Wilcox Chapter2R

2.1 Chapter 2 : Boundary Value Problems in Electrostatics 2.1 Method of images: flat conducting surface Going to solve some specific problems in this chapter using: 1) reduced Green function technique, 2) method of images, 3) expansion in orthogonal functions, and 4) conformal mapping. First problem: (Methods 1 & 2 illustrated) qregion of interest is equivalent to -q q This says the Dirichlet Green function is just GD(x≥ ,x≥') = 1 |x≥ - x≥'| - 1 |x≥ - x≥"| , (2.1) (x≥' = source pt.; x≥ = field pt.) where x≥' = x'i^ + y'j^ + z'k^ (z'>0) 2.2 locates the +unit charge and x≥" = x'i^ + y'j^ - z'k^ locates the image charge. Notice that G D(x≥ ,x≥')|z=0 = 0 and that G D(x≥ ,x≥') = GD(x≥',x≥). We can write down the solution by inspection, but let us confirm it from a differentialequation point of view. Want to solve ∇ 2 GD(x≥ ,x≥') = -4πδ(x≥ - x≥'), (2.2) subject to G D(x≥ ,x≥')|z=0 = 0. (2.3) First, however, just to get some experience, let us solve the easier problem of a point charge in free space from the diff. eqn point of view. Obviously, the solution to ∇ 2 Gf(x≥ ,x≥') = -4πδ(x≥ - x≥') in free space is Gf(x≥ ,x≥') = 1 |x≥ - x≥'|. (2.4) This can be derived from the diff. eqn as follows using the representations, δ(x≥ - x'≥) = ∫ d3k (2π)3 eik≥.(x≥-x≥'), (2.5) Gf(x≥ ,x≥') = ∫ d3k (2π)3 G(k) eik≥.(x≥-x≥'). (2.6) (k = |k≥|) It is easy to see that G(k) = 4π k2; the fact that G(k) depends only on k = |k≥| is due to the fact that the geometry is translationally invariant in x,y & z directions.Let's now do the integrals: 2.3 Gf = ∫ d3k (2π)3 4π k2 eik≥.(x≥-x≥')-kε (ε>0) (2.7) ÷convergence factor (sometimes suppressed) Gf = 4π (2π)3 ∫ dφ dcosθ k2dk k2 eik|x≥-x≥'|cosθ(2.8) Gf = 8π2 (2π)3 ∫ dk ∫ -1 1 dcosθ eik|x≥-x≥'|cosθ (2.9) 2 k|x≥ - x≥'| sin k|x≥ - x≥'| => Gf = 8π2 (2π)3 2 |x≥ - x≥'| ∫ 0∞ dk sin k|x≥ - x≥'| k. (2.10) |consider ∫ dz eiz z y x ∫ 0∞ dx sinx x = π 2 (complex variables) => G f(x≥,x≥') = 1 |x≥ - x≥'|. [Many other ways of evaluating this, including inserting a convergence factor, e -kε, in (2.10), doing the integral, and then taking ε->0 limit. Can't resist an alternate evaluation, which also uses a convergence factor earlier in the evaluation and inverts the order of the k and angularintegrals: 2.4 Gf = 4π (2π)3 ∫ dΩ ∫ 0∞ dk eik|x≥ - x≥'|cosθ-kε (below) 1 i|x≥ - x≥'|cosθ-ε e k|x≥-x≥'|cosθ-kε |0∞ = -1 i|x≥ - x≥'|cosθ-ε => Gf = -4π (2π)3 2π 1 i|x≥ - x≥'| ∫ -1 1 dcosθ 1 cosθ + iε (ε has been re-scaled, but is still positive) Gf = -1 iπ|x≥ - x≥'| ∫ -11 dcosθ cosθ-iε cos2θ + ε2 "Magic": δ(x) = 1 π lim ε .0+ ε x2 + ε2 => G f = 1 |x≥ - x≥'| ∫ -11 dx δ(x) (x=cos θ) = 1 |x≥ - x≥'| . ] 2.2 Reduced Green function technique applied to flat conductor Now go back to our half-plane geometry. The geometry here is translationally invariant also, but only in the x andy directions. Therefore, assume G D(x≥,x≥') = 4π ∫ d2k (2π)2 eik≥.(x≥-x≥')⊥ g(z,z',k), (2.11) where 2.5 d2k = dk xdky, k≥.(x≥-x≥')⊥ = kx(x-x') + ky(y-y'). [k≥ = kxi^ + kyj^]. Let's call g(z,z',k) the "reduced Green function". We then have -∇2 G(x≥,x≥') = 4π ∫ d2k (2π)2 eik≥.(x≥-x≥')⊥ [k2 - ∂2 ∂z2] g(z,z',k), = 4π ∫ d2k (2π)2 eik≥.(x≥-x≥')⊥ δ(z-z'), (2.12) which implies [k2 - ∂2 ∂z2] g(z,z',k) = δ(z-z'). (2.13) The BC on GD(x≥,x≥') now read g(z,z',k)|z=0 = 0. (2.14) Let's try to solve this simpler eqn. For z ≠ z' we have [k2 - ∂2 ∂z2] g(z,z',k) = 0, (2.15) the solution for which is g(z,z',k) = C(z')e+-kz. (2.16) Our solution must be bounded as z .∞, so g(z,z',k) = ⎩⎨⎧C1 e-kz, z > z' C2ekz + C3e-kz, z < z'.(2.17) The above BC now gives C2 = -C3. (2.18) 2.6 The diff. eqn implies a discontinuity in g(z,z',k) at z = z': ∫ z'+ z'- dz [k2 - ∂2 ∂z2] g(z,z',k) = ∫ z'+ z'- δ(z-z')dz. (2.19) Since g(z,z',k) is continuous, g(z,z',k) | z'+ z'- = 0 as well as - ∂ ∂z g(z,z',k) |z'+ z'- = 1 from the above. These imply C 1e-kz' = C2(ekz'- e-kz'), (continuity of g) (2.20) C 1ke-kz' + C2k(ekz' + e-kz') = 1. ( discontinuity (2.21) of ∂g ∂z ) The solution is C1 = 1 2k (ekz'- e-kz'), (2.22) C2 = 1 2k e-kz'. (2.23) This means that g(z,z',k) = 1 2k x ⎩⎨⎧ek(z'-z) - e-k(z'+z), z>z' ek(z-z') - e-k(z'+z), z<z'. (2.24) which we can also write as g(z,z',k) = 1 2k (e-k|z-z'|- e-k|z+z'|), ¢z',z. (2.25) Therefore GD(x≥,x≥') = 4π ⌡⎮⎮⎮⌠d2k (2π)2 eik≥.