Equipotential Cylinder: Can Method of Images Work?

[neworder1]

An infinite string with linear charge density \lambda is put parallel to the axis of an infinite conducting grounded cylinder (V=0) of radius R, the distance between the string and the center of the cylinder is l. Find the potential outside the cylinder.

Is it possible to solve this problem using the method of images, i.e. placing an image string with some charge density \lambda_{1} inside the cylinder so that the cylinder will be an equipotential surface? If V weren't 0, I could place a string of equal charge (i.e. -\lambda) somwehere inside and it should work, but for V=0, the charges can't be equal.

 

[Meir Achuz]

Yes, it is the 2D example of a charge outside a sphere.

 

[neworder1]

Ok, but how do I calculate this? If I place an image string with charge \lambda_{1} inside the cylinder, at distance a from the center,I get the following equation for equipotential surfaces:
\lambda ln(r)-\lambda_{1} ln(r_{1})=C, where r, r_{1} are distances form resp. 1st and 2nd string. Now if I put C=0 (since on the surface of the cylinder V=0, and I want the cylinder to be an equipotential surface), I get:
\lambda ln(r)=\lambda_{1} ln(r_{1}). Now I need to calculate \lambda_{1} and a from this. Any help?

 

[pam]

It's done like the 3D problem.
Write (R\hat{r}-{\vec L})e^\lambda=(R\hat{r}-{\vec L}')e^{-\lambda'}.
Then let L'=R^2/L.