# 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.

• I'm fairly certain that your expression for the electric field is incorrect, since it has dimensions of $q/\epsilon_0$. (Note that $\sigma$ has dimensions of charge per area.) – Michael Seifert Oct 9 at 18:49
• If the plates can be considered to be at uniform potential (true if they are conductors and the situation is static or slowly changing) then it suffices to find the answer along any single path between them. So pick an easy one... – dmckee Oct 9 at 19:25
• @MichaelSeifert In fact, the electric field was wrong. – Mathieu Rousseau Oct 9 at 19:29

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 $$C$$.