a=重-9e.95j (2.6.2) mmm Suppose the particle is at rest (vo=0)when it is first released from the positive plate. The final speed v of the particle as it strikes the negative plate is v,=2la,ly=1 (2.6.3) m where y is the distance between the two plates.The kinetic energy of the particle when it strikes the plate is K=m=95,y (2.6.4) 21 2.7 Electric Dipole An electric dipole consists of two equal but opposite charges,+q and -g,separated by a distance 2a,as shown in Figure 2.7.1. P(xy.0) Figure 2.7.1 Electric dipole The dipole moment vector p which points from-g to +g (in the +y-direction)is given by p=2gaj (2.7.1) The magnitude of the electric dipole is p=2ga,where g>0.For an overall charge- neutral system having N charges,the electric dipole vector p is defined as (2.7.2) 2-11
ˆ e y q qE m m m = = = − F E a j G G G (2.6.2) Suppose the particle is at rest ( 0 v = 0 ) when it is first released from the positive plate. The final speed v of the particle as it strikes the negative plate is 2 2 | | y y y yqE v a y m = = (2.6.3) where is the distance between the two plates. The kinetic energy of the particle when it strikes the plate is y 1 2 2 K m y y = v = qE y (2.6.4) 2.7 Electric Dipole An electric dipole consists of two equal but opposite charges, +q and , separated by a distance , as shown in Figure 2.7.1. −q 2a Figure 2.7.1 Electric dipole The dipole moment vector p which points from G −q to +q (in the + y - direction) is given by ˆ p = 2qa j G (2.7.1) The magnitude of the electric dipole is p = 2qa , where . For an overall chargeneutral system having N charges, the electric dipole vector q > 0 p G is defined as 1 i N i i q = = p ≡ ∑ i r G G (2.7.2) 2-11
where r is the position vector of the charge g,.Examples of dipoles include HCL,CO, H2O and other polar molecules.In principle,any molecule in which the centers of the positive and negative charges do not coincide may be approximated as a dipole.In Chapter 5 we shall also show that by applying an external field,an electric dipole moment may also be induced in an unpolarized molecule. 2.7.1 The Electric Field of a Dipole What is the electric field due to the electric dipole?Referring to Figure 2.7.1,we see that the x-component of the electric field strength at the point P is E 女 +0-a[+0+a严 (2.7.3) where r2=r2+a22racos0=x2+(ya)2 (2.7.4) Similarly,the y-component is q sine sine E,1 y-a y+a 2 [r+0-a2[x+0+ar] (2.7.5) TEo 4πe0 In the "point-dipole"limit wherera,one may verify that (see Solved Problem 2.13.4) the above expressions reduce to insinocoso ,3p (2.7.6) and E(c1) (2.7.7) where sin=x/r and cos=y/r.With3prcos 3p.r and some algebra,the electric field may be written as )=( +3p加 4π6r3 (2.7.8) Note that Eq.(2.7.8)is valid also in three dimensions where r=xi+yj+k.The equation indicates that the electric field E due to a dipole decreases with r as 1/r3, 2-12
where is the position vector of the charge . Examples of dipoles include HCL, CO, H i r G i q 2O and other polar molecules. In principle, any molecule in which the centers of the positive and negative charges do not coincide may be approximated as a dipole. In Chapter 5 we shall also show that by applying an external field, an electric dipole moment may also be induced in an unpolarized molecule. 2.7.1 The Electric Field of a Dipole What is the electric field due to the electric dipole? Referring to Figure 2.7.1, we see that the x-component of the electric field strength at the point P is 2 2 3/ 2 3/ 2 2 2 2 2 0 0 cos cos 4 4 ( ) ( ) x q q x E r r x y a x y a θ θ πε πε + − + − ⎛ ⎞ ⎛ ⎞ = − ⎜ ⎟ = ⎜ ⎟ − ⎝ ⎠ ⎡ ⎤ + − ⎡ + + ⎝ ⎠ ⎣ ⎦ ⎣ x ⎤ ⎦ 2 ∓ (2.7.3) where (2.7.4) 2 2 2 2 r r a 2 c ra osθ x ( y a) ± = + ∓ = + Similarly, the y -component is 2 2 3/ 2 3/ 2 2 2 2 2 0 0 sin sin 4 4 ( ) ( ) y q q y a y E r r x y a x y a θ θ πε πε + − + − ⎛ ⎞ ⎛ ⎞ − + = − ⎜ ⎟ = ⎜ ⎟ − ⎝ ⎠ ⎡ ⎤ + − ⎡ + + ⎝ ⎠ ⎣ ⎦ ⎣ a ⎤ ⎦ (2.7.5) In the “point-dipole” limit where , one may verify that (see Solved Problem 2.13.4) the above expressions reduce to r a 3 0 3 sin cos 4 x p E r θ θ πε = (2.7.6) and ( 2 3 0 3cos 1 4 y p E r θ πε = − ) (2.7.7) where sinθ = x /r and cosθ = y /r. With3pr cosθ = 3p ⋅ r G G and some algebra, the electric field may be written as 3 5 0 1 3( ( ) 4πε r r ⎛ ⋅ = −⎜ + ⎝ ⎠ p p r r E r ) ⎞ ⎟ G G G G G G (2.7.8) Note that Eq. (2.7.8) is valid also in three dimensions where ˆ ˆ ˆ r i = + x yj + zk G . The equation indicates that the electric field E G due to a dipole decreases with r as , 3 1/r 2-12
