Electric Fields Around Conductors Part 2, Solution Manual

Capacitance coefficients for shells We can write in general, Q1 = C11 ϕ1 + C12ϕ2 , Q2 = C21 ϕ1 + C22ϕ2 . (275) With charge Q1 on the inner shell, the field between the shells equals Q1 /4πϵ0 r2 , so the potential difference is ( ) ∫ a ∫ a Q1 1 1 Q1 dr ϕ2 − ϕ1 = − E dr = − = − . (276) 2 4πϵ0 a b b b 4πϵ0 r Hence, Q1 = Comparing this with Eq. (275) gives C11 = 4πϵ0 ab a−b 4πϵ0 ab (ϕ1 − ϕ2 ). a−b and C12 = − (277) 4πϵ0 ab . a−b (278) In the region external to the outer shell of radius a, both shells look like point charges. So the potential at radius a is simply ϕ2 = (Q1 + Q2 )/4πϵ0 a, which yields Q2 = 4πϵ0 aϕ2 − Q1 . Using the Q1 from Eq. (277) allows us to write Q2 in terms of ϕ1 and ϕ2 : 4πϵ0 a2 4πϵ0 ab 4πϵ0 ab ϕ1 + Q2 = 4πϵ0 aϕ2 − (ϕ1 − ϕ2 ) = − ϕ2 . (279) a−b a−b a−b Comparing this with Eq. (275) gives C21 = − 4πϵ0 ab a−b and C22 = 4πϵ0 a2 . a−b (280) As expected, C12 = C21 . In the event that we have a standard capacitor with charge Q on the inner shell and −Q on the outer, the field is zero outside the outer shell. So ϕ2 = 0, and ϕ1 equals the ∆ϕ between the shells. Both of the equations in Eq. (275) then reduce to Q = [4πϵ0 ab/(a − b)]∆ϕ, which agrees with the result in Eq. (3.18) for a two-sphere capacitor. Capacitance-coefficient symmetry (a) We can write in general, Q1 = C11 ϕ1 + C12ϕ2 , Q2 = C21 ϕ1 + C22ϕ2 . (281) Step 1: Let us add charge to conductor 1, while holding ϕ2 constant at zero. During this process, we will have to remove charge from conductor 2 to maintain ϕ2 = 0. (Equivalently, charge will naturally flow off conductor 2 if it is grounded, because charge will be repelled from the charge added to conductor 1.) So conductor 2 will become negatively charged. But this process doesn’t involve any work, as we will see. From above, if ϕ2 = 0 then dQ1 = C11 dϕ1 and dQ2 = C21 dϕ1 (so evidently C21 is negative). The total work done in raising ϕ1 from 0 to ϕ1f equals the work done in adding charge to conductor 1 and removing charge from conductor 2. Therefore, ∫ (ϕ1 dQ1 + ϕ2 dQ2 ) Wstep 1 = ∫ ϕ1f [ ] 1 ϕ1 (C11 dϕ1 ) + (0)(C21 dϕ1 ) = C11 ϕ21f , = (282) 2 0 where we have used ϕ2 = 0. As promised, no work is involved in the flow of charge off conductor 2. Step 2: Now let us add charge to conductor 2, while holding ϕ1 constant at ϕ1f . If ϕ1 is constant then dQ1 = C12 dϕ2 and dQ2 = C22 dϕ2 . Therefore, ∫ (ϕ1 dQ1 + ϕ2 dQ2 ) (283) Wstep 2 = ∫ ϕ2f [ ] 1 = ϕ1f (C12 dϕ2 ) + ϕ2 (C22 dϕ2 ) = C12 ϕ1f ϕ2f + C22 ϕ22f . 2 0 The total work done is 1 1 Wtotal = C11 ϕ21f + C22 ϕ22f + C12 ϕ1f ϕ2f . 2 2 (284) (b) In this process, the roles of 1 and 2 are interchanged, so we simply need to switch the 1’s and 2’s in the result in part (a). Therefore, the total work done is 1 1 Wtotal = C22 ϕ22f + C11 ϕ21f + C21 ϕ2f ϕ1f . 2 2 (285) Since the final state is the same, and since there is no dissipation of energy in the charging process, the total work done must be the same. Hence C12 = C21 . Capacitor energy As usual, the charges on the two conductors (label them as C1 and C2 ) of the capacitor are Q and −Q. If the potential difference between the conductors is ϕ, then we can write the potentials in the general forms of ϕ0 + ϕ and ϕ0 , for some value of ϕ0 (which may be zero). The integral in Eq. (2.32) breaks up into two separate