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Consider a ring shaped surface of the shell between $x$
and $x+dx$.\\
Radius of sphere $=a$\\
Radius of the ring $=a\sin\theta$\\
$x=a\cos\theta$\\
Area of the ring =$(ad\theta)\times a\sin\theta\times
2\pi$\qquad\quad $dx=a\sec^2\theta d\theta$\\
Charges density $=\frac{Q}{4\pi a^2}=\sigma$
\begin{align*}
\therefore~d\phi=&\left(\frac{1}{4\pi\epsilon_0}\right)
\frac{\sigma\times2\pi a^2\sin\theta d\theta}{R}\qquad\qquad\qquad{QP=r-x}\\
=&\left({1}{4\pi\epsilon_0}\right)\left({Q}{2}\right)
\frac{\sin\theta d\theta}{\sqrt{r^2+a^2-2ar\cos\theta}}\qquad{AP=\sqrt{a^2\sin^2\theta+r^2+x^2+2xr}}\\
\phi=&\left(\frac{1}{4\pi\epsilon_0}\right) \left(\frac{Q}{2}\right)
\int_0^\pi \frac{\sin\theta d\theta}{\sqrt{r^2+a^2-2ar\cos\theta}}\qquad{=\sqrt{a^2+r^2-2ar\cos\theta}}\\
=&\frac{Q}{8\pi\epsilon_0}\int_{-1}^1\frac{dt}{(r^2+a^2-2art)^{1/2}}\\
=&\frac{Q}{8\pi\epsilon_0}\left[(r^2+a^2-2art)^{1/2}\left|_{t=1} -
(r^2+a^2-2art)^{1/2}\right|_{t=-1}\right]\\
=&\frac{Q}{8\pi\epsilon_0}\left[\frac{1}{ar}\left((r+a)^2\right)^{1/2}-\left((r-a^2)\right)^{1/2}\right]\\
\end{align*}
\noindent \underline{$r>a$}
$$
\phi = \frac{Q}{8\pi\epsilon_0r}~2a = \frac{Q}{4\pi\epsilon_0r}
$$
\noindent \underline{$r<a$}
$$
\phi = \frac{Q}{8\pi\epsilon_0}~2r~ =\frac{Q}{4\pi\epsilon_0a}
$$