unlike the 1/r2behavior for a point charge.This is to be expected since the net charge of a dipole is zero and therefore must fall off more rapidly than 1/2 at large distance.The electric field lines due to a finite electric dipole and a point dipole are shown in Figure 2.7.2. Figure 2.7.2 Electric field lines for (a)a finite dipole and(b)a point dipole. Animation 2.3:Electric Dipole Figure 2.7.3 shows an interactive Shock Wave simulation of how the dipole pattern arises. At the observation point,we show the electric field due to each charge,which sum vectorially to give the total field.To get a feel for the total electric field,we also show a "grass seeds"representation of the electric field in this case.The observation point can be moved around in space to see how the resultant field at various points arises from the individual contributions of the electric field of each charge. Figure 2.7.3 An interactive ShockWave simulation of the electric field of an two equal and opposite charges. 2.8 Dipole in Electric Field What happens when we place an electric dipole in a uniform field E=Ei,with the dipole moment vector p making an angle with the x-axis?From Figure 2.8.1,we see that the unit vector which points in the direction of p is cosi+sinj.Thus,we have p=2ga(cos0i+sine j) (2.8.1) 2-13
unlike the behavior for a point charge. This is to be expected since the net charge of a dipole is zero and therefore must fall off more rapidly than at large distance. The electric field lines due to a finite electric dipole and a point dipole are shown in Figure 2.7.2. 2 1/r 2 1/r Figure 2.7.2 Electric field lines for (a) a finite dipole and (b) a point dipole. Animation 2.3: Electric Dipole Figure 2.7.3 shows an interactive ShockWave simulation of how the dipole pattern arises. At the observation point, we show the electric field due to each charge, which sum vectorially to give the total field. To get a feel for the total electric field, we also show a “grass seeds” representation of the electric field in this case. The observation point can be moved around in space to see how the resultant field at various points arises from the individual contributions of the electric field of each charge. Figure 2.7.3 An interactive ShockWave simulation of the electric field of an two equal and opposite charges. 2.8 Dipole in Electric Field What happens when we place an electric dipole in a uniform field , with the dipole moment vector p making an angle with the x-axis? From Figure 2.8.1, we see that the unit vector which points in the direction of E ˆ E = G i G p G is ˆ cos sin ˆ θ i + θ j. Thus, we have ˆ 2 ( qa cos sin )ˆ p = θ i + θ j G (2.8.1) 2-13
Figure 2.8.1 Electric dipole placed in a uniform field. As seen from Figure 2.8.1 above,since each charge experiences an equal but opposite force due to the field,the net force on the dipole is F=F+F=0.Even though the net force vanishes,the field exerts a torque a toque on the dipole.The torque about the midpoint O of the dipole is i=F.xF+xF=(acos0i+asin0j)x(F.i)+(-acos0i-asin0j)x(-Fi) =asin0F(-k)+asin0F(-k) (2.8.2) =2aF sin0(-k) where we have usedF=F=F.The direction of the torque is-k,or into the page. The effect of the torquet is to rotate the dipole clockwise so that the dipole moment pbecomes aligned with the electric field E.With F=gE,the magnitude of the torque can be rewritten as T=2a(gE)sin=(2ag)Esin0=pE sin and the general expression for toque becomes i=pxE (2.8.3) Thus,we see that the cross product of the dipole moment with the electric field is equal to the torque. 2.8.1 Potential Energy of an Electric Dipole The work done by the electric field to rotate the dipole by an angle de is dw =-rde=-pEsinede (2.8.4) 2-14
