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?

  • $\begingroup$ Is it a conducting disk? A uniformly charged insