Fuente TeX:

V=\frac{k}{8\pi\varepsilon_0}\int_0^{L^2}\frac{du}{\sqrt{y^2+u}}=\frac{k}{4\pi\varepsilon_0}\sqrt{y^2+u}|^{L^2}_0=\frac{k}{4\pi\varepsilon_0}\left[\sqrt{y^2+L^2}-y\right]