integrals, one for each conductor (which are the only places where ρ is nonzero; more precisely, the surface density σ is nonzero). Since the potential takes on a constant value on each conductor, these potentials can be taken out of the integrals, yielding ∫ ∫ 1 1 1 1 1 U = (ϕ0 + ϕ) ρ dv + ϕ0 ρ dv = (ϕ0 + ϕ)Q + ϕ0 (−Q) = Qϕ, (286) 2 2 2 2 2 C1 C2 as desired. Adding a capacitor Let the two capacitors be labeled 1 and 2. If the initial charge on capacitor 1 is Q, then Q = C 1 Vi , (287) where C1 = 100 pF and Vi = 100 volts. So Q = (10−10 F)(100 V) = 10−8 C. When capacitor 2 is connected in parallel, the charge Q is shared between the two capacitors, that is, Q = Q1 + Q2 . But the voltages across the two capacitors are equal because they are connected in parallel. This voltage is Vf = 30 volts. So we have Q1 = C1 Vf and Q2 = C2 Vf . Adding these relations gives Q = (C1 + C2 )Vf , (288) which is the statement that capacitances in parallel simply add. Equating the righthand sides of Eqs. (287) and (288) gives C1 (100 V) = (C1 + C2 )(30 V) =⇒ C2 = C1 · 70 = 233 pF. 30 (289) The initial energy stored is 1 1 QVi = (10−8 C)(100 V) = 5 · 10−7 J. 2 2 (290) The final energy stored is 1 1 1 1 Q1 Vf + Q2 Vf = QVf = (10−8 C)(30 V) = 1.5 · 10−7 J. 2 2 2 2 (291) (The final energy is smaller than the initial energy by the factor Vf /Vi .) Therefore, 3.5 · 10−7 J of energy is lost. This much energy has to go somewhere before the system can settle down to static equilibrium. If it is not stored anywhere else (for instance, in a weight lifted by a motor driven by the current from C1 to C2 ) it will eventually be dissipated in circuit resistance, no matter how small that resistance may be. (If the circuit is superconducting, the current will keep sloshing back and forth. We’ll talk about LC circuits in Chapter 8.) Energy in coaxial tubes We’ll solve this exercise first by using the energy density in the electric field, and then by using the capacitance. If λ is the charge per unit length on the inner cylinder (with −λ on the outer cylinder), then the field between the cylinders (ignoring end effects) is λ/2πϵ0 r. The energy stored in the field is therefore (with ℓ = 0.3 m being the length) ) )2 ( ∫ ( ∫ r2 ∫ λ dr λ2 ℓ ϵ0 r2 λ2 ℓ ϵ 0 r2 U= . = ln 2πrℓ dr = E 2 dv = r1 2 r1 4πϵ0 r1 r 4πϵ0 2πϵ0 r 2 (292) To write this in terms of the (magnitude of the) potential difference ϕ between the tubes, instead of in terms of λ, note that ( ) ∫ r2 ∫ r2 r2 λ λ dr ln E dr = ϕ= = . (293) 2πϵ r 2πϵ r1 0 0 r1 r1 Solving for λ and plugging the result into Eq. (292) gives ( )2 2πϵ0 ϕ πϵ0 ℓϕ2 ℓ U = ln(r2 /r1 ) = ln(r2 /r1 ) 4πϵ0 ln(r2 /r1 ) ) ( s2 C 2 2 π 8.85 · 10−12 kg m3 (0.3 m)(45 V) = = 5.9 · 10−8 J. ln(4/3) (294) Alternatively: We can solve the problem using capacitance. Using the value of ϕ we found above, the capacitance of the tubes is given by C= λℓ 2πϵ0 ℓ Q = = , ln(r2 /r1 ) ϕ (λ/2πϵ0 ) ln(r2 /r1 ) (295) which is independent of λ, as it should be. (The capacitance depends only on the geometry of the system, and not on the charge that is placed on the conductors.) The energy stored is then ( ) 1 2πϵ0 ℓ 1 2 πϵ0 ℓϕ2 Cϕ = , (296) ϕ2 = ln(r2 /r1 ) 2 2 ln(r2 /r1 ) in agreement with the first solution. Maximum energy storage between cylinders First, note that the stored energy should indeed achieve a maximum for some