(x≥-x≥')⊥ 1 2k (e-k|z-z'|- e-k|z+z'|). (2.26) We will actually do the integrals to verify this is a correct result. Brute force. Consider (first term above): 2.7 I ≡ 4π 2(2π)2 ∫ 02π dφ ∫ 0∞ dk eikρcosφ e-k|z-z'| I = -1 2π ∫ 02π dφ 1 iρcosφ - |z-z'| I = - 1 2π ∫ 02π dφ -iρcosφ - |z-z'| ρ2cos2φ + (z-z')2 I = |z-z'| 2π ∫ 02π dφ 1 ρ2cos2φ + (z-z')2 Look it up in a table of indefinite integrals (CRC, 22nd ed., #375, p.442). Get I = |z-z'| 2π [1 |z-z'|√⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ρ2 + (z-z')2 tan-1(|z-z'|tan φ √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ρ2 + (z-z')2)]|2π 0 2π = 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ρ2 + (z-z')2 . Similarly for the 2nd term. Replacing ρ . |(x≥-x≥')⊥|, we recover that G(x≥,x≥') = 1 |x≥ - x≥'| - 1 |x≥ - x≥"|, where x≥" has the same meaning as before. Explicitly G(x≥,x≥') = 1 [ (x-x')2 + (y-y')2 + (z-z')2 ]1/2 ÷* - 1 [ (x-x')2 + (y-y')2 + (z+z')2 ]1/2 (2.27) ÷* 2.8 We can confirm the presence of the surface delta function mentioned earlier. We have n^ z' z,z' > 0 ∂G ∂n' |s = -∂G ∂z' |z'=0 = - { 1 2 . 2(z-z') [(x-x')2 + (y-y')2 + (z-z')2]3/2 - -1 2 . 2(z+z') [(x-x')2 + (y-y')2 + (z+z')2]3/2 }|z'=0(2.28) => ∂G ∂n' |s = -2z [ρ2 + z2]3/2 [ρ2 = (x-x')2 + (y-y')2](2.29) Notice ρ ≠ 0, z = 0, ∂G ∂n'|z'=0 = 0. (2.30) ρ = 0, z .0+, ∂G ∂n'|z'=0 = -∞. (2.31) Also |dx'dy' 1 4π lim z .0+ ∫s da'∂G ∂n' = 1 4π lim z .0+ ∫ ∫ρdρdφ -2z [ρ2 + z2]3/2 , (2.32) (take origin at x=y=0) = lim z .0+ ⎝⎜⎛ ⎠⎟⎞-z 2 ∫∞ 0 2ρdρ [ρ2 + z2]3/2, (2.33) = lim z .0+ ⎝⎜⎜⎛ ⎠⎟⎟⎞- z 2 ( 1 -1/2) 1 √⎯⎯⎯⎯⎯⎯ρ2 + z2|∞ 0 = lim z .0+ ⎝⎜⎛ ⎠⎟⎞- z |z| = -1, (2.34) => lim z .0+ ∂G ∂n'|z'=0 = -4πδ(s)(x≥ ⊥-x≥'⊥). (2.35) 2.9 (See Eq.(1.89) of the last Chapter.) We see that the surface delta function arises from the image charge which has beenplaced on the surface for z'=0. [We can also see the surface delta function from GD(x≥,x≥') = 4π∫ d2k (2π)2 eik≥.(x≥,x≥')⊥ 1 2k (e-k|z-z'| - e-k|z+z'| ) => -∂ ∂z' GD(x≥,x≥')|z'=0 = 4π∫ d2k (2π)2 eik≥.(x≥,x≥)⊥ 1 2 (-e-kz - e-kz) => lim z .0+ 1 4π ∂ ∂n' GD(x≥,x≥')|z'=0 = -∫ d2k (2π)2 eik≥.(x≥-x≥')⊥ . ] -(s)≥≥(x⊥⊥-x') δ Of course, the total surface charge induced (when z' is not on the surface, say) is -1 because of the nature of our Green function. [Since ßsurface = 1 4π ∂G ∂n|s we see that ßsurface = -2z' 4π[ρ2 + z2]3/2. ] 2.3 Method of images: conducting sphere Can do other problems with the method of images (usually need an infinite number of them, however). Point chargeoutside sphere: 2.10 radius=aP q x" n n≥q'x'≥ ^ ^x≥ ' Can find q',x≥ as follows. Guess: Φ(x≥) = q |x≥-x≥'| + q' |x≥-x≥"|, (2.36) or Φ(x≥) = q |xn^-x'n^'| + q' |xn^-x"n^'| . (2.37) Now we want Φ(x≥)|x=a = q a|n^ - x' a n^'| + q' x"|n^' - a x" n^| = 0. (2.38) Notice q a = -q' x" , x' a = a x" works. => q' = -a x'q, x" = a2 x', (2.39) => Φ(x≥) = q |x≥-x≥'| - qa x'|x≥ - a2 x'2 x≥'|. (2.40) 2.11 Essentially the Green function for this geometry. More explicitly GD(x≥,x≥') = 1 (x2 + x'2 - 2xx'cos γ)1/2 - 1 (x2x'2 a2 + a2 - 2xx'cos γ)1/2 . (2.41) (cosγ = n^.n'^ = cos ¥cos ¥' + sin ¥sin ¥'cos(φ - φ')) Can see explicitly that G(x≥,x≥') = G(x≥',x≥) and can show that the gradient on the surface ( inward normal to sphere) is given by ∂G ∂n'|x'=a = - ∂G ∂x'|x'=a = - (x2 - a2) a(x2 + a2- 2ax cos γ)3/2. (2.42) Can also verify that as x .a+, ∂G ∂n'|s acts as a surface delta function. I am not going to do this now, but this problem can also be done using the reduced Green function approach.