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.