value of b between 0 and a, due to the following reasoning. The energy is zero when b = a, because there is zero volume containing a nonzero field (a nonzero field exists only between the cylinders). And the energy is essentially zero when b ≈ 0, because in that case the charge per unit length on the inner cylinder will have to be very small (otherwise the field at the surface, which is proportional to 1/b, would exceed E0 ). This then means that the field is very small in the region between the cylinders (except very close to the inner cylinder, where it is E0 ). Therefore, since the stored energy is zero at both b = a and b ≈ 0, it must achieve a maximum at some intermediate value. We’ll solve this exercise first by using the energy density in the electric field, and then by using the capacitance. For convenience, let b ≡ ka, where k is a numerical factor to be determined. If E0 is the field at radius ka, then since E ∝ 1/r for a cylinder, the field equals E0 (ka/r) at larger values of r (but less than a). The energy stored in the field in a length ℓ of the capacitor is then U= ϵ0 2 ∫ E 2 dv = = )2 ka 2πrℓ dr r ka ∫ a dr πϵ0 ℓk2 a2 E02 = πϵ0 ℓa2 E 20 k 2 ln(1/k). ka r ϵ0 2 ∫ a ( E0 (297) As noted above, this equals zero when k = 0 or k = 1. (At k = 0, the smallness of k 2 wins out over the largeness of ln(1/k).) Taking the derivative of −k 2 ln k to find the maximum gives −k 2 (1/k) − 2k ln k = 0 =⇒ ln k = −1/2 =⇒ k = e−1/2 . Hence b = e−1/2 a ≈ (0.607)a. The stored energy is then ( )2 πϵ0 ℓa2 E 20 U = πϵ0 ℓa2 E02 e−1/2 (1/2) = . 2e (298) The energy per unit length is obtained by erasing the ℓ. Alternatively: We can solve the problem using capacitance. The (magnitude of the) potential difference ϕ between the tubes is ∫ a ∫ a (a ) λ λ dr = ln . (299) ϕ= E dr = r 2πϵ b 2πϵ 0 0 b b The capacitance is then C= Q λℓ 2πϵ0 ℓ = = , ln(a/b) ϕ (λ/2πϵ0 ) ln(a/b) (300) which is independent of λ, as it should be. (The capacitance depends only on the geometry of the system, and not on the charge that is placed on the conductors.) With b ≡ ka this becomes C = 2πϵ0 ℓ/ ln(1/k). If λ is the charge per unit length on the inner cylinder, then E0 = λ/2πϵ0 (ka) =⇒ λ = 2πϵ0 (ka)E0 . The stored energy is then ( )2 Q2 2πϵ0 (ka)E0 ℓ (λℓ)2 U= (301) = = = πϵ0 ℓa2 E02k 2 ln(1/k). 2C 2C 2 · 2πϵ0 ℓ/ ln(1/k) in agreement with Eq. (297). The solution proceeds as above. Force, and potential squared (a) In Gaussian units, (potential)2 ∼ ( charge distance )2 ∼ charge2 ∼ force . distance2 (302) In the case of 1 statvolt, we have (1 statvolt)2 = (1 esu)2 = 1 dyne. (1 cm)2 (b) 1 volt equals 1/300 statvolts, so )2 ( 6 10 statvolt ∼ 107 dynes = 100 newtons. (1 megavolt)2 = 300 (303) (304) But 9.8 newtons equals 2.2 pounds (these are both the weight of 1 kilogram). So the desired force is (100 N)(2.2 pounds/9.8 N) ≈ 20 pounds. Force and energy for two plates From Eq. (1.49), the force per unit area on one of the plates is σ times the average of the fields on either side of the plate. (Equivalently, it is σ times the field from the other plate.) This average field is E/2, where E is the field between the plates. But E equals σ/ϵ0 , so σ = ϵ0 E (it will be more useful to write the field in terms of E than σ). The force per unit area is therefore F ϵ0 E 2 E E =⇒ F = A = σ = (ϵ0 E) . 