Arturo O. answered • 04/17/17

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Start with Gauss' law. For the Gaussian surface, use a hollow cylinder of radius r (a < r < b) centered at the same axis. (It looks like you tried to use a sphere for the Gaussian surface, which will not work here. You need to use a cylindrical surface.)

EA = Q/ε

_{0}A = surface area of the Gaussian surface cylinder without the end surfaces (radial electric field)

Let L = length of the cylinder. It is actually infinity, but you will find that L drops out of the math later.

A = 2πrL

Q = (the charge on the line) + (the charge enclosed inside the shell)

Q = λL + ρV = λL + ρ[π(r

^{2}- a^{2})L],where V = π(r

^{2}- a^{2})L is the volume of the shell with charge in its interior over a radius range from a to r.Put this all together in Gauss' law and get

E(r)(2πrL) = (1/ε

_{0}){λL + ρ[π(r^{2}- a^{2})L]}Note that L cancels out, so we do not have to worry about an infinity. Then

E(r) = [1/(2πε

_{0}r)] [λ + ρπ(r^{2}- a^{2})]Be sure to carefully check my math.