Problem Set 4 Solutions

Massachusetts Institute of Technology Department of Physics

Physics 8.07 Fall 2005

Problem 1: Conducting Sphere in a Gradient Field a) As shown in Griﬃths Problem 3.45,

pirˆi Qijrˆirˆj Vind(~x)= 2 + 3 . 4π0 r 8π0 r

b) The external electric ﬁeld Ei = Ci + Dijxj corresponds to a potential 1 Vext(~x )= −C x − D x x . i i 2 ij i j

(Diﬀerentiating this gives Ei = −∂iV provided Dij = Dji. If Dij =6 Dji, ∇~ × E~ =6 0, which is forbidden in electrostatics.) The total potential is V = Vext + Vind. Requiring V to vanish at r = R gives 3 5 pi = 4π0R Ci , Qij = 4π0R Dij .

c) Using the results of part (b), the total potential for r > R may be written 3 5 R 1 2 R V (~x )= −Cirˆi r − 2 − Dijrˆirˆj r − 3 . r ! 2 r !

Using σ = 0[ˆr·E~ ]= −0(∂V/∂r)(R, θ) (withr î being held fixed during ∂/∂r becauser î depends on angles but not on radius) now gives 5 σ(ˆr) = 30C rˆ + 0RD rˆ rˆ . i i 2 ij i j

Noting that Cirî = C cos θ where C = |C~ | and θ is the angle betweenr ˆ and C~ = E~0, the first term agrees with Griffiths eq. (3.77).

1 d) For V = − 2 Dijxixj the Poisson equation gives

∇~ · E~ = −Dii = −ρ/0 .

A potential with Dii =6 0 is created by a uniform charge density ρ = 0Dii. This is certainly incompatible with a conducting sphere for r < R, but in principle a uniform charge density could exist in some volume.

1 Figure 1: Electric ﬁeld lines for the quadrupole of problem 1(e).

e) The electric ﬁeld created by one line charge at ~s0 ≡ (x0, y0) is

~ λ(~s − ~s0) E = 2 2π0|~s − ~s0|

where ~s ≡ (x, y). For ~s0 = (0,d), to ﬁrst order in ~x/d,

~ λ(x, y − d) E = 2 2 2 2π0(x + y − 2dy + d ) λ(x, y − d) ≈ 2 2π0d (1 − 2y/d) λ(x, y − d)(1+2y/d) ≈ 2 2π0d λ(x, −y − d) ≈ 2 . 2π0d Superposing the four contributions as mentioned in the problem set gives

~ 2λ(x, −y) E ≈ 2 π0d corresponding to 2λ Dxx(0) = −Dyy(0) = 2 . π0d The ﬁeld lines are shown in the ﬁgure above.

2 Problem 2: Molecular Dipoles a) Griﬃths p. 163 gives p = 6.1 × 10−30 C−m for water, and the Boltzmann constant is k = 1.38 × 10−23 J/K. Room temperature is T ≈ 300 K, and with E = 106 V/m, we get

pE (6.1 × 10−30)(106) a = = = 1.5 × 10−3 . kT (1.38 × 10−23)(300)

The average dipole moment is

pa 1 2 −4 hpzi = 1+ a + ··· ≈ 4.9 × 10 p . 3 15

3 −10 b) Using ~p = αE~ with α = 4π0(d/2) and a bond length d = 1.1 × 10 m, we get 1.35 × 10−28 J U = αE2 = = 0.84 × 10−9 eV 1.602 × 10−19 J/eV This is far less than the ionization energy (14.5 eV for atomic nitrogen) and far less than electronic excitation energies. Nitrogen cannot be ionized by this electric ﬁeld. Instead, any free charges in the atmosphere (such as those liberated by cosmic rays, very energetic particles from space) get accelerated in the electric ﬁeld, thereby acquiring enough kinetic energy to collisionally ionize more molecules, liberating more charge, leading to an avalanche. Ex- periments have shown that cosmic ray showers can initiate lighting strikes!

Problem 3: Griﬀths Problems 4.10 (p. 169) and 4.11 (p. 179)

a) In Problem 4.10, P~ = k~x = kr~er within a sphere of radius R implying σb = P~ ·~er = kR and ρb = −∇~ · P~ = −3k. We thus have a uniformly charged sphere surrounded by a thin charged spherical shell. The electric ﬁeld inside the sphere is E~ = ρb~x/(30) = −k~x/0 = −P~ /0. By Gauss’s law, the ﬁeld 3 2 outside the sphere vanishes, since the total charge is (4π/3)ρbR +4πR σb = 0.

b) In Problem 4.11, ∇~ · P~ = 0 inside the cylinder while σb = ±P (with opposite bound charges at the two ends of the cylinder). The result is like a dipole (if L a) or capacitor (if L a); see the figure above for a qualitative sketch of the fields. In the case L a, the electric field inside the capacitor has magnitude E = σb/0 = P/0 and points opposite to P~ , so E~ = −P~ /0.

3 (i) (ii) (iii)

Figure 2: Electric ﬁeld lines for Problem 3(b). The polarization is uniform and vertical between the bound charges (red and blue blobs).

c) Griffiths eq. (4.18) is E~in = −P~ /(30). It gives the average field inside a sphere due to all the charges within that sphere, assuming that the sphere is small enough so that P~ does not vary significantly over its volume. In part (a), the result was E~ = −P~ /0 but P~ = kr~er varies significantly. If a small sphere were made about the origin, the mean polarization vanishes, so Griffiths eq. (4.18) correctly yields E~ = 0 at that point. It isn’t useful elsewhere, though. As for the cylinder in part (b), although P~ is uniform, it is not spherically symmetric, implying that the field produced by charges outside a small sphere cannot be neglected, so E~ =6 E~in. In general, Griffiths eq. (4.18) is not very useful and it should not be used to estimate the electric field produced by polarization.

Problem 4: Displacement Field a) From Griﬃths Example 4.2,

−P/(30)~ez , r < R , E~ = 3 3 PR /(30r )(2~er cos θ + ~eθ sin θ) , r > R .

Combining this with a uniform P~ = P~ez = P (~er cos θ − ~eθ sin θ) inside the sphere, we get

(2P/3)(~er cos θ − ~eθ sin θ) , r < R , D~ = 3 3 (PR )/(3r )(2~er cos θ + ~eθ sin θ) , r > R .

It is easy to see that [Dr] = 0at r = R (no free surface charge) while [Dθ]= P sin θ = [P~ · ~eθ] where [A] ≡ lim→0[A(R + ) − A(R − )].

4 (a) (c)

Figure 3: Displacement field lines for Problem 4; the polarized material is colored yellow. The displacement field vanishes in part (b) so only (a) and (c) are shown. Note that in the absence of free charges the displacement fields form closed loops.

b) For the radially polarized sphere (Griffiths Problem 4.10), E~ = −P~ /0 inside the sphere and E~ = 0 outside. Since P~ is zero outside the sphere, we conclude D~ = 0 everywhere. This is consistent with [Dr]=0 and [Dθ] = [Pθ]=0at r = R. c) For the bar electret in the shape of a capacitor with wide plates, E~ = −P~ /0 in the interior of the capacitor while E~ → 0 away from the capacitor where P~ = 0. Thus, D~ ≈ 0 away from the region of the fringing fields, consistent with the boundary conditions [Dn] = [Dt] = 0 across the capacitor plates. Without an exact expression for D~ it is impossible to test these conditions at the sides of the capacitor where the fringing field is present, however the absence of free surface charge implies that the normal component of D~ is continuous while the tangential component is discontinuous due to the bound charge. d) Sketches of the D-field lines are shown in the figure on the next page. In no case can D~ be written as the gradient of a potential, unless one counts the trivial case D~ = 0 of part (b). Note that while D~ -field lines can only begin and end on free charges, they can also make closed loops (unlike static E~ fields). These close loops are associated with changes in the polarization vector because ∇~ × D~ = ∇~ × P~ .

5 Problem 5: Griffths Problem 4.18 (p. 184) Neglecting the fringing fields, we know from Problem 4(c) that the induced polarization causes no displacement field in a parallel plate capacitor. Thus, for this case, the only source for D~ is the free charges.

3 a) From Gauss’s law for the displacement ﬁeld, D~ · d~a = ρf d x. The only free charges are on the top and bottom plates,H with surfaceR charge density ±σf . Since ∇~ × D~ = 0 if we neglect the fringing ﬁeld, D~ ≈ −σf ~ez in both slabs 1 and 2, where ~ez points normally from slab 2 into slab 1.

b) The electric ﬁeld in a linear dielectric follows from the constitutive relation E~ = D/~ (κ0) where κ is the dielectric constant. We’re given κ1 = 2 and

κ2 = 1.5 so E~1 = −(σf /20)~ez and E~2 = −(2σf /30)~ez. ~ ~ ~ ~ ~ 1 ~ ~ 1 ~ c) We have P = D − 0E = D(1 − 1/κ) yielding P1 = 2 D and P2 = 3 D.

d) The electric ﬁeld is uniform in each slab of thickness a, so ∆V =(E1 +E2)a = (7/6)(aσf /0).

e) The bound charge density at surfaces is σb = P~ · ~n. At the top of slab 1, 1 1 ~n = ~ez so σb = − 2 σf . At the bottom of slab 1, σb = + 2 σf . At the top of 1 1 slab 2, σb = − 3 σf and at the bottom of slab 2, σb =+ 3 σf . ~ −1 1 f) The electric ﬁeld in slab 1 is E = −0 (σf + σb)~ez where σb = − 2 σf is the bound charge at the top of slab 1. Thus, E~ = −(σf /20)~ez in slab 1. In 1 slab 2, we use σb = − 3 σf , the bound charge at the top of slab 2 to get E~ = −(2/3)(σf /0)~ez in slab 2. These results agree with part (b).