2 2 2 A Since E is given by ϕ/s, we can write F in terms of the potential as ) ( 2 −12 s2 C2 2 Aϵ0 ϕ2 (0.2 m) 8.85 · 10 kg m3 (10 V) = 2.0 · 10−8 N. = F = 2 2 2(0.03 m) 2s (305) (306) If the charge is held constant as the plates come together, then the electric field is independent of the separation, so we see from Eq. (305) that the force is also independent of the separation. (Equivalently, ϕ is proportional to s in Eq. (306), so F is independent of s.) The total work done by the electric force (which could be used to lift an external object, etc.) is then W = F · s = (2.0 · 10−8 N)(0.03 m) = 6 · 10−10 J. Note that the work can be written symbolically as W =F ·s= Aϵ0 E 2 ϵ0 E 2 ϵ0 E 2 . · s = (As) = (volume) 2 2 2 (307) Since ϵ0 E 2 /2 is the energy density, the work does indeed equal the energy initially stored in the field. Alternatively, the work can be written in terms of ϕ as (using C = ϵ0 A/s for a parallel-plate capacitor) Aϵ0 ϕ2 1 ϵ0 A 2 1 ·s= ϕ = Cϕ2 , 2 s 2s2 2 which is the energy stored in the capacitor. W =F ·s= (308) What is the work done if the plates remain connected to the 10 volt battery? In this case, since ϕ is constant, the force of Aϵ0 ϕ2 /2s2 in Eq. (306) grows like 1/s2 as s goes to zero. The integral of this diverges near zero, so the work is theoretically infinite. However, eventually the battery won’t be able to supply the necessary charge to the plates, so ϕ will inevitably decrease. Conductor in a capacitor (a) The initial energy stored in the capacitor is ϵ0 E 2 /2 times the volume As. The field is E = σ/ϵ0 , so the initial energy is Ui = ϵ0 2 ( σ ϵ0 )2 (As) = σ 2 As . 2ϵ0 (309) Alternatively, we get the same result if we use Ui = Q2 /2C, with Q = Aσ and C = ϵ0 A/s. Now consider the moment when the conducting slab is completely inside the capacitor. Since the charge on each plate remains fixed, the final charge densities are still ±σ. (The charges shift around at intermediate times, but this doesn’t concern us.) The field in the vacuum half of the capacitor is therefore still σ/ϵ0 . And the field inside the conductor is zero, of course. What happens is that the conductor basically becomes two sheets of charge: −σ on top, and σ on bottom. The σ bottom sheet neutralizes the −σ bottom plate of the capacitor, so there is effectively no charge in the bottom half of the original capacitor, which means we can ignore that part. The stored energy U therefore decreases, simply because the volume of nonzero field decreases. The final energy is ( ) ϵ0 σ 2 σ 2 As Uf = . (310) (A · s/2) = 2 ϵ0 4ϵ0 Uf is correctly smaller than Ui . (If Uf turned out to be larger than Ui in an isolated system, we would have a problem.) The decrease in U shows up as kinetic energy, so K = σ 2 As/4ϵ0 . Note that when the slab is completely inside, since the effective width of the capacitor decreases to s/2, the voltage decreases to (σ/ϵ0 )(s/2) = σs/2ϵ0 . But we didn’t need this fact when finding K . σ' +++ +++ +++ +++ +++ +++ + −−−−−−−−−−−−−−−−−−− +++ +++ +++ +++ +++ +++ + −−−−−−−−−−−−−−−−−−− -σ' Figure 83 -σ' σ' (b) As in part (a), the initial energy stored in the capacitor is Ui = σ 2 As/2ϵ0 . Now consider the moment when the conducting slab is completely inside the capacitor. The field inside the conductor is zero, so if the final charge densities on the capacitor are ±σ ′ , we must have the opposite densities on