Electric Fields in Matter, Solution Manual

Electric fields in matter Densities on a capacitor The charge density σ2 on the right part of each plate is κ times the charge density σ1 on the left part. So σ1 (b − x)a + σ2 xa = Q =⇒ σ1 (b − x)a + (κσ1 )xa = Q. (639) The two charge densities, σ1 and σ2 = κσ1 , are therefore given by σ1 = Q/a , b + (κ − 1)x σ2 = κQ/a . b + (κ − 1)x (640) Since κ > 1, both of these densities decrease as x increases. It is possible for both densities to decrease while the total charge remains at the given value Q, because the charge in the right region increases (while the charge in the left region decreases), but it does so at a slower rate than the area increases; so the density decreases. We would have a paradox if the areas stayed the same. An analogy: 10 people each have the same amount of money. 20 other people each have the same amount, but it is smaller than what the first 10 have. One of these 20 people takes some money from each of the first 10, and also from each of the other 19, so that she now has the same amount as the first 10. The total amount of money held by all 30 people is still the same, but the average amounts in the two groups (now with 11 and 19 people) have both decreased. Leyden jar Assume that the jar is cylindrical, with the height being twice the diameter d (the result will depend somewhat on the proportions assumed). Then the volume is π(d/2)2 · (2d). Setting this equal to 10−3 m3 gives d = 0.086 m. The area of the capacitor (assuming it has no top) is A = π(d/2)2 + πd(2d) = 9πd2 /4 = 0.052 m2 . So the capacitance is ) ( −12 s2 C2 2 κϵ0 A (4) 8.85 · 10 kg m3 (0.052 m ) C= = 9.2 · 10−10 F. (641) = 0.002 m s If we had chosen the height to instead be four times the diameter, then the capacitance would be about 20% larger. As long as the jar isn’t too squat (in which case it would be better called a tray) or too tall (in which case it would be better called a tube), the dependence of the capacitance on the exact dimensions is fairly weak. (If the height is h = nd, then you can show that the capacitance is proportional to (n + 1/4)/n2/3 .) The capacitance of a sphere is 4πϵ0 r, so a sphere will have a capacitance of 9.2·10−10 F if r = 8.3 m. The diameter is then 16.6 m, or about 54 feet. Maximum energy storage The maximum field is 14 kilovolts/mil, which in volts/meter equals Emax = 1.4 · 104 V = 5.5 · 108 V/m. 2.54 · 10−5 m (642) The capacitance of the Mylar-filled capacitor is κϵ0 A/s. The energy stored is still Cϕ2 /2, so the maximum possible energy density is energy volume = = 1 1 1 κϵ0 A 1 1 (643) Cϕ2 = (Es)2 = κϵ0 E 2 2 2 s V 2 As ( 2 2 ) s C 1 (3.25) 8.85 · 10−12 (5.5 · 108 V/m)2 = 4.4 · 106 J/m3 . kg m3 2 The maximum energy per kilogram of Mylar is therefore 4.4 · 106 J/m3 = 3100 J/kg. 1400 kg/m3 (644) To determine how high the capacitor could lift itself, let the entire mass of the capacitor be m. Then 3m/4 of this is Mylar, so conservation of energy gives E = mgh =⇒ (3100 J/kg)(3m/4) = mgh =⇒ h = (3/4)((3100 J/kg) = 240 m. 9.8 m/s2 (645) The D cell in Exercise 4.41 had an energy storage of 1.8 · 105 J/kg, which is about 60 times as much as the Mylar capacitor. However, the capacitor can deliver all the stored energy in less than a microsecond! Partially filled capacitors The second capacitor in the figure consists of two capacitors in series; you can imagine the boundary between them to be two plates with charge Q and −Q superposed. Both of these capacitors have plate separation s/2 and area A, so the capacitances are (with the two halves labeled by “v” for vacuum and “d” for dielectric) Cv = ϵ0 A/(s/2) and Cd = κϵ0 A/(s/2). Since C0 = ϵ0 A/s, we have Cv = 2C0 and Cd = 2κC0 . Problem 3.18 gives the rule for adding capacitors in series, so the desired capacitance is (with “S” for series) 1 1 1 1 1 2κ = + = + =⇒ CS = C0 . (646) CS Cv Cd 2C0 2κC0 κ+1 The third capacitor in the figure consists of two capacitors in parallel. They both have plate separation s and area A/2, so the capacitances are Cv = ϵ0 (A/2)/s and Cd = κϵ0 (A/2)/s. These can be written as Cv = C0 /2 and Cd = κC0 /2. Problem 3.18 gives the rule for adding capacitors in parallel, so the desired capacitance is (with “P” for parallel) κC0 1+κ C0 + = C0 . (647) CP = Cv + Cd = 2 2 2 If κ = 1, then both CS and CP are equal to C0 , as they should be. For any other value of κ (greater