Suggestion then is to do a couple of things: Solve it as if it were a dielectric cylinder in a uniform field without any additional free electric charge and then take the limit as the dielectric constant goes to infinity. In addition add the charge Q=it should be a charge per unit length, but in any case, let the charge be uniformly distributed on the surface and superimpose this solution with the previous one. ## \\ ## The Legendre method in a uniform field for a cylinder is somewhat lengthy, but there is a shortcut that should give the same answer which I will show you here: Given an applied electric field ## E_o ## perpendicular to the axis of the cylinder. There will be some uniform polarization ## P_i ## that occurs for this geometry, with ## E_p=-(1/2)P_i/\epsilon_o ##. (The (1/2) factor is a result of the geometry=it comes out of the detailed Legendre analysis=I'm simply giving a result that I have seen previously stated. The ## E_p ## is the electric field that occurs from the surface charge density ## \sigma_p ## that results in the case of uniform polarization ## P_i ##. The Legendre solution will show that this is in fact the case). Meanwhile ## E_i=E_o+E_p ## and ## P_i=\epsilon_o \chi E_i ## where ## \chi ## is the dielectric susceptibility. Then the dielectric constant ## \epsilon=\epsilon_o(1+\chi) ##. A little algebra gives (assuming my algebra is correct) ## E_i=\frac{E_o}{1+(1/2)(\frac{\epsilon}{\epsilon_o}-1)} ##. This just gives the electric field inside the cylinder, and the surface polarization charge density is ## \sigma_p=P \cdot \hat{n} =P cos(\phi) ## where ## P=P_i=(\epsilon-\epsilon_o) E_i ##. You should be able to compare this result to the Legendre result and they should agree. For the limit as ## \epsilon ## gets large (a conductor), I get ## P=2 \epsilon_o E_o ## , and also that ## E_i=0 ##. ## \\ ## Anyway, I'd be interested in seeing your complete Legendre solution, but I recommend first doing it without any free charge on the cylinder, and just superimposing that part afterwards. ## \\ ## One additional input is I believe the free charge per unit length is going to give you a ## A_o' ln(r) ## term which will be absent until you include the free (surface) charge . ## \\ ## And an additional input: I believe you need a potential term of the form ## -E_o cos(\phi) r ## as ## r ## gets large. To satisfy the boundary condition of equal potential everywhere at ## r=a ##, you need an ## A_1'cos(\phi)/r ## term to offset the previous term at ## r=a ##. This means ## A_1'=+E_o a^2 ##. It remains to calculate the electric fields, along with the resulting surface charge to show that this solution for the potential works everywhere. (It appears to me that the ## B ## coefficients might be equal to zero and all you need is ## A_1 ## and ## A_1' ## along with a subsequent ## A_o' ##. This solution for the potential is of course for ## r>a ##. For ## r<a ## the potential is zero everywhere. ## A_o ## can be chosen to make ## V=0 ## for ## r=a ## in the solution for ## r \geq a ##.) Additional comment: This last method is easier than my previous suggestion above to solve it as a dielectric, etc. Using the fact that the potential must be the same everywhere on the surface makes for a simple solution for ## A_1' ##.