0 1 i X X ¹ind(input) ® @xij 3xij ind ¡ j;x A = 3 qj + 5 rij;¯¹j;¯ (input) 3 (33) 4¼²0 r r r j(=6 i) ij ij ¯ ij The last equations should be solved self-consistently.
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  • Magnetic Dipoles Magnetic Field of Current Loop I

    Magnetic Dipoles Magnetic Field of Current Loop I

    PHY2061 Enriched Physics 2 Lecture Notes Magnetic Dipoles Magnetic Dipoles Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor. Magnetic Field of Current Loop z B θ R y I r x For distances R r (the loop radius), the calculation of the magnetic field does not depend on the shape of the current loop. It only depends on the current and the area (as well as R and θ): ⎧ μ cosθ B = 2 μ 0 ⎪ r 4π R3 B ==⎨ where μ iA is the magnetic dipole moment of the loop μ sinθ ⎪ B = μ 0 ⎩⎪ θ 4π R3 Here i is the current in the loop, A is the loop area, R is the radial distance from the center of the loop, and θ is the polar angle from the Z-axis. The field is equivalent to that from a tiny bar magnet (a magnetic dipole). We define the magnetic dipole moment to be a vector pointing out of the plane of the current loop and with a magnitude equal to the product of the current and loop area: μ K K μ ≡ iA i The area vector, and thus the direction of the magnetic dipole moment, is given by a right-hand rule using the direction of the currents. D. Acosta Page 1 10/24/2006 PHY2061 Enriched Physics 2 Lecture Notes Magnetic Dipoles Interaction of Magnetic Dipoles in External Fields Torque By the FLB=×i ext force law, we know that a current loop (and thus a magnetic dipole) feels a torque when placed in an external magnetic field: τ =×μ Bext The direction of the torque is to line up the dipole moment with the magnetic field: F μ θ Bext i F Potential Energy Since the magnetic dipole wants to line up with the magnetic field, it must have higher potential energy when it is aligned opposite to the magnetic field direction and lower potential energy when it is aligned with the field.