[YMP/EM-03002] Potential Due to Uniformly Charged Disk

Charge on the annular ring between $x$ and $x+dx=2\pi
xdx\sigma$
\begin{align*}
d\phi=&\frac{1}{4\pi\epsilon_0}~\frac{2\pi
xdx\sigma}{\sqrt{x^2+r^2}}\\
\phi=&\int_0^a \frac{1}{4\pi\epsilon_0}~\frac{2\pi xdx\sigma}
{\sqrt{x^2+r^2}}\\
=&\frac{\sigma}{2\epsilon_0}\int_0^a\frac{xdx}{\sqrt{x^2+a^2}}\\
=&\frac{\sigma}{2\epsilon_0}\left[\sqrt{a^2+r^2}-r\right]\\
\sqrt{a^2+r^2} = & r(1+a^2/r^2)^{1/2} \approx r + \frac{a^2}{2r}
\end{align*}
For large distances $r\gg a$
\begin{align*}
\phi \approx& ~\frac{\sigma}{2\epsilon_0}~\frac{a^2}{2r} =
\frac{\sigma\pi a^2}{4\pi\epsilon_0 r}\\
=&\frac{Q}{4\pi\epsilon_0r}
\end{align*}