charge in sphere part 2 REVIEWED

Charge in Sphere Problem part 2 PhL 11.6.13 This stuff all ended up in Appendix L! I want to do a more general problem than I did in part 1. Problem: For this spherical picture with positive q at the center, compute everything. None of the media is conducting. We can add the layers at the two boundaries to say Q23 = Q2 - Q3 Q12 = Q1 - Q2 The charge on the great sphere will be -Q1. We will compute all these charges below as part of the solution to this problem. Solution: We know that φ = φ(r) and E = Er in all three regions. The first issue is what are the general forms of the solution in each region? In my first go at this, I allowed linear terms Ar in φ3 and Dr in φ2, but I now realize they cannot be present because the field must always have the form K/r2 due to spherical symmetry. But I leave them in since that is how I started, and then A = D = 0 pretty soon. One BC I got wrong the first time was the first term in φ3. If you are close to q in region 3, then you can use the result of the LAST document to conclude that the first term must be (1/4πε3)q/r, and when I did this problem the first time I wrongly used ε0 there. You HAVE to do that other problem first in order to know about this BC. Using obvious atomic forms we can write in each region φ3 = (1/4πε3) q/r + Ar + B // has right small r limit, I think A = 0 but lets see φ2 = C q/r + Dr + E // just trying out a few known atomic forms. φ1 = F q/r // since very far away we expect to have 1/r drop-off The continuity conditions on φ are at r = b: F q/b = C q/b + Db + E 1 meets 2 at r = a: C q/a + Da + E = (1/4πε3) q/a + Aa + B 2 meets 3 This is only 2 conditions on 6 parameters. But we now have to do field continuity. Compute the three fields -E = φ : -E3 = -(1/4πε3)q/r2 + A -E2 = -C q/r2 + D -E1 = -Fq/r2 The field conditions are at r = b: 1:2 ε1En1 = ε2En2 -ε1 Fq/b2 = ε2[ -C q/b2 + D] at r = a: 2:3 ε3En3 = ε2En2 ε3[-(1/4πε3) q/a2 + A] = ε2[ -C q/a2 + D] We now have 4 conditions on 6 constants. There are no other conditions I can think of. I suspect that A and D both vanish, so lets assume A = 0 and D = 0 and solve the four resulting equations which are then F q/b = C q/b + E (1) C q/a + E = (1/4πε3) q/a + B (2) -ε1 Fq/b2 = ε2[ -C q/b2] => -ε1F = -ε2C (3) ε3[-(1/4πε3) q/a2] = ε2[ -C q/a2] => ε3 [(1/4πε3)] = ε2C (4) Variables are C, F, E, B. The last equation tells us C, and then (3) tells us F C = [(1/4πε3)] (ε3/ε2) = (1/4πε2) -F = - (ε2/ε1)C = - [(1/4πε3)] (ε3/ε2) (ε2/ε1) = - (1/4πε3) (ε3/ε1) = - (1/4πε1) which I summarize as C = + (1/4πε3) (ε3/ε2) = (1/4πε2) F = + (1/4πε3) (ε3/ε1) = (1/4πε1) Then equation (1) says E = (q/b)(F-C) = - (q/b)(C-F) = - (q/b) (1/4πε3) [ (ε3/ε2) - (ε3/ε1)] = - (q/b) (1/4π)(1/ε2-1/ε1) Finally we can solve equation (3) for B B = - (1/4πε3) q/a + C q/a + E = - (1/4πε3) q/a + (1/4πε3) (ε3/ε2) q/a - (q/b) (1/4πε3) [ (ε3/ε2)- (ε3/ε1)] = (1/4πε3){ - q/a + (ε3/ε2) q/a - (q/b) [ (ε3/ε2)- (ε3/ε1)] } = (1/4πε3){ (q/a) [(ε3/ε2) - 1] - (q/b) [ (ε3/ε2)- (ε3/ε1)] } = (1/4π){ (q/a) [1/ε2- 1/ε3] - (q/b) [ 1/ε2 -1/ε1] } So here are all the constants A = 0 B = (1/4πε3){ (q/a) [(ε3/ε2) - 1] - (q/b) [ (ε3/ε2)- (ε3/ε1)] } C = (1/4πε3) (ε3/ε2) D = 0 E = - (q/b) (1/4πε3) [ (ε3/ε2)- (ε3/ε1)] F = + (1/4πε3) (ε3/ε1) Now define κ = (1/4πε3) Then we have A = 0 B = κ{ (q/a) [(ε3/ε2) - 1] - (q/b) [ (ε3/ε2)- (ε3/ε1)] } = κ (q/a) [(ε3/ε2) - 1] + E C = κ (ε3/ε2) D = 0 E = - κ (q/b) [ (ε3/ε2)- (ε3/ε1)] F = + κ (ε3/ε1) So write the non-trivial ones again C = κ (ε3/ε2) F = κ (ε3/ε1) E = - κ (q/b) [ (ε3/ε2)- (ε3/ε1)] B = κ (q/a) [(ε3/ε2) - 1] + E c = (ε3/ε2) κc = (1/4πε2) f = (ε3/ε1) κf = (1/4πε1) e = - (q/b) [ (ε3/ε2)- (ε3/ε1)] κe = -(q/b) (1/4π) [1/ε2 - 1/ε1] β = (q/a) [(ε3/ε2) - 1] + e