Answer:
[tex]\Delta V=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})[/tex]
Explanation:
Using the Gauss Law, we obtain the electric Field for a uniform large line of charge:
[tex]2\pi r L*E=\lambda *L/\epsilon_{o}[/tex]
[tex]E=\lambda /\(2 \pi* r *\epsilon_{o})[/tex]
We calculate the potential difference from the electric field:
[tex]\Delta V=-\int\limits^{r_{1}}_{r_{2}} E \, dr =-\int\limits^{r_{1}}_{r_{2}} \lambda dr/ (2\pi*r*\epsilon_{o})=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})[/tex]