(Don't have the mathematical machinery yet - Ch.3!) Now that we have the Green function, we can solve otherproblems for this geometry. In particular, consider: V (const.)q The solution is Φ(x≥) = ∫ d3x'GD(x≥,x≥') ρ(x≥') -1 4π ∫o s da'Φ(x≥')∂G ∂n', (2.43) ÷÷ qδ(x≥ - x≥') V so all we need is the surface integral, ∫o s ∂G ∂n'|x'=a da' = -a2 a ∫ dΩ' (x2 - a2) (x2 + a2 - 2ax cos γ)3/2. (2.44) 2.12 Take z'-axis along x≥ : (dΩ' = sin γdγdφ) => ∫o s ∂G ∂n'|x'=a da' = -a(x2 - a2)2π 2ax ∫ 2ax sin γdγ (x2 + a2- 2ax cos γ)3/2, = -(x2 - a2) 2x 2π (1 -1/2) 1 (x2 + a2- 2ax cos γ)1/2 |π 0, = (2 πx2 - a2){1 |x+a| - 1 |x-a|}, ÷1 x-awhen x>a, = -4πa x (x = |x≥|). (2.45) Therefore Φ(x≥) = q G D(x≥,x≥') + Va |x≥| . (2.46) The extra term has a simple interpretation: due to surface charge on the sphere, uniformly distributed. The totalcharge on the surface is Q = -q a |x≥'| + Va. (2.47) Q is determined because V is given. Turn this around. Specify Q => V is determined. => Va = Q + q a |x≥'|. (2.48) Gives solution with a given charge Q on the surface: Φ(x≥) = q G D(x≥,x≥') + (Q + q a |x≥'|) |x≥|. (2.49) Another situation: 2.13 az=-Rx≥ z=R E≥ 0θ +Q -Q Specify Φ |s = 0, (2.50) ρ(x≥) = Q[δ((x≥ - (0,0,-R)) - δ(x≥ - (0,0,R))]. (2.51) Gives Φ(x≥) = Q (r2 + R2 + 2rRcos θ)1/2 - aQ R(r2 + a4/R2 + 2a2r/R cos θ)1/2 - Q (r2 + R2 - 2rRcos θ)1/2 + aQ R(r2 + a4/R2 - 2a2r/R cos θ)1/2. (2.52) Now 1 (R2 + r2 ¶2rRcosθ)1/2 ≈ R>>r 1 R (1 • r R cosθ), (2.53) 1 (r2 + a4/R2 ¶ 2a2r/R cos θ)1/2 ≈ R>>r 1 r (1 •a2 Rr cosθ), (2.54) so Φ(x≥) ≈ -2Q/R2 (r - a3 r2) cosθ. (2.55) Identify: 2Q/R2 . E0. => Φ(x≥) = -E0 (r - a3 r2)cosθ = -E0z + d≥.r≥ r3, 2.14 d≥ = a3 E≥ 0. (2.56) With this interpretation, result is exact. Of course, if we were to specify a potential on the surface, we would just get a term Va |x≥| as before. 2.4 Normal force on a charged surface We need the force/area on conductor surfaces. conductorn^ da Vß 1 2Analog of F=qE≥ ≥ is F=≥ßE≥ (" f" is now force/area). In this case f≥ = ß(E≥ 2total - E≥ 2disk). (2.57) Now we know that (E≥ 2total - E≥ 1total).n^ = 4π ß, => E≥ 2ntotal.n^ = 4π ß. (2.58) Remember E2ndisk = -∫sß(x≥)dΩ, (2.59) (flat) => E 2ndisk = 2π ß(outside surface, d Ω < 0), (2.60) 2.15 => f≥.n^ = ß(4π ß- 2π ß) = 2π ß2. (2.61) Notice fn > 0, so the force is directed outward (likes repel). What has happened is just: cancel{(remainder) (disk) 2π ß. -2π ß , .2π ß .2π ß (remainder) (disk)}add 2.5 Charged conducting sphere force example Now can do a problem using our result (2.56) above. Picture: Fz radius=a gap ??≥E=E0k^ext We just found that Φ = -E0(r-a3 r2)cosθ. (2.62) Therefore the induced surface charge density is ß = - 1 4π∂Φ ∂r|a = 3 4π E0cosθ (total charge=0). (2.63) Get total upward force by integrating over the upper surface: Fz = 2π∫upper ß2cosθda = 2π∫ 0π/2 ∫ 02π 9 16π2 E02cos3θsinθ a2dθ dφ, (2.64) ÷by symmetry ÷!! 2.16 Fz = 9(2π)2 16π2 E02a2 ∫ 0π/2 cos3θsinθdθ , (2.65) - cos4θ 4 ∫ 0π/2 = 1 4 «upper sphere Fz = 9E02a2 16 (this is the force we must counteract). (2.66) Now, imagine some total charge Q placed on the sphere before it is cut into two. Then, the external forces on the twopieces will no longer be equal in magnitude and opposite indirection. Now, the potential is Φ = Φ uncharged + Q r, (2.67) since the charge Q will distribute itself uniformly over a spherical equipotential surface. This gives (magnitude only) ß = 3 4π E0cosθ + 1 4π Q a2, (2.68) add in [0, / 2]π Fupper z = (2π)2a2 ∫ 0π/2 dθ sinθ cosθ (3 4π E0cosθ + 1 4π Q a2)2, (2.69) Flower z = (2π)2a2 ∫ π/2π dθ sinθcosθ (3 4π E0cosθ + 1 4π Q a2)2. (2.70) subtract in [/2,]ππ (the "add in..." and "subtract in..." comments are true if E 0 and Q have the same sign.) Get 2.17 FU Z = 9E02a2 16 + 1 2 E0Q + Q2 8a2, (2.71) Fl Z = - 9E02a2 16 + 1 2 E0Q - Q2 8a2. (2.72) Notice FU Z + Fl Z = E0Q (as it should), (2.73) FU Z - Fl Z 2 = 9E02a2 16 + Q2 8a2. (2.74) ÷ This is the force that must be exerted on each shpere to keep the spheres together, but they still accelerate in the E≥ field. 