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First up, I needed to compute the potential at some external point due to charge stuffed inside the region between two concentric cylinders, the volume charge density being given.
Two methods came to mind but they are apparently yielding different answers(maybe I goofed up somewhere)
The first is to solve the Poisson's equation considering that the potential doesn't vary with $\theta$ and $z$ and setup and solve a differential equation.
With that said, what if I consider the composite object as two cylinders, the bigger one with a volume charge density $\rho_v$, radius $a$ and a smaller one inside it (along the common axis) with density $-\rho_v$, radius $b$
(In this way the stuff along the axis cancels out leaving the same resultant object) and calculate the potentials due to them at some point $r$. The resultant potential would then be the sum. Though this doesn't apparently yield the same result.

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up vote 1 down vote accepted

It does give the same result--- it would be good to give your steps, if you want to find the error. The easiest way to see this is to just use Gauss's law: if you make a cylinder of radius R, the electric field outward times the surface area is the charge inside. If the length of the Gaussian cylinder is L, and the inner and outer radius of the physical cylindrical annulus is $R_0$,$R_1$, then for the exterior

$$L 2\pi R E(R) = L \rho (\pi R_1^2 - \pi R_0^2) $$

Which gives you the standard 1/R field, log(R) potential, outside a charged cylinder and by its form is equivlent to the difference-of-two-cylinders solution.

For inside the inner cylinder, the field is zero, and between the two, it is that sum of a 1/R and constant which matches the other two regions. The field in all regions can be found by Gaussian cylinders, and the potential is found by integrating in R. The result satisfies Laplace's equation with source, since you can prove Gauss's law for solutions of this equation directly.

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