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Solução de problemas do livro do Jackson, tomara que ajude alguém

Tipologia: Exercícios

2018

1 / 163

Baixe Exercícios: Problemas do livro Jackson e outras Exercícios em PDF para Física Clássica, somente na Docsity! Solutions to Problems in Jackson, Classical Electrodynamics, Third Edition Homer Reid December 8, 1999 Chapter 2 Problem 2.1 A point charge q is brought to a position a distance d away from an infinite plane conductor held at zero potential. Using the method of images, find: (a) the surface-charge density induced on the plane, and plot it; (b) the force between the plane and the charge by using Coulomb’s law for the force between the charge and its image; (c) the total force acting on the plane by integrating σ2/2ε0 over the whole plane; (d) the work necessary to remove the charge q from its position to infinity; (e) the potential energy between the charge q and its image (com- pare the answer to part d and discuss). (f) Find the answer to part d in electron volts for an electron originally one angstrom from the surface. (a) We’ll take d to be in the z direction, so the charge q is at (x, y, z) = (0, 0, d). The image charge is −q at (0, 0,−d). The potential at a point r is Φ(r) = q 4πε0 [ 1 |r − dk| − 1 |r + dk| ] The surface charge induced on the plane is found by differentiating this: 1 Homer Reid’s Solutions to Jackson Problems: Chapter 2 2 σ = −ε0 dΦ dz ∣ ∣ z=0 = − q 4π [−(z − d) |r + dk|3 + (z + d) |r + dk|3 ] ∣ ∣ z=0 = − qd 2π(x2 + y2 + d2)3/2 (1) We can check this by integrating this over the entire xy plane and verifying that the total charge is just the value −q of the image charge: ∫ ∞ −∞ ∫ ∞ −∞ σ(x, y)dxdy = − qd 2π ∫ ∞ 0 ∫ 2π 0 rdψdr (r2 + d2)3/2 = −qd ∫ ∞ 0 rdr (r2 + d2)3/2 = −qd 2 ∫ ∞ d2 u−3/2du = −qd 2 ∣ ∣ ∣ −2u−1/2 ∣ ∣ ∣ ∞ d2 = −q √ (b) The point of this problem is that, for points above the z axis, it doesn’t matter whether there is a charge −q at (0, 0, d) or an infinite grounded sheet at z = 0. Physics above the z axis is exactly the same whether we have the charge or the sheet. In particular, the force on the original charge is the same whether we have the charge or the sheet. That means that, if we assume the sheet is present instead of the charge, it will feel a reaction force equal to what the image charge would feel if it were present instead of the sheet. The force on the image charge would be just F = q2/16πε0d 2, so this must be what the sheet feels. (c) Total force on sheet = 1 2ε0 ∫ ∞ 0 ∫ 2π 0 σ2dA = q2d2 4πε0 ∫ ∞ 0 rdr (r2 + d2)3 = q2d2 8πε0 ∫ ∞ d2 u−3du = q2d2 8πε0 ∣ ∣ ∣ ∣ −1 2 u−2 ∣ ∣ ∣ ∣ ∞ d2 = q2d2 8πε0 [ 1 2 d−4 ] Homer Reid’s Solutions to Jackson Problems: Chapter 2 5 r23 = [(x − x0) 2 + (y + y0) 2] r24 = [(x+ x0) 2 + (y + y0) 2]. From this you can see that • when x = 0, r1 = r2 and r3 = r4 • when y = 0, r1 = r3 and r2 = r4 and in both cases the argument of the logarithm in (2) is unity. (b) σ = −ε0 d dy Φ = − λ 2π ( 1 r2 dr2 dy + 1 r3 dr3 dy − 1 r1 dr1 dy − 1 r4 dr4 dy ) ∣ ∣ y=0 We have dr1/dy = (y − y0)/r1 and similarly for the other derivatives, so σ = − λ 2π ( y − y0 r22 + y + y0 r23 − y − y0 r21 − y + y0 r24 ) ∣ ∣ y=0 = −y0λ π ( 1 (x− x0)2 + y2 0 − 1 (x+ x0)2 + y2 0) ) (c) Total charge per unit length in z Qx = ∫ ∞ 0 σdx = −y0λ π [ ∫ ∞ 0 dx (x− x0)2 + y2 0 − ∫ ∞ 0 dx (x+ x0)2 + y2 0 ] For the first integral the appropriate substitution is (x − x0) = y0 tanu, dx = y0 sec2 udu. A similar substitution works in the second integral. = −λ π [ ∫ π/2 tan−1 − x0 y0 du− ∫ π/2 tan−1 x0 y0 du ] = −λ π [ π 2 − tan−1 −x0 y0 − π 2 + tan−1 x0 y0 ] = −2λ π tan−1 x0 y0 . (3) The calculations are obviously symmetric with respect to x0 and y0. The total charge on the plane x = 0 is (3) with x0 and y0 interchanged: Qy = −2λ π tan−1 y0 x0 Since tan−1 x− tan−1(1/x) = π/2 the total charge induced is Q = −λ Homer Reid’s Solutions to Jackson Problems: Chapter 2 6 which is, of course, also the sum of the charge per unit length of the three image charges. (d) We have Φ = λ 4πε0 ln r22r 2 3 r21r 2 4 Far from the origin, r21 = [(x − x0) 2 + (y − y0) 2] = [ x2(1 − x0 x )2 + y2(1 − y0 y )2 ] ≈ [ x2(1 − 2 x0 x ) + y2(1 − 2 y0 y ] = [ x2 − 2x0x+ y2 − 2y0y) ] = (x2 + y2) [ 1 − 2 xx0 + yy0 x2 + y2 ] Similarly, r22 = (x2 + y2) [ 1 − 2 −xx0 + yy0 x2 + y2 ] r23 = (x2 + y2) [ 1 − 2 xx0 − yy0 x2 + y2 ] r24 = (x2 + y2) [ 1 − 2 −xx0 − yy0 x2 + y2 ] Next, r21r 2 4 = (x2 + y2)2 [ 1 − 4 (xx0 + yy0) 2 (x2 + y2)2 ] r22r 2 3 = (x2 + y2)2 [ 1 − 4 (xx0 − yy0) 2 (x2 + y2)2 ] so Φ = λ 4πε0 ln 1 − 4 (xx0−yy0) 2 (x2+y2)2 1 − 4 (xx0+yy0)2 (x2+y2)2 . The (x2 + y2) term in the denominator grows much more quickly than the (xx0 + yy0) term, so in the asymptotic limit we can use ln(1 + ε) ≈ ε to find Φ = λ 4πε0 [ −4 (xx0 − yy0) 2 (x2 + y2)2 + 4 (xx0 + yy0) 2 (x2 + y2)2 ] = λ 4πε0 [−4(x2x2 0 + y2y2 0 − 2xyx0y0) + 4(x2x2 0 + y2y2 0 + 2xyx0y0) (x2 + y2)2 ] Homer Reid’s Solutions to Jackson Problems: Chapter 2 7 = λ 4πε0 [ 16xyx0y0 (x2 + y2)2 ] = 4λ πε0 (xy)(x0y0) (x2 + y2)2 . √ Problem 2.4 A point charge is placed a distance d > R from the center of an equally charged, isolated, conducting sphere of radius R. (a) Inside of what distance from the surface of the sphere is the point charge attracted rather than repelled by the charged sphere? (b) What is the limiting value of the force of attraction when the point charge is located a distance a(= d−R) from the surface of the sphere, if a R? (c) What are the results for parts a and b if the charge on the sphere is twice (half) as large as the point charge, but still the same sign? Let’s call the point charge q. The charged, isolated sphere may be replaced by two image charges. One image charge, of charge q1 = −(R/d)q at radius r1 = R2/d, is needed to make the potential equal at all points on the sphere. The second image charge, of charge q2 = q − q1 at the center of the sphere, is necessary to recreate the effect of the additional charge on the sphere (the “additional” charge is the extra charge on the sphere left over after you subtract the surface charge density induced by the point charge q). The force on the point charge is the sum of the forces from the two image charges: F = 1 4πε0 [ qq1 [ d− R2 d ]2 + qq2 d2 ] (4) = q2 4πε0 [ −dR [d2 −R2]2 + d2 + dR d4 ] (5) As d → R the denominator of the first term vanishes, so that term wins, and the overall force is attractive. As d → ∞, the denominator of both terms looks like d4, so the dR terms in the numerator cancel and the overall force is repulsive. (a) The crossover distance is found by equating the two bracketed terms in (5): Homer Reid’s Solutions to Jackson Problems: Chapter 2 10 its image is PE = 1 4πε0 ( qq′ |r − r′| ) = 1 4πε0 ( −q2a r(r − a2/r) ) = 1 4πε0 ( −q2a r2 − a2 ) (8) Result (7) is only half of (8). This would seem to violate energy conservation. It would seem that we could start with the point charge at infinity and allow it to fall in to a distance r from the sphere, liberating a quantity of energy (8), which we could store in a battery or something. Then we could expend an energy equal to (7) to remove the charge back to infinity, at which point we would be back where we started, but we would still have half of the energy saved in the battery. It would seem that we could keep doing this over and over again, storing up as much energy in the battery as we pleased. I think the problem is with equation (8). The traditional expression q1q2/4πε0r for the potential energy of two charges comes from calculating the work needed to bring one charge from infinity to a distance r from the other charge, and it is assumed that the other charge does not move and keeps a constant charge during the process. But in this case one of the charges is a fictitious image charge, and as the point charge q is brought in from infinity the image charge moves out from the center of the sphere, and its charge increases. So the simple expression doesn’t work to calculate the potential energy of the configuration, and we should take (7) to be the correct result. (b) In this case there are two image charges: one of the same charge and location as in part a, and another of charge Q − q′ at the origin. The work needed to remove the point charge q to infinity is the work needed to remove the point charge from its image charge, plus the work needed to remove the point charge from the extra charge at the origin. We calculated the first contribution above. The second contribution is − ∫ ∞ r q(Q− q′)dy 4πε0y2 = − 1 4πε0 ∫ ∞ r [ qQ y2 + q2a y3 ] dy = − 1 4πε0 ∣ ∣ ∣ ∣ −qQ y − q2a 2y2 ∣ ∣ ∣ ∣ ∞ r = − 1 4πε0 [ qQ r + q2a 2r2 ] so the total work done is W = 1 4πε0 [ q2a 2(r2 − a2) − q2a 2r2 − qQ r ] . Homer Reid’s Solutions to Jackson Problems: Chapter 2 11 Review of Green’s Functions Some problems in this and other chapters use the Green’s function technique. It’s useful to review this technique, and also to establish my conventions since I define the Green’s function a little