Massachusetts Institute of TechnologyDepartment of Physics

Physics 8.07 Fall 2005

Problem Set 4 Solutions

Problem 1: Conducting Sphere in a Gradient Field

a) As shown in Griffiths Problem 3.45,

Vind(~x) =pir̂i

4πε0 r2+

Qij r̂ir̂j

8πε0 r3.

b) The external electric field Ei = Ci + Dijxj corresponds to a potential

Vext(~x ) = −Cixi −1

2Dijxixj .

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

pi = 4πε0R3Ci , Qij = 4πε0R

5Dij .

c) Using the results of part (b), the total potential for r > R may be written

V (~x ) = −Cir̂i

(

r −R3

r2

)

−1

2Dij r̂ir̂j

(

r2 −R5

r3

)

.

Using σ = ε0[r̂ · ~E ] = −ε0(∂V/∂r)(R, θ) (with r̂i being held fixed during ∂/∂rbecause r̂i depends on angles but not on radius) now gives

σ(r̂) = 3ε0Cir̂i +5

2ε0RDij r̂ir̂j .

Noting that Cir̂i = C cos θ where C = |~C | and θ is the angle between r̂ and~C = ~E0, the first term agrees with Griffiths eq. (3.77).

d) For V = −1

2Dijxixj the Poisson equation gives

~∇ · ~E = −Dii = −ρ/ε0 .

A potential with Dii 6= 0 is created by a uniform charge density ρ = ε0Dii.This is certainly incompatible with a conducting sphere for r < R, but inprinciple a uniform charge density could exist in some volume.

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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

~E =λ(~s − ~s0)

2πε0|~s − ~s0|2

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

~E =λ(x, y − d)

2πε0(x2 + y2 − 2dy + d2)

≈λ(x, y − d)

2πε0d2(1 − 2y/d)

≈λ(x, y − d)(1 + 2y/d)

2πε0d2

≈λ(x,−y − d)

2πε0d2.

Superposing the four contributions as mentioned in the problem set gives

~E ≈2λ(x,−y)

πε0d2

corresponding to

Dxx(0) = −Dyy(0) =2λ

πε0d2.

The field lines are shown in the figure above.

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Problem 2: Molecular Dipoles

a) Griffiths p. 163 gives p = 6.1 × 10−30 C−m for water, and the Boltzmannconstant is k = 1.38×10−23 J/K. Room temperature is T ≈ 300 K, and withE = 106 V/m, we get

a =pE

kT=

(6.1 × 10−30)(106)

(1.38 × 10−23)(300)= 1.5 × 10−3 .

The average dipole moment is

〈pz〉 =pa

3

(

1 +1

15a2 + · · ·

)

≈ 4.9 × 10−4 p .

b) Using ~p = α~E with α = 4πε0(d/2)3 and a bond length d = 1.1× 10−10 m, weget

U = αE2 =1.35 × 10−28 J

1.602 × 10−19 J/eV= 0.84 × 10−9 eV

This is far less than the ionization energy (14.5 eV for atomic nitrogen) andfar less than electronic excitation energies. Nitrogen cannot be ionized bythis electric field. Instead, any free charges in the atmosphere (such as thoseliberated by cosmic rays, very energetic particles from space) get acceleratedin the electric field, thereby acquiring enough kinetic energy to collisionallyionize 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 chargedsphere surrounded by a thin charged spherical shell. The electric field insidethe sphere is ~E = ρb~x/(3ε0) = −k~x/ε0 = −~P/ε0. By Gauss’s law, the fieldoutside the sphere vanishes, since the total charge is (4π/3)ρbR

3+4πR2σb = 0.

b) In Problem 4.11, ~∇ · ~P = 0 inside the cylinder while σb = ±P (with oppositebound charges at the two ends of the cylinder). The result is like a dipole (ifL � a) or capacitor (if L � a); see the figure above for a qualitative sketchof the fields. In the case L � a, the electric field inside the capacitor hasmagnitude E = σb/ε0 = P/ε0 and points opposite to ~P , so ~E = −~P/ε0.

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(i) (ii) (iii)

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

c) Griffiths eq. (4.18) is ~Ein = −~P/(3ε0). It gives the average field inside asphere due to all the charges within that sphere, assuming that the sphereis small enough so that ~P does not vary significantly over its volume. Inpart (a), the result was ~E = −~P/ε0 but ~P = kr~er varies significantly. Ifa 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 chargesoutside a small sphere cannot be neglected, so ~E 6= ~Ein. In general, Griffithseq. (4.18) is not very useful and it should not be used to estimate the electricfield produced by polarization.

Problem 4: Displacement Field

a) From Griffiths Example 4.2,

~E ={

−P/(3ε0)~ez , r < R ,PR3/(3ε0r

3)(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 thesphere, we get

~D ={

(2P/3)(~er cos θ − ~eθ sin θ) , r < R ,(PR3)/(3r3)(2~er cos θ + ~eθ sin θ) , r > R .

It is easy to see that [Dr] = 0 at r = R (no free surface charge) while

[Dθ] = P sin θ = [~P · ~eθ] where [A] ≡ limε→0[A(R + ε) − A(R − ε)].

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(a) (c)

Figure 3: Displacement field lines for Problem 4; the polarized material iscolored 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 fieldsform closed loops.

b) For the radially polarized sphere (Griffiths Problem 4.10), ~E = −~P/ε0 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] = 0 atr = R.

c) For the bar electret in the shape of a capacitor with wide plates, ~E = −~P/ε0

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, consistentwith the boundary conditions [Dn] = [Dt] = 0 across the capacitor plates.

Without an exact expression for ~D it is impossible to test these conditionsat the sides of the capacitor where the fringing field is present, however theabsence of free surface charge implies that the normal component of ~D iscontinuous while the tangential component is discontinuous due to the boundcharge.

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 countsthe trivial case ~D = 0 of part (b). Note that while ~D-field lines can onlybegin 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 .

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Problem 5: Griffths Problem 4.18 (p. 184)

Neglecting the fringing fields, we know from Problem 4(c) that the induced polarizationcauses no displacement field in a parallel plate capacitor. Thus, for this case, the onlysource for ~D is the free charges.

a) From Gauss’s law for the displacement field,∮ ~D · d~a =

∫

ρf d3x. The onlyfree charges are on the top and bottom plates, with surface charge density±σf . Since ~∇ × ~D = 0 if we neglect the fringing field, ~D ≈ −σf ~ez in bothslabs 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/(κε0) where κ is the dielectric constant. We’re given κ1 = 2 and

κ2 = 1.5 so ~E1 = −(σf/2ε0)~ez and ~E2 = −(2σf/3ε0)~ez.

c) We have ~P = ~D − ε0~E = ~D(1 − 1/κ) yielding ~P1 = 1

2

~D and ~P2 = 1

3

~D.

d) The electric field is uniform in each slab of thickness a, so ∆V = (E1+E2)a =(7/6)(aσf/ε0).

e) The bound charge density at surfaces is σb = ~P · ~n. At the top of slab 1,~n = ~ez so σb = −1

2σf . At the bottom of slab 1, σb = +1

2σf . At the top of

slab 2, σb = −1

3σf and at the bottom of slab 2, σb = +1

3σf .

f) The electric field in slab 1 is ~E = −ε−1

0 (σf + σb)~ez where σb = −1

2σf is the

bound charge at the top of slab 1. Thus, ~E = −(σf/2ε0)~ez in slab 1. Inslab 2, we use σb = −1

3σf , the bound charge at the top of slab 2 to get

~E = −(2/3)(σf/ε0)~ez in slab 2. These results agree with part (b).

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