2.6 Expansion in orthogonal series: conducting box Next problem (new technique): z=c V(x,y) x=a xy=byƒ=0 on other surfaces Find the inside solution. Since there is no charge in V, need to solve ∂2φ ∂x2 + ∂2φ ∂y2 + ∂2φ ∂z2 = 0. (2.75) Assume 2.18 φ(x,y,z) = X (x) Y(y) Z(z). (2.76) Substituting above yields (divide through by φ=XYZ) 1 X d2X dx2 + 1 Y d2Y dy2 + 1 Z d2Z dz2 = 0. (2.77) (Notice total derivatives have replaced partials). Must be that 1 X d2X dx2 = -α2, 1 Y d2Y dy2 = -β2, 1 Z d2Z dz2 = α2 + β2. (2.78) Nothing to determine these constants yet. Keep our minds open as to their values ( α,β can be real or imag. at this point). Solution will be unique => BC's will tell us their values. Assuming α,β ≠ 0: X(x) ~ C1sinαX + C2cosαX, (2.79) Y(y) ~ C3sinβy + C4cosβy, (2.80) Z(z) ~ C5sinhγz + C6coshγz, (2.81) γ ≡ √⎯⎯⎯⎯⎯⎯α2 + β2. BC's: φ|x=0 = 0 => C 2 = 0, (2.82) φ|x=a = 0 => α = nπ a , n =1,2,3,... (2.83) « φ|y=0 = 0 => C 4 = 0, γ=π√⎯⎯n2 a2+m2 b2(2.84) ÷ φ|y=b = 0 => β = mπ b , m=1,2,3,... (2.85) Must also have φ|z=0 = 0 => C 6 = 0. (2.86) 2.19 Most general solution: φ(x,y,z) = ∑ n,m=1∞ Amnsin(αnx) sin( βmy) sinh( γmnz). (2.87) Last BC: V(x,y) = ∑ n,m Amnsin(αnx) sin( βmy) sinh( γmnc). (2.88) Use orthogonality to invert this (n,n' ≥ 1) ∫ -aa dx sin(nπx a) sin(n'πx a) = aδnn', (2.89) ∫ -aa dx cos(nπx a) cos(n'πx a) = aδnn'. (2.90) [Can get these by using sinx siny = -1 2 (cos(x+y) - cos(x-y)) cosx cosy = 1 2 (cos(x+y) + cos(x-y)). ] => Amn = 4 ab sinh( γmnc) ∫ 0a dx∫ 0b dy V(x,y)sin( αnx)sin(βmy). (2.91) 2.7 Fourier series problems: conducting box Another way to solve this problem: reduced G.F. method. However, we have to proceed differently from before since weno longer have translational invariance. It really combinesthe image, reduced G.F. and Fourier series techniques. Start with fundamental theorem for Fourier series (f(x) piecewise continuous in -a <- x <- a ): f(x) = a0 2 + ∑ n=1∞ ( an cos(πnx a) + bnsin(πnx a) ), (2.92) 2.20 where (from orthogonality) a n = 1 a ∫ -aa f(x) cos(πnx a)dx, (2.93) bn = 1 a ∫ -aa f(x) sin(πnx a)dx. (2.94) Now consider Q(x,x') ≡ 1 a ∑ n=1∞ [sin(πnx a)sin(πnx' a)]. (2.95) We have that ∫ -aa dx f(x)Q(x,x') = ∫ -aa dx∑ n=1∞ bn sin(πnx a)Q(x,x') = 1 a ∑ n,n' bn sin(πn'x' a) ∫ -aa dx sin(πnx a)sin(πn'x a) = ∑ n bn sin(πnx' a) = f(x) - f(-x) 2. (2.96) ÷ Since this is the odd part of f(x) Since this holds for any f(x) => Q(x,x') = 1 2 [δ(x-x') - δ(x+x')] , -a <- x,x' <- a. (2.97) Likewise, for P(x,x') = 1 a [12 + ∑ n=1∞ cos(nπx a) cos(nπx' a)], (2.98) one can show ∫ -aa f(x) P(x,x') dx = f(x) + f(-x) 2, (2.99) => P(x,x') = 1 2 [δ(x-x') + δ(x+x')], -a <- x,x' <- a. (2.100) 2.21 Explicit BC on G(x≥,x≥'): GD(x≥,x≥')|x=0,a = 0, (2.101) GD(x≥,x≥')|y=0,b = 0, (2.102) GD(x≥,x≥')|z=0,c = 0. (2.103) My claim: Solving ∇2 GD(x≥,x≥') =-4π[δ(x-x')- δ(x+x')] [ δ(y-y') - δ(y+y')] δ(z-z') (2.104) for G D|big box = 0 in -a <- x,x' <- a -b <- y,y' <- b 0 <- z,z' <- c gives the same solution in the small box as ∇2 GD(x≥,x≥') = -4 π ∂(x≥ - x≥'). (2.105) for G D|small box = 0 in 0 <- x,x' <- a 0 <- y,y' <- b 0 <- z,z' <- c Point: Fourier series can only represent periodic functions. Three image charges(in big box)y xoriginal box (2-D cutout)-1 +1-1 2.22 By the way, this means we can formally write down the form of GD here as (an infinite number of them, however) GD(x≥,x≥') = 1 |x≥-x≥'| - 1 |x≥-x≥"| + 1 |x≥-x≥"'| - 1 |x≥-x≥'IV| + ... (2.106) ÷ where rest of periodic images in x,y, and z. x≥' = (x',y',z'), x≥" = (x',-y',z'), x≥"' = (-x',-y',z'), x≥'IV = (-x',y',z'), etc. Since G D(x≥,x≥') is now an odd function in x' and y', let's expand (also G D(x≥,x≥') = G D(x≥',x≥)) GD(x≥,x≥') = 16π ab ∑ n,m=1 gnm(z,z')sin(nπx a) sin(nπx' a) sin(mπy b) x sin(mπy' b). (2.107) Then -∇2GD(x≥,x≥') = 16π ab ∑ n,m=1 sin(nπx a) sin(nπx' a)sin(mπy b) sin(mπy' b) x[n2π2 a2 + m2π2 b2 - ∂2 ∂z2] gnm (z,z'). (2.108) On the other hand 4π[ ∂(x-x') - ∂(x+x') ][ ∂(y-y') - ∂(y+y')] ∂(z-z') = 16π ab ∑ n',m'=1 sin(n'πx a) sin(n'πx' a) sin(m'πy b) sin(m'πy' b) ∂(z-z'). (2.109) Must be that [n2π2 a2 + m2π2 b2 - ∂2 ∂z2] gnm(z,z') = ∂(z-z'). (2.110) 2.23 As before, we can solve for g nm(z,z'). Solutions: gnm(z,z') = ⎩⎨⎧C1(z')sinh γnmz, z <- z' C2(z')sinh γnmz + C3(z')cosh γnmz, z >- z'. (2.111) where γnm = π√⎯⎯⎯n2 a2 + m2 b2. The above diff. eqn and BC's require gnm(z,z')|z'+ z'- = 0, (2.112) - ∂ ∂z gnm(z,z')|z'+ z'- = 1. (2.113) gnm(z,z')| z=c = 0. (2.114) Solution is gnm(z,z') = sinh(γnm z<)sinh[γnm(c-z>)] γsinhγnmc, (2.115) where z< is the lesser of z,z'; z > is the greater of z,z'. Full