We assume that the sphere of radius $R$ centered at 0. Let us assume an observation
point $o$ above the north pole of the sphere (by symmetry this should provide a
good answer). We consider rings from the bottom up to the north polo $(0,0,R)$.
The ring at a high $z$, $-R \le z \le R$ has a radius
$\rho=\sqrt{R^2 - z^2}$. We prefer to see the problem as a function
of the polar angle from $-\pi/2$ to $\pi/2$. We have that $\rho=R \sin \theta$ with $\theta$
the polar angle. It is well known that for a ring with uniform charge density
$\sigma$ , radius $r$ and an observation point in the axis of the ring at a
distance $d$ from the center (in the direction of the axis of the ring) produces the field
\begin{eqnarray*}
E(d,\rho)= \frac{\sigma \rho \, d}{2 \epsilon_0(\rho^2 + d^2)^{3/2}}.
\end{eqnarray*}
The distance between the observation point $o$ and the ring at $z$
height is $d=o-z$, and $z=R \cos \theta$, then we find
\begin{eqnarray*}
E(\theta)=
\frac{\sigma (o-R \cos \theta)
R \sin \theta}{2 \epsilon_0(R^2 \sin^2 \theta + (o-R \cos \theta)^2)^{3/2}}.
\end{eqnarray*}
We need to integrate along $\theta$ between $0$ and $\pi$.
Along the polar axis the element of integration is $d \ell = R \, d \theta$,
so we will need to multiply by $R \, d \theta$.
\begin{eqnarray*}
E = \frac{\sigma}{2 \epsilon_0} \int_{-\pi/2}^{\pi/2}
\frac{(o-R \cos \theta)
R^2 \sin \theta}{(R^2 \sin^2 \theta + (o-R \cos \theta)^2)^{3/2}}
d \theta.
\end{eqnarray*}
Let us perform the following substitution
\begin{eqnarray*}
u= \cos \theta \quad , \quad du=- \sin \theta d \theta \\
\theta=0 \implies u = 1\\
\theta=\pi \implies u = -1 ,
\end{eqnarray*}
then
\begin{eqnarray*}
E = \frac{\sigma R^2}{2 \epsilon_0} \int_{-1}^1
\frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du
\end{eqnarray*}
We split the integrand in two fractions (forget the coefficient for now).
\begin{eqnarray*}
\int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du
\quad \mathrm{and} \quad
-\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du
\end{eqnarray*}
For the first integral, let us make $x=R^2 + o^2 - 2 o R u$,
then $dx=-2 o R du$, and in terms of $x$,
\begin{eqnarray*}
-\frac{1}{2 R} \int \frac{dx}{x^{3/2}} =
\frac{1}{ R \sqrt{x} },
\end{eqnarray*}
Then the first integral is
\begin{eqnarray*}
\int_{-1}^1 \frac{o} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
\left . \frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}} \right |_0^1 \\
\end{eqnarray*}
Let us do the second integral usig integration by parts.
We write
\begin{eqnarray*}
-\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
-\frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}} \\
&& + \int
\frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du
\end{eqnarray*}
Now,
\begin{eqnarray*}
\int \frac{1}{o \sqrt{R^2 + o^2 - 2 o R u}} du =
-\frac{1}{o^2 R} \sqrt{R^2 + o^2 - 2 o R u},
\end{eqnarray*}
then
\begin{eqnarray*}
\int_{-1}^1 \frac{u R} {(R^2 + o^2 - 2 o R u)^{3/2}} du &=&
\frac{u }{o \sqrt{R^2 + o^2 - 2 o R u}}
+ \frac{ \sqrt{R^2 + o^2 - 2 o R u}}{o^2 R} \\
&=& \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
\end{eqnarray*}
Putting the first and the second integrals back together we get
\begin{eqnarray*}
\frac{1}{R \sqrt{R^2 + o^2 - 2 o R u}}
- \frac{ R^2 + o^2 - o R u}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
=
\frac{-R^2 + o Ru}{o^2 R \sqrt{R^2 + o^2 - 2 o R u}}
\end{eqnarray*}
Hence we found that
\begin{eqnarray*}
\int \frac{o - u R}{(R^2 + o^2 - 2 o R u)^{3/2}} du
= \frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}},
\end{eqnarray*}
and so
\begin{eqnarray*}
\left .
\frac{o u - R}{o^2 \sqrt{o^2 - 2 o R u + R^2}} \right |_{-1}^1
&=& \frac{o - R}{o^2 \sqrt{o^2 - 2 o R + R^2}}
+ \frac{o + R}{o^2 \sqrt{o^2 + 2 o R + R^2}} \\
&=&
\frac{o - R}{o^2 |o - R|}
+ \frac{o + R}{o^2 |o + R|}
\end{eqnarray*}
\begin{eqnarray*}
E= \frac{\sigma R^2}{2 \epsilon_0}
\left [
\frac{o - R}{o^2 |o - R|}
+ \frac{o + R}{o^2 |o + R|}
\right ].
\end{eqnarray*}
That is
\begin{eqnarray*}
E = \left \{
\begin{array}{cc}
\frac{\sigma R^2}{o^2 \epsilon_0} & o > R \\
\\
0 & o < R
\end{array}
\right .
\end{eqnarray*}
but
\begin{eqnarray*}
\frac{\sigma R^2}{o^2 \epsilon_0}
= \frac{4 \pi \sigma R^2}{4 \pi o^2 \epsilon_0}
= \frac{Q}{4 \pi o^2 \epsilon_0}
\end{eqnarray*}
where $4 \pi R^2 \sigma$ is the total charge in the sphere.
Then
\begin{eqnarray*}
E = \left \{
\begin{array}{cc}
\frac{Q}{4 \pi o^2 \epsilon_0} & o > R \\
\\
0 & o < R
\end{array}
\right .
\end{eqnarray*}
What if $o=R$?
