potential of a ring of charge REVIEWED

Electrostatics: Potential inside a ring of charge PhL 11.12.13 I think my treatments below are viable. It is not clear to me now 2/9/14 why I was interested in this problem at the time. I guess if you think of the mag surface current Kz on a piece of round conductor, it is a like a ring of charge??? I have totally lost the connection and the edit log for this date is also confusing. But I think the work here regarding this Jackson problem is probably all OK. I might file a copy of this in electrostatics somewhere. My interest in this problem arose recently when doing the transmission line Chapter 4 work where I mistakenly used the 3D propagator instead of the 2D one. I was reminded of my GR7 errata with the two integrals, and it all has a bearing on this simple problem. We want to look at this 3D problem solved a two different ways. Using spherical atoms, Jackson blue provides a solution at the top of page 104. I see he has dropped equation numbers on equations that perhaps are just applications. I will quote this result in a while. [ This is also in green in cgs.] In Part 1 I do it my way and get a K function In Part 2 I do it Jackson's way and get a spherical coordinate solution. // I never show that the two solutions in general are equivalent. // But I do show that they agree at the center of the ring In Part 3 I plot my solution, it is a very simple plot in the plane only, log diverge at the ring In Part 4 I provide "atomic support" for the Jackson method In Part 5 I do the 2D ring charge problem which is a completely different animal. 1. The Poisson Integral Solution to the 3D ring charge problem. 1 2. The Jackson Solution to the 3D ring charge problem 3 3. More on My Solution to the 3D ring charge problem 7 4. Jackson's method written out 9 5. The 2D ring charge problem 9 1. The Poisson Integral Solution to the 3D ring charge problem. My interest however is the Poisson Integral solution. Start with 2φ = -ρ/ε0 ρ = ρ0δ(r-a)δ(z) = ring of charge Then I expect to have, using cylindrical coordinates, φ(x) = (1/4π) (1/ε0) ∫dV' ρ(x')/R = (1/4π) (ρ0/ε0) !Syntax Error, Ir'dr'∫dθ'!Syntax Error, Idz' δ(r'-a)δ(z')/R = (a/4π) (ρ0/ε0) ∫dθ' /R where R2 = [r2 + a2 + z2] - 2arcos(θ-θ') = A2 - 2arcos(x) Our result is then φ(x) = (a/4π) (ρ0/ε0) !Syntax Error, Idx 1/ = 2 (a/4π) (ρ0/ε0) !Syntax Error, Idx 1/ This integral appears wrong TWICE in GR7, I did not notice that until today. Here are the correct integrals from my note: !Syntax Error, Idx(1 /) = (2 /) K() a > b > 0 !Syntax Error, Idx = (2 ) E() a > b > 0 Now in my application of this second integral, I have a = A2 = r2 + a2 + z2 and b = 2ar. This means a+b = r2 + a2 + z2+ 2ar = (r+a)2 + z2 The solution is then φ(x = r,θ,z) = 2 (a/4π) (ρ0/ε0) K() or φ(x = r,θ,z) = (a/π) (ρ0/ε0) K() Dimensions of ρ0 are C/m (linear charge density) while ε0 = farad/m so dim(φ) = m * C/m * m/farad * m-1 = C/F = volt I am now going to change the symbols in my cylindrical coordinates solution to be φ(x = ρ,φ,z) = (a/π) (ρ0/ε0) K() Now I have 2πaρ0 = q, the total charge of my ring, so (a/π) (ρ0/ε0) = (a/π) (q/2πaε0) = q/(2π2ε0) and then my solution is φ(x = ρ,φ,z) = q/(2π2ε0) K() // ring at z = 0 As expected, things are independent of φ. Now I want to make one more change: I