Electric Field Outside a Round Wire at DC REVIEWED

Electric Field Outside a Round Wire at DC PhL 12.11.13 My conclusion is shown in the picture below, I think this issue is resolved. I found a physical situation which gives fields which satisfy the Max equations. The Jackson paper reference is noted. Question 1. Current I flows in a DC round wire of radius a. (a) What is the Ez field just outside the surface? (b) What is Ez outside the wire? (a) I think the current profile is uniform and just drops to zero as a step function at r = a. If so, then we have Ez = J/σ at r = a-ε where J = I/(πa2). If Ez is continuous across the boundary as I think it is, then the answer is that just outside the wire, Ez = I(πa2σ). [ I still think this is correct ] (b) There are perhaps several approaches here. We seem to know that dive E = ∂zEz(r) = 0 if there are no other field components in existence, but that fact just says Ez does not vary with z which we already know. We know that (2 - με ∂t2)E = μ∂tJ + (1/ε) grad ρ (1.2.1) In our case, outside the wire, I assume J = 0 since perfect insulator say, and ρ = 0 and E is static, so 2Ez(r) = 0 or 22DEz(r) = 0 This means that we must have Ez(r) = A + B ln (r/a) ok Then the boundary condition says Ez(a) = A = I(πa2σ) ok So then we have Ez(r) = I(πa2σ) + B ln (r/a) What sets the other constant? Maybe one of the Maxwell equations? curl E = 0 curl E = [ r-1∂θEz - ∂zEθ] + [∂zEr - ∂rEz] + [ r-1∂r(rEθ) - r-1∂θEr ] = + [- ∂rEz] This seems to say that ∂rEz(r)= 0 everywhere and that would set B = 0. ok Well, we know there is an H field outside the wire given by 2πrHθ = I Hθ(r) = I/2πr. We should have curl H = 0 since ∂t = 0 and J = 0 outside, so then curl H = [ r-1∂θHz - ∂zHθ] + [∂zHr - ∂rHz] + [ r-1∂r(rHθ) - r-1∂θHr ] = 0 + [ r-1∂r(rHθ) ] = 0 // and this works properly since Hθ(r) = I/2πr. So my tentative conclusion is rather astounding. It says Ez(r) = I(πa2σ) everywhere outside the wire, all the way out to r = ∞. Conclusion: I think this is correct, that Ez = constant in all 3D space for this problem. Here is my interpretation of this solution: This uniform Ez field is provided by a huge parallel plate capacitor at the ends of the round wire (at ±∞). The capacitor makes Ez uniform everywhere between the plates. It just happens that in the round wire it causes a current to flow. This is certainly A possible Maxwell equation solution, and I "found it" above but in retrospect it is quite obvious. "Surface charges on circuit wires and resistors play three roles" by J. D. Jackson, in American Journal of Physics -- July 1996 -- Volume 64, Issue 7, pp. 855 . Scribd has it but I am too tired to upload something. There are lots of fancy papers on this subject addressing induced charge, relativistic effects and so on. But for me, I think the answer is this: For an idealized isolated infinite round wire, Ez = constant everywhere in space. This is consistent with Maxwell's Equations. Notice above how the curl E equation killed off the lnr term. In the more realistic situation of two oppositely directed wires, if they were forced to have a uniform current density even, things are messier. Now we have surface charges and external fields etc. The notion of fields outside a uniform wire is slippery I feel. agreed Question 2. Can a round wire have an Eφ ≠ 0 field at the surface? I keep coming back to this painful question. If I argue that Eφ = 0 due to "quasi-static", then why doesn't that mean Ez = 0 as well at the surface. Hold on this question. [ Section D.8 at least faces this question ] Question 3. In the round wire with wave going down it, there is Ez at the surface I think. I know that Ez creates a current Jz inside the wire and pushes those conduction electrons along. Does this Ez push on the surface charge density and make it do something? Does it make a surface current in the z direction? If this surface charge is really in a little channel at the surface, then yes, there should be a surface current. The free electrons then oscillate back and forth in the z direction along with z. [ Final Reader Exercise ]