Green function for the box: GD(x≥,x≥') = 16π ab ∑ n,m sin(nπx a) sin(nπx' a)sin(mπy b)sin(mπy' b) x [sinh(γnmz<)sinh[γnm(c-z>)] γsinhγc]. (2.115) Now recover result for φ|z=c = V(x,y): φ(x≥) = -1 4π ∫o s da'φ(x≥) ∂G ∂n' = - 1 4π ∫ dx'dy' V(x',y') ∂G ∂n'|z'=c . (2.116) For z' > z: ∂gnm(z,z') ∂z' |z'=c = - sinh γnmz sinh γnmc, (2.118) 2.24 so, as before φ(x≥) = ∑ n,m=1∞ 4 sinhγnmz ab sinh γnmc∫ 0a dx'∫ 0b dy' V(x',y') sin(nπx' a) x sin(mπy' b) sin(nπx a) sin(mπy b). (2.119) 2.8 Separation of variables in cylindrical coordinates Many, many ways of solving 2-D electrostatics problems with simple geometry. First, some background: Separation of variables for ®,φ. Want to solve Laplace eqn. In polar coordinates it reads 1 ® ∂ ∂ ® ( ®∂Φ ∂ ®) + 1 ®2 ∂2Φ ∂φ2 = 0. (2.120) Assume Φ( ®,φ) = R( ®)ψ(φ). (2.121) Substitute & multiply by ®2 Φ: ® R d d ® ( ® dR d ®) + 1 ψ d2ψ dφ2 = 0. (2.122) Must have ® R d d ® ( ® dR d ®) = ν2, (2.123) 1 ψ d2ψ dφ2 = -ν2. (2.124) ν ≠ 0 => R( ®) = a ®ν + b ®-ν, ψ(φ) = Acos νφ + Bsin νφ, (2.125) ν = 0 => R( ®) = a0 + b0 ln ® ®0, ψ(φ) = A0 + B0φ. (2.126) 2.25 When solving for a problem where the full azimuthal range of φ is allowed, must have ( φ = 0, 2 π are the same) ν ≠ 0: sin(0) = sin(2 πν ) => ν = ¶1, ¶2,..., (2.127) ν = 0: B0.0 = B0.2π => B0 = 0. (2.128) Full solution for this case Φ(ρ,φ) = a0 + b0 ln ρ ρ0 + ∑ n=1∞ an ρnsin(nφ + αn) + ∑ n=1∞ bn ρ-nsin(nφ + βn). (2.129) Becomes a Fourier series problem. 2.9 Corner problem in cylindrical coordinates Solve the corner problem: Example: p ® βφxy Φ=V ρ=0 is included => b n=0. Want Φ|φ=0,β = V => ψ|ν≠0 φ=0,β = 0, ψ|ν=0 φ=0,β = V. (2.130) ν ≠ 0: Acos(0) + Bsin (0) = 0, ( φ=0) (2.131) 2.26 => A = 0. Bsin ν β = 0, ( φ=β) => ν = mπ β, m=1,2,3... . (2.132) ν =0: a 0 (A0 + B0φ) = V, ÷ 0,β => B0 = 0, a0A0 = V . (2.133) General solution is «undetermined Φ(ρ,φ) = V + ∑ m=1∞ amρmπ/βsin(mπφ β). (2.134) Notice that if the surfaces have Φ=V for ρ .∞, then the solution degenerates to Φ= V everywhere. Near ρ=0, the term with the lowest power will dominate. Assuming a 1≠0, Φ(ρ,φ) ≈ V + a1ρπ/βsin(πφ β). (2.135) Electric fields are Eρ = - ∂Φ ∂ρ ~- - πa1 β ρπ/β− 1sin(πφ β), (2.136) Eφ = - 1 ρ ∂Φ ∂φ ~- - πa1 β ρπ/β− 1cos(πφ β). (2.137) Notice that √⎯⎯⎯⎯⎯⎯⎯Eρ2 + Eφ2 ≈ π|a1| β ρπ/β− 1. (2.138) It looks like: 2.27 β=7/4 ~ρ-3/7 a10β=/ 4 ~ρ3 a01π π > > The Green functions for this geometry can be found, but I will resist this temptation until the next chapter. 2.10 Cylindrical halves at different potentials Do this problem as an example: gap φ VV2 1VV1 2+ 2pot. in gap =radius=b ÷ (don't mix up w/b ) n This is a standard BV problem; however, the resulting Fourier series can be summed exactly. Start off: Φ = a0 + b0 lnρ/ρ0 + ∑ n=1∞ anρnsin(nφ + αn) + ∑ n=1∞ bnρ−nsin(nφ + βn) (2.139) 2.28 Origin is included => b n = 0, all n. By integrating around the perimeter, we can immediately establish that a 0 represents the average value of the potential on the surface, so a0 = V1 + V2 2. (2.140) The BC are: Φ(b,φ = -π 2 < φ < π 2) = V1, (2.141) Φ(b,φ = π 2 < φ < 3π 2) = V2. (2.142) There are two sets of constants, the a n and αn, that we need to evaluate. The standard, long, way of doing this would be to rewrite the ∑ n=1∞ anρnsin(nφ + αn) term in (2.139) as ∑ n=1∞ ()anρn sin(nφ) + an'ρn cos(nφ) and evaluate the two sets of constants, a n and a n' using (2.141) and (2.142). However, whenever we have a symmetry in theproblem, we should exploit it to simplify the work. We note that the function ⎝⎜⎛ ⎠⎟⎞Φ - V 1 + V2 2 on the inner surface is odd in the variable φ + π 2 about φ = - π 2 . The symmetry of the problem implies that it is odd everywhere in the interior. Introducing φ' = φ + π 2 , we have ⎝⎜⎛ ⎠⎟⎞Φ - V1 + V2 2 = ∑ n=1∞ anρn sin(nφ'- nπ 2 + αn). (2.143) This means αn = nπ 2. (2.144) Thus, 2.29 (2.141) => V 1 = V1 + V2 2 + ∑ n=1 anbn sin n( φ + π 2), - π 2 < φ < π 2. (2.145) This is a Fourier series problem for a n: => anbn = 2 π ∫ -π/2π/2 (V1-V2 2) sin n( φ + π 2) dφ, => anbn = 2 π [V1-V2 2] [-cosnπ n + 1 n], => an = ⎩⎪⎪⎨⎪⎪⎧0, n even b-n 2(V1-V2) πn, n odd . (2.146) Check by using (2.142): (2.142) => V 2 = V1 + V2 2 + ∑ n=1 anbn sin n( φ + π 2), π 2 < φ < 3π 2 . (2.147) => anbn = - 2 π ∫ π/23π/2 (V1-V2 2) sin n( φ + π 2) dφ, (2.148) => an = ⎩⎪⎪⎨⎪⎪⎧0, n even b-n 2(V1-V2) πn, n odd . (2.149) (Same as before) => Φ = V1 + V2 2 + 2(V1-V2) π ∑ n odd∞ (ρ b)n sin[n(φ+π/2)] n. (2.150) Now show it can be summed. Let Z = ρ b ei(φ+π/2), (2.151) Im Z = ρ b sin(φ+π/2) = ρ b cosφ, (2.152) Φ = V1 + V2 2 + 2(V1-V2) π Im∑ n odd Zn n. (2.153) 2.30 By comparing infinite series representations and using properties of the log, ∑ n odd Zn n = 1 2ln (1+Z 1-Z), (2.154) => Φ = V1 + V2 2 + 2(V1-V2) π 1 2 Im[ ln(1+Z 1-Z)]. (2.155) However Im [ ln(Z')] = tan-1(Im Z' Re Z'), (2.156) and (1+Z 1-Z) = 1 - |Z|2 + 2i ImZ |1 - Z|2. (2.157) => Re(1+Z 1-Z) = 1-|Z|2 |1-Z|2, Im(1+Z 1-Z) = 2ImZ |1 - Z|2. (2.158) => Φ(ρ,φ) = V1 + V2 2 + V1-V2 π tan-1(2ρ/b cosφ 1 - (ρ/b)2), (2.159) => ß = 1 4π ∂Φ ∂ρ |ρ=b = V1 - V2 4π2 1 bcosφ. (2.160) 2.11 Conformal mapping techniques Another important technique is conformal mapping. Conformal mapping techniques are good for (essentially) 2-Dproblems with Dirichlet or Neumann BC's. z = x + iy, (2.161) f(z) = u(x,y) + iv(x,y). (2.162) Cauchy: ∂u ∂x = ∂v ∂y, ∂u ∂y = - ∂v ∂x, (2.163) => ∇2u = ∂2u ∂x + ∂2u ∂y = ∂ ∂x (∂v ∂y) - ∂ ∂y (∂v ∂x) = 0. (2.164) 2.31 ∇2v = 0 also. (2.165) Also notice, if E≥ = -∇≥u, (2.166) ∇≥u .∇≥v = ∂u ∂x ∂v ∂x + ∂u ∂y ∂v ∂y = - ∂u ∂x∂u ∂y + ∂u ∂y ∂u ∂x = 0. (2.167) so lines of v = const. lay along electric field lines. Basic theorem (Churchill, "Complex Variables and Applications", 3rdEd., p.206): Suppose that the analytic function f(z) = u(x,y) + iv(x,y) maps an arc C in the z plane onto an arc ΓΓΓΓ in the w plane. Let f(z) be conformal (f(z) analytic and f'(z) ≠≠≠≠ 0 at z.)on C and let a function h(u,v) be differentiable on ΓΓΓΓ. If the function h(u,v) satisfies either of the conditions h = c or dh dn = 0 (not dh dn = c) along ΓΓΓΓ, then H(x,y) = h [u(x,y), v(x,y) ] satisfies the corresponding condition along C. Do an example. Here is a mapping: z w y v-V V -V +VuOxπ ÷ ÷ ÷ ÷ ÷ ÷ w=ez > (E0=2V/π) f(z) = iE 0z + V, E 0,V > 0. (2.168) We see that u(x,y) = -E 0y + V, (2.169) 2.32 => E x = 0, Ey = -∇yu = E0. (2.170) on the other hand f[z(w)] = iE 0 ln w + V, = iE 0 ln(u+iv) + V, = iE0[ ln√⎯⎯⎯⎯u2+v2 + itan-1(v/u)] + V. (2.171) use principle branch, if youwish Identify Φ(u,v) = -E 0tan-1 (v/u) + V. (2.172) => E u = - ∂Φ ∂u = - E0v u2+v2 , (2.173) Ev = - ∂Φ ∂v = E0u u2+v2 . (2.174) Notice E u2 + Ev2 = E02 ρ2 ,ρ2 = u2+v2. (2.175) There are many ways of confirming this result. One way is to use our Green function for the geometry: y (was z) x That Green function was GD(x≥ ,x≥') = 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯(x-x')2+(y-y')2+(z-z')2 - 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯(x-x')2+(y+y')2+(z-z')2 , 2.33 Then Φ(x≥) = - 1 4π ∫o s da'Φ(x≥') ∂G ∂n', (2.176) ∂G ∂n' |s = - ∂G ∂y' |y'=0 = -2y ((x-x')2+y2+(z-z')2)3/2. (2.177) Φ(x≥) = - yV 2π {∫ -∞0 dx'∫ -∞∞ dz' 1 ((x-x')2+y2+(z-z')2)3/2 -∫ 0∞ dx'∫ -∞∞ dz' 1 ((x-x')2+y2+(z-z')2)3/2 }(2.178) Now ∫ -∞∞ dz' 1 ((x-x')2+y2+(z-z')2)3/2 = ∫ -∞∞ dz" 1 ((x-x')2+y2+z"2)3/2 = (z" (x-x')2+y2) 1 √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯(x-x')2+y2+z"2 |-∞∞ = (2 (x-x')2+y2). (2.179) We also have ∫-∞0 dx' (1 (x-x')2+y2) = -∫ -∞x dx" 1 x"2+y2 = - 1 y tan-1 x y |x ∞ = - 1 y (tan-1 (x y) - π/2). (2.180) Likewise ∫ 0∞ dx' (1 (x-x')2+y2) = - 1 y (-π/2 - tan-1(x y)), (2.181) => Φ(x≥) = 2V π tan-1(x y) = 2V π (π/2 - tan-1(y x)) = V - 2V π tan-1(y x). (2.182) 2.34 Same result as before if we realize y <=> v, (2.183) x <=> u, (2.184) E0 <=> 2V π . (2.185) Conformal methods can also give us some very nontrivial answers. Example: z wy v V -Vu -π -÷E≥x =>-1 w=e +zzππ πV -V0 f(z) = iV π z, Re (f(z)) = - V π y. (2.186) Complicated in general, but let's just consider the electric field along v=0 (v=0 <=> y=0 here) u = excosy + x, (2.187) v = exsiny + y. (2.188) Must try to express x(u,v) and y(u,v). Now Ev|v=0 = - ∂ ∂v(- V π y(u,v))|v=0 = V π ∂y ∂v|v=0. (2.189) (2.187) => 0 = excosy ∂x ∂v - exsiny ∂y ∂v + ∂x ∂v, (2.190) 2.35 (2.188) => 1 = exsiny ∂x ∂v + excosy ∂y ∂v + ∂y ∂v. (2.191) For y=0 the second eqn becomes 1 = ex ∂y ∂v|y=0 + ∂y ∂v|y=0, (2.192) => ∂y ∂v|y=0 = 1 ex + 1 . (2.193) From (2.187) at y = 0 we have u = ex + x, (2.194) so that 1 ex + 1 . ⎩⎪⎨⎪⎧ 1/u , u .