differently than Jackson. The whole technique is based on the divergence theorem. Suppose A(x) is a vector valued function defined at each point x within a volume V . Then ∫ V (∇ · A(x′)) dV ′ = ∮ S A(x′) · dA′ (9) where S is the (closed) surface bounding the volume V . If we take A(x) = φ(x)∇ψ(x) where φ and ψ are scalar functions, (9) becomes ∫ V [ (∇φ(x′)) · (∇ψ(x′)) + φ(x′)∇2ψ(x′) ] dV ′ = ∮ S φ(x′) ∂ψ ∂n ∣ ∣ ∣ ∣ x ′ dA′ where ∂ψ/∂n is the dot product of ~∇ψ with the outward normal to the surface area element. If we write down this equation with φ and ψ switched and subtract the two, we come up with ∫ V [ φ∇2ψ − ψ∇2φ ] dV ′ = ∮ S [ φ ∂ψ ∂n − ψ ∂φ ∂n ] dA′. (10) This statement doesn’t appear to be very useful, since it seems to require that we know φ over the whole volume to compute the left side, and both φ and ∂φ/∂n on the boundary to compute the right side. However, suppose we could choose ψ(x) in a clever way such that ∇2ψ = δ(x − x0) for some point x0 within the volume. (Since this ψ is a function of x which also depends on x0 as a parameter, we might write it as ψx0 (x).) Then we could use the sifting property of the delta function to find φ(x0) = ∫ V [ ψx0 (x′)∇2φ(x′) ] dV ′ + ∮ S [ φ(x′) ∂ψx0 ∂n ∣ ∣ ∣ ∣ x ′ − ψx0 (x′) ∂φ ∂n ∣ ∣ ∣ ∣ x ′ ] dA′. If φ is the scalar potential of electrostatics, we know that ∇2ψ(x′) = −ρ(x′)/ε0, so we have φ(x0) = − 1 ε0 ∫ V ψx0 (x′)ρ(x′)dV ′ + ∮ S [ φ(x′) ∂ψx0 ∂n ∣ ∣ ∣ ∣ x ′ − ψx0 (x′) ∂φ ∂n ∣ ∣ ∣ ∣ x ′ ] dA′. (11) Equation (11) allows us to find the potential at an arbitrary point x0 as long as we know ρ within the volume and both φ and ∂φ/∂n on the boundary. boundary. Usually we do know ρ within the volume, but we only know either φ or ∂φ/∂n on the boundary. This lack of knowledge can be accommodated by choosing ψ such that either its value or its normal derivative vanishes on the boundary surface, so that the term which we can’t evaluate drops out of the surface integral. More specifically, Homer Reid’s Solutions to Jackson Problems: Chapter 2 12 • if we know φ but not ∂φ/∂n on the boundary (“Dirichlet” boundary con- ditions), we choose ψ such that ψ = 0 on the boundary. Then φ(x0) = − 1 ε0 ∫ V ψx0 (x′)ρ(x′)dV ′ + ∮ S φ(x′) ∂ψx0 ∂n ∣ ∣ ∣ ∣ x ′ dA′. (12) • if we know ∂φ/∂n but not φ on the boundary (“Neumann” boundary conditions), we choose ψ such that ∂ψ/∂n = 0 on the boundary. Then φ(x0) = − 1 ε0 ∫ V ψx0 (x′)ρ(x′)dV ′ + ∮ S φx0 (x′) ∂φ ∂n ∣ ∣ ∣ ∣ x ′ dA′. (13) Again, in both cases the function ψx0 (x) has the property that ∇2ψx0 (x) = δ(x − x0). Homer Reid’s Solutions to Jackson Problems: Chapter 2 3 Multiplying the first term by R2/b2 on top and bottom yields σ = − τ 2π [ R2 b − b R2 + b2 − 2bR cosφ ] = − τ 2πb [ R2 − b2 R2 + b2 − 2bR cosφ ] (d) To find the force on the charge, we note that the potential of the image charge is Φ(x) = − τ 4πε0 ln C2 |x − R′ î|2 . with C some constant. We can differentiate this to find the electric field due to the image charge: E(x) = −∇Φ(x) = − τ 4πε0 ∇ ln |x − R′ î|2 = − τ 4πε0 2(x− R′ î) |x − R′̂i|2 . The original line charge is at x = R, y = 0, and the field there is E = − τ 2πε0 1 R − R′ î = − τ 2πε0 R R2 − b2 î. The force per unit width on the line charge is F = τE = − τ2 2πε0 R R2 − b2 tending to pull the original charge in toward the cylinder. Problem 2.12 Starting with the series solution (2.71) for the two-dimensional potential problem with the potential specified on the surface of a cylinder of radius b, evaluate the coefficients formally, substitute them into the series, and sum it to obtain the potential inside the cylinder in the form of Poisson’s integral: Φ(ρ, φ) = 1 2π ∫ 2π 0 Φ(b, φ′) b2 − ρ2 b2 + ρ2 − 2bρ cos(φ′ − φ) dφ′ What modification is necessary if the potential is desired in the region of space bounded by the cylinder and infinity? Homer Reid’s Solutions to Jackson Problems: Chapter 2 4 Referring to equation (2.71), we know the bn are all zero, because the ln term and the negative powers of ρ are singular at the origin. We are left with Φ(ρ, φ) = a0 + ∞ ∑ n=1 ρn {an sin(nφ) + bn cos(nφ)} . (1) Multiplying both sides successively by 1, sin n′φ, and cosn′φ and integrating at ρ = b gives a0 = 1 2π ∫ 2π 0 Φ(b, φ)dφ (2) an = 1 πbn ∫ 2π 0 Φ(b, φ) sin(nφ)dφ (3) bn = 1 πbn ∫ 2π 0 Φ(b, φ) cos(nφ)dφ. (4) Plugging back into (1), we find Φ(ρ, φ) = 1 π ∫ 2π 0 Φ(b, φ′) { 1 2 + ∞ ∑ n=1 (ρ b )n [sin(nφ) sin(nφ′) + cos(nφ) cos(nφ′)] } dφ′ = 1 π ∫ 2π 0 Φ(b, φ′) { 1 2 + ∞ ∑ n=1 (ρ b )n cosn(φ − φ′) } . (5) The bracketed term can be expressed in closed form. For simplicity define x = (ρ/b) and α = (φ − φ′). Then 1 2 + ∞ ∑ n=1 xn cos(nα) = 1 2 + 1 2 ∞ ∑ n=1 [ xneinα + xne−inα ] = 1 2 + 1 2 [ 1 1 − xeiα + 1 1 − xe−iα − 2 ] = 1 2 + 1 2 [ 1 − xe−iα − xeiα + 1 1 − xeiα − xe−iα + x2 − 2 ] = 1 2 + [ 1 − x cosα 1 + x2 − 2x cosα − 1 ] = 1 2 + x cosα − x2 1 + x2 − 2x cosα = 1 2 [ 1 − x2 1 + x2 − 2x cosα ] . Plugging this back into (5) gives the advertised result. Homer Reid’s Solutions to Jackson Problems: Chapter 2 5 Problem 2.13 (a) Two halves of a long hollow conducting cylinder of inner radius b are separated by small lengthwise gaps on each side, and are kept at different potentials V1 and V2. Show that the potential inside is given by Φ(ρ, φ) = V1 + V2 2 + V1 − V2 π tan−1 ( 2bρ b2 − ρ2 cosφ ) where φ is measured from a plane perpendicular to the plane through the gap. (b) Calculate the surface-charge density on each half of the cylinder. This problem is just like the previous one. Since we are looking for an expression for the potential within the cylinder, the correct expansion is (1) with expansion coefficients given by (2), (3) and (4): a0 = 1 2π ∫ 2π 0 Φ(b, φ)dφ = 1 2π [ V1 ∫ π 0 dφ + V2 ∫ 2π π dφ ] = V1 + V2 2 an = 1 πbn [ V1 ∫ π 0 sin(nφ)dφ + V2 ∫ 2π π sin(nφ)dφ ] = − 1 nπbn [ V1 |cosnφ|π0 + V2 |cosnφ|2π π ] = − 1 nπbn [V1(cosnπ − 1) + V2(1 − cosnπ)] = { 0 , n even 2(V1 − V2)/(nπbn) , n odd } bn = 1 πbn [ V1 ∫ π 0 cos(nφ)dφ + V2 ∫ 2π π cos(nφ)dφ ] = 1 nπbn [ V1 |sinnφ|π0 + V2 |sin nφ|2π π ] = 0. With these coefficients, the potential expansion becomes Φ(ρ, φ) = V1 + V2 2 + 2(V1 − V2) π ∑ n odd 1 n (ρ b )n sinnφ. (6) Homer Reid’s Solutions to Jackson Problems: Chapter 2 8 We can add these together and use the differential equation satisfied by gn to find ∇2G = δ(y − y′) · 2 ∞ ∑ n=1 sin(nπx) sin(nπx′) = δ(y − y′) · δ(x − x′) since the infinite sum is just a well-known representation of the δ function. (b) The suggestion is to take gn(y, y′) = { An1 sinh(nπy′) + Bn1 cosh(nπy′), y′ < y; An2 sinh(nπy′) + Bn2 cosh(nπy′), y′ > y. (9) The idea to use hyperbolic sines and cosines comes from the fact that sinh(nπy) and cosh(nπy) satisfy a homogeneous version of the differential equation for gn (i.e. satisfy that differential equation with the δ function replaced by zero). Thus gn as defined in (9) satisfies its differential equation (at all points except y = y′) for any choice of the As and Bs. This leaves us free to choose these coefficients as required to satisfy the boundary conditions and the differential equation at y = y′. First let’s consider the boundary conditions. Since y is somewhere between 0 and 1, the condition that gn vanish for y′ = 0 is only relevant to the top line of (9), where it requires taking Bn1 = 0 but leaves An1 undetermined for now. The condition that gn vanish for y′ = 1 only affects the lower line of (9), where it requires that 0 = An2 sinh(nπ) + Bn2 cosh(nπ) = (An2 + Bn2)e nπ + (−An2 + Bn2)e −nπ (10) One way to make this work is to take An2 + Bn2 = −e−nπ and − An2 + Bn2 = enπ. Then Bn2 = enπ + An2 → 2An2 = −enπ − e−nπ so An2 = − cosh(nπ) and Bn2 = sinh(nπ). With this choice of coefficients, the lower line in (9) becomes gn(y, y′) = − cosh(nπ) sinh(nπy′)+sinh(nπ) cosh(nπy′) = sinh[nπ(1−y′)] (11) for (y′ > y). Actually, we haven’t completely determined An2 and Bn2; we could multiply (11) by an arbitrary constant γn and (10) would still be satisfied. Next we need to make sure that the two halves of (9) match up at y′ = y: An1 sinh(nπy) = γn sinh[nπ(1 − y)]. (12) Homer Reid’s Solutions to Jackson Problems: Chapter 2 9 0 10000 20000 30000 40000 50000 60000 70000 0 0.2 0.4 0.6 0.8 1 g( yp rim e) yprime Figure 1: gn(y, y′) from Problem 2.15 with n=5, y=.41 This obviously happens when An1 = βn sinh[nπ(1 − y)] and γn = βn sinh(nπy) where βn is any constant. In other words, we have gn(y, y′) = { βn sinh[nπ(1 − y)] sinh(nπy′), y′ < y; βn sinh[nπ(1 − y′)] sinh(nπy), y′ > y. = βn sinh[nπ(1 − y>)] sinh(nπy<) (13) with y< and y> defined as in the problem. Figure 1 shows a graph of this function n = 5, y = .41. The final step is to choose the normalization constant βn such that gn sat- isfies its differential equation: ( ∂2 ∂2y′2 − n2π2 ) gn(y, y′) = δ(y − y′). (14) To say that the left-hand side “equals” the delta function requires two things: • that the left-hand side vanish at all points y′ 6= y, and • that its integral over any interval (y1, y2) equal 1 if the interval contains the point y′ = y, and vanish otherwise. The first condition is clearly satisfied regardless of the choice of βn. The second condition may be satisfied by making gn continuous, which we have already done, but giving its first derivative a finite jump of unit magnitude at y′ = y: Homer Reid’s Solutions to Jackson Problems: Chapter 2 10 ∂ ∂y′ gn(y, y′) ∣ ∣ ∣ ∣ y′=y+ y′=y− = 1. Differentiating (13), we find this condition to require nπβn [− cosh[nπ(1 − y)] sinh(nπy) − sinh[nπ(1 − y)] cosh(nπy)] = −nπβn sinh(nπ) = 1 so (14) is satisfied if βn = − 1 nπ sinh(nπ) . Then (13) is gn(y, y′) = − sinh[nπ(1 − y>)] sinh(nπy<) nπ sinh(nπ) and the composite Green’s function is G(x, y; x′, y′) = 2 ∞ ∑ n=1 gn(y, y′) sin(nπx) sin(nπx′) = −2 ∞ ∑ n=1 sinh[nπ(1 − y>)] sinh(nπy<) sin(nπx) sin(nπx′) nπ sinh(nπ) .