will put my ring of charge at z = b to match Jackson below. Then φ(x = ρ,φ,z) = q/(2π2ε0) K() 2. The Jackson Solution to the 3D ring charge problem For some reason Jackson writes his z value as r. His solution is actually quite unclear because he doesn't really say what "r" is for his general solution I quoted above. I presume it is the distance of the observation point from his picture's origin. I am guessing that r,θ are two spherical coordinates since I see spherical atoms in his formula, and then there is no φ dependence. My solution is more geared to cylindrical coordinates. Now do we relate Jackson's r and θ to my ρ and z ? r2 = ρ2 + z2 z = rcosθ = cosθ => cosθ = z/ Then Jackson also has cosα = b/c = b/ which is just a constant. Then r< = min(r, ) = min(, ) r> = max(r, ) = max(, ) Then I can write his solution as φJ = [q/4πε0] Σn=0∞ [min(, )]n [max(, )]-n-1 Pn(b/) Pn(z/) which I can compare to my solution φL(x = ρ,φ,z) = q/(2π2ε0) K() Let's examine both solutions in the plane of the ring. For me, z = b. For him, we still have to do the min max depending no whether we are inside or outside the ring. So φJ = [q/4πε0] Σn=0∞ [min(, )]n [max(, )]-n-1 Pn(b/) Pn(b/) φL = q/(2π2ε0) K() Now at ρ = a, I have K(2a/2a) = K(1) which is no doubt a singular point for K. Now let's restrict interest to "inside the ring". Then < and we get φJ = [q/4πε0] Σn=0∞ ()n ()-n-1 Pn(b/) Pn(b/) φL = q/(2π2ε0) K() Now how about right at the center of the ring: φJ = [q/4πε0] Σn=0∞ bn ()-n-1 Pn(b/) Pn(1) φL = q/(2π2ε0) K(0) Now look up K(0) in AS special values (I hope) So then my solution becomes φL = q/(2π2ε0) (π/2) = [ 1/4πε0 ] q/a which is of course the correct result since all parts of the ring are distance a away. Jackson meanwhile gets φJ = [q/4πε0] Σn=0∞ bn ()-n-1 Pn(b/) which seems to depend on b, but as I will show below, it does not in fact depend on b. First, for b = 0 it gives the right answer since then only the n=0 term survives and φJ = [q/4πε0] 1/a P0(0) = [q/4πε0] 1/a In order for this result to be correct for any b, one would have to show that Σn=0∞ bn ()-n-1 Pn(b/) = 1/a for any b !! (*) Maple seems to indicate that this is true! Rewrite differently, x ≡ b/< 1 => = b/x a2 + b2 = b2/x2 => a2 = b2(1/x2-1) = b2(1-x2)/x2 => b2 = a2x2/(1-x2) => (b/a)2 = x2/(1-x2) => (b/a) = x/ Σn=0∞ bn (b/x)-n-1 Pn(x) = 1/a ? Σn=0∞ bn b-n-1xn+1 Pn(x) = 1/a ? Σn=0∞ b-1xn+1 Pn(x) = 1/a ? b-1Σn=0∞ xn+1 Pn(x) = 1/a ? Σn=0∞ xn+1 Pn(x) = b/a = x/ ? Σn=0∞ xn Pn(x) = 1/ ? Now there is the following generating function sum rule: If I set t = z = x this says 1/ = Σn=0∞ xnPn(x) |x| < min | x ± | = min | x ± i | (**) But | x ± i |2 = (x ± i ) (x ∓ i ) = x2 - i2(1-x2) = x2+(1-x2) = 1 Thus I have shown that | x ± i | = 1 => min | x ± i | = 1 => |x| < 1 and therefore the sum (**) is valid in my application. Therefore the last ? line above is true, and therefore (*) is true in fact for any real b. 3. More on My Solution to the 3D ring charge problem If the ring is at z = 0, my solution is this: φ(x = ρ,φ,z) = q/(2π2ε0) K() // ring lies at z = 0 If we are interested in φ inside the ring itself, we set z = 0 to get φ = q/(2π2ε0) K() 0 < ρ < a where the argument of K runs 0 to 1. Rewrite this as φ = q/(4πε0) [ (2/π) K() ] If we plot the argument of K versus ρ we find which tells us