Point charge in two dielectrics, Part 2 REVIEWED

Point charge in two dielectrics, Part 2 PhL 10.24.13 See notes in the Part 3 version. There is more stuff I need to know about this problem before I can attack my two King logjam question. A Wrong Assumption: First of all, since div D = ρ = the free charge only, you would think that the field D would be D1 = D2 = q (1/4π) (1/r) since D does not "see" the polarization charge at the boundary. Here we place the Jackson region 1 point charge at a spherical system origin. We then know that P1 = ε0χ1 E1 in region 1 P2 = ε0χ2 E2 in region 2 D1 = ε1E1 in region 1 D2 = ε2E2 in region 2 But since D1 = D2 = D = the same in both regions, we seem to get q (1/4π) (1/r) = ε1E1 in region 1 q (1/4π) (1/r) = ε2E2 in region 2 But this is NOT what we get. The reason is that in region 1 we don't get E1 = [...], but instead we know the E field lines are curved. So what have I done wrong here? What is wrong is this. Knowing that div D = ρ does not determine D. This is just a condition on D. The other major condition is curl E = 0 for a static situation. And we have to make sure things are OK at boundaries, and THEN we have a solution. I said that D does not see polarization charge at a boundary. That is true, but we have to incorporate boundary conditions. Recall the integral form of our two rules, div D = ρ ∫V ρ dV = ∫S D dA (1.1.13) curl E = 0 E ds = 0 (1.1.17) This means that E|| is the same on both sides of a boundary, while D is the same on both sides. For our little Jackson problem that means E||(1) = E||(2) ε1E(1) = ε2E(2) or (1/ε1)D||(1) = (1/ε2)D||(2) D(1) = D(2) Thus, even thought D does not see polarization charge at a boundary, it does not simply "pass through" a boundary as I suggested above. There is refraction of D at the boundary just as with E ! Now how would you approach solving the D problem? I don't really know! I only know how to approach the φ problem and E = -φ so I am better at doing the E problem. My "cylindrical atoms" solution to the problem was this: φ1(ρ,z) = (1/4πε1) !Syntax Error, Idk e-k|z|J0(kρ) + !Syntax Error, I dk Ak ekz J0(kρ) φ2(ρ,z) = !Syntax Error, I dk Bk e-kz J0(kρ) R = Ez1 = ∂zφ1 = (1/4πε1) !Syntax Error, Idk (∂ze-k|z|)J0(kρ) + !Syntax Error, I dk Ak k ekz J0(kρ) Ez2 = ∂zφ2 = !Syntax Error, I dk Bk (-k) e-kz J0(kρ) Eρ1 = ∂ρφ1 = (1/4πε1) !Syntax Error, Idk k e-k|z|J0'(kρ) + !Syntax Error, I dk kAk ekz J0'(kρ) Eρ2 = ∂ρφ2 = !Syntax Error, I dk kBk e-kz J0'(kρ) Ak = (1/4πε1)e-2ks [(ε1-ε2)/ (ε1+ε2)] Bk = (1/4πε1) [2ε1 / (ε1+ε2)] ≡ B Ratio. For fun, let's look at the ratio of Ez2/Eρ2 Ez2/Eρ2 = !Syntax Error, I dk Bk (-k) e-kz J0(kρ) / !Syntax Error, I dk k Bk e-kz J0'(kρ) = - !Syntax Error, I dk k e-kz J0(kρ) / !Syntax Error, I dk k e-kz J0'(kρ) where the Bk just cancel, being constants as shown above. If we use ρ2 + z2 = r2 to eliminate either ρ or z, the solution claims that this ratio is a constant independent of r ! That result seems completely non-obvious looking at the integrals. Maybe write this ratio this way Ez2/Eρ2 = ∂zφ2 / ∂ρφ2 where φ2 = B!Syntax Error, I dk e-kz J0(kρ) For z > 0, we can see that !Syntax Error, I dk e-kz J0(kρ) = 1/R for a unit point charge at the origin, but this is true only for z > 0 !! That is why we get the lines there all zeroing in on the location of q. So it is quite easy to obtain the image charge solution from my results above. Free charge on the boundary. What about the surface charge distribution? We get that from div D = ρ ∫V ρ dV = ∫S D dA (1.1.13) If we apply this to a tiny pillbox straddling the boundary, we find that n dA = Dz2dA - Dz1dA => n = Dz2 - Dz1 => n = ε2Ez2 - ε1 Ez1 where from above, at z = s, Ez1 = ∂zφ1 = (1/4πε1) !Syntax Error, Idk (-ke-ks)J0(kρ) + !Syntax Error, I dk Ak k eks J0(kρ) Ez2 = ∂zφ2 = B !Syntax Error, I dk (-k) e-ks J0(kρ) Ak = (1/4πε1)e-2ks [(ε1-ε2)/ (ε1+ε2)] = (1/4πε1)e-2ks f f = [(ε1-ε2)/ (ε1+ε2)] B = (1/4πε1) [2ε1 / (ε1+ε2)] = (1/4πε1) g g = [2ε1 / (ε1+ε2)] Therefore Ez1 = ∂zφ1 = - (1/4πε1) !Syntax Error, Idk k e-ksJ0(kρ) + (1/4πε1) f !Syntax Error, I dk k e-ks J0(kρ) = (1/4πε1) [ f - 1] !Syntax Error, I dk k e-ks J0(kρ) Ez2 = (1/4πε1) [-g] !Syntax Error, I dk k e-ks J0(kρ) Then the surface charge density seems to be this n = ε2Ez2 - ε1 Ez1 = (1/4πε1) [-ε2g - ε1(f-1)] !Syntax Error, I dk k e-ks J0(kρ) = - (1/4πε1) [ε2g + ε1(f-1)] !Syntax Error, I dk k e-ks J0(kρ) Meanwhile, Maple tells us that [ε2g + ε1(f-1)] = 0 so our result is that n = 0. This is the correct result for the free charge at the boundary! I think the integral shown is a simple elementary function but hold on that for a moment. Polarization charge on the boundary. Now let's find the charge that E sees! div E = ρ/ε ∫V (ρ/ε) dV = ∫S E dA [****] (1.1.14) If we apply this to a tiny pillbox straddling the boundary, what do we do about ε? Suppose we put half the charge into each half space, just as something to try. Then let εav be the average of the two, so (n/ε1 + n/ε2 )dA = (Ez2 - Ez1)dA but this is baloney, you could use any fraction you wanted. So I guess the problem is that you cannot find the total charge ρ from the above conditions! You have to involve the P vector somehow. So let's accept Jackson's claim that where ρ is the free charge, and therefore ρpol = - div P when then leads to div P = -ρpol -∫V ρpol dV = ∫S P dA Now apply this to our pillbox to get (see Attempt #2 picture) - npoldA = [P2z - P1z] dA or - npol = P2z - P1z Now we know that P1 = ε0χ1E1 = ε0(ε1/ε0- 1) E1 = (ε1-ε0) E1 P2 = ε0χ2E2 and therefore - npol = ε0χ2E2z - ε0χ1E1z or - npol/ε0 = χ2E2z - χ1E1z Meanwhile, our boundary condition on E is this: ε1E(1) = ε2E(2) => ε1E1z = ε2E2z Since Ez2 is my simpler known form, I will keep it so that then - npol/ε0 = χ2E2z - χ1(ε2/ε1)E2z = [ χ2 - χ1(ε2/ε1)] E2z But we know that so χ1 = ε1/ε0- 1 χ2 = ε2/ε0- 1 Then -npol/ε0 = [ χ2 - χ1(ε2/ε1)] E2z = [- (ε2- ε1)/ε1] E2z where Maple did this little piece of algebra My conclusion then is that npol/ε0 = [(ε2- ε1)/ε1] E2z = [(ε2- ε1)/ε1] B !Syntax Error, I dk (-k) e-ks J0(kρ) = [(ε2- ε1)/ε1] (1/4πε1) [2ε1 / (ε1+ε2)] !Syntax Error, I dk (-k) e-ks J0(kρ) = - (ε2-ε1)] (1/2πε1) [1 / (ε1+ε2)] !Syntax Error, I dk k e-ks J0(kρ) = (1/2πε1) (ε1-ε2)/(ε1+ε2) !Syntax Error, I dk k e-ks J0(kρ) Now let's go do this integral. Maple cannot do it, so I will go look it up in GR7. So set α = s, ν=0, β = ρ and we meet the conditions and the result is then 2s(2ρ)0 Γ(3/2)/[(s2+ρ2)3/2] = 2s{/2)/[(s2+ρ2)3/2] = s / (s2+ρ2)3/2 and therefore I conclude that npol/ε0 = (q/2πε1) (ε1-ε2)/(ε1+ε2) s / (s2+ρ2)3/2 and finally npol = - (qε0/2πε1) (ε2-ε1)/(ε2+ε1) s/ (s2+ρ2)3/2 Jackson's result was this so we are in perfect agreement. Hurray! Question: Given some polarization charge ρpol , how do you compute its electric field? Answer: Look back at Jackson's claim If we are in a medium of ε ≠ ε0, the claim is that we have div E = ρ/ε0 + ρpol/ε0 where ρpol = - div P. Thus, we must propagate both the charges with 1/4πε0 , not with 1/4πε !! It's just that when we add up these two sources, we find div D = ρ and div E = ρ/ε where ρ is the total charge. Let's prove this claim. Start with the above div E = ρ/ε0 + ρpol/ε0 = ρ/ε0 – 1/ε0 div P or ε0div E = ρ – div P or div( ε0E + P) = ρ Then define as Jackson does, , so that div( D) = ρ Then since D = ε E we get div(E) = ρ/ε where ρ is only the free charge. So if you ignore the polarization charge, then you compute a potential by using 1/4πεR as the propagator. But if don't ignore it, they you propagate each piece as 1/4πε0R . This is an important point that I missed earlier. So, when we talk about the charge on our boundary as being just polarization charge, it has to propagate with 1/4πε0R in either region 1 or region 2! Compute E field due to polarization charge. Here is my claim, words follow: E = Ez Ez= ∫[2πρdρ]n(ρ) { 1/ (z2+