Horizontal $E$-field for a charged conducting disk For part of a simulation I am writing, I need to know the electric field emitted from a  charged conducting disk. If the disk was laid out in the $x$-$y$ plane, I am interested in the field in that same plane, not vertically.
The method for getting it in the axis of the disk (the $z$-axis) is easy, but I can't figure out how to do this.
Does anyone know the equation or how to calculate it?
 A: Actually the conducting disk problem is solved very easily in the so-called oblate spheroidal coordinates.
First, alter the coordinates so that your disc is centered at the origin and is orthogonal to the $z$-direction. I will follow the notation of the Wiki article:
$$
x=a\cosh\mu\cos\nu\cos\phi\\
y=a\cosh\mu\cos\nu\sin\phi\\
z=a\sinh\mu\sin\nu
$$
where $a$ is the radius of the disc. Then the condition of the potential $\Phi=\Phi_0$ on the disc is written as
$$
\Phi(\mu=0)=\Phi_0.
$$
Now, the coordinates are orthogonal, so we have a pretty nice Laplace equation:
$$
0=a^2\Delta\Phi=\\\frac{1}{\sinh^2\mu+\sin^2\nu}\left[\frac{1}{\cosh\mu}\frac{\partial}{\partial\mu}\left(\cosh\mu\frac{\partial\Phi}{\partial\mu}\right)+\frac{1}{\cos\nu}\frac{\partial}{\partial\nu}\left(\cos\nu\frac{\partial\Phi}{\partial\nu}\right)\right]+\frac{1}{\cosh^2\mu+\cos^2\nu}\frac{\partial^2\Phi}{\partial\phi^2}=0
$$
Very involved, but let us guess that (look at the picture of the coordinates on Wiki, $\mu=const$ surfaces look very appeling to the role of equipotential surfs) $\Phi=\Phi(\mu)$ is the function of $\mu$ only. Then the Laplace equation is trivially reduced to
$$
\frac{\partial}{\partial\mu}\left(\cosh\mu\frac{\partial\Phi}{\partial\mu}\right)=0.
$$
Very nice indeed, we can solve it! Solution is:
$$
\Phi=\Phi_0+C\int_0^{\mu}\frac{d\mu}{\cosh\mu}=\Phi_0+2C\left(\arctan e^\mu-\pi/4\right)
$$
where we fix $C$ by requiring $\Phi(+\infty)=0$. One funny thing is that the integral we have encountered in the solution is called the Gudermannian function
$$
\mathrm{gd}(\mu)=2\left(\arctan e^\mu-\pi/4\right)=\arctan(\sinh\mu).
$$
This way or another the final solution is:
$$
\Phi(\mu)=\Phi_0\left(1-\frac{2}{\pi}\arctan\sinh\mu\right)=\frac{2}{\pi}\Phi_0\arcsin\frac{1}{\cosh\mu}
$$
Now, one may want to express it through the usual cylindrical coordinates $r,\phi,z$. According to Wiki, we have:
$$
\cosh\mu=\frac{\sqrt{(r+a)^2+z^2}+\sqrt{(r-a)^2+z^2}}{2a}
$$
This is enough to write $\Phi$ in cylindrical coordinates (just substitute). However, it is rather a complicated formula. We stil have to relate $\Phi_0$ to $Q$ the charge of the disk. For the case $z=0$ and $r>a$ outside the disc we have:
$$
\cosh\mu=r/a
$$
that is
$$
\Phi(r)=\frac{2\Phi_0}{\pi}\arcsin\frac{a}{r}.
$$
Now we know that for $r\rightarrow\infty$ we have $\Phi\simeq\frac{Q}{r}$. At the same time
$$
\Phi(r)\simeq\frac{2\Phi_0}{\pi}\frac{a}{r}=\frac{Q}{r}
$$
so the answer is
$$
\Phi_0=\frac{\pi Q}{2a}\\
\Phi(r)=\frac{Q}{a}\arcsin\frac{a}{r}.
$$
For convenience, in general case:
$$
\Phi(r,z)=\frac{Q}{a}\arcsin\frac{2a}{\sqrt{(r+a)^2+z^2}+\sqrt{(r-a)^2+z^2}}
$$
Edit:
Alec S answer states that:
$$
\Phi(r,\theta=\pi/2)=\frac{2\Phi_0}{\pi}\sum_{l=0}^{\infty}\frac{(-1)^l}{2l+1}\left(\frac{a}{r}\right)^{2l+1}\frac{(-1)^l(2l)!}{2^{2l}(l!)^2}=\\
=\frac{2\Phi_0}{\pi}\sum_{l=0}^{\infty}\frac{1}{4^l(2l+1)}\binom{2l}{l}\left(\frac{a}{r}\right)^{2l+1}=\frac{2\Phi_0}{\pi}\arcsin\frac{a}{r},
$$
see the series for the $\arcsin$ function.
