750
Solutions to the problems
entries are zero from symmetry, and the diagonal elements are zero due to the example in Section 10.2 combined with the previous paragraph. Alternatively, the average value of, say, x1 2 over the surface of a sphere equals r 2 /3, because it has the same average value as x2 2 and x3 2 , and the sum of all three averages is r 2 . If you want to get some practice with Q, Exercise 10.26 deals with the quadrupole arrangement in Fig. 10.5. 10.7 Force on a dipole Let the dipole consist of a charge −q at position r and a charge q at position r + s. Then the dipole vector is p = qs. If the dipole is placed in an electric field E, the net force on it is F = (−q)E(r) + qE(r + s).
(12.470)
The x component of this is Fx = (−q)Ex (r)+qEx (r+s). Now, the change in a function f due to a small displacement s is ∇f · s, by the definition of the gradient (or at least that’s one way of defining it). So we can write Fx as ! Fx = q Ex (r + s) − Ex (r) = q∇Ex · s = (qs) · ∇Ex ≡ p · ∇Ex ,
(12.471)
as desired. Likewise for the other two components. 10.8 Force from an induced dipole If q is the charge of the ion, then the magnitude of the electric field of the ion at the location of the atom is E = q/4π 0 r2 . If the polarizability of the atom is α, then the induced dipole moment of the atom is p = αE = αq/4π 0 r2 . This dipole moment points along the line from the ion to the atom (see Fig. 12.130), so the magnitude of the field of the induced dipole at the location of the ion is Edipole = 2p/4π 0 r3 . The magnitude of the force on the ion is therefore F = qEdipole =
2pq 2(αq/4π 0 r2 )q 2αq2 = = . 3 3 4π 0 r 4π 0 r (4π 0 )2 r5
(12.472)
You can quickly show that the force is attractive for either sign of q. The potential energy relative to infinity is r r 2αq2 dr
αq2 F(r )dr = − − =− . U(r) = − 2
5 2(4π 0 )2 r4 ∞ ∞ (4π 0 ) r (12.473) Atom
Ion r
p
q
The polarizability of sodium is given by α/4π 0 = 27 · 10−30 m3 . If the magnitude of the potential energy equals |U| = 4 · 10−21 J, then solving for r and setting q = e gives ⎤1/4 ⎡ 1/4
2 −30 3 −19 2 (27 · 10 m )(1.6 · 10 C) (α/4π 0 )q ⎥ ⎢ =⎣ r= ⎦ 2 C2 s 2(4π 0 )|U| −12 −21 (4 · 10 2 · 4π 8.85 · 10 J) kg m3
Eion
Figure 12.130.
Edipole
= 9.4 · 10−10 m.
(12.474)
If r is larger than this, then (on average) the thermal energy is sufficient to kick the ion out to infinity.