(15) Problem 2.16 A two-dimensional potential exists on a unit square area (0 ≤ x ≤ 1, 0 ≤ y ≤ 1) bounded by “surfaces” held at zero potential. Over the entire square there is a uniform charge density of unit strength (per unit length in z). Using the Green function of Problem 2.15, show that the solution can be written as Φ(x, y) = 4 π3ε0 ∞ ∑ m=0 sin[(2m + 1)πx] (2m + 1)3 { 1 − cosh[(2m + 1)π(y − (1/2))] cosh[(2m + 1)π/2] } . Referring to my Green’s functions review above, the potential at a point x0 within the square is given by Φ(x0) = − 1 ε0 ∫ V G(x0;x ′)ρ(x′)dV ′ + ∮ S [ Φ(x′) ∂G ∂n ∣ ∣ ∣ ∣ x ′ − G(x0;x ′) ∂Φ ∂n ∣ ∣ ∣ ∣ x ′ ] dA′. (16) In this case the surface integral vanishes, because we’re given that Φ vanishes on the boundary, and G vanishes there by construction. We’re also given that Homer Reid’s Solutions to Jackson Problems: Chapter 2 13 = ln (Z2 + a2)1/2 + Z (Z2 + a2)1/2 − Z = ln (1 + (a2/Z2))1/2 + 1 (1 + (a2/Z2))1/2 − 1 ≈ ln 2 + a2 2Z2 a2 2Z2 = ln 4Z2 + a2 a2 = ln[4Z2 + a2] − ln a2. Since Z is much bigger than a, the first term is essentially independent of a and is the ’nonessential constant’ Jackson is talking about. The remaining term is the 2D Green’s function: G = − lna2 = − ln[(x − x′)2 + (y − y′)2] in rectangular coordinates = − ln[ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′)] in cylindrical coordinates. (b) The 2d Green’s function is defined by ∫ ∇2G(ρ, φ; ρ′, φ′)ρ′dρ′dφ′ = 1 but ∇2G = 0 at points other than (ρ, φ). These conditions are met if ∇2G(ρ, φ; ρ′, φ′) = 1 ρ′ δ(ρ − ρ′)δ(φ − φ′). (20) You need the ρ′ on the bottom there to cancel out the ρ′ in the area element in the integral. The Laplacian in two-dimensional cylindrical coordinates is ∇2 = 1 ρ′ ∂ ∂ρ′ ( ρ′ ∂ ∂ρ′ ) − 1 ρ′2 ∂ ∂φ′2 . Applying this to the suggested expansion for G gives ∇2G(ρ, φ; ρ′, φ′) = 1 2π ∞ ∑ −∞ { 1 ρ′ ∂ ∂ρ′ ( ρ′ ∂gm ∂ρ′ ) − m2 ρ′2 gm } eim(φ−φ′). If gm satisfies its differential equation as specified in the problem, the term in brackets equals δ(ρ−ρ′)/ρ′ for all m and may be removed from the sum, leaving ∇2G(ρ, φ; ρ′, φ′) = ( δ(ρ − ρ′) ρ′ ) · 1 2π ∞ ∑ −∞ eim(φ−φ′) = ( δ(ρ − ρ′) ρ′ ) δ(φ − φ′). Homer Reid’s Solutions to Jackson Problems: Chapter 2 14 (c) As in Problem 2.15, we’ll construct the functions gm by finding solutions of the homogenous radial differential equation in the two regions and piecing them together at ρ = ρ′ such that the function is continuous but its derivative has a finite jump of magnitude 1/ρ. For m ≥ 1, the solution to the homogenous equation { 1 ρ′ ∂ ∂ρ′ ( ρ′ ∂ ∂ρ′ ) − m2 ρ′2 } f(ρ′) = 0 is f(ρ′) = Amρ′m + Bmρ′−m. Thus we take gm = { A1mρ′m + B1mρ′−m , ρ′ < ρ A2mρ′m + B2mρ′−m , ρ′ > ρ. In order that the first solution be finite at the origin, and the second solution be finite at infinity, we have to take B1m = A2m = 0. Then the condition that the two solutions match at ρ = ρ′ is A1mρm = B2mρ−m which requires A1m = γmρ−m B2m = ρmγm for some constant γm. Now we have gm = γm ( ρ′ ρ )m , ρ′ < ρ γm ( ρ ρ′ )m , ρ′ > ρ The finite-derivative step condition is dgm dρ′ ∣ ∣ ∣ ∣ ρ′=ρ+ − dgm dρ′ ∣ ∣ ∣ ∣ ρ′=ρ− = 1 ρ or −mγm ( 1 ρ + 1 ρ ) = 1 ρ so γm = − 1 2m . Then gm = − 1 2m ( ρ′ ρ )m , ρ′ < ρ − 1 2m ( ρ ρ′ )m , ρ′ > ρ = − 1 2m ( ρ< ρ> )m . Homer Reid’s Solutions to Jackson Problems: Chapter 2 15 Plugging this back into the expansion gives G = − 1 4π ∞ ∑ −∞ 1 m ( ρ< ρ> )m eim(φ−φ′) = − 1 2π ∞ ∑ 1 1 m ( ρ< ρ> )m cos[m(φ − φ′)]. Jackson seems to be adding a ln term to this, which comes from the m = 0 solution of the radial equation, but I have left it out because it doesn’t vanish as ρ′ → ∞. Problem 2.18 (a) By finding appropriate solutions of the radial equation in part b of Problem 2.17, find the Green function for the interior Dirichlet prob- lem of a cylinder of radius b [gm(ρ, ρ′ = b) = 0. See (1.40)]. First find the series expansion akin to the free-space Green function of Problem 2.17. Then show that it can be written in closed form as G = ln [ ρ2ρ′2 + b4 − 2ρρ′b2 cos(φ − φ′) b2(ρ2 + ρ′2 − 2ρρ′ cos(φ − φ′)) ] or G = ln [ (b2 − ρ2)(b2 − ρ′2) + b2|ρ − ρ′|2 b2|ρ − ρ′|2 ] . (b) Show that the solution of the Laplace equation with the potential given as Φ(b, φ) on the cylinder can be expressed as Poisson’s inte- gral of Problem 2.12. (c) What changes are necessary for the Green function for the exterior problem (b < ρ < ∞), for both the Fourier expansion and the closed form? [Note that the exterior Green function is not rigorously correct because it does not vanish for ρ or ρ′ → ∞. For situations in which the potential falls of fast enough as ρ → ∞, no mistake is made in its use.] (a) As before, we write the general solution of the radial equation for gm in the two distinct regions: gm(ρ, ρ′) = { A1mρ′m + B1mρ′−m , ρ′ < ρ A2mρ′m + B2mρ′−m , ρ′ > ρ. (21) The first boundary conditions are that gm remain finite at the origin and vanish on the cylinder boundary. This requires that B1m = 0 Homer Reid’s Solutions to Jackson Problems: Chapter 2 18 Evaluated at ρ′ = b this is ∂G ∂ρ′ ∣ ∣ ∣ ∣ ρ′=b = − 1 2π { ρ2 − b2 b(ρ2 + b2 − 2ρb cos(φ − φ′)) } . In the surface integral, the extra factor of b on the bottom is cancelled by the factor of b in the area element dA′, and (24) becomes just the result of Problem 2.12. (c) For the exterior problem we again start with the solution (21). Now the boundary conditions are different; the condition at ∞ gives A2m = 0, while the condition at b gives A1m = γmb−m B1m = −γmbm. From the continuity condition at ρ′ = ρ we find A2m = γmρm [ (ρ b )m − ( b ρ )m] . The finite derivative jump condition gives −mγm [ (ρ b )m − ( b ρ )m] 1 ρ − mγm [ (ρ b )m + ( b ρ )m] 1 ρ = 1 ρ or γm = − 1 2m ( b ρ )m . Putting it all together we have for the exterior problem gm = 1 2m [( b2 ρρ′ )m − ( ρ< ρ> )m] . This is the same gm we came up with before, but with b2 and ρρ′ terms flipped in first term. But the closed-form expression was symmetrical in those two expressions (except for the mysterious ln term) so the closed-form expression for the exterior Green’s function should be the same as the interior Green’s function. Solutions to Problems in Jackson, Classical Electrodynamics, Third Edition Homer Reid June 15, 2000 Chapter 3: Problems 1-10 Problem 3.1 Two concentric spheres have radii a, b(b > a) and each is divided into two hemi- spheres by the same horizontal plane. The upper hemisphere of the inner sphere and the lower hemisphere of the outer sphere are maintained at potential V . The other hemispheres are at zero potential. Detemine the potential in the region a ≤ r ≤ b as a series in Legendre polynomials. Include terms at least up to l = 4. Check your solution against known results in the limiting cases b → ∞ and a → 0. The expansion of the electrostatic potential in spherical coordinates for prob- lems with azimuthal symmetry is Φ(r, θ) = ∞ ∑ l=0 [ Alr l + Blr −(l+1) ] Pl(cos θ). (1) We find the coefficients Al and Bl by applying the boundary conditions. Mul- tiplying both sides by Pl′(cos θ) and integrating from -1 to 1 gives ∫ 1 −1 Φ(r, θ)Pl(cos θ)d(cos θ) = 2 2l + 1 [ Alr l + Blr −(l+1) ] . At r = a this yields V ∫ 1 0 Pl(x)dx = 2 2l + 1 [ Ala l + Bla −(l+1) ] , 1 Homer Reid’s Solutions to Jackson Problems: Chapter 3 2 and at r = b, V ∫ 0 −1 Pl(x)dx = 2 2l + 1 [ Alb l + Blb −(l+1) ] . The integral from 0 to 1 vanishes for l even, and is given in the text for l odd: ∫ 1 0 Pl(x)dx = (−1 2 )(l−1)/2 (l − 2)!! 2 ( l+1 2 ) ! . The integral from -1 to 0 also vanishes for l even, and is just the above result inverted for l odd. This gives V (−1 2 )(l−1)/2 (l − 2)!! 2 ( l+1 2 ) ! = 2 2l + 1 [ Ala l + Bla −(l+1) ] −V (−1 2 )(l−1)/2 (l − 2)!! 2 ( l+1 2 ) ! = 2 2l + 1 [ Alb l + Blb −(l+1) ] . or αl = Ala l + Bla −(l+1) −αl = Alb l + Blb −(l+1) with αl = V (−1 2 )a(l−1)/2 (2l + 1)(l − 2)!! 