κβ = (q/a) (1/4π) [1/ε2 - 1/ε3] + κe The potentials and fields are then E3 = κ q/r2 = (1/4πε3) (q/r2) E2 = κc q/r2 = (1/4πε2) (q/r2) E1 = κf q/r2 = (1/4πε1) (q/r2) φ3 = κ q/r + κβ = (1/4πε3) (q/r) + (q/a) (1/4π) [1/ε2 - 1/ε3] + κe φ2 = κc q/r + κe = (1/4πε2) (q/r) -(q/b) (1/4π) [1/ε2 - 1/ε1] φ1 = κf q/r = (1/4πε1) (q/r) In all three regions, the field goes as constant/r2 as you might guess. Fact: In each region, the E field thinks all space is filled with its ε. That is a simple and fascinating result. You can see this would be true with any number of spherical layers. In Maple I enter all the stuff and then check BC's: (Here κ = (1/4πε3) = 1 ) and all is well. I then generate the complete φ and E functions and set in some constants And finally here are some plots And here you see the expected two kinks in φ and the expected two jumps in E. Now what is the charge Q mentioned earlier? Start with div E = ρpol/ε0 . // since there is no free charge in this problem other than q. Now use area pointing radially outward for dA and then use (1/ε0)∫V ρpol dV = ∫S E dA Apply this with a pillbox across the 1,2 boundary to get (r = b) (1/ε0)npol(12) = (E1- E2) = κf q/b2 - κc q/b2 = κ (q/b2)(f-c) = (1/4πε3) (q/b2) [(ε3/ε1) - (ε3/ε2)] = (1/4π) (q/b2) [(1/ε1) - (1/ε2)] and therefore npol(12) = (1/4π) [(ε0/ε1) - (ε0/ε2)] (q/b2) The total charge on this boundary is then Q12 = 4πb2 * npol(12) = [(ε0/ε1) - (ε0/ε2)] q . Next, put a pillbox across the 2,3 boundary to get (1/ε0)npol(23) = (E2 - E3) = κc q/a2 - κ q/a2 = κ (q/a2)[ (ε3/ε2) - 1] = (1/4πε3) (q/a2)[ (ε3/ε2) - 1] = (1/4π) (q/a2)[ (1/ε2) - (1/ε3) ] so that npol(23) = (1/4π) (q/a2) [ (ε0/ε2) - (ε0/ε3) ] and then Q23 = 4πa2* npol(23) = q [ (ε0/ε2) - (ε0/ε3) ] So here are the polarization charge results: npol(12) = (1/4π) [(ε0/ε1) - (ε0/ε2)] (q/b2) npol(23) = (1/4π) [ (ε0/ε2) - (ε0/ε3) ] (q/a2) Q12 = [(ε0/ε1) - (ε0/ε2)] q = Q1 - Q2 (*) Q23 = [(ε0/ε2) - (ε0/ε3)] q = Q2 - Q3 (**) Now in region 3 we have E = (1/4πε3) q/r = (1/4πε0) [(ε0/ε3)q]/r so E there is seeing a total vacuum charge of [(ε0/ε3)q]. But this must be q + Q3 which tells us that Q3 = (ε0/ε3)q - q = [(ε0/ε3) - 1 ] q which we see is negative as expected (as Q3 defined) Then from (**) we get Q2 = Q23 + Q3 = [(ε0/ε2) - (ε0/ε3)] q + [(ε0/ε3) - 1 ] q = [(ε0/ε2) - 1 ] q < 0 and then from (*) we get Q1 = Q12 + Q2 = [(ε0/ε1) - (ε0/ε2)] q + [(ε0/ε2) - 1 ] q = [(ε0/ε1) - 1 ] q < 0 So here are the complete charge results: npol(12) = (1/4π) [(ε0/ε1) - (ε0/ε2)] (q/b2) npol(23) = (1/4π) [ (ε0/ε2) - (ε0/ε3) ] (q/a2) Q12 = [(ε0/ε1) - (ε0/ε2)] q = Q1 - Q2 (*) Q23 = [(ε0/ε2) - (ε0/ε3)] q = Q2 - Q3 (**) Q1 = [(ε0/ε1) - 1 ] q < 0 Q2 = [(ε0/ε2) - 1 ] q < 0 Q3 = [(ε0/ε3) - 1 ] q < 0 Here is an alternative way to determine the charges. If you are in region 3, you have E = (1/4πε3)(q/r2) = (1/4πε0)([(ε0/ε3)q]/r2), so region 3 is seeing a total effective vacuum point charge of (ε0/ε3)q. We can equate this to q +Q3 if we add up the one pol charge that observer sees from region 3. Thus, q + Q3 = (ε0/ε3)q and so Q3 = [(ε0/ε3) - 1 ] q < 0 If you are in region 2, you have E = (1/4πε2)(q/r2) = (1/4πε0)([(ε0/ε2)q]/r2), so region 2 is seeing a total effective vacuum point charge of (ε0/ε2)q. We can equate this to q +Q3 -Q3 + Q2 if we include the three pol charges that observer sees from region 2. Thus, q + Q2 = (ε0/ε2)q and so Q2 = [(ε0/ε2) - 1 ] q < 0 If you are in region 1, you have E = (1/4πε1)(q/r2) = (1/4πε0)([(ε0/ε1)q]/r2), so region 1 is seeing a total effective vacuum point charge of (ε0/ε1)q. We can equate this to q +Q3 -Q3 + Q2_Q2+Q1 if we include the five pol charges that observer sees from region 1. Thus, q + Q1 = (ε0/ε1)q and so Q1 = [(ε0/ε1) - 1 ] q Thus we replicate the earlier Qi results.