the two faces of the slab, as shown in Fig. 83. This creates zero field inside the slab. The field in the vacuum half of the capacitor is σ ′ /ϵ0 , so the voltage difference between the plates is (σ ′ /ϵ0 )(s/2). But we are told that the voltage is held constant by the battery, so it must still take on the original value of ϕ = (σ/ϵ0 )s. Hence σ ′ = 2σ. (In short, half the separation means twice the field.) The final energy stored in the capacitor is therefore ( ) ϵ0 2σ 2 σ 2 As Uf = . (311) (A · s/2) = 2 ϵ0 ϵ0 This is larger than the initial energy, in contrast with the situation in part (a). So if there weren’t anything else going on, we would have a violation of energy conservation. But there is indeed something else going on; the battery is doing work. It must dump more charge onto the plates to increase the density from σ to σ ′ = 2σ. It does this by moving charges from the negative plate to the positive plate, through the constant potential difference ϕ = σs/ϵ0 . The amount of charge moved is q = (σ ′ − σ )A = σA, so the work done is W = qϕ = (σA)(σs/ϵ0 ) = σ 2 As/ϵ0 . Conservation of energy therefore gives the final kinetic energy of the slab as W + Ui = Uf + K =⇒ σ 2 As σ 2 As σ 2 As σ 2 As . (312) + K =⇒ K = = + 2ϵ0 ϵ0 2ϵ0 ϵ0 Basically, of the σ 2 As/ϵ0 work done, half goes into increasing U , and half goes into K. It makes sense that the K here is larger than the K in part (a), because the present case ends up with more charge on the plates, so the forces involved are larger. Force on a capacitor sheet We will neglect the edge fields on the assumption that the gap s is much smaller than y and b. The charge Q on sheet A will be split between the area yb on each side of the sheet; so the surface density on each side is σ = (Q/2)/yb. Likewise the charge −Q on sheet B will be split between the area yb on each “wing” of the bent sheet. The electric field on either side of sheet A is therefore E= σ (Q/2)/yb Q . = = 2ϵ0 yb ϵ0 ϵ0 (313) The voltage difference between the sheets is V = Es = Qs/2ϵ0 yb, so the capacitance is C = Q/V = 2ϵ0 yb/s. (This is just the standard parallel-plate expression C = ϵ0 A/s with area A = 2yb.) Equation 3.32 then gives1 ) ( ( ) Q2 s Q2 d Q2 d 1 s =− = F = . (314) 2 dy 2ϵ0 yb 2 dy C 4ϵ0 by2 The negative sign indicates that the energy of the capacitor decreases as y increases. So the direction of the force on A is downward, because the decrease in energy will show up as kinetic energy, or work done on some other object. (Equivalently, the F in Eq. (3.32) was defined as the force that some other object must apply to A to keep it at rest. This force is upward, in the direction of decreasing y; hence the negative sign.) We can write y in terms of V via the above relation, V = Qs/2ϵ0 yb =⇒ y = Qs/2ϵ0 V b. Hence, the magnitude of the force is √ ( )2 Fs Q2 s 2ϵ0 V b ϵ0 V 2 b =⇒ V = F = = . (315) s 4ϵ0 b Qs ϵ0 b Note that the force F is independent of y. This is consistent with the discussion at the end of the solution to Problem 3.26. Although we (justifiably) ignored the edge effects in computing the capacitance, it is in fact these edge effects that produce the force. These edge effects depend on the charge density on the plates, and for a given voltage V , this density is independent of y . Alternatively: We can find the force by directly calculating how the stored energy changes with y. The stored energy is Q2 s 1 U = QV = . 