than 1, of course), both CS and CP are larger than C0 . This makes sense because the effect of a dielectric is to partially cancel the existing charge, which means that more charge must be added if the same potential is to be maintained. If κ = ∞, then CS = 2C0 and CP = ∞. The former makes sense because the dielectric is actually a conductor in this case, so we effectively have a vacuum capacitor with separation s/2. The latter makes sense because the capacitance of a conductor is infinite, since any charge you dump on it will be neutralized by the shifting of charges within the conductor. So the left half of the third capacitor has infinite capacitance. Capacitor roll Let w and s stand for the width and thickness of the materials. Then the various constants in the problem are (after converting to meters) κ = 2.3, wp = 5.72 · 10−2 m, sp = 2.54 · 10−5 m, wa = 5.08 · 10−2 m, and sa = 1.27 · 10−5 m. We want the capacitance to be C = 5 · 10−8 F. The capacitance of a parallel-plate capacitor (with width w, length ℓ, and separation s) in the presence of a dielectric is C= κϵ0 wℓ κϵ0 A = s s =⇒ ℓ = Cs . κϵ0 w (648) If we have the tape stretched out straight, with the polyethylene sandwiched between two strips of aluminum, then we need the length of this linear aluminum capacitor to be (5 · 10−8 F)(2.54 · 10−5 m) Csp = ( ℓ= ) = 1.23 m. (649) s 2 C2 −2 m) κϵ0 wa (2.3) 8.85 · 10−12 kg m3 (5.08 · 10 If we simply rolled up the capacitor into a spiral, then one aluminum strip would touch the other. This would ruin the capacitor, because we need the two strips to have opposite charge. So we need to add on a second layer of polyethylene, as shown in Fig. 153. (We don’t need a third layer of polyethylene on bottom.) So it appears that the total length of each kind of tape should be 2(1.23 m) = 2.46 m. However, from Problem 3.21 and Exercise 3.57, we effectively have twice as much area in the capacitor, because the two sides of each strip of aluminum tape act like two different sheets. So ℓ only needs to be half of the above 1.23 m. Hence ℓ = 0.61 m, and the total length of each kind of tape is 2(0.61 m) = 1.23 m. To calculate the diameter of the rolled-up capacitor, note that the area (in the plane of the page) of the capacitor in Fig. 153 is (2sp + 2sa )(ℓ) = (7.62 · 10−5 m)(0.61 m) = 4.7 · 10−5 m2 . This area doesn’t change when we roll up the capacitor, so the radius of the roll is given by πr2 = 4.7 · 10−5 m2 =⇒ r = 3.9 · 10−3 m. (650) The diameter is therefore a little less than 0.8 cm. If √ you missed the above factor of 1/2 in ℓ, the diameter would be larger by a factor of 2, but it would still be in the right ballpark. Work in a dipole field From Eq. (10.15) the potential energy, per unit charge, in the field of a dipole is ϕ = p cos θ/4πϵ0 r2 . Hence √ √ p 2p p(1/ 2) √ ϕA = = . (651) and ϕ = B 4πϵ0 a2 4πϵ0 a2 4πϵ0 (a/ 2)2 √ The work done per unit charge is therefore ϕB − ϕA = ( 2 − 1)(p/4πϵ0 a2 ). polyethylene aluminum Figure 153 A few dipole moments (a) Let the origin be located at the −2q charge (although the choice doesn’t affect the result, because the net √ √ charge is zero). Then p points downward with magnitude p = 2 · q( 3 d/2) = 3 qd. (b) p = 0, by symmetry. (c) Let the origin be located at the upper left corner (although, again, the choice doesn’t matter). Then px √ = qd + q(2d) = 3qd, and py = (−q )(−d) + (2q )(−d) = −qd. So p has magnitude 10 qd and points diagonally rightward and downward at an angle of tan θ = −1/3 =⇒ θ = −18.4◦ below the horizontal. Fringing field from a capacitor The capacitor can be thought of as a large number of dipoles situated next to each other, and these dipoles all produce essentially the same field at a given point far away. We can therefore treat the capacitor like one effective dipole. The charge on it is Q = CV = (2.5 · 10−10 F)(2000 V) = 5 · 10−7 C, so the dipole moment is p = Qs = (5 · 10−7 C)(0.015 m) = 7.5 · 10−9 C m. (a) From Eq. (10.18) the field at a point 3 m away in the plane of the plates is E= 7.5 · 10−9 C m p ) = 2.5 V/m. ( = s 2 C2 3 4πϵ0 r3 4π 8.85 · 10−12 kg m3 (3 m) (652) If the upper plate is the positively charged one, this field points downward. (b) From Eq. (10.18) the field at a point in the direction perpendicular to the plates is E = p/2πϵ0 r3 . This is just twice the field in part (a), so at 3 m away it equals 5 V/m. If the upper plate is