∞ 1, u .-∞ (2.195) Tells us, for example, that (E0 = V π) Ev . ⎩⎪⎨⎪⎧V π 1 u , u .∞ V π, u .-∞ (2.196) 2.36 Problems 2.1.1 Consider the infinite parallel conducting plate interior problem. + 1 z=0 z=a(x',y',z') Find the image charge locations and values of all the image charges needed to satisfy the BC's for the Dirichlet Greenfunction. Write the Green function formally as a sum overthese charges. Show that your expression convergesabsolutely. 2.1.2 (a) Find the Dirichlet electrostatic Green function, G(x≥;x≥'), for the (3 dimensional) geometry shown in Cartesian coordinates. The walls are perfect conductors. conductor +1 Ox y' ' 2.37 The angle between the infinitely long conducting planes is exactly 90 °. (Take your coordinate origin at O.) (b) Find the work done, W, in removing the +1 charge to spatial infinity.(c) Using the Green's function from the (a) part, find thepotential for the situation where the vertical wall is atpotential V and the horizontal wall is at -V, as shown. OΦ=V Φ=-V 2.1.3 An infinitely long unit line charge is located with cylindrical coordinates ( ρ',φ') parallel to the line at which perpendicular conducting planes intersect. Using the image method, find the Dirichlet Green function. φ'ρ'line charge Express your answer in cylindrical coordinates. Show that G(x≥,x≥') = ln ⎣⎢⎢⎡ ⎦⎥⎥⎤ ρ4 + ρ'4 - 2ρ'2ρ2 cos(2(φ+φ')) ρ4 + ρ'4 - 2ρ'2ρ2 cos(2(φ-φ')) . [Beware of the point raised in prob. 1.3.1.] 2.38 2.1.4 A half-cylinder of conducting material of radius R is formed on top of an infinite conducting plane. A long line charge, λ, is located in vacuum above the conductors. λ RPicture: Cut- through section 90 °x≥x'≥ Find the potential at an arbitrary position, x≥, in vacuum. Show explicitly that all the boundary conditions are met. 2.1.5 A particle of charge e is attached to a massless string of length L and oscillates above a semi-infinite conductingplane that fills the entire region below the origin, O. Thedistance from the point of attachment to the plane is D.(Gravity is not operating in this problem and D > L.) θ m, eDL (t) O By computing the potential energy of the system to lowest order in the angle θ(t) (or other means), find the frequency of small oscillations of the charge. 2.1.6 By using the image method, find the Green function for a line charge of unit linear density inside a cylinder ofradius b and parallel to it's axis. Assume two line charges 2.39 of equal and opposite density and require that the potential vanish on a circular cylinder containing the positive linecharge. Given the radius b, the distance from the exterior image line charge to the cylinder's center as ρ", and the distance from the interior line charge to the cylinder's center as ρ', one can show that ρ" = b 2 ρ'. [Hint: the answer is G(ρ,φ;ρ',φ') = ln ⎝⎜⎛ ⎠⎟⎞ b4 + ρ2ρ'2 - 2b2ρρ'cos(φ-φ') b2(ρ2 + ρ'2 - 2ρρ'cos(φ-φ')).] 2.1.7 Consider a long line of charge, λ, located near a rounded conductor corner, as shown. The corner is 1/4 of a circle of radius a. a origin ’ ’ a Using cylindrical coordinates, ρ and φ, find the Dirichlet Green function ( λ=1) by the method of images. Take your origin at the hypothetical corner, which is also the circle center, as shown. 2.2.1 An arbitrary patch of area on the z = 0 plane is raised to a constant potential V, as shown. The rest of the plane(extending to infinity) is specified to have V=0. 2.40 0V z=0 planeV Show that the electric potential, Φ, at an arbitrary position above or below the plane can be written Φ(x≥) = V 2π |Ω(x≥)|, where Ω(x≥) is the solid angle subtended by the patch at the observation position, x≥. [Hint: Use the Green function for this geometry.] 