Let us situate this on the $$x$$ axis, and let our uniform line of charge be positioned on the interval $$(-L, 0]$$ for some large number $$L$$. The voltage $$V$$ as a function of $$x$$ on the interval $$(0,\infty)$$ is given by integrating the contributions from each bit of charge. Let the charge density be $$\lambda$$. Thus, for an infinitesimal length element $$dx'$$, we have $$\lambda = \frac{dq}{dx'}$$.$V(x) = \frac{1}{4\pi \epsilon_0} \int\limits_{\text{line}} \frac{dq}{r}=\frac{\lambda}{4\pi \epsilon_0} \int\limits_{-L}^0 \frac{dx'}{x-x'}=\frac{\lambda}{4\pi \epsilon_0} \left( \ln|x+L| -\ln|x|\right)$Try using this now to find an approximate voltage drop over your interval. I don't really think you can do it exactly without knowing the value of $$L$$.