4 ( l+1 2 ) ! . The solution is Al = αl [ bl+1 + al+1 a2l+1 − b2l+1 ] Bl = −αl [ al+1bl+1(bl + al) a2l+1 − b2l+1 ] The first few terms of (1) are Φ(r, θ) = 3 4 V [ (a2 + b2)r a3 − b3 − a2b2(a + b) r2(a3 − b3) ] P1(cos θ)− 7 16 [ (a4 + b4)r3 a7 − b7 − a4b4(a3 + b3) r4(a7 − b7) ] P3(cos θ)+· · · In the limit as b → ∞, the problem reduces to the exterior problem treated in Section 2.7 of the text. In that limit, the above expression becomes Φ(r, θ) → 3 4 V (a r )2 P1(cos θ) − 7 16 V (a r )4 P3(cos θ) + · · · in agreement with (2.27) with half the potential spacing. When a → 0, the problem goes over to the interior version of the same problem, as treated in section 3.3 of the text. In that limit the above expression goes to Φ(r, θ) → −3 4 V (r b ) P1(cos θ) + 7 16 V (r b )3 P3(cos θ) + · · · This agrees with equation (3.36) in the text, with the sign of V flipped, because here the more positive potential is on the lower hemisphere. Homer Reid’s Solutions to Jackson Problems: Chapter 3 5 Problem 3.3 A thin, flat, conducting, circular disk of radius R is located in the x − y plane with its center at the origin, and is maintained at a fixed potential V . With the information that the charge density on a disc at fixed potential is proportional to (R2 − ρ2)−1/2, where ρ is the distance out from the center of the disc, (a) show that for r > R the potential is Φ(r, θ, φ) = 2V π R r ∞ ∑ l=0 (−1)l 2l + 1 ( R r 2l ) P2l(cos θ) (b) find the potential for r < R. (c) What is the capacitance of the disk? We are told that the surface charge density on the disk goes like σ(r) = K(R2 − r2)−1/2 = K R [ 1 + 1 2 ( r R )2 + 3 · 1 (2!)(2 · 2) ( r R )4 + 5 · 3 · 1 (3!)(2 · 2 · 2) ( r R )6 + · · · ] = K R ∞ ∑ n=0 (2n − 1)!! n! · 2n ( r R )2n (3) for some constant K. From the way the problem is worded, I take it we’re not supposed to try to figure out what K is explicitly, but rather to work the problem knowing only the form of (3). At a point infinitesimally close to the surface of the disk (i.e., as θ → π/2), the component of ∇Φ in the direction normal to the surface of the disk must be proportional to the surface charge. At the surface of the disk, the normal direction is the negative θ̂ direction. Hence 1 r ∂ ∂θ Φ(r, θ) ∣ ∣ ∣ ∣ θ=(π/2) = ± σ ε0 . (4) with the plus (minus) sign valid for Φ above (below) the disc. For r < R the potential expansion is Φ(r, θ) = ∞ ∑ l=0 Alr lPl(cos θ). (5) Combining (3), (4), and (5) we have ∞ ∑ l=0 Alr l−1 d dθ Pl(cos θ) ∣ ∣ ∣ ∣ cos θ=0 = ± K Rε0 ∞ ∑ n=0 (2n − 1)!! n! · 2n ( r R )2n . (6) Homer Reid’s Solutions to Jackson Problems: Chapter 3 6 For l even, dPl/dx vanishes at x = 0. For l odd, I used some of the Legendre polynomial identities to derive the formula d dx P2l+1(x) ∣ ∣ ∣ ∣ x=0 = (−1)l(2l + 1) (2l − 1)!! l! · 2l . This formula reminds one strongly of expansion (3). Plugging into (6) and equating coefficents of powers of r, we find A2l+1 = ± (−1)lK (2l + 1)R2l+1ε0 so Φ(r, θ) = A0 ± K ε0 ∞ ∑ l=1 (−1)l 2l + 1 ( r R )2l+1 P2l+1(cos θ). I wrote A0 explicitly because we haven’t evaluated it yet–the derivative condition we used earlier gave no information about it. To find A0, observe that, on the surface of the disk (cos θ = 0), all the terms in the above sum vanish ( because Pl(0) is 0 for odd l) so Φ = A0 on the disk. But Φ = V on the disk. Therefore, A0 = V . We have Φ(r, θ) = V ± K ε0 ∞ ∑ l=1 (−1)l 2l + 1 ( r R )2l+1 P2l+1(cos θ) (7) where the plus (minus) sign is good for θ less than (greater than)π/2. Note that the presence of that ± sign preserves symmetry under reflection through the z axis, a symmetry that is clearly present in the physical problem. (a) For r > R, there is no charge. Thus the potential and its derivative must be continuous everywhere–we can’t have anything like the derivative discontinuity that exists at θ = π/2 for r < R. Since the physical problem is symmetric under a sign flip in cos θ, the potential expansion can only contain Pl terms for l even. The expansion is Φ(r, θ) = ∞ ∑ l=0 B2lr −(2l+1)P2l(cos θ). At r = R, this must match up with (7): V ± K ε0 ∞ ∑ l=1 (−1)l 2l + 1 P2l+1(cos θ) = ∞ ∑ l=0 B2lR −(2l+1)P2l(cos θ). Multiplying both sides by P2l(cos θ) sin(θ) and integrating gives B2l 2R−(2l+1) 4l + 1 = V ∫ 1 −1 Pl(x)dx + K ε0 ∞ ∑ l=1 (−1)l 2l + 1 { − ∫ 0 −1 P2l+1(x)P2l(x)dx + ∫ 1 0 P2l+1(x)Pl(x)dx } = 2V δl,0 + 2K ε0 ∞ ∑ l=1 (−1)l 2l + 1 ∫ 1 0 P2l+1(x)P2l(x)dx. Homer Reid’s Solutions to Jackson Problems: Chapter 3 7 but I can’t do this last integral. Problem 3.4 The surface of a hollow conducting sphere of inner radius a is divided into an even number of equal segments by a set of planes; their common line of intersection is the z axis and they are distributed uniformly in the angle φ. (The segments are like the skin on wedges of an apple, or the earth’s surface between successive meridians of longitude.) The segments are kept at fixed potentials ±V , alternately. (a) Set up a series representation for the potential inside the sphere for the general case of 2n segments, and carry the calculation of the coefficients in the series far enough to determine exactly which coefficients are different from zero. For the nonvanishing terms exhibit the coefficients as an integral over cos θ. (b) For the special case of n = 1 (two hemispheres) determine explicitly the poten- tial up to and including all terms with l = 3. By a coordinate transformation verify that this reduces to result (3.36) of Section 3.3. (a) The general potential expansion is Φ(r, θ, φ) = ∞ ∑ l=0 l ∑ m=−l [ Almrl + Blmr−(l+1) ] Ylm(θ, φ). (8) For the solution within the sphere, finiteness at the origin requires Blm = 0. Multiplying by Y ∗ l′m′ and integrating over the surface of the sphere we find Alm = 1 al ∫ Φ(a, θ, φ) Y ∗ lm(θ, φ) dΩ = V al n ∑ k=1 (−1)k ∫ π 0 ∫ 2kπ/n 2(k−1)π/n Y ∗ lm(θ, φ) sin θ dφ dθ = V al [ 2l + 1 4π (l − m)! (l + m)! ]1/2{∫ 1 −1 P m l (x) dx } n ∑ k=1 (−1)k { ∫ 2kπ/n 2(k−1)π/n e−imφ dφ } . (9) The φ integral is easy: ∫ 2kπ/n 2(k−1)π/n e−imφ dφ = − 1 im [ e−2imkπ/n − e−2im(k−1)π/n ] . This is to be summed from k = 1 to n with a factor of (−1)k thrown in: ∑ = − 1 im [ (e−2mπi(1/n) − 1) − (e−2mπi(2/n) − e−2mπi(1/n)) + · · · − (1 − e−2mπi((n−1)/n)) ] = 2 im { 1 − e−2mπi/n + e2(−2mπi/n) − e3(−2mπi/n) + · · · + e(n−1)(−2mπi/n) } . (10) Homer Reid’s Solutions to Jackson Problems: Chapter 3 10 (c) When we put the grounded sphere around the two charges, a surface charge distribution forms on the sphere. Let’s denote by Φs the potential due to this charge distribution alone (not including the potential of the dipole) and by Φd the potential due to the dipole. To calculate Φs, we pretend there are no charges within the sphere, in which case we have the general expansion (1), with Bl = 0 to keep us finite at the origin. The total potential is just the sum Φs + Φd: Φ(r, θ) = p 4πε0r2 cos θ + ∞ ∑ l=0 Alr lPl(cos θ). The condition that this vanish at r = b ensures, by the orthogonality of the Pl, that only the l = 1 term in the sum contribute, and that A1 = − p 4πε0b3 . The total potential inside the sphere is then Φ(r, θ) = p 4πε0b2 ( 1 − r b ) P1(cos θ). Problem 3.7 Three point charges (q,−2q, q) are located in a straight line with separation a and with the middle charge (−2q) at the origin of a grounded conducting spherical shell of radius b, as indicated in the figure. (a) Write down the potential of the three charges in the absence of the grounded sphere. Find the limiting form of the potential as a → 0, but the product qa2 = Q remains finite. Write this latter answer in spherical coordinates. (b) The presence of the grounded sphere of radius b alters the potential for r < b. The added potential can be viewed as caused by the surface-charge density induced on the inner surface at r = b or by image charges located at r > b. Use linear superposition to satisfy the boundary conditions and find the potential everywhere inside the sphere for r < a and r > a. Show that in the limit a → 0, Φ(r, θ, φ) → Q 2πε0r3 ( 1 − r5 b5 ) P2(cos θ). (a) On the z axis, the potential is Φ(z) = q 4πε0 [ −2 z + 1 |z − a| + 1 z + a ] = q 4πε0r [ −2 + ( 1 + (a z ) + (a z )2 · · · ) + ( 1 − (a z ) + (a z )2 + · · · )] = q 2πε0z [ (a z )2 + (a z )4 + · · · ] . Homer Reid’s Solutions to Jackson Problems: Chapter 3 11 As before, from this result we can immediately infer the expression for the potential at all points: Φ(r, θ) = q 2πε0r [ (a r )2 P2(cos θ) + (a r )4 P4(cos θ) + · · · ] = qa2 2πε0r3 [ P2(cos θ) + (a r )2 P4(cos θ) + · · · ] → Q 2πε0r3 P2(cos θ) as a → 0 (11) (b) As in the previous problem, the surface charges on the sphere produce an extra contribution Φs to the potential within the sphere. Again we can express Φs with the expansion (1) (with Bl = 0), and we add Φs to (11) to get the full potential within the sphere: Φ(r, θ) = Q 2πε0r3 P2(cos θ) + ∞ ∑ l=0 Alr lPl(cos θ) From the condition that Φ vanish at r = b, we determine that only the l = 2 term in the sum contributes, and that A2 = − Q 2πε0b5 . Then the potential within the sphere is Φ(r, θ) = Q 2πε0r3 [ 1 − (r b )5 ] P2(cos θ). Problem 3.9 A hollow right circular cylinder of radius b has its axis coincident with the z axis and its ends at z = 0 and z = L. The potential on the end faces is zero, while the potential on the cylindrical surface is given as V (φ, z). Using the appropriate sepa- ration of variables in cylindrical coordinates, find a series solution for the potential