the positively charged one, this field points upward (both above and below the capacitor). We should check that our far-field dipole approximation is in fact valid. The capacitance of a parallel plate capacitor is C = ϵ0 A/s, so the area is A= Dipole field plus uniform field θ r p (653) If the plates are square, then they are about 0.65 m on a side. The largest distance √ from the center to a point on the plates is therefore (0.65 m)/ 2 = 0.46 m. The given distance of 3 m is reasonably large compared with this, so our dipole approximation is a fairly good one. However, if s or C were increased enough, then the length scale of the capacitor would be on the order of 3 m. z P1 (0.015 m)(250 · 10−12 F) sC = = 0.42 m2 . s 2 C2 ϵ0 8.85 · 10−12 kg m3 y r P2 Figure 154 We want the field of the dipole to point in the −ŷ direction with magnitude 1.5 · 105 V/m. So we are interested in two points in the yz plane like the ones shown in 2 Fig. 154. Since √ (10.17) tells us that 3 cos θ − 1 = 0. Hence √ we need Ez = 0, Eq. cos θ = ±1/ 3, and then sin θ = ± 2/3. We want the points with opposite signs in these two trig relations (the other two points yield fields in the +ŷ direction). From Eq. (10.17) we have Ey = 3p sin θ cos θ/4πϵ0 r3 . Therefore, r3 = √ √ √ 3p sin θ cos θ 3(6 · 10−10 C m)(± 2/ 3 )(∓1/ 3 ) ) . ( = 2 C2 5 s −12 4πϵ0 Ey 4π 8.85 · 10 kg m3 (−1.5 · 10 V/m) (654) This yields r = 0.037 m. The coordinates of the lower right point are then (y, z) = (r sin θ, −r cos θ) = (0.030, −0.021) m. (655) The upper left point has the negative of these coordinates. Field lines Let the dipole point in the z direction. If r = r0 sin2 θ, then x = r sin θ = r0 sin3 θ, z = r cos θ = r0 sin2 θ cos θ. (656) Therefore, dx dθ dz dθ = 3r0 sin2 θ cos θ, = r0 (2 sin θ cos2 θ − sin3 θ) = r0 sin θ(2 cos2 θ − sin2 θ) = r0 sin θ(3 cos2 θ − 1). (657) The ratio of these derivatives gives the slope of the tangent to the r = r0 sin2 θ curve as dz 3 cos2 θ − 1 . (658) = 3 sin θ cos θ dx This is the same as the ratio Ez /Ex as given by Eq. (10.17). The tangent to the r = r0 sin2 θ curve therefore points in the same direction as the E field, as we wanted to show. Alternatively, we can work with polar coordinates, as we did in Section 2.7.2. With respect to the local r̂-θˆ basis, the slope of the field-line curve is 1 2 cos θ 1 dr = . · 2r0 sin θ cos θ = r dθ sin θ r0 sin2 θ (659) But from Eq. (10.18) this equals Er /Eθ . So again, the tangent to the r = r0 sin2 θ curve points in the same direction as the E field. Average dipole field on a sphere From Fig. 10.6, we quickly see that the average Ex value is zero, because for every field line pointing one way, there is another field line with the opposite Ex value. This holds for any vertical plane containing the z axis, so it holds for Ey too. Hence the averages of both Ex and Ey over the whole surface of a sphere are zero. It isn’t obvious from Fig. 10.6 that the average Ez value is zero. We can’t use a quick symmetry argument, so we need to actually integrate Ez over a spherical shell. The area element of a horizontal ring on the sphere is da = (2πr sin θ)(r dθ). Using the form of Ez given in Eq. (10.17), and ignoring the r’s and all the other constant factors, we have π ∫ π  Ezavg ∝ (3 cos2 θ − 1) sin θ dθ = (− cos3 θ + cos θ) = 0, (660) 0 as desired. 0 Quadrupole for a square The y ≡ x2 coordinate of all four of the given charges is zero, so the quadrupole matrix from Problem 10.6 becomes   2 0 3x1 x3 3x 1 − r2 ∑ 2   Q= qi −r 0 0 . (661) all charges 0 3x23 − r2 3x3 x1 We haven’t bothered to tack on the index i (which labels each of the four charges) to the coordinates, and we’ve dropped the primes on the coordinates. All four charges have either x1 or x3 equal to zero, so the x1 x3 terms in the matrix vanish. Also, all of the charges have the same value of r2 , so the sum of r2 over all the charges is zero, because the sum of the charges itself is zero. We are therefore left with only the 3x21 and 3x23 terms. Each of these picks up a contribution of 3a2 from each of two charges, so the complete quadrupole matrix is   −6 0 0 (662) Q = ea2  0 0 0  . 