Figure 2.8.1 Electric dipole placed in a uniform field. As seen from Figure 2.8.1 above, since each charge experiences an equal but opposite force due to the field, the net force on the dipole is net 0 F F = + − + F = G G G . Even though the net force vanishes, the field exerts a torque a toque on the dipole. The torque about the midpoint O of the dipole is ˆ ˆ ˆ ˆ ˆ ( cos sin ) ( ) ( cos sin ) ( ) ˆ ˆ sin ( ) sin ( ) ˆ 2 sin ( ) a a F a a F a F a F aF θ θ θ θ θ θ θ + + − − + − + − = × + × = + × + − − × − = − + − = − τ r F r F i ˆ j i i j i k k k G G G G G (2.8.2) where we have used F F F . The direction of the torque is + − = = −kˆ , or into the page. The effect of the torque is to rotate the dipole clockwise so that the dipole moment becomes aligned with the electric field E τ G p G G . With F = qE , the magnitude of the torque can be rewritten as τ = = 2 ( a qE)sinθ θ (2aq)Esin = pEsinθ and the general expression for toque becomes τ = p×E G G G (2.8.3) Thus, we see that the cross product of the dipole moment with the electric field is equal to the torque. 2.8.1 Potential Energy of an Electric Dipole The work done by the electric field to rotate the dipole by an angle dθ is dW = − = τ dθ − pEsinθ dθ (2.8.4) 2-14
The negative sign indicates that the torque opposes any increase in.Therefore,the total amount of work done by the electric field to rotate the dipole from an angle 0 to 0 is (-pEsin0yd0-pE(cos0-cos0) (2.8.5) The result shows that a positive work is done by the field when coscos.The change in potential energy AU of the dipole is the negative of the work done by the field: AU=U-Uo=-W=-pE(cos0-cos) (2.8.6) where U=-PE cos is the potential energy at a reference point.We shall choose our reference point to be=z/2 so that the potential energy is zero there,Uo=0.Thus,in the presence of an external field the electric dipole has a potential energy U=-pE cos0=-p·E (2.8.7) A system is at a stable equilibrium when its potential energy is a minimum.This takes place when the dipole p is aligned parallel to E,making U a minimum with Uin=-pE.On the other hand,when p and E are anti-parallel,U=+pE is a maximum and the system is unstable. If the dipole is placed in a non-uniform field,there would be a net force on the dipole in addition to the torque,and the resulting motion would be a combination of linear acceleration and rotation.In Figure 2.8.2,suppose the electric field E at +g differs from the electric field E at-g. E(x-a)x E(x+a) Figure 2.8.2 Force on a dipole Assuming the dipole to be very small,we expand the fields about x: E(x+a)≈E(x)+a dE (2.8.8) dx E(x-a)≈E(x)-a d The force on the dipole then becomes 2-15
The negative sign indicates that the torque opposes any increase inθ . Therefore, the total amount of work done by the electric field to rotate the dipole from an angle θ0 to θ is ( 0 0 W p ( s E in )d pE cos cos θ θ = − θ θ = θ − ) ∫ θ (2.8.5) The result shows that a positive work is done by the field when 0 cosθ > cosθ . The change in potential energy of the dipole is the negative of the work done by the field: ∆U ∆ = U U −U0 = −W = − pE (cosθ − cosθ0 ) (2.8.6) where 0 U PE cos = − θ0 is the potential energy at a reference point. We shall choose our reference point to be 0 θ = π 2 so that the potential energy is zero there, . Thus, in the presence of an external field the electric dipole has a potential energy 0 U = 0 U p = − E cosθ = −p⋅E G G (2.8.7) A system is at a stable equilibrium when its potential energy is a minimum. This takes place when the dipole p is aligned parallel to E G G , making U a minimum with Umin = − pE . On the other hand, when p G and E G are anti-parallel, is a maximum and the system is unstable. Umax = + pE If the dipole is placed in a non-uniform field, there would be a net force on the dipole in addition to the torque, and the resulting motion would be a combination of linear acceleration and rotation. In Figure 2.8.2, suppose the electric field E+ G at differs from the electric field at . +q E− G −q Figure 2.8.2 Force on a dipole Assuming the dipole to be very small, we expand the fields about x : ( ) ( ) , ( ) ( ) dE dE E x a E x a E x a E x a dx dx + − ⎛ ⎞ ⎛ + ≈ + ⎜ ⎟ − ≈ − ⎜ ⎝ ⎠ ⎝ ⎠ ⎞ ⎟ (2.8.8) The force on the dipole then becomes 2-15