Difference of potential between 2 disks of radius $R$ where $d << R$ Well,
We have 2 disks of radius R where the distance between these 2 disks : $d << R$.
These disks are uniformly charged.
I have calculated the electric field near one disk, I have: 
$$E(z) = \frac{\sigma}{2 \epsilon_0}\left(1 - \frac{1}{\sqrt{1 + \frac{R^2}{z^2}}}\right)$$ 
Now I'm stuck to find the potential between these 2 disks.
Because one disk has a positive charge and another one has a negative charge, the total electric field is
$$E_{tot} = 2E(z)$$
We know that
$$\Delta V = -\int_{+\infty}^{d}E_{tot} \cdot \mathrm dr$$
However, I get an infinite potential. 
I suppose that I'm doing the wrong calculation.
The teacher explained that because $d << R$, we can approximate them as infinite planes. So, should I calculate the potential between 2 infinite planes?
Note: $z$ is the axis perpendicular to the surface of the disk.
 A: As commented by Michael Seifert, your expression of the electric field is indeed incorrect. It should read
    $$E_z=-{\sigma\over 2\varepsilon_0}\left[{z\over |z|}-{z\over\sqrt{R^2+z^2}}\right]$$
at a distance $z$ on the axis of a disk of radius R carrying a density $\sigma$. The potentiel is given by
   $$E_z=-{\partial V\over\partial z}\ \Leftrightarrow\ V(z)=-\int E_z(z)dz$$
To perform the integration, consider separately the cases $z>0$ and $z<0$. When $z>0$,
   $$V(z)=-{\sigma\over 2\varepsilon_0}\Big[z-\sqrt{R^2+z^2}\Big]+C$$
where $C$ is constant. Imposing the vanishing of $V$ at infinity leads to $C=0$. When
$z<0$,
   $$V(z)={\sigma\over 2\varepsilon_0}\Big[z+\sqrt{R^2+z^2}\Big]+C'$$
The continuity of $V(z)$ at $z=0$ imposes $C=C'=0$.
Now, consider a point very close to the disk or equivalently a very large disk $R\gg 1$. As expected for an infinite plane, the expression of the electric field becomes constant
   $$E_z=-{\sigma\over 2\varepsilon_0}$$
when $z>0$ and
   $$E_z={\sigma\over 2\varepsilon_0}$$
when $z<0$. The potential is therefore linear
   $$V(z)=-E_zz+C={\sigma\over 2\varepsilon_0}|z|+C$$
However, it is not possible to impose the vanishing of $V(z)$ in both limits $z\rightarrow +\infty$ and $z\rightarrow -\infty$ only by fixing the constant $C$.