0 0 6 The monopole and dipole moments are zero, so Eq. (12.469) gives the potential at 2 position r as (1/4πϵ0 )(r̂ · Qr̂/2r3 ). If r̂ = (0, 0, 1), √ then we quickly find rˆ· Qr̂ = 6ea , so ϕ = 3ea2 /4πϵ0 r3 , as desired. If r̂ = (1, 0, 1)/ 2, then r̂ · Qr̂ = 0, so ϕ = 0. This makes sense, because the given point is equidistant from the upper +e and right −e charges, yielding zero potential. Likewise for the lower +e and left −e charges. Pascal’s triangle and the multipole expansion (a) If we make a copy of one of the configurations, negate all of its charges, shift it one unit to the left relative to the original, and then add its charges to the original, then we end up with the next configuration in the series. This is demonstrated in Fig. 155. (b) For the octupole, the potential at a point P on the axis is (ignoring the factor of q/4πϵ0 ) ) ( 3 1 3 3 1 1 1 3 + − + − = ϕP = − . 1− 1 + a/r 1 + 2a/r r r + a r + 2a r + 3a 1 + 3a/r r (663) We can expand this with the Taylor series 1/(1 + ϵ) ≈ 1 − ϵ + ϵ2 − ϵ3 . With ϵ ≡ a/r we have 1[ ϕP ≈ 1 − 3(1 − ϵ + ϵ2 − ϵ3 ) r + 3(1 − 2ϵ + 22 ϵ2 − 23 ϵ3 ) ] − 1(1 − 3ϵ + 32 ϵ2 − 33 ϵ3 ) . (664) Collecting terms with the same power of ϵ, the sum in brackets can be written as 1+ϵ0 (−3 · 10 + 3 · 20 − 1 · 30 ) − ϵ 1 ( −3 · 1 1 + 3 · 2 1 − 1 · 3 1 ) +ϵ2 (−3 · 12 + 3 · 22 − 1 · 32 ) −ϵ3 (−3 · 13 + 3 · 23 − 1 · 33 ). (665) monopole 1 dipole -1 1 -1 1 quadrupole 1 -1 1 -2 2 -1 1 octupole -1 1 -1 3 -3 -3 3 -1 1 24-pole Figure 155 You can quickly check that only the ϵ3 term has a nonzero coefficient; the sum equals 6ϵ3 . Remembering the 1/r out front in Eq. (664), bringing back in the factor of q/4πϵ0 , and using ϵ ≡ a/r, the potential at P is ϕP ≈ q 1 3 q 6a3 (6ϵ ) = . 4πϵ0 r 4πϵ0 r4 (666) As promised, this is proportional to 1/r4 . And the units are (charge)/[ϵ0 ·(length)], which are correct. ( ) ∑3 You can see that each line in Eq. (665) takes the form of (−ϵ)m k=0 (−1)k k3 k m , for m = 0, 1, 2, 3. (The first line takes this form if we let 00 equal 1.) The sum is nonzero only for m = 3. To prove the theorem mentioned in the problem, start with the “(1 − x)N = [binomial expansion]” equation, and then repeat the process m times of alternately taking derivatives and multiplying by x (which will generate the mth power of the k’s in the above sum), and then set x = 1. The left-hand side will vanish as long as we haven’t taken more than N − 1 derivatives. If we take N derivatives, the left-hand side (and hence the right-hand side) will equal (−1)N N !. It’s easy to see how all of this works out if you do the procedure for, say, N = 4. Force on a dipole Let the middle dipole be located at the origin. Intuitively, the downward field at the origin arising from the left dipole is slightly stronger at the (negative) left end of the middle dipole than at its (positive) right end. The left end therefore feels a larger force upward than the right end feels downward. So there is a net upward force due to the field of the left dipole. Similar reasoning shows that there is a net rightward force due to the field of the right dipole. So the total force on the middle dipole is upward and rightward. Let’s be quantitative. From Eq. (10.26) the x component of the force on a dipole is Fx = p · ∇Ex , and likewise for the other components. Let’s first look at the y force due to the field from the left dipole. This field is Ey = −p/4πϵ0 (b + x)3 at points on the x axis near the origin. The gradient of this has only an x component, and it is  ∂Ey  3p = . (667) ∂x x=0 4πϵ0 b4 Therefore, Fy = p · ∇Ey = px (∇Ey )x = p 3p2 3p . = 4πϵ0 b4 4πϵ0 b4 (668) Now consider the x force due to the field from the right dipole. This field is Ex = 2p/4πϵ0 (b − x)3 at points on the x axis near the origin. The gradient of this has only an x component, and it is  6p ∂Ex  = . (669) ∂x  4πϵ b4 x=0 0 Therefore, Fx = p · ∇Ex = px (∇Ex )x = p 6p2 6p = . 4πϵ0 b4 4πϵ0 b4 (670) Since Fy /Fx = 1/2, the field points up to the right at an angle tan √ θ = 1/2 =⇒ θ = 26.6◦ with respect to the x axis. The magnitude is F = 3 5p2 /4πϵ0 b4 = 2 4 (6.71)p /4πϵ0 b . Alternatively, you could work out the force from scratch, by letting the dipole consist of two charges ±q at positions x = ±ℓ/2, with qℓ = p. If you explicitly calculate the forces on the two charges due to the left and right dipoles, to leading order in ℓ, you will end up with the above values of Fx and Fy . The differences in the forces on the two charges will effectively give the above gradients of the E’s. Energy of dipole pairs (a) If we bring in the right dipole from infinity, with it pointing leftward, this requires