f1

Point charge in two dielectrics, Part 2 PhL 10.24.13 There is more stuff I need to know about this problem before I can attack my two King logjam question. 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 curl E = 0 does see polarization charge at a boundary. 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 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. 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 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 e-ks J0(kρ) = - (1/4πε1) [ε2g + ε1(f-1)] !Syntax Error, I dk e-ks J0(kρ) Meanwhile, we have ε2g + ε1(f-1) = ε22ε1 / (ε1+ε2) + ε1 [