2 4ϵ0 yb (316) 1 The parameter y need not represent the distance between the plates, for Eq. (3.32) to hold. That equation is valid for any y describing the relative position of the conductors in a capacitor; it gives the force in the direction corresponding to the parameter y . This U can also be found via: U = (volume) ϵ0 E 2 ϵ0 = (2syb) 2 2 ( Q 2ϵ0 yb )2 = Q2 s . 4ϵ0 yb (317) This decreases as y increases, that is, as sheet A moves downward. (The volume increases with y, but E decreases with y, and E is squared in the expression for U ). Taking the differential of Eq. (316) gives the decrease in U as Q2 s dy dU = − . 4ϵ0 b y 2 (318) If A is attached to some other object D, this decrease in energy equals the increase in the energy of D, due to the work F dy that sheet A does on D. Therefore, F = Q2 s/4ϵ0 by2 . (We’re now defining F to be the force that A applies to D, instead of the other way around, as it is defined in Eq. (3.32).) Since the sheets are isolated, Q remains constant as y increases, whereas V does not. On the other hand, if the sheets are connected to a constant voltage source, Q will increase with increasing y, and U will also increase. But in this case the voltage source will supply energy for both the increase in U and the external work. The expressions for the force F (given in Eqs. (314) and (315)) will be exactly the same; so F is now constant. See the solution to Problem 3.26 for a more complete discussion of this point. Force on a coaxial capacitor We’ll need to know the capacitance of the cylinders. Let ℓ be the distance of overlap. And let ±λ = ±Q/ℓ be the charge densities per unit length on the overlap region of the cylinders. The (magnitude of the) potential difference ϕ between the cylinders is ( ) ∫ r2 ∫ r2 λ dr r2 λ . (319) ln ϕ= E dr = = r1 2πϵ0 r1 r1 2πϵ0 r The capacitance is then C= Q λℓ 2πϵ0 ℓ = = , ln(r2 /r1 ) ϕ (λ/2πϵ0 ) ln(r2 /r1 ) (320) which is independent of λ, as it should be. (The capacitance depends only on the geometry of the system, and not on the charge that is placed on the conductors.) Consider how the energy changes when there is a downward displacement of the inner cylinder, so that the overlap distance increases by ∆ℓ. The capacitance increases by ∆C = 2πϵ0 ∆ℓ/ ln(r2 /r1 ). With constant potential difference, the stored electrical energy Cϕ2 /2 increases by (∆C)ϕ2 /2. At the same time, an amount of charge ∆Q = (∆C)ϕ flows onto the capacitor. The battery thereby does work, in amount (∆Q)ϕ = (∆C)ϕ2 . This is twice the increase in stored energy in the field. The difference is the work done against the external force F that balances the electrical attraction of the cylinders. That is, the work done by the battery shows up as energy in the capacitor plus energy of an external object: 1 (∆C)ϕ2 = (∆C)ϕ2 + F ∆ℓ. 2 (321) Hence F ∆ℓ = (∆C)ϕ2 /2, from which we obtain F = (1/2)ϕ2 ∆C/∆ℓ. This is a quite general formula. In the case at hand, ∆C/∆ℓ = 2πϵ0 / ln(r2 /r1 ). With r2 /r1 = 3/2 and ϕ = 5000 volts, we find ( 2π 8.85 · 10−12 2πϵ0 1 1 = (5000 V)2 F = ϕ2 2 ln(r2 /r1 ) ln(3/2) 2 s2 C 2 kg m3 ) = 1.71 · 10−3 N. (322) Equipotentials for two pipes Consider the potential ϕ due to a single line charge with density λ. If ϕ is chosen to be zero at a point a distance