no work, because the leftward displacement is always perpendicular to the vertical field from the left dipole. But when we rotate the right dipole 90◦ to the desired orientation, this requires pE worth of work; see Eq. (10.22). Since E = p/4πϵ0 d3 , the work required is W = p2 /4πϵ0 d3 . Note that if we bring in the right dipole from infinity, with it pointing upward, then we cannot say that the leftward displacement is always perpendicular to the field from the left dipole, because this field is not vertical at points off the x axis. If we imagine the right dipole to consist of two point charges slightly above and slightly below the x axis, then at the location of these charges, the field from the left dipole has a slight horizontal component (rightward above the x axis, and leftward below). Positive work must be done to move each charge against this field. You can be quantitative about this if you want. (b) From reasoning similar to that in part (a), the work required is W = −p2 /4πϵ0 d3 . (c) If we bring in the right dipole from infinity, with it pointing upward, this requires no work, because the works for the charges on the two ends of the dipole cancel. But when we rotate it 90◦ to the desired orientation, this requires −pE worth of work, from Eq. (10.22). Since E = 2p/4πϵ0 d3 along the axis of the left dipole, the work required is W = −2p2 /4πϵ0 d3 . (d) From reasoning similar to that in part (c), the work required is W = 2p2 /4πϵ0 d3 . The task of Problem 10.3 is to directly calculate the above potential energies by looking at the point charges that make up the dipoles. Polarized hydrogen Since volume is proportional to r3 , the negative charge inside radius ∆z is q = −e(∆z/a)3 . Gauss’s law therefore gives the field due to the inner part of the electron cloud as ∫ q −e(∆z)3 −e∆z Ee · da = =⇒ Ee · 4π (∆z)2 = . (671) =⇒ Ee = ϵ0 4πϵ0 a3 ϵ 0 a3 This field pulls the proton downward. In equilibrium, it must be equal and opposite to the applied field E that pushes the proton upward. Hence ∆z is given by ∆z = 4πϵ0 Ea3 , e (672) which agrees with Eq. (10.27). The hydrogen atom won’t actually remain spherically symmetric, but that won’t affect the rough size of ∆z . Mutually induced dipoles Consider the setup shown in Fig. 156 . The field at B due to pA is 2pA /4πϵ0 r3 . Hence the induced dipole at B is pB = α(2pA /4πϵ0 r3 ). In a similar manner we find pA = α(2pB /4πϵ0 r3 ). Substituting pB from the first of these relations into the second gives pA = 4α2 pA /(4πϵ0 )2 r6 . This is satisfied by pA = 0, or by any value of pA provided that )1/3 ( 4α2 α r6 = =⇒ r = ≡ rc , (673) 2πϵ0 (4πϵ0 )2 where the “c” stands for critical (or cutoff). If r > rc , and if both dipole moments are nonzero at a given instant, they will decay to zero. But if r < rc , and if both dipole moments are nonzero at a given instant, they will increase until limited by the nonlinearity of polarizability. Atomic polarizabilities α/4πϵ0 are typically, in order of magnitude, an atomic volume; see Section 10.5. So rc = (2·α/4πϵ0 )1/3 is on the order of an atomic radius. The object we are concerned with therefore might look something like what is shown in Fig. 157. Whether the lowest state of this system is a spontaneously polarized structure cannot be decided by considering only the interactions of dipoles. Ordinarily the lowest state of two similar atoms would be symmetrical with pA + pB = 0. But we cannot exclude the possibility that the symmetry is “spontaneously broken.” Hydration In a water molecule, the side with the two hydrogen atoms is positively charged, and the side with the oxygen atom is negatively charged. So if the given ion is negative, the hydrogen side will be closer to the ion. The field of a dipole along its axis equals 2p/4πϵ0 r3 , so the force on the negative ion will be attractive with magnitude 2ep/4πϵ0 r3 . The work required to move the ion to infinity by applying an equal and oppositive force is therefore ∫ ∞ ep 2ep dr W = = . (674) 3 4πϵ0 r20 r0 4πϵ0 r polarizable atoms A B α α PA PB r Figure 156 A Figure 157 B Alternatively, from Eq. (10.15) we know that the initial energy of the negative ion is (−e)ϕ = (−e)(p/4πϵ0 r20 ). The work that must be done in bringing the ion to infinity where ϕ = 0 is the negative of this, in agreement with the above result. With p = 6.13 · 10−30 C-m and r0 = 1.5 · 10−10 m, we find