anywhere inside the cylinder. The general solution of the Laplace equation for problems in cylindrical coordi- nates consists of a sum of terms of the form R(ρ)Q(φ)Z(z). The φ function is of the form Q(φ) = A sin νφ + B cos νφ Homer Reid’s Solutions to Jackson Problems: Chapter 3 12 with ν an integer. The z function is of the form Z(z) = Cekz + De−kz . In this case, Z must vanish at z = 0 and z = L, which means we have to take k imaginary, i.e. Z(z) = C sin(knz) with kn = πn L , n = 1, 2, 3, · · · With this form for Z, R must be taken to be of the form R(ρ) = EIν(knρ) + FKν(knρ). Since we’re looking for the potential on the inside of the cylinder and there is no charge at the origin, the solution must be finite as ρ → 0, which requires F = 0. Then the potential expansion becomes Φ(ρ, φ, z) = ∞ ∑ n=1 ∞ ∑ ν=0 [Anν sin νφ + Bnν cos νφ] sin(knz)Iν(knρ). (12) Multiplying by sin ν′φ sin kn′z and integrating at r = b, we find ∫ L 0 ∫ 2π 0 V (φ, z) sin νφ sin(knz) dφ dz = πL 2 Iν(knb)Anν so Anν = 2 πLIν(knb) ∫ L 0 ∫ 2π 0 V (φ, z) sin(νφ) sin(knz) dφ dz. (13) Similarly, Bnν = 2 πLIν(knb) ∫ L 0 ∫ 2π 0 V (φ, z) cos(νφ) sin(knz) dφ dz. (14) Problem 3.10 For the cylinder in Problem 3.9 the cylindrical surface is made of two equal half- cylinders, one at potential V and the other at potential −V , so that V (φ, z) = { V for −π/2 < φ < π/2 −V for π/2 < φ < 3π/2 (a) Find the potential inside the cylinder. (b) Assuming L >> b, consider the potential at z = L/2 as a function of ρ and φ and compare it with two-dimensional Problem 2.13. The potential expansion is (12) with coefficients given by (13) and (14). The relevant integrals are ∫ L 0 ∫ 2π 0 V (φ, z) sin(νφ) sin(knz) dφ dz Solutions to Problems in Jackson, Classical Electrodynamics, Third Edition Homer Reid June 15, 2000 Chapter 3: Problems 11-18 Problem 3.11 A modified Bessel-Fourier series on the interval 0 ≤ ρ ≤ a for an arbitrary function f(ρ) can be based on the ”homogenous” boundary conditions: At ρ = 0, ρJν(kρ) d dρ Jν(k′ρ) = 0 At ρ = a, d dρ ln[Jν(kρ)] = −λ a (λ real) The first condition restricts ν. The second condition yields eigenvalues k = yνn/a, where yνn is the nth positive root of x dJν(x)/dx + λJν(x) = 0. (a) Show that the Bessel functions of different eigenvalues are orthogonal in the usual way. (b) Find the normalization integral and show that an arbitrary function f(ρ) can be expanded on the interval in the modified Bessel-Fourier series f(ρ) = ∞ ∑ n=1 AnJν (yνn a ) with the coefficients An given by An = 2 a2 [ ( 1 − ν2 y2 νn ) J2 ν (yνn) + ( dJν(yνn) dyνn )2 ] −1 ∫ a 0 f(ρ)ρJν (yνnρ a ) dρ. 1 Homer Reid’s Solutions to Jackson Problems: Chapter 3 2 (a) The function Jν(kρ) satisfies the equation 1 ρ d dρ [ ρ d dρ Jν(kρ) ] + ( k2 − ν2 ρ2 ) Jν(kρ) = 0. (1) Multiplying both sides by ρJν(k′ρ) and integrating from 0 to a gives ∫ a 0 { Jν(k′ρ) d dρ [ ρ d dρ Jν(kρ) ] + ( k2ρ − ν2 ρ ) Jν(k′ρ)Jν(kρ) } dρ = 0. (2) The first term on the left can be integrated by parts: ∫ a 0 Jν(k′ρ) d dρ [ ρ d dρ Jν(kρ) ] dρ = ∣ ∣ ∣ ∣ ρJν(k′ρ) d dρ Jν(kρ) ∣ ∣ ∣ ∣ a 0 − ∫ a 0 ρ [ d dρ Jν(k′ρ) ] [ d dρ Jν(kρ) ] dρ. (3) One of the conditions we’re given is that the thing in braces in the first term here vanishes at ρ = 0. At ρ = a we can invoke the other condition: d dρ ln[Jν(kρ)] ∣ ∣ ∣ ∣ ρ=a = 1 Jν(kρ) d dρ Jν(kρ) ∣ ∣ ∣ ∣ ρ=a = −λ a → a d dρ Jν(ka) = −λJν(ka). Plugging this into (3), we have ∫ a 0 Jν(k′ρ) d dρ [ ρ d dρ Jν(kρ) ] dρ = −λJν(k′ρ)Jν(kρ) − ∫ a 0 ρ [ d dρ Jν(k′ρ) ] [ d dρ Jν(kρ) ] . (4) This is clearly symmetric in k and k′, so when we write down (2) with k and k′ switched and subtract from (2), the first integral (along with the ν2/ρ term) vanishes, and we are left with (k′2 − k2) ∫ a 0 ρJν(k′ρ)Jν(kρ) dρ = 0 proving orthogonality. (b) If we multiply (1) by ρ2J ′(kρ) and integrate, we find ∫ a 0 ρJ ′ ν(kρ) d dρ [ρJ ′ ν(kρ)]dρ+k2 ∫ a 0 ρ2Jν(kρ)J ′ ν(kρ)dρ−ν2 ∫ a 0 Jν(kρ)J ′ ν(kρ)dρ = 0. (5) Homer Reid’s Solutions to Jackson Problems: Chapter 3 3 The first and third integrals are of the form ∫ f(x)f ′(x)dx and can be done immediately. In the second integral we put f(ρ) = ρ2Jν(kρ), g′(ρ) = J ′ ν(kρ) and integrate by parts: ∫ a 0 ρ2Jν(kρ)J ′ ν(kρ)dρ = ∣ ∣ρ2J2 ν (kρ) ∣ ∣ a 0 − 2 ∫ a 0 ρJ2 ν (kρ)dρ − ∫ a 0 ρ2Jν(kρ)J ′ ν(kρ)dρ → ∫ a 0 ρ2Jν(kρ)J ′ ν(kρ)dρ = 1 2 a2J2 ν (ka) − ∫ a 0 ρJ2 ν (kρ)dρ. Using this in (5), a2 2 J ′2 ν (ka) + (ak)2 2 aJ2 ν (ka) − k2 ∫ a 0 ρJ2 ν (kρ)dρ − ν2 2 J2 ν (ka) = 0 so ∫ a 0 ρJ2 ν (kρ)dρ = ( a2 2 − ν2 2k2 ) J2 ν (ka) + a2 2k2 J ′2 ν (ka) = a2 2 { ( 1 − ν2 (ka)2 ) J2 ν (ka) + [ d d(ka) Jν(ka) ]2 } This agrees with what Jackson has if you note that k is chosen such that ka = ynm. Problem 3.12 An infinite, thin, plane sheet of conducting material has a circular hole of radius a cut in it. A thin, flat, disc of the same material and slightly smaller radius lies in the plane, filling the hole, but separated from the sheet by a very narrow insulating ring. The disc is maintained at a fixed potential V , whilc the infinite sheet is kept at zero potential. (a) Using appropriate cylindrical coordinates, find an integral expression involving Bessel functions for the potential at any point above the plane. (b) Show that the potential a perpendicular distance z above the center of the disc is Φ0(z) = V ( 1 − z√ a2 + z2 ) (c) Show that the potential a perpendicular distance z above the edge of the disc is Φa(z) = V 2 [ 1 − kz πa K(k) ] where k = 2a/(z2 + 4a2)1/2, and K(k) is the complete elliptic integral of the first kind. Homer Reid’s Solutions to Jackson Problems: Chapter 3 6 with Rl(r; r ′) = 1 [ 1 − ( a b )]2l+1 ( rl < − a2l+1 rl+1 < )( 1 rl+1 > − rl > b2l+1 ) . (11) Actually in this case the potential cannot have any Φ dependence, so all terms with m 6= 0 in (10) vanish, and we have G(x;x′) = − 1 4π ∞ ∑ l=0 Pl(cos θ)Pl(cos θ′)Rl(r; r ′). In this case the boundary surfaces are spherical, which means the normal to a surface element is always in the radial direction: ∂ ∂n G(x;x′) = − 1 4π ∞ ∑ l=0 Pl(cos θ)Pl(cos θ′) ∂ ∂n Rl(r; r ′). The surface integral in (9) has two parts: one integral S1 over the surface of the inner sphere, and a second integral S2 over the surface of the outer sphere: S1 = − 1 4π ∞ ∑ l=0 Pl(cos θ) ∂Rl ∂n ∣ ∣ ∣ ∣ r′=a { ∫ π 0 ∫ 2π 0 Φ(a, θ′)Pl(cos θ′)a2 sin θ′ dφ dθ′ } = −V 2 ∞ ∑ l=0 a2Pl(cos θ) ∂Rl ∂n ∣ ∣ ∣ ∣ r′=a { ∫ 1 0 Pl(x) dx } = −V 2 ∞ ∑ l=0 a2γlPl(cos θ) · ∂Rl ∂n ∣ ∣ ∣ ∣ r′=a where γl = ∫ 1 0 Pl(x) dx = (−1 2 )(l−1)/2 (l − 2)!! 2[(l + 1)/2]! , l odd = 0, l even. A similar calculation gives S2 = −V 2 ∞ ∑ l=0 b2Pl(cos θ) ∂Rl ∂n ∣ ∣ ∣ ∣ r′=b { ∫ 0 −1 Pl(x) dx } = V 2 ∞ ∑ l=0 b2γlPl(cos θ) ∂Rl ∂n ∣ ∣ ∣ ∣ r′=b because Pl is odd for l odd, so its integral from -1 to 0 is just the negative of the integral from 0 to 1. The final potential is the sum of S1 and S2: Φ(r, θ) = V 2 ∞ ∑ l=0 γlPl(cos θ) ∣ ∣ ∣ ∣ r′2 ∂Rl ∂n ∣ ∣ ∣ ∣ r′=b r′=a (12) Homer Reid’s Solutions to Jackson Problems: Chapter 3 7 Since the point of interest is always between the two spheres, to find the normal derivative at r = a we differentiate with respect to r<, and at r = b with respect to r>. Also, at r = a the normal is in the +r direction, while at r = b the normal is in the negative r direction. a2 ∂ ∂n Rl(r; r ′) ∣ ∣ ∣ ∣ r′=a = (2l + 1)a2 al−1 [ 1 − ( a b )]2l+1 ( 1 rl+1 − rl b2l+1 ) b2 ∂ ∂n Rl(r; r ′) ∣ ∣ ∣ ∣ r′=b = (2l + 1)b2 b−(l+2) [ 1 − ( a b )]2l+1 ( rl − a2l+1 rl+1 ) Combining these with some algebra gives Φ(r, θ) = V 2 ∞ ∑ l=0 (2l + 1)γlPl(cos θ) [ (ab)l+1(bl + al)r−(l+1) − (al+1 + bl+1)rl b2l+1 − a2l+1 ] in agreement with what we found in Problem 3.1. Problem 3.14 A line charge of length 2d with a total charge Q has a linear charge density varying as (d2 − z2), where z is the distance from the mid- point. A grounded, conducting spherical shell of inner radius b > d is centered at the midpoint of the line charge. (a) Find the potential everywhere inside the spherical shell as an expansion in Legendre polynomials. (b) Calculate the surface-charge density induced on the shell. (c) Discuss your answers to parts a and b in the limit that d << b. First of all, we are told that the charge density ρ(z) = λ(d2 − z2), and that the total charge is Q, whence Q = 2λ ∫ d 0 (d2 − z2)dz = 4 3 d3λ → λ = 3Q 4d3 . In this case we have azimuthal symmetry, so the Green’s function is G(x;x′) = − 1 4π ∞ ∑ l=0 Pl(cos θ′)Pl(cos θ)Rl(r; r ′) (13) Homer Reid’s Solutions to Jackson Problems: Chapter 3 8 with Rl(r; r ′) = rl < ( 1 rl+1 > − rl > b2l+1 ) . Since the potential vanishes on the boundary surface, the potential inside the sphere is given by Φ(r, θ) = − 1 ε0 ∫ V G(r, θ; r′, θ′)ρ(r′, θ′)dV. In this case ρ is only nonzero on the z axis, where r = z. Also, Pl(cos θ)=1 for z > 0, and (−1)l for z < 0. This means that the contributions to the integral from the portions of the line charge for z > 0 and z < 0 cancel out for odd l, and add constructively for even l: Φ(r, θ) = 1 4πε0 ∞ ∑ l=0,2,4,... Pl(cos θ) [ 2 ∫ d 0 Rl(r; z)ρ(z) dz ] We have ∫ d 0 Rl(r; z)ρ(z) dz = λ ∫ d 0 rl < ( 