r0 from the line, then ϕ is zero on the entire circle (or cylinder) of radius r0 around the line. To find the potential at a point at a different radius r1 , we need only find the change in potential along a radial line from r0 to r1 . This change is ( ) ∫ r1 r1 λ λ ln ϕ(r1 ) = − dr = − . (323) 2πϵ0 r0 r0 2πϵ0 r If r1 > r0 then ϕ(r1 ) is negative (assuming λ is positive), which is correct. With two line charges with densities ±λ located at positions (±x0 , 0), the potential (relative to the origin) at an arbitrary point located r1 from the positive line and r2 from the negative line is ( ) ( ) ( ) λ r1 r2 r2 (−λ) λ ϕ=− ln ln ln . (324) − = 2πϵ0 2πϵ0 2πϵ0 r1 x0 x0 If r1 < r2 (so the point is closer to the positive line) then ϕ is positive, which is correct. We see that the potential is constant on a curve for which r2 /r1 = k, for some constant k. So our goal is to show that the curve defined by r2 /r1 = k is a circle. Since r12 = (x − x0 )2 + y 2 and r22 = (x + x0 )2 + y 2 , we can rewrite the relation r22 = k 2 r12 as [ ] (x + x0 )2 + y 2 = k 2 (x − x0 )2 + y 2 (k 2 − 1)(x2 + y 2 + x20 ) − 2(k 2 + 1)x0 x = 0 ( ) k2 + 1 =⇒ x2 − 2 2 x0 x + y 2 = −x20 . k −1 =⇒ (325) The fact that the coefficients of x2 and y 2 here are the same means that the curve is a circle. But let’s finish the calculation anyway. Letting b ≡ (k 2 + 1)x0 /(k 2 − 1) and completing the square yields (x − b)2 + y 2 = b2 − x02 . (326) √ This describes a circle with radius r = b2 − x02 centered at the point (b, 0). b is positive if k > 1 (relevant to the right half-plane), and b is negative if k < 1 (relevant to the left half-plane). Note that since b > x0 , r is indeed real. If k → ∞, then b → x0 and r → 0, as expected (we have a small circle around the positive line). And if k → 1 from the positive side, then b → ∞ and r → b → ∞, also as expected (we have a large circle with its leftmost point at the origin). You can show that r = 2k/|k 2 − 1|. If k is replaced with 1/k (that is, r1 and r2 interchange their values), then b becomes −b, and r remains the same, as expected. If you also want to demonstrate that the field lines are circles, you can show that any circle described by x2 + (y − h)2 = x02 + h2 intersects any circle described by Eq. (326) at right angles, for any values of b and h. (The parameter h gives the y value of the center of the field-line circle; any such circle must pass through the pipes and have its center on the y axis.) There are various ways to show this. One is to take the differential of each circle equation to find the slopes of the curves, and to then show that these slopes are the negative reciprocals of each other, by using the difference of the circle equations, namely bx − hy = x02. Average of six points The Taylor expansions are ϕ(x0 + δ, y0 , z0 ) = ϕ(x0 − δ, y0 , z0 ) = δ2 ∂ 2 ϕ δ3 ∂ 3 ϕ ∂ϕ + ··· , + + 3! ∂x3 2! ∂x2 ∂x δ2 ∂ 2 ϕ δ3 ∂ 3 ϕ ∂ϕ ϕ(x0 , y0 , z0 ) − δ − + + ··· , ∂x 3! ∂x3 2! ∂x2 ϕ(x0 , y0 , z0 ) + δ (327) and likewise for the y0 ± δ and z0 ± δ points. When we add up all six terms and divide by 6 to take the average, the terms with odd powers of δ cancel in pairs, and we are left with ) ] [ ( 2 ∂ ϕ ∂2ϕ ∂2ϕ 1 6ϕ(x0 , y0 , z0 ) + δ 2 ϕavg = (328) + + 2 + O(δ 4 ) . 