W = (1.6 · 10−19 C)(6.13 · 10−30 C-m) = 3.9 · 10−19 J. ) ( s2 C 2 −10 m)2 4π 8.85 · 10−12 kg m3 (1.5 · 10 (675) Field from hydrogen chloride From Eq. (10.18) the magnitude of the field is p/2πϵ0 r3 at points on the z axis, and p/4πϵ0 r3 at points on the y axis. From Fig. 10.14, p points in the direction from the Cl to the H, which is downward here. So the field points downward on the z axis and upward on the y axis. An angstrom equals 10−10 m, so the magnitude of the field at z = 10 angstroms is E= 3.43 · 10−30 C m p ) = 6.2 · 107 V/m. ( = s2 C2 −9 m)3 2πϵ0 z 3 2π 8.85 · 10−12 kg m3 (10 (676) And the magnitude at y = 10 angstroms is just half of this, or 3.1 · 107 V/m. Hydrogen chloride dipole moment If the electron distribution is spherically symmetric around the chlorine nucleus, then we effectively have one proton and one electron separated by 1.28 angstroms. The dipole moment is therefore p = qd = (1.6 · 10−19 C)(1.28 · 10−10 m) = 2.05 · 10−29 C m, center of negative charge x 17e -18e 1.28.10-10 m Figure 158 which is about 6 times the actual dipole moment, 3.43 · 10 e −30 (677) C m. To determine where the charge really is, we can treat the entire group of electrons like a point charge of −18e∫located at their “center of gravity” (defined to be the origin with respect to which rρ dv = 0). So we have the distribution shown in Fig. 158. Let x be the distance from the chlorine nucleus to the electron center. Then with the chlorine nucleus as our origin, we want x to satisfy (17e)(0) + (−18e)x + (e)(1.28 · 10−10 m) = 3.43 · 10−30 C m 1.28 · 10−10 m 3.43 · 10−30 C m =⇒ x = − = 5.9 · 10−12 m, 18 18(1.6 · 10−19 C) (678) or 0.059 angstroms, which is about 1/22 of the way from the chlorine nucleus to the proton. If it were 1/18 of the way (at 0.071 angstroms), which is the location of the center of the positive charge, then p would be zero. This would be the case if the electron from the hydrogen atom remained spherically symmetric around the hydrogen nucleus (the proton). Some electric susceptibilities The susceptibility is given by χ = κ − 1, so χ = CN p2 /ϵ0 kT yields C = (κ − 1)ϵ0 kT /N p2 . The κ values for water and methanol in Table 10.1 are given for room temperature (20◦ C), whereas the κ for ammonia is given for −34◦ C. The values of kT at these temperatures are 4.0 · 10−21 J and 3.3 · 10−21 J, respectively. Knowing the values of κ, kT , N , and p (and ϵ0 , of course), we can compute C = (κ − 1)ϵ0 kT /N p2 . So we just need to find N for the various substances. A mole of something with molecular weight w has a mass of w grams. Equivalently, since the proton mass is 1.67 · 10−24 g, it takes 1/1.67 · 10−24 = 6 · 1023 protons to make a gram. This number is essentially Avogadro’s number. Water has a molecular weight of 18, so the number of water molecules in 1 gram is (6 · 1023 /mole)/(18 g/mole) = 3.33 · 1022 g−1 . The number of molecules per cm3 is then obtained by multiplying by the mass density, ρ = 1.00 g/cm3 (which doesn’t change the number in the case of water). Finally, to obtain the number of molecules per m3 , N , we must multiply by 106 . The resulting N ’s for the three substances are shown in the table below. The dipole moments, p, from Fig. 10.14 are also listed. The calculated values of C = (κ − 1)ϵ0 kT /N p2 are listed in the righthand column. κ 80 H2 O 23 NH3 CH3OH 34 kT 4.0 · 10−21 J 3.3 · 10−21 J 4.0 · 10−21 J N 3.3 · 1028 m−3 2.9 · 1028 m−3 2.5 · 1028 m−3 p 6.13 · 10−30 C-m 4.76 · 10−30 C-m 5.66 · 10−30 C-m C 2.3 1.0 1.5 Discontinuity in E⊥ The internal field is E = −P/3ϵ0 , so E⊥in = P cos θ/3ϵ0 . (As a double check, the factor here is indeed cos θ because the field at the north pole is P/3ϵ0 , pointing downward.) This component points inward in the upper hemisphere (and outward in the lower hemisphere), because E points downward. The perpendicular external field is the radial field from a dipole, E⊥out = Er = p0 cos θ/2πϵ0 r30 . This component points outward in the upper hemisphere (and inward in the lower hemisphere). The effective out = 2P cos θ/3ϵ0 . Due to the different dipole moment p0 equals (4πr03/3)P . Hence E⊥ inward/outward dir e ct i o ns of the ve ctors, th e discontinuity in E⊥ is 2P cos θ/3ϵ0 −− ( P c o s θ / 3 ϵ ) = P c o s θ / ϵ = P /ϵ , a s d es i r ed . 