1 rl+1 > − rl > b2l+1 ) (d2 − z2) dz This is best split up into two separate integrals: = λ ∫ d 0 rl < rl+1 > (d2 − z2) dz − λ b2l+1 ∫ d 0 rl <rl >(d2 − z2) dz The second integral is symmetric between r and r′, so we may integrate it directly: − λ b2l+1 ∫ d 0 rl <rl >(d2 − z2) dz = − λrl b2l+1 ∫ d 0 zl(d2 − z2) dz = − λrl b2l+1 [ dl+3 l + 1 − dl+3 l + 3 ] = − λrldl+3 (l + 1)(l + 3)b2l+1 (14) The first integral must be further split into two: λ ∫ d 0 rl < rl+1 > (d2 − z2) dz Homer Reid’s Solutions to Jackson Problems: Chapter 3 11 for r < a, Φ(r, θ) = Φin(r, θ) = ∞ ∑ l=0 Alr lPl(cos θ) for r > a, Φ(r, θ) = Φout(r, θ) = ∞ ∑ l=0 Blr −(l+1)Pl(cos θ) Continuity at r = a requires that Ala l = Bla −l+1 → Bl = a2l+1Al so Φ(r, θ) = { Φin(r, θ) = ∑ ∞ l=0 Alr lPl(cos θ), r < a Φout(r, θ) = ∑ ∞ l=0 Ala 2l+1r−(l+1)Pl(cos θ), r > a. (16) Now, in the steady state there can be no discontinuities in the current den- sity, because if there were than there would be more current flowing into some region of space than out of it, which means charge would pile up in that region, which would be a growing source of electric field, which would mean we aren’t in steady state. So the current density is continuous everywhere. In particular, the radial component of the current density is continuous across the boundary of the sphere, i.e. Jr(r = a−, θ) = Jr(r = a+, θ). (17) Outside of the sphere, Ohm’s law says that J = σ′E = −σ′∇Φout. Inside the sphere, there is an extra term coming from the chemical force: J = σ(E + F k̂) = σ(−∇Φin + F k̂). Applying (17) to these expressions, we have σ ( − ∂ ∂r Φin ∣ ∣ ∣ ∣ r=a + F cos θ ) = −σ′ ∂ ∂r Φout ∣ ∣ ∣ ∣ r=a Using (16), this is FP1(cos θ) − ∞ ∑ l=0 lAla l−1Pl(cos θ) = ( σ′ σ ) ∞ ∑ l=0 (l + 1)Ala l−1Pl(cos θ). Multiplying both sides by Pl′ (cos θ) and integrating from −π to π, we find F − A1 = ( σ′ σ ) 2A1 (18) Homer Reid’s Solutions to Jackson Problems: Chapter 3 12 for l=1, and −lAl = ( σ′ σ ) (l + 1)Al (19) (20) for l 6= 1. Since the conductivity ratio is positive, the second relation is impos- sible to satisfy unless Al = 0 for l 6= 1. The first relation becomes A1 = σ σ + 2σ′ F. Then the potential is Φ(r, θ) = { σ σ+2σ′ Fr cos θ, r < a σ σ+2σ′ Fa3r−2 cos θ, r > a (21) The dipole moment p is defined by Φ(r, θ) → 1 4πε0 p · r r3 as r → ∞. (22) The external portion of (21) can be written as Φ(r, θ) = σ σ + 2σ′ Fa3z r3 and comparing this with (22) we can read off p = 4πε0 σ σ + 2σ′ Fa3k̂. The electric field is found by taking the gradient of (21): E(r, θ) = { − σ σ+2σ′ F k̂, r < a σ σ+2σ′ F ( a r )3 (2 cos θr̂ + sin θθ̂), r > a The surface charge σs(θ) on the sphere is proportional to the discontinuity in the electric field: σs(θ) = ε0[Er(r = a+) − Er(r = a−)] = 3ε0σ σ + 2σ′ F cos θ. (b) The current flowing out of the upper hemisphere is just ∫ J · dA = σ ∫ (Ein + F k̂) · dA = σ ( 1 − σ σ + 2σ′ ) F ∫ π/2 0 ∫ 2π 0 cos θ sin θ a2 dφ dθ = 2 σσ′ σ + 2σ′ · πa2F (23) Homer Reid’s Solutions to Jackson Problems: Chapter 3 13 The Ohmic power dissipation in a volume dV is dP = σE2dV (24) To see this, suppose we have a rectangular volume element with sides dx, dy, and dz. Consider first the current flowing in the x direction. The current density there is σEx and the cross-sectional area is dydz, so I = σExdydz. Also, the voltage drop in the direction of current flow is V = Exdx. Hence the power dissipation due to current in the x direction is IV = σE2 xdV . Adding in the contributions from the other two directions gives (24). For the power dissipated outside the sphere we use the expression for the electric field we found earlier: Pout = σ′ ∫ ∞ a ∫ π 0 ∫ 2π 0 E2(r, θ, φ)r2 sin θ dφ dθ dr = 2πσ′ ( σ σ + 2σ′ )2 F 2a6 ∫ ∞ a ∫ π 0 1 r4 (4 cos2 θ + sin2 θ) sin θ dθ dr = 8π 3 σ′ ( σ σ + 2σ′ )2 F 2a3 Dividing by (23), we find the effective external voltage Ve: Ve = Pout/I = 4 3 aF · σ σ + 2σ′ and the effective external resistance: Re = Pout/I2 = 2 3πaσ′ . (c) The power dissipated inside the sphere is Pin = σ ∫ (E + F k̂)2dV = 4σσ2′ (σ + 2σ′)2 F 2 ∫ dV = 16σσ2′ 3(σ + 2σ′)2 πa3F 2 Since we’re in steady state, the current flowing out through the upper hemi- sphere of the sphere must be replenished by an equal current flowing in through the lower half of the sphere, so to find the internal voltage and resistance we can just divide by (23): Vi = Pin/I = 8 3 σ′ σ + 2σ′ aF Ri = Pin/I2 = 4 3πaσ . Homer Reid’s Solutions to Jackson Problems: Chapter 3 16 (27), because Km is singular at the origin, while Im is singular at infinity, and there is no linear combination of these functions that will be finite over the whole range of ρ′. Hence we must use terms of the form (26). To ensure finiteness at the origin we must exlude the Nm term, so D = 0. To ensure vanishing at z′ = 0 we must take A = −B, so the z′ function in the region 0 ≤ z′ ≤ z is proportional to sinh(kz′). To ensure vanishing at z′ = L we must take A = −Be−2kL, so the z′ function in the region z ≤ z′ ≤ L is proportional to sinh[k(z′ − L)]. With these restrictions, the differential equation and the boundary conditions are satisfied for all terms of the form (25) with no limitation on k. Hence the Green’s function will be an integral, not a sum, over these terms: G(x′;x) = { ∑ ∞ m=0 ∫ ∞ 0 Am(k, ρ, φ, z)eimφ′ sinh(kz′)Jm(kρ′) dk, 0 ≤ z′ ≤ z ∑ ∞ m=0 ∫ ∞ 0 Bm(k, ρ, φ, z)eimφ′ sinh[k(z′ − L)]Jm(kρ′) dk, z ≤ z′ ≤ L Problem 3.18 The configuration of Problem 3.12 is modified by placing a conducting plane held at zero potential parallel to and a distance L away from the plane with the disc insert in it. For definiteness put the grounded plane at z = 0 and the other plane with the center of the disc on the z axis at z = L. (a) Show that the potential between the planes can be written in cylindrical coor- dinates (z, ρ, φ) as Φ(z, ρ) = V ∫ ∞ 0 dλJ1(λ)J0(λρ/a) sinh(λz/a) sinh(λL/a) . (b) Show that in the limit a → ∞ with z, ρ, L fixed the solution of part a reduces to the expected result. Viewing your result as the lowest order answer in an expansion in powers of a−1, consider the question of corrections to the lowest order expression if a is large compared to ρ and L, but not infinite. Are there difficulties? Can you obtain an explicit estimate of the corrections? (c) Consider the limit of L → ∞ with (L − z), a and ρ fixed and show that the results of Problem 3.12 are recovered. What about corrections for L a, but not L → ∞? (a) The general solution of the Laplace equation in cylindrical coordinates with angular symmetry that vanishes at z = 0 is Φ(ρ, z) = ∫ ∞ 0 A(k)J0(kρ) sinh(kz) dk. (28) Homer Reid’s Solutions to Jackson Problems: Chapter 3 17 Multiplying both sides by ρJ0(k ′ρ) and integrating at z = L yields ∫ ∞ 0 ρJ0(k ′ρ)Φ(ρ, L) dρ = ∫ ∞ 0 A(k) sinh(kL) { ∫ ∞ 0 ρJ0(k ′ρ)J0(kρ) dρ } dk = ∫ ∞ 0 A(k) sinh(kL) { 1 k δ(k − k′) } dk = 1 k′ A(k′) sinh(k′L) so A(k) = k sinh(kL) ∫ ∞ 0 ρJ0(kρ)Φ(ρ, L) dρ = V k sinh(kL) ∫ a 0 ρJ0(kρ) dρ = V k sinh(kL) ∫ ka 0 uJ0(u) du. (29) I worked out this integral earlier, in Problem 3.12: ∫ x 0 uJ0(u) du = xJ1(x). Then (29) becomes A(k) = V k sinh(kL) · (ka)J1(ka) and (28) is Φ(ρ, z) = V ∫ ∞ 0 aJ1(ka)J0(kρ) sinh(kz) sinh(kL) dk = V ∫ ∞ 0 J1(λ)J0(λρ/a) sinh(λz/a) sinh(λL/a) dλ. (30) (b) For x 1, J0(x) → 1 − 1 4 x2 + · · · and for x 1 and y 1, sinh(x) sinh(y) = x + 1 6x3 + · · · y + 1 6y3 + · · · = x y [ 1 + 1 6 (x2 − y2) ] + O(x4) With these approximations we may expand the terms containing a in (30): J0(λρ/a) sinh(λz/a) sinh(λL/a) ≈ ( 1− 1 4 ( λρ a )2 ) ( z L ) ( 1 + 1 6 ( λ a )2 (x2 − y2) ) (31) = z L [ 1 − ( λ a )2( 1 6 (L2 − z2) + 1 4 ρ2 ) + · · · ] (32) Homer Reid’s Solutions to Jackson Problems: Chapter 3 18 Then the potential expansion (30) becomes Φ(ρ, z) = V z L [ ∫ ∞ 0 J1(λ) dλ − 1 a2 [ 1 6 (L2 − z2) + 1 4 ρ2 ] ∫ ∞ 0 λ2J1(λ) dλ + · · · ] The first integral evaluates to 1, so for a infinite the potential becomes simply Φ(z) = V z/L. This is just what we expect to get for the potential between two infinite sheets, one grounded and the other at potential V. The second integral, unfortunately, has a bit of an infinity problem. It’s not hard to see where the problem comes: I derived the expansion above based on the premise that λ/a is small, but the integral goes over all λ up to ∞, so for any finite a the expansions eventually become invalid in the integral. I’m still trying to work out a better procedure for estimating corrections for finite a. (c) In this part we’re interested in taking L → ∞ and looking at the potential a fixed distance away from the plane with the circular insert. Calling the fixed distance z′, the z coordinate of the point we’re interested in is L− z′. We have sinh k(L − z′) sinh kL = sinh(kL) cosh(−kz′) + cosh(kL) sinh(−kz′) sinh kL = cosh(kz′) − coth(kL) sinh(kz′) (33) Now, coth(kL) differs significantly from 1 only for kLa . 