2 2 ∂z 6 ∂x ∂y If ∇2 ϕ = 0, then the δ 2 term here is zero, so we have ϕavg = ϕ(x0 , y0 , z0 ) + O(δ 4 ), (329) which equals ϕ(x0 , y0 , z0 ) through terms of order δ 3 , as desired. The relaxation method After four iterations, the values of a through g, as they would appear in the array, are (at least for the order of my path running through the array): 29.2 26.2 18.4 9.2 61.9 56.5 37.5 These values are close to the true values (which settle down after about 25 iterations), obtained from the computer program in Exercise 3.77: 29.2135 25.8427 17.9775 8.98876 61.7978 56.1798 37.0787 Using either of the above sets of ϕ values, we see that the ϕ = 25 and ϕ = 50 equipotentials will look something like the ones shown in Fig. 84. Relaxation method, numerical Here is a Mathematica program that gets the job done for an 18 × 18 array instead of the 9 × 9 array that appeared in Exercise 3.76: ITS=150; (* number of iterations *) n=6; (* width of box in middle; 1/3 size of whole array *) NN=3n+1; (* number of lattice points in each direction *) T=Table[100, {i,1,NN},{j,1,NN}]; (* create matrix with all entries = 100 *) (* create an initial array with equipotential squares varying linearly from 0 on the outer boundary to 100 on the inner square: *) φ = 100 50 25 φ =0 Figure 84 Do[Do[T[[i,j]]=(100./n)*(k-1), {i,k,NN-(k-1)},{j,k,NN-(k-1)}], {k,1,n+1}]; (* now do the iterative averaging: *) Do[ (* averaging for rows above middle box: *) Do[Do[T[[i,j]]=(T[[i,j-1]]+T[[i,j+1]]+T[[i-1,j]]+T[[i+1,j]])/4., {j,2,NN-1}],{i,2,n}]; (* averaging for rows left of middle box: *) Do[Do[T[[i,j]]=(T[[i,j-1]]+T[[i,j+1]]+T[[i-1,j]]+T[[i+1,j]])/4., {j,2,n}],{i,n+1,2n+1}]; (* averaging for rows right of middle box: *) Do[Do[T[[i,j]]=(T[[i,j-1]]+T[[i,j+1]]+T[[i-1,j]]+T[[i+1,j]])/4., {j,2n+2,NN-1}],{i,n+1,2n+1}]; (* averaging for rows below middle box: *) Do[Do[T[[i,j]]=(T[[i,j-1]]+T[[i,j+1]]+T[[i-1,j]]+T[[i+1,j]])/4., {j,2,NN-1}],{i,2n+2,NN-1}], {it,1,ITS}] (* repeat averaging process ITS times *) PaddedForm[MatrixForm[T], {6, 3}] (* print matrix *) The results for the “triangle” of entries analogous to those in Exercise 3.76 are: 14.357 14.124 13.416 12.227 10.601 8.666 6.561 4.386 2.193 29.179 28.722 27.313 24.892 21.512 17.503 13.192 8.789 44.917 44.271 42.224 38.515 33.052 26.641 19.917 61.945 61.221 58.796 53.892 45.541 36.091 80.423 79.872 77.846 72.717 59.129 The bold entries correspond to the seven entries in Exercise 3.76. The agreement is reasonably good. If the middle box is 48 × 48 instead of the above 6 × 6 or the 3 × 3 we had in Exercise 3.76, then, for example, the 36.091 entry becomes 35.2961. By looking at how this number changes with the size of the box, it appears to converge to approximately 35.2 in the continuum limit involving an infinite number of lattice points. Interestingly, the computing time doesn’t appear to be helped much by our choice of initial equipotentials that varied linearly from the outer boundary to the central square. If we had instead picked ϕ = 0 on the outer boundary and ϕ = 100 at every other point (all the other points, not just the ones in the middle square), then the computing time would be only slightly longer. The computing time increases by a factor of 16 for every doubling of the array’s width, because there are 4 times as many points that each iteration needs to run through, and also it turns out that we need to do 4 times as many iterations to achieve a given accuracy.