00⊥0 E at the center of a polarized sphere Consider a horizontal ring at an angle θ down from the top of the sphere, with angular span dθ. The area of this ring is 2π(R sin θ)(R dθ). Since the density is σ = P cos θ, the charge in the ring is q = 2πP R2 sin θ cos θ dθ. A little bit of charge dq in a ring in the upper hemisphere creates a diagonally downward field of dq/4πϵ0 R2 at the center of the sphere. But by symmetry we are concerned only with the vertical component, which brings in a factor of cos θ. Integrating over all the dq’s in a ring simply gives the total charge q in the ring. The net field from the ring therefore points downward with magnitude P sin θ cos2 θ dθ 2πP R2 sin θ cos θ dθ . cos θ = 4πϵ0 R2 2ϵ0 (679) You can verify that this expression is valid for rings in the lower hemisphere too; all contributions to the field point downward. Integrating over θ from 0 to π gives a total magnitude of  ∫ π P cos3 θ π P sin θ cos2 θ dθ  = P , =− (680) E=  2ϵ 3ϵ 6ϵ 0 0 0 0 0 as desired. The direction is downward. Uniform field via superposition (a) Applying Gauss’s law to a sphere of radius r inside the given sphere yields 4πr2 E = q ϵ0 =⇒ 4πr2 E = (4πr 3 /3)ρ ϵ0 =⇒ E = ρr . 3ϵ0 (681) positive r1 C1 s C2 P r2 The field points radially, so we have E = (ρ/3ϵ0 )r. (b) Let s be the vector from C2 to C1 in Fig. 159. The field at an arbitrary point P inside both distributions is (using r1 − r2 = −s) E = E1 + E 2 = negative Figure 159 ρr1 (−ρ)r2 ρ ρs = (r − r2 ) = − . + 3ϵ0 3ϵ0 3ϵ0 1 3ϵ0 (682) The vector s is the displacement of the positive charge distribution with respect to the negative, so the polarization density is P = ρs. (This is true because if the dipoles consist of charges q separated by a distance s, then the magnitude of P is given by P = N p = N qs = ρs.) This is the numerator of the result in Eq. (682), so we have ρs P E=− =− , (683) 3ϵ0 3ϵ0 in agreement with Eq. (10.47). (c) For a cylindrical distribution, applying Gauss’s law to an internal cylinder of radius r and length ℓ gives 2πrℓE = q ϵ0 =⇒ 2πrℓE = (πr 2 ℓ)ρ ϵ0 =⇒ E = ρr . 2ϵ0 (684) The field points radially, so we have E = (ρ/2ϵ0 )r. From the same reasoning as in part (b) (the picture looks exactly the same, when viewing along the axis), the field at an arbitrary point P inside both distributions is E = E1 + E 2 = ρr1 (−ρ)r2 ρ ρs = (r1 − r2 ) = − . + 2ϵ0 2ϵ0 2ϵ0 2ϵ0 (685) We again have P = ρs, so the field inside a cylinder with uniform transverse polarization is E = −P/2ϵ0 . Remark: A similar result holds if we kick things down another dimension and consider a slab. For a slab, applying Gauss’s law to an internal box centered on the center-plane of the slab, with thickness 2x and end-face area A, gives E(2A) = q ϵ0 =⇒ 2EA = (2x · A)ρ ϵ0 =⇒ E = ρx . ϵ0 (686) The field points perpendicular to the faces of the slab, so we have E = (ρ/ϵ0 )x. The same reasoning holds again, with two slightly displaced slabs. It’s even easier in this case, because everything takes place in one dimension: E = E1 + E 2 = ρx1 ρ ρs (−ρ)x2 = (x1 − x2 ) = − . + 2ϵ0 ϵ0 ϵ0 ϵ0 (687) We again have P = ρs, so the desired field is E = −P/ϵ0 . The only difference in the E’s for the various objects we have considered is the numerical factor (3, 2, or 1) in the denominator. Conducting-sphere limit Equation (10.54) gives the polarization of a dielectric sphere as P = 3ϵ0 E0 (κ − 1)/(κ + 2). In the κ → ∞ limit, this becomes P = 3ϵ0 E0 . Let’s check that this produces an external field that is perpendicular to the conductor, which must be the case for a perfect conductor. The total dipole moment of the sphere is p = (4πa3 /3)P = 4πϵ0 a3 E0 . And the sphere does indeed act like a dipole, as far as the external field is concerned, from the reasoning near the beginning of Section 10.9. From Eq. (10.18) the tangential field of a dipole is p sin θ/4πϵ0 a3 , which gives E0 sin θ here. But this is exactly what is needed to cancel the tangential component of the original uniform field E0 (you can check that the direction is correct). The tangential component of the total external field is therefore zero, as desired. The field strength inside the dielectric sphere, which from Eq. (10.53) is 3E0 /(2 + κ), goes to zero as κ → ∞. The sphere therefore becomes an equipotential, which is correct for a conducting sphere. A sketch of the field lines outside the sphere is shown in Fig. 160. Using the above value of p, we see that the polarizability α, defined by p = αE0 , of a perfectly conducting