that as ρ ranges from ρ= 0 to ρ = ∞, the K argument is always ≤ 1. It is only = 1 at the point where ρ = a. So we don't need any fancy K transformation formulas. For small ρ we have already shown that [...] → 1/a as expected. For large ρ we have K() ≈ K() = K() ≈ K(0) = π/2 and then [ (2/π) K() ] ≈ (2/π) (1/ρ) (π/2) = 1/ρ and then φ = q/(4πε0) (1/ρ) // as expected. Here is a Maple plot for ρ in (0,a): At ρ = 0 we know the result is [... ] = 1/a which in our plot is just 1. At ρ→a we expect logarithmic divergence since it is like a point next to an infinite line charge which has the general behavior (1/2π) ln(1/(a-ρ)) where a-ρ = r, and the 2D propagator in Stak is (1/2π)ln(1/r). I can in fact plot right through the singular point like so and then you see the usual 1/r drop off. Conclusion: I think my expression for φ of the ring change in terms of the K function is correct. 4. Jackson's method written out For this azisym problem, we know what the atoms are, so φ(r,θ) = Σn=0∞ An r-n-1 Pn(cosθ) large r He computes things for θ = 0 and finds there that φ(r,θ=0) = Σn=0∞ [ (q/4πε0)cn Pn(cosα)] r-n-1 Pn(1) large r where Pn(1) = 1. This then determines that An = (q/4πε0)cn Pn(cosα) and then the full result must be φ(r,θ) = Σn=0∞ (q/4πε0)cn Pn(cosα) r-n-1 Pn(cosθ) large r He does a similar thing for small r and there you have it. I think his presentation is a little strange. 5. The 2D ring charge problem I don't recall ever doing this problem before. I scanned Stak Chapter 6 and did not notice it. The result I get here seems reasonable in various ways. So here is the derivation: Consider the general 2D situation 22D φ = -ρ/ε0 The potential of a unit positive point charge is φ = (1/4πε0)(1/R) in 3D. For Stak, the ε0 is just missing in the bottom, as on his page 50 (5.103). So I would say that in 2D the potential of a unit point charge at the origin is given by φ = (1/2πε0)ln(1/r). (*) This diverges logarithmically at both r = 0 and r = ∞. I think Stak gets this by assuming the divergence theorem div E = -ρ/ε0 in two dimensions and maintaining E = -φ. This is also the Green's Function, so our solution is then φ(x) = (1/2πε0)∫dA' ρ(x') ln(1/R) R = |x - x'| R2 = (x-x')2 + (y-y')2 Writing this in polar coordinates φ(x) = (1/2πε0)∫r'dr' ∫dθ' ρ(r',θ') ln(1/R) R2 = r2+r'2 - 2rr'cos(θ-θ') Now install our ring charge ρ(r',θ') = ρ0δ(r'-a) to get φ(x) = ρ0(a/2πε0) ∫dθ' (1/2)ln(1/R2) R2 = r2+a2 - 2racos(θ-θ') = - ρ0(a/4πε0) !Syntax Error, Idx ln[r2+a2 - 2racosx ] = - ρ0(a/2πε0) !Syntax Error, Idx ln[r2+a2 - 2racosx ] = - ρ0(a/2πε0) !Syntax Error, Idx ln[(a2)(1 + (r/a)2 - 2(r/a) cosx )] = - ρ0(a/2πε0) !Syntax Error, Idx { ln(a2) + ln(1 + (r/a)2 - 2(r/a) cosx ) } = - ρ0(a/2πε0)[ ln(a2)π + !Syntax Error, Idx ln(1 + (r/a)2 - 2(r/a) cosx ) ] I cannot find this second integral in GR7 where it ought to be. So I am now looking for I ≡ !Syntax Error, Idx ln(1 + A2 - 2A cosx ) ] Can Maple do this? It gives a big mess including lots of dilogs. But I remember this in Stak and (6.22) claims that ln(1 + A2 - 2A cosx ) = -2 Σn=1∞ (An/n)cos(nx) // where |A| < 1 Then my integral becomes I ≡ !Syntax Error, Idx {-2 Σn=1∞ (An/n)cos(nx)} = -2 Σn=1∞ (An/n) !Syntax Error, Idx cos(nx) = -2 Σn=1∞ (An/n) [sin(nπ)/n] // Maple But sin(nπ) = 0 for n = 1,2,3... and so this integral is I = 0. My answer is then φ(x) = - ρ0(a/2πε0)[ ln(a2)π ] // inside which, if correct, says φ = a constant. This says that φ = this constant everywhere inside the ring of charge in 2D! I guess that is possible. Then for the outside region I will swap a and r and get φ(x) = - ρ0(a/2πε0)[ ln(r2)π ] // outside Meanwhile, write ρ02πa = q for the total ring. Then ρ0(a/2πε0) = (q/2πa) (a/2πε0) = (q/4π2ε0) and then our result is φ(x) = - (q/4π2ε0) [ ln(a2)π ] // inside φ(x) = - (q/4π2ε0) [ ln(r2)π ] // outside or φ(x) = - (q/2πε0) ln(a) // inside (***) φ(x) = - (q/2πε0) ln(r) // outside At least things match at the boundary. If I take a→0 so the ring becomes a point charge, the result is φ(x) = - (q/2πε0) ln(r) and this agrees with my starting form (*) for such a point charge, so that is encouraging. Here is a Maple plot with a = 2 and I omit the leading factor but keep the minus sign: Recall that in 3D, inside a sphere of uniform charge you get φ = constant, and outside the sphere the charge appears as if it were q at the sphere center. Here, in 2D we find that inside a circle of uniform charge we also get φ = constant. Outside the ring, the ring behaves as a point charge at the center of charge q. So very similar result. Let's now go back and look at the integral again. I will do the integral and then we don't need that Stak 6.22 expansion. I ≡ !Syntax Error, Idx ln(1 + α2 - 2α cosx ) Then ∂αI = !Syntax Error, Idx (2α - 2cosx) = 2 !Syntax Error, Idx = 2α !Syntax Error, Idx - 2 !Syntax Error, Idx I would probably make this a contour integral, but let's try to look it up in GR7. The second integral seems to be this so with n = 1 and a = α we get !Syntax Error, Idx = πα/(1-α2) α < 1 = -(π/α)/(1-α2) α > 1 !Syntax Error, Idx = π/(1-α2) α < 1 = -(π)/ (1-α2) α > 1 So we then find ∂αI = 1/(1-α2) { 2α π - 2 πα } = 0 α < 1 ∂αI = -1/(1-α2) { 2α π - 2 π/α} α > 1 This second is ∂αI = -2π/(1-α2) { α - 1/α} = -2π/(1-α2)* (α2- 1)/α = 2π/α So we summarize ∂αI = 0 α < 1 ∂αI = 2π/α α > 1 Now integrate to get I = C1 α < 1 I = 2π lnα + C2 α > 1 Now match at α = 1 to get C1 = 2πln1 + C2 = C2 so we then have I = C1 α < 1 I = 2π lnα + C1 α > 1 where recall I ≡ !Syntax Error, Idx ln(1 + α2 - 2α cosx ) How do we get the constant C1? If α is very large we get I = !Syntax Error, Idx ln(α2) = π lnα2 = 2πlnα large α Equate this to I = 2π lnα + C1 to find that C1 = 0. Thus I = 0 α < 1 I = 2π lnα α > 1 or I = 2πlnα θ(α>1) // here is the integral all done! Now go back to φ(x) = - ρ0(a/2πε0)[ ln(a2)π + !Syntax Error, Idx ln(1 + (r/a)2 - 2(r/a) cosx ) ] = - ρ0(a/2πε0)[ ln(a2)π + I(α = r/a)] = - ρ0(a/2πε0)[ ln(a2)π +2πln(r/a) θ((r/a)>1)] = - ρ0(a/2πε0)[ 2 π ln(a) +2πln(r/a) θ(r>a)] If r < a, the result is just φ(x) = - ρ0(a/2πε0)[ 2 π ln(a) r < a If r > a then φ(x) = - ρ0(a/2πε0) [ 2 π ln(a) +2πln(r/a) ] = - ρ0(a/2πε0) 2πln(r) so then solution is then φ(x) = - ρ0(a/2πε0) 2π { either ln(a) or ln(r) } Then recall ρ0(a/2πε0) = (q/2πa) (a/2πε0) = (q/4π2ε0) so get φ(x) = - (q/4π2ε0) 2π { either ln(a) or ln(r) } = - (q/2πε0) { either ln(a) or ln(r) } which then replicates our solution (***) above. This solution also applies to an infinite cylinder with a uniform surface charge! I think I am going to make use of this fact when I look at the curl J around a cylinder in which J = constant!