Finally, the radial (and the only nonzero) component of $E$ is given in $x-y$ plane by (outside the disk)
$$
E_r(r)=\frac{Q}{r^2}\frac{1}{\sqrt{1-\frac{a^2}{r^2}}}
$$ 
A: The exact solution is $${\bf E}(R<r, \theta =\pi/2)=\frac{Q}{4 \pi \epsilon_0 }\left(\frac{1}{r^2}\right)\sum_{l=0}^{\infty}\frac{(2l)!}{2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l}\hat{{\bf r}}.$$
Clearly the field inside the conductor (that is, for $r<R$) vanishes. Here $Q$ is the total charge on the disk. The field, for large values of $r$, looks essentially like a point charge (due to the fact that the series tapers off rather quickly) but closer to the disk there are terms of all orders in $r$.
This was obtained by solving a similar problem, that is, finding the potential everywhere for a charged conducting disk. Most of my solution to that problem is reproduced here.
We assume that the azimuthally-symmetric potential $\Phi(r,\theta,\phi)$
  is separable. That is, it can be written $$\Phi(\boldsymbol{r})=\Phi(r,\theta)=\sum_{l=0}^{\infty}\left(A_{l}r^{l}+\frac{B_{l}}{r^{l+1}}\right)P_{l}(\cos\theta).$$
 Along the $z$
 -axis, using the fact that $P_{l}(1)=1$
 ,$$\Phi(r,\theta=0)=\sum_{l=0}^{\infty}A_{l}r^{l}+\frac{B_{l}}{r^{l+1}}$$
 The coefficients $A_{l}$
  and $B_{l}$
  form the coefficients for a unique power series expansion in $r$
  while the legendre polynomials $P_{l}$
  form a complete orthonogonal set. If a separable solution exists, then finding each of these coefficients should be enough to reproduce the potential everywhere. Far away from the disc the potential must limit to zero. Then the $A_{l}$
  are zero and $$\Phi(r,\theta=0)=\frac{1}{r}\sum_{l=0}^{\infty}\frac{B_{l}}{r^{l}}.$$
 All that remains is to find the coefficients $B_{l}$
 .
First, we need to find the appropriate normalization factor for the surface charge density $\sigma(\rho)$
 :$$\int\sigma(\rho')da'=\int_{0}^{2\pi}d\phi'\int_{0}^{R}\frac{\rho'd\rho'}{\sqrt{R^{2}-\rho'^{2}}}=2\pi R$$.
 Since $\int\sigma da=Q$
 , this implies that $$\sigma(\rho)=\frac{Q}{2\pi R}\frac{1}{\sqrt{R^{2}-\rho^{2}}}.$$
 Next, we integrate over the charge density to find the potential everywhere along the axis $\theta=0$
 , and find that the potential along the entire $z$
 -axis can be written$$\Phi(r,\theta=0) = \frac{1}{4\pi\epsilon_{0}}\int\frac{\sigma(\boldsymbol{r}')da'}{|\boldsymbol{r}-\boldsymbol{r}'|}
 =... $$ $$...=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2\pi R}\int_{0}^{2\pi}d\phi'\int_{0}^{R}\frac{1}{\sqrt{R^{2}-\rho'^{2}}}\frac{1}{\sqrt{r^{2}+\rho'^{2}}}\rho'd\rho'$$
    $$= \frac{1}{4\pi\epsilon_{0}}\frac{Q}{R}\arctan\left(\frac{R}{r}\right)$$
 where the last step was done by computer. In addition, we have to demand that in the limit that $r\rightarrow0
 , \Phi(r,\theta)\rightarrow V: $ $$
 \Phi(r\rightarrow0)=\frac{1}{8\epsilon_{0}}\frac{Q}{R}=V.$$
 In other words $Q=8\epsilon_{o}$ and$$\Phi(r,\theta=0)=\frac{2V}{\pi}\arctan\left(\frac{R}{r}\right).$$
We can write this solution as a power series using the well known expansion for $\arctan\left(\frac{R}{r}\right)$
  for $\left|\frac{R}{r}\right|\leq1$
 ,
$$\Phi(r,\theta=0)=\frac{2V}{\pi}\left(\frac{R}{r}\right)\sum_{l=0}^{\infty}\frac{(-1)^{l}}{2l+1}\left(\frac{R}{r}\right)^{2l}.$$
 By comparing these power series coefficients to the $B_{l}$
  we find:$$B_{l}=\begin{cases}
\frac{2V}{\pi}\frac{(-1)^{l}R^{2l+1}}{2l+1} & l\in\mbox{even}\\