sphere equals 4πϵ0 a3 . The quantity that we normally work with, α/4πϵ0 , therefore equals a3 for our conducting sphere of radius a. Since α/4πϵ0 = 0.66 · 10−30 m3 for hydrogen, a conducting sphere of equal polarizability has a radius of (0.66 · 10−30 m3 )1/3 = 8.7 · 10−11 m. This is very close to the Bohr radius, 5.3 · 10−11 m. Continuity of D The E field due to the slab is the same as the E field due to two capacitor plates with surface charge densities ±P . Both E and P are zero outside the slab, so the external D is likewise zero. Our task is therefore to show that D is zero inside the slab. And indeed, E = −P/ϵ0 (this is the field between two plates with densities ±P ), +0P = ϵ ( P/ϵ ) + P = 0. so D = ϵ0 E 0− Discontinuity in D∥ Inside the sphere, we have E = −P/3ϵ0 , so the displacement vector is D ≡ ϵ0 E + P = ϵ0 (−P/3ϵ0 ) + P = 2P/3. The tangential component of this is D∥in ≡ Dθin = − 2P sin θ . 3 (688) The minus sign here comes from the fact that P points toward the north pole, whereas the positive θˆ direction is defined to point away from the north pole. Outside the sphere, E is the field due to a dipole with p0 = (4πR3 /3)P. From Eq. (10.18) the tangential component of the dipole field is Eθ = p0 sin θ/4πϵ0 R3 . In terms of P this becomes Eθ = P sin θ/3ϵ0 . Since P = 0 outside the sphere, the external D is obtained by simply multiplying the external E by ϵ0 . Therefore Dout ≡ Dθout = ∥ P sin θ . 3 (689) Comparing this with the D∥in in Eq. (688), we see that D∥ has a discontinuity of P sin θ. D∥ increases by P sin θ in going from inside to outside. This makes sense, because we know that Eθ is continuous across the boundary, so the discontinuity in the tangential component of D ≡ ϵ0 E + P is simply the discontinuity in the tangential component of P. Energy density in a dielectric With a dielectric present, the capacitance of a parallel-plate capacitor is C = κϵ0 A/s ≡ ϵA/s. The energy stored is still Cϕ2 /2, because it equals Qϕ/2 for all the same reasons as in the vacuum case (imagine a battery doing work in transferring charge from one plate to the other). So the energy density is 1 1 1 ϵA 1 energy ϵE 2 = Cϕ2 = (Es)2 , = V 2 volume 2 As 2 s (690) E= 0 Figure 160 E0 as desired. Since ϵ ≡ κϵ0 , this energy density is κ times the ϵ0 E 2 /2 energy density without the dielectric. Basically, since C is κ times larger, so is the energy, and hence the energy density. To see physically why the energy is larger, consider the case of induced dipole moments, discussed in Section 10.5. The stretched atoms and molecules are effectively little springs that are stretched, so they store potential energy. This makes the total energy larger than it would be for the same equivalent charge on/at the capacitor plates (free charge plus bound-charge layer). In an electromagnetic wave in a dielectric, the energy density of the magnetic field is still B 2 /2µ0 . (It would be B 2 /2µ in a magnetized material, but we’re assuming that the material here is only electrically polarizable.) But from Eq. (10.83) the amplitudes √ of the E and B fields are related by B0 = µ0 ϵ E0 . So B 2 /2µ0 = ϵE2 /2. The E and B energy densities are therefore equal, just as they are in vacuum. Reflected wave The incident, transmitted, and reflected waves are, respectively (using E × B ∝ v to find the direction of the B’s), Ei = ẑEi sin(ky − ωt), Er = ẑEr sin(ky + ωt), Bi = x̂Bi sin(ky − ωt), Br = −x̂Br sin(ky + ωt), Bt = x̂Bt sin(k ′ y − ωt). Et = ẑEt sin(k ′ y − ωt), (691) The total wave in the empty space y < 0 is the sum of the incident and reflected waves. Let’s apply continuity of E and B at y = 0. After setting y = 0, we can cancel all the sin ωt terms (which means that our results will hold for all t), but we must be careful about the extra minus sign in sin(−ωt). We obtain − E i + Er = − Et and − B i − Br = − Bt . (692) Et = (c/n)Bt . (693) We also have Ei = cBi , Er = cBr , Given the “i” quantities, we have four equations in four unknowns (the “r” and “t” quantities). Eliminating the B’s quickly turns Eq. (692) into Ei − E r = E t and Ei + Er = nEt . (694) n−1 Ei n+1 and Et = 2 Ei . n+1 (695) Solving these yields Er = So Er /Ei = (n − 1)/(n + 1), and Et /Ei = 2/(n + 1). The energy is proportional to the square of E, so the faction of the incident energy that is reflected, with n = 1.6, is ( ) n−1 2 Er2 = 0.053. = Ei2 n+1 (696)