2.2.2 Obtain the Neumann Green function for the half-infinite geometry: G+1 ∂ n∂ z=0 = 0 z=0n3 Solve it: 2.41 (a) By the method of images. (b) By the reduced Green function method [You can leave the result in integral form.] (c) Confirm the presence of a surface delta function in ∂G ∂n' |z=0 . That is, show ρ ≠ 0 , z'= 0, ∂G ∂n' |z=0 = 0 ρ = 0 , z' . 0+, ∂G ∂n' |z=0 = +∞ ( ρ2 = (x-x')2 + (y-y')2 ) and that 1 4π lim z' .0+ ∫ S da ∂G ∂n' |z=0 = 1. 2.2.3 Consider again the Dirichlet Green function for infinite parallel conducting plates (see above). (a) Show that G0(x≥ ,x≥') = 4π ⌡⎮⎮⌠d2k (2π)2 eik≥.(x≥ -x'≥)⊥g(z,z'k), where g(z,z',k)= sinh(kz <)sinh(k(a-z >)) ksinh(ka) , and where z < (z>) is the lesser (greater) of z and z'. (b) Using the result of part (a), verify the result (which can be shown using the reciprocity theorem of Ch.1), ⎭⎪⎬⎪⎫Qz=0 = -(1- z' a) Qz=a = - z' a. z' is the source position 2.42 (c) Using the explicit expression for the capcitance (derived in Ch. 1) and the result of (b) to show that the capcitanceis C = A 4πa . [Hint: Take one of the surface integrals to include only a finite area, A. We have used a sledgehammer to crack a nut.] 2.3.1 Find the Green's function solution for a unit point charge located inside a hollow, grounded, perfectly conducting shperical shell of radius "a". Use the imagemethod and find the potential both inside and outside theconducting shell. 2.3.2 Consider a point charge, q, located inside a thin spherical conducting sphere of radius R. In this case theconductor is neutral (no net charge on it). R qconductor (sphere) x'≥ Find the potential due to the charge q everywhere inside and outside the sphere. Can one tell where the charge is insidethe sphere from the outside field? Generalize your conclusionto a point charge inside an arbitrarily shaped conductingneutral hollow object. 2.43 2.3.3 (a) A half-sphere of radius R is placed atop an infinite two dimensional plane. All surfaces are conducting. half sphere of radius R R Using a set of image charges, find the Dirichlet Green function at all points above the conducting surfaces.(b) Find the Dirchlet Green function for the vacuum regioninside the hemispherical conducting half-sphere. 2.3.4 Will an image charge solution for the Neumann Green function for a sphere, either inside or outside, of radius Rwork? If so, find them; if not, explain why. 2.6.1 Solve the Laplace equation, ∂ 2Φ ∂x2 + ∂2Φ ∂y2 = 0, in two dimensions by separation of variables and find the potential, expressed as an infinite series, everywhere inside a two dimensional box which has Φ=0 on it's borders except along y=b where Φ=V. (V is a constant). 2.44 y xb aΦ(x,y) ΦΦΦΦ=V =0=0 =0 Ans.: Φ(x,y) = 4V π ∑ n=1∞ sinh(mπy a) m sinh(mπb a) sin(mπx a), m=2n-1. 2.10.1 Using the Dirichlet Green function derived in prob. 2.1.6, rederive the expression (2.159) for the potentialinside a long cylinder of radius b with potentials V 1 and V 2 as in the cylindrical halves section above. (Make sure you set up the angle φ relative to the potentials exactly as in the example.) 2.10.2 (a) Using Fourier series methods, find the potential, Φ(ρ,φ), of an infinitely long cylinder (radius b) raised to two separate potentials, V 1 (for -π 2<φ<π 2) and V 2 (for π 2<φ<π and -π<φ<-π 2) in the region exterior to the cylinder ( ρ>b). 2.45 V V b y x2 1gap gap (b) By summing or other means produce a closed form solution. Compare with the interior solution and comment. 2.11.1 (Adapted from Churchhill, Ch.9) The two dimensional Dirichlet problem for the geometry shown in prob. 2.6.1 isconformally mapped to the space between the semicircularregions shown, all surfaces having zero potential except thesurface between v=0, -r 0<u<-1, which again has potential V. =VΦ Φ ΦΦ =0=0 =0v u -1 1r0r θP (a) Find the mapping, w = f(z), which takes the geometry in prob. 2.6.1 to the above. 2.46 (b) Using this mapping, show that the potential everywhere inside the semicircular region is given by Φ(r,θ) = 4V π ∑ n=1∞ sinh(mπθ ln(r0)) m sinh(mπ2 ln(r0))) sin(mπln(r) ln(r0)), m=2n-1, where θ and r are cylindrical coordinates which locate a given point P in the semicircular region.