[Physics] 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?

Best Answer

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