[Physics] Horizontal $E$-field for a charged conducting disk

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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?

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) $$E_r(r)=\frac{Q}{r^2}\frac{1}{\sqrt{1-\frac{a^2}{r^2}}}$$