1, in which region kz′ . z/L 1, so cosh(kz′) ≈ 1 and sinh(kz′) ≈ 0. By the time k gets big enough that kz′ is starting to get significant, coth(kL) has long since started to look like 1, so the two terms in (33) add directly. The result is that, for all k, (33) can be approximated as exp(−kz′). Then (30) becomes Φ(ρ, z) = aV ∫ ∞ 0 J1(ka)J0(kρ)e−kz′ dk as we found in Problem 3.12. Homer Reid’s Solutions to Jackson Problems: Chapter 3 3 so (3) becomes σ′(a, L) = − q 2π ∫ ∞ 0 dk kJ0(ak) sinh(kz0) sinh(kL) (4) (c) At a = 0, (4) becomes σ′(0, L) = −q 2π ∫ ∞ 0 k sinh(kz0) sinh(kL) . I have no idea how to do this integral. Problem 3.22 The geometry of a two-dimensional potential problem is defined in polar coordinates by the surfaces φ = 0, φ = β, and ρ = a, as indicated in the sketch. Using separation of variables in polar coordinates, show the the Green function can be written as G(ρ, φ; ρ′, φ′) = ∞ ∑ m=1 − 1 mπ ρ mπ/β < ( 1 ρ mπ/β > − ρ mπ/β > a2mπ/β ) sin ( mπφ β ) sin ( mπφ′ β ) Problem 2.25 may be of use. As before, the procedure for determining the Green’s function is to split the region of interest into two parts (one on each ’side’ of the observation point), find separate solutions of the Laplace equation that satisfy the boundary conditions in each region, and then join the two solutions at the source point such that their values match up but the first derivative (in whichever dimension we chose ’sides’) has a finite discontinuity. Suppose the observation point is (ρ, φ). Let’s break the region into two subregions, defined by 0 ≤ ρ′ ≤ ρ and ρ ≤ ρ′ ≤ a. The general solution of the Laplace equation in two-dimensional polar coordinates is Φ(ρ′, φ′) =A0 + B0 ln ρ′ + ∑ n ρ′n[An sin nφ′ + Bn cosnφ′] + ρ′−n[Cn sin nφ′ + Dn cosnφ′]. The solution in the first region must be admissible down to ρ′ = 0, which excludes the ln term and the negative powers of ρ. However, these terms may be included in the solution for the second region. In both regions, the solution must vanish at φ = 0, which excludes the cos terms (i.e. Bn = Dn = 0). The solution must also vanish at φ = β, which requires that n = mπ/β, m = 1, 2, · · · . With these considerations we may write down the solutions for G in the two regions: Homer Reid’s Solutions to Jackson Problems: Chapter 3 4 G(ρ, φ; ρ′, φ′) = ∞ ∑ m=1 Amρ′mπ/β sin ( mπφ′ β ) , 0 ≤ ρ′ ≤ ρ (5) = ∞ ∑ m=1 [ Bmρ′mπ/β + Cmρ′−mπ/β ] sin ( mπφ′ β ) , ρ ≤ ρ′ ≤ a (6) The solution in the second region must vanish at ρ′ = a for all φ′, i.e. Bmamπ/β + Cma−mπ/β = 0 so Bm = γma−mπ/β and Cm = −γmamπ/β where γm can be anything. Then (6) becomes G(ρ, φ; ρ′, φ′) = ∞ ∑ m=1 γm [ ( ρ′ a )mπ/β − ( a ρ′ )mπ/β ] sin ( mπφ′ β ) ρ ≤ ρ′ ≤ a. The solutions in the two regions must agree on the boundary between the two regions, i.e. at ρ′ = ρ. This determines Am and γm: Am = λm [ (ρ a )mπ/β − ( a ρ )mπ/β ] γm = λmρmπ/β where λm can be anything. Using these expressions for Am, Bm, and Cm we can write G(ρ, φ; ρ′, φ′) = ∞ ∑ m=1 λm [ (ρ a )mπ/β − ( a ρ )mπ/β ] ρ′mπ/β sin ( mπφ′ β ) 0 ≤ ρ′ ≤ ρ = ∞ ∑ m=1 λm [ ( ρ′ a )mπ/β − ( a ρ′ )mπ/β ] ρmπ/β sin ( mπφ′ β ) ρ ≤ ρ′ ≤ a. This may be more succintly written as G(ρ, φ; ρ′, φ′) = ∑ m λmfm(ρ; ρ′) sin ( mπφ′ β ) (7) where fm(ρ; ρ′) = [ (ρ> a )mπ/β − ( a ρ> )mπ/β ] ρ mπ/β < . Homer Reid’s Solutions to Jackson Problems: Chapter 3 5 The final step is to choose the constant λm in (7) such as to make ∇2G(ρ, φ; ρ′, φ′) = 1 ρ δ(ρ′ − ρ)δ(φ′ − φ). (8) The Laplacian of (7) is ∇2G = [ ∂2 ∂ρ′2 + 1 ρ′2 ∂2 ∂φ′2 ] G = ∑ m λm [ d2 dρ′2 fm(ρ; ρ′) − ( mπ ρ′β )2 fm(ρ; ρ′) ] sin ( mπφ′ β ) This is equal to (8) if λm = κm 1 β sin ( mπφ β ) (9) and κm [ d2 dρ′2 fm(ρ; ρ′) − ( mπ ρ′β )2 fm(ρ; ρ′) ] = 1 ρ δ(ρ′ − ρ). At all points ρ′ 6= ρ, the latter condition is already satisfied by f as we con- structed it earlier. At ρ′ = ρ, the condition is achieved by choosing κm to satisfy κm d dρ′ fm(ρ; ρ′) ∣ ∣ ∣ ∣ ρ′=ρ+ ρ′=ρ− = 1 ρ . (10) Referring to (7), we have d dρ′ fm ∣ ∣ ∣ ∣ ρ′+ρ+ = mπ β [ (ρ a )mπ/β + ( a ρ )mπ/β ] ρmπ/β−1 (11) d dρ′ fm ∣ ∣ ∣ ∣ ρ′+ρ− = mπ β [ (ρ a )mπ/β − ( a ρ )mπ/β ] ρmπ/β−1. (12) Subtracting (12) from (11) we obtain dfm dρ′ ∣ ∣ ∣ ∣ ρ′=ρ+ ρ′=ρ− = 2mπ β amπ/β · 1 ρ . Then from (10) we read off κm = β 2mπ a−mπ/β and plugging this into (9) gives λm = 1 2mπ a−mπ/β sin ( mπ β ) φ. Plugging this into (7) we obtain finally G(ρ, φ; ρ′, φ′) = ∑ m 1 2mπ [ (ρ<ρ> a2 )mπ/β − ( ρ< ρ> )mπ/β ] sin ( mπφ β ) sin ( mπφ′ β ) I seem to be off by a factor of 2 here, but I can’t find where. ! #"%$& ')(*+&,-.0/2134 6578 9 õ Ä¼ÕÆ ¢&J ¡-ª/¡¢¬Ï»¹¯7 ¡ªL:[¢«¬;ªJ-Jµ¨·ªJ ¡ ¡ ¡®× £/6¯»[¯7Â ¡®;©¢¨· ¡ª/N -¢[Ï¢JÐÒ¢ ` å é Ý øù îba K ì Ñ ï ¤ ò ï ¤å þ ò Ô þ é üýJþ ÷Ý ¤ï îba K Ñø ò ø ò ó Ô ö þ üýJþ Ý ¤ï îba K Ñø ò øCÑ øï ó Ô öþ òØõ ó Ô ö Ñ >6>?> Ý ¤ï îba K ß ¥ed ü å Ñø é þ ¥ef ü å ï º ò ø éûûï ¥ º û í Ô á ñ þ ¥ ¢¶¡«¡ªJ-!-¢[7¯C¶[ ¡¢ å ï é ³ ~+!m
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The medium inside and outside the cylinder has a dielectric constant of unity. (a) Determine the potential and electric fields in the three regions, neglecting end effects. (b) Sketch the lines of force for a typical case of b ≈ 2a. (c) Discuss the limiting forms of your solution appropriate for a solid dielectric cylinder in a uniform field, and a cylindrical cavity in a uniform dielectric. We will take the axis of the cylinder to be the z axis and the electric field to be aligned with the x axis: E0 = E0̂i. Since the cylinder is very long and we’re told to neglect end effects, we can ignore the z direction altogether and treat this as a two-dimensional problem. (a) The general solution of the Laplace equation in two dimensional polar co- ordinates is Φ(r, ϕ) = ∑ [Anrn + Bnr−n][Cn sin(nϕ) + Dn cos(nϕ)] For the region inside the shell (r < a), the B coefficients must vanish to keep the potential from blowing up at the origin. Also, in the region outside the shell 1 Homer Reid’s Solutions to Jackson Problems: Chapter 4 2 (r > b), the only positive power of r in the sum must be that which gives rise to the external electric field, i.e. −E0r cosϕ with An = 0 for n > 1. With these observations we may write expressions for the potential in the three regions: Φ(r, ϕ) = ∑ rn[An sinnϕ + Bn cosnϕ], r < a ∑ rn[Cn sin nϕ + Dn cosnϕ] + r−n[En sin nϕ + Fn cosnϕ], a < r < b −E0r cosϕ + ∑ r−n[Gn sin nϕ + Hn cosϕ], r > b The normal boundary condition at r = a is ε0 ∂Φ ∂r ∣ ∣ ∣ ∣ x=a− = ε ∂Φ ∂r ∣ ∣ ∣ ∣ x=a+ or ε0 ε ∑ nan−1[An sin nϕ + Bn cosnϕ] = ∑ nan−1[Cn sin nϕ + Dn cosnϕ] − na−(n+1)[En sin nϕ + Fn cosnϕ] From this we obtain two equations: ε0 ε An = Cn − Ena−2n (1) ε0 ε Bn = Dn − Fna−2n (2) Next, the tangential boundary condition at r = a is ∂Φ ∂ϕ ∣ ∣ ∣ ∣ x=a+ = ∂Φ ∂ϕ ∣ ∣ ∣ ∣ x=a− or ∑ nan[An cosnϕ − Bn sin nϕ] = ∑ nan[Cn cosnϕ − Dn sinnϕ] + na−n[En cosnϕ − Fn sin nϕ] from which we obtain two more equations: An = Cn + Ena−2n (3) Bn = Dn + Fna−2n (4) Similarly, from the normal boundary condition at r = b we obtain −ε0 ε E0 cosϕ − ε0 ε ∑ nb−(n+1)[Gn sinnϕ + Hn cosϕ] = ∑ nbn−1[Cn sin nϕ + Dn cosnϕ] − nb−(n+1)[En sin nϕ + Fn cosϕ] Homer Reid’s Solutions to Jackson Problems: Chapter 4 5 Figure 1: Field lines in Problem 4.8 for b = 2a, ε = 5ε0. b → ∞, in which case the field becomes E(r, ϕ) = 4εε0 (ε + ε0)2 E0̂i, r < a 2ε0 (ε + ε0) E0̂i− 2ε0(ε − ε0) (ε + ε0)2 (a r )2 E0[cos ϕr̂ + sin ϕϕ̂], r > a. Homer Reid’s Solutions to Jackson Problems: Chapter 4 6 Problem 4.9 A point charge q is located in free space a distance d away from the center of a dielectric sphere of radius a (a < d) and dielectric constant ε/ε0. (a) Find the potential at all points in space as an expansion in spherical harmonics. (b) Calculate the rectangular components of the electric field near the center of the sphere. (c) Verify that, in the limit ε/ε0 → ∞, your result is the same as that for the conducting sphere. We will take the origin of coordinates at the center of the sphere, and put the point charge on the z axis at z = +h. Then the problem has azimuthal symmetry. (a) Since there is no free charge within the sphere, ∇·D = 0 there. But since the permittivity is uniform within the sphere, we may also write ∇·(D/ε) = ∇·E = 0 there. This means that polarization charge only exists on the surface of the sphere, so within the sphere the potential satisfies the normal Laplace equation, whence Φ(r, θ) = ∑ l Alr lPl(cos θ) (r < a). Now, in the region r > a, the potential may be written as the sum of two components Φ1 and Φ2, where Φ1 comes from the polarization charge on the surface of the sphere, while Φ2 comes from the external point charge. Since Φ1 satisfies the Laplace equation for r > a, we may expand it in Legendre polynomials: Φ1(r, θ) = ∑ l Blr −(l+1)Pl(cos θ) (r > a). On the other hand, Φ2 is just the potential due to a point charge at z = d: Φ2(r, θ) = q 4πε0 ∑ rl dl+1 Pl(cos θ), r < d q 4πε0 ∑ dl rl+1 Pl(cos θ), r > d. (9) Putting this all together we may write the potential in the three regions as Φ(r, θ) = ∑ Alr lPl(cos θ), r < a ∑ [ Blr −(l+1) + q 4πε0 rl dl+1 ] Pl(cos θ), a < r < d ∑ [ Bl + qdl 4πε0 ] r−(l+1)Pl(cos θ), r > d. Homer Reid’s Solutions to Jackson Problems: Chapter 4 7 The normal boundary condition at r = a is ε ∂Φ ∂r ∣ ∣ ∣ ∣ r=a− = ε0 ∂Φ ∂r ∣ ∣ ∣ ∣ r=a+ → ε ε0 lAla l−1 = −(l + 1)Bla −(l+2) + lqal−1 4πε0dl+1 → Al = ε0 ε [−(l + 1) l Bla −(2l+1) + q 4πε0dl+1 ] (10) The tangential boundary condition at r = a is ∂Φ ∂θ ∣ ∣ ∣ ∣ r=a− = ∂Φ ∂θ ∣ ∣ ∣ ∣ r=a+ → Ala l = Bla −(l+1) + q 4πε0 al d(l+1) → Bl = Ala 2l+1 − q 4πε0 a2l+1 dl+1 (11) Combining (10) and (11), we obtain Al = 1 ε ε0 + l+1 l ( 2l + 1 l ) q 4πε0dl+1 Bl = 1 ε ε0 + l+1 l ( 1 − ε ε0 ) qa2l+1 4πε0dl+1 In particular, as ε/ε0 → ∞ we have Al → 0 as must happen, since the field within a conducting sphere vanishes; and Bl → − qa2l+1 4πε0dl+1 . (12) With the coefficients (12), the potential outside the sphere due to the polariza- tion charge at the sphere boundary is Φ1(r, θ) = 1 4πε0 ( −qa d ) ∑ ( a2 d )l 1 rl+1 Pl(cos θ). Comparing with (9) we see that this is just the potential of a charge −qa/d on the z axis at z = a2/d. This is just the size and position of the image charge we found in Chapter 2 for a point charge outside a conducting sphere. Homer Reid’s Solutions to Jackson Problems: Chapter 4 10 cylinders. For our Gaussian pillbox we take a disk of thickness dz and radius r, a < r < b centered on the axis of the cylinders. By symmetry there is no component of E normal to the top or bottom boundary surfaces, and the com- ponent normal to the side surfaces (the radial component) is uniform around the disc. Hence ∮ E · dA = 2π r dzEρ = q ε0 = 1 ε0 (2π a dz)σ → Eρ(ρ) = aσ ε0r where σ is the surface charge on the inner conductor. This must integrate to give the correct potential difference between the conductors: V = − ∫ b a Eρ(ρ)dρ = −aσ ε0 ln b a which tells us that, to establish a potential difference V between the conductors, the battery has to flow enough charge to establish a surface charge of magnitude σ = ε0V a ln(b/a) (17) on the cylinder faces (the surface charges are of opposite sign on the two cylin- ders). It is useful to figure out the energy per unit length stored in the electric field between the cylinder plates here. This is just Wv = 1 2 ∫ b a ∫ 2π 0 E ·D ρ dρ dφ = πε0 ∫ b a E2(ρ)ρ dρ = π a2σ2 ε0 ln(b/a) = πε0V 2 ln(b/a) (18) where the v subscript stands for ’vacuum’, since (18) is the energy per unit length stored in the field between the cylinders with just vacuum between them. Now suppose we introduce a dielectric material between the cylinders. If the voltage between the cylinders is kept at V , then the E field must be just the same as it was in the no-dielectric case, because this field integrated from a to b must still give the same potential difference. However, in order to establish this same E field in the presence of the retarding effects of the dielectric, the battery now has to establish a surface charge that is greater that it was before by a factor (ε/ε0). With this greater charge on the electrodes, the D field will now be bigger by a factor (ε/ε0) than it was in our above calculation. So the Homer Reid’s Solutions to Jackson Problems: Chapter 4 11 energy per unit length stored in the field between the cylinders increases by a factor (ε/ε0 − 1) over the result (18): ∆Wd = (ε − ε0) πV 2 ln(b/a) . On the other hand, to get to this point the battery has had to flow enough charge to increase the surface charges to be of magnitude (ε/ε0) times greater than (17). In doing this the internal energy of the battery decreases by an amount equal to the work it had to do to flow the excess charge, namely ∆Wb = −V dQ = V (2π a dσ) = (ε − ε0) 2πV 2 ln(b/a) (per unit length). The energy lost by the battery is twice that gained by the dielectric, so the system with dielectric between the cylinders has lower overall energy than the system with vacuum between the cylinders by a factor ∆W = (ε − ε0) πV 2 ln(b/a) (19) (per unit length). Turning now to the situation in this problem, we’ll take the axis of the cylinders as the z axis, so that the surface of the liquid is parallel to the xy plane. We’ll take the boundary between the liquid and the air above it to be at z = 0. With no potential between the cylinder plates, the liquid between the cylinders is at the same height as the liquid outside. Now suppose a battery of fixed potential V is connected between the two cylinder plates. As we showed earlier, the combined system of battery and di- electric can lower its energy by having more of the dielectric rise up between the cylinders. However, at some point the energy win we get from this is balanced by the energy hit we take from the gravitational potential energy of having the excess liquid rise higher between the cylinders. The height at which we no longer gain by having more liquid between the cylinders is the height to which the system will settle. So suppose that, with a battery keeping a voltage V between the electrodes, the liquid between the electrodes rises to a height h above the surface of the liquid outside the electrodes. The decrease in electrostatic energy this affords over the case with just vacuum filling that space is just (19) times the height, i.e. Ee = −h(ε − ε0) πV 2 ln(b/a) (20) This must be balanced by the gravitational potential energy Eg of the excess liquid. Eg is easily calculated by noting that the area between the cylinders is π(b2 − a2), so the mass of liquid contained in a height dh between the cylinders is dm = ρπ(b2 −a2)dh, and if this mass is at a height h above the liquid surface its excess gravitational energy is dEg = (dm)gh = πgρ(b2 − a2)hdh. Homer Reid’s Solutions to Jackson Problems: Chapter 4 12 Integrating over the excess height of liquid between the cylinders, Eg = πgρ(b2 − a2) ∫ h 0 h′ dh′ = 1 2 πgρ(b2 − a2)h2. (21) Comparing (20) to (21), we find that the gravitational penalty of the excess liquid just counterbalances the electrostatic energy reduction when h = 2(ε − ε0)V 2 ρg(b2 − a2) ln(b/a) = 2χeε0V 2 ρg(b2 − a2) ln(b/a) Solving for χe, χe = ρgh(b2 − a2) ln(b/a) 2ε0V 2 . So I seem to be off by a factor of 2 somewhere. Actually we should note one detail here. When the surface of the liquid between the cylinders rises, the surface of the liquid outside the cylinders must fall, since the total volume of the liquid is conserved. Hence there are really two other contributions to the energy shift, namely, the change in gravitational and electrostatic energies of the thin layer of liquid outside the cylinders that falls away when the liquid rises between the cylinders. But if the surface area of the vessel containing the liquid is sufficiently larger than the area between the cylinders, the difference layer will be thin and its energy shifts negligible. Homer Reid’s Solutions to Jackson Problems: Chapter 4 15 fprintf(g,"e\n"); for(phi=0; phi<=2*M_PI; phi+=(2*M_PI/100)) fprintf(g,"%g %g\n",B*cos(phi),B*sin(phi)); fprintf(g,"e\n"); /* * Draw field lines. */ for (i=1.0; i<=NUMLINES; i+=1.0) { /* * Compute starting x and y coordinates and initiate plot. */ x=-2.0*B; y=2.0*B * ((NUMLINES - 2.0*i)/NUMLINES); fprintf(g,"plot ’-’ t ’’ with lines\n"); /* * Plot NUMPOINTS points for this field line. */ for (j=0.0; j<NUMPOINTS; j+=1.0) { /* * compute polar coordinates of present location */ r=sqrt(x*x + y*y); if (x==0.0) phi=(y>0.0) ? M_PI/2.0 : -M_PI/2.0; else phi=atan(y/x); /* * compute rise and run of electric field */ RComp=Er(r,phi); PhiComp=Ephi(r,phi); dx=cos(phi)*RComp - sin(phi)*PhiComp; dy=sin(phi)*RComp + cos(phi)*PhiComp; /* * bump x coordinate forward a fixed amount, and y * coordinate up or down by an amount depending on * the direction of the electric field at this point */ x+=DELTAX; y+=DELTAX * (dy/dx); fprintf(g,"%g %g\n",x,y); Homer Reid’s Solutions to Jackson Problems: Chapter 4 16 }; fprintf(g,"e\n"); }; printf("Thank you for your support.\n"); } Solutions to Problems in Jackson, Classical Electrodynamics, Third Edition Homer Reid November 8, 2000 Chapter 5: Problems 1-10 Problem 5.1 Starting with the differential expression dB = µ0I 4π dl′ × x − x′ |x − x′|3 for the magnetic induction at the point P with coordinate x produced by an incre- ment of current I dl′ at x′, show explicitly that for a closed loop carrying a current I the magnetic induction at P is B = µ0I 4π ∇Ω where Ω is the solid angle subtended by the loop at the point P . This corresponds to a magnetic scalar potential, ΦM = −µ0IΩ/4π. The sign convention for the solid angle is that Ω is positive if the point P views the “inner” side of the surface spanning the loop, that is, if a unit normal n to the surface is defined by the direction of current flow via the right-hand rule, Ω is positive if n points away from the point P , and negative otherwise. This is the same convention as in Section 1.6 for the electric dipole layer. I like to change the notation slightly: the observation point is r1, the coordi- nate of a point on the current loop is r2, and the displacement vector (pointing to the observation point) is r12 = r1 − r2. The solid angle subtended by the current loop at r1 is given by a surface integral over the loop: Ω = ∫ S cos γ dA r2 12 1