0 & l\in\mbox{odd }
\end{cases}$$
 so that, outside a sphere of radius $R$
  from the origin:$$\Phi(r,\theta)=\frac{2V}{\pi}\left(\frac{R}{r}\right)\sum_{l=0}^{\infty}\frac{(-1)^{l}}{2l+1}\left(\frac{R}{r}\right)^{2l}P_{2l}(\cos\theta).$$
Expanding $\arctan(\frac{R}{r})$
  when $\left(\frac{R}{r}\right)\geq1$
  , we can do the same procedure to get the inner coefficients:$$\left.\arctan\left(\frac{R}{r}\right)\right|_{r=0}=\frac{\pi}{2}-\sum_{l=0}^{\infty}\frac{(-1)^{l}}{2l+1}\left(\frac{r}{R}\right)^{2l+1}$$
 which has the appropriate behavior in the limit $\theta\rightarrow\pi$
 . So the potential becomes$$\Phi(r,\theta)\biggr|_{r=0}=\frac{2V}{\pi}\left(\frac{\pi}{2}-\sum_{l=0}^{\infty}\frac{(-1)^{l}}{2l+1}\left(\frac{r}{R}\right)^{2l+1}P_{2l+1}(\cos\theta)\right).$$
Now that you know the potential, it is a simple matter to compute the fields, which are given by $${\bf E}=-\nabla \Phi.$$ Use the fact that the field is symmetric about the $z$-axis to ignore the $\theta$ derivative, and finally that $$P_{2l}(0)=\frac{(-1)^l(2l)!}{2^{2l}(l!)^2}$$ to obtain the final result.
Addendum: Peter Kravchuk is correct that there is a closed form expression for this result. All credit goes to him for pointing this out, I am simply going to add it to the end of this answer for completeness. For simulational purposes I'm not sure whether closed form or series is more useful. 
$$\Phi(r,\theta=\pi/2)=\frac{2V}{\pi}\sum_{l=0}^{\infty}\frac{(2l)!}{(2l+1)2^{2l}(l!)^2}\left(\frac{R}{r}\right)^{2l+1}$$
which equals
$$\Phi(r,\theta=\pi/2)=\frac{2V}{\pi}\arcsin(R/r).$$
Differentiating and substituting gives $${\bf E}(R<r, \theta=\pi/2)=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\frac{1}{\sqrt{1-(R/r)^2}}\hat{{\bf r}}.$$
One will notice that electric field diverges at the edges of the disk -- a consequence of the fact that the charge density there also diverges.
A: I'll suppose your disk (radius $R$) is uniformly charged with surfacic charge density $\sigma$ (your disk being conductor, any charged Q deposed on it will be distributed all over it so that $Q=\sigma\times\pi R^2$). The field outside the disk is necessarily radial so we can place the point $M$ where we would like to compute the field on the $x$ axis so that $\vec{OM}=x\,\vec{e_x}$.
We'll try to compute first the electric potential $V$ in order to find the field $\vec{E}(M) = -\frac{dV}{dx}\,\vec{e_x}$ in $M$.
Let's imagine some point $P$ on the disk of polar coordinates $(r,\theta)$. The surface $dS=rd\theta\,dr$ around it is filled by a charge $dq=\sigma\,dS$. The potential $dV$ created by this charge in $M$ will then be, using the Al Kashi theorem,
$$
dV = \frac{dq}{4\pi\varepsilon_0\,PM} 
= \frac{\sigma\,rd\theta\,dr}{4\pi\varepsilon_0\,\sqrt{r^2+x^2 - 2rx\cos\theta}}
$$
To get the entire potential, you just have to integrate over $\theta$ from 0 to $2\pi$ and $r$ from $0$ to $R$:
$$
V(x) = \int_{r=0}^R\int_{\theta=0}^{2\pi} dV
     = \int_{r=0}^R\int_{\theta=0}^{2\pi} \frac{\sigma\,rd\theta\,dr}{4\pi\varepsilon_0\,\sqrt{r^2+x^2 - 2rx\cos\theta}}
$$
Once you got this integral, deriving it with respect to $x$ will give you the electric field:
$$\vec{E} = - \frac{dV}{dx}\,\vec{e_x}$$
Note: if you want to integrate numerically, you can first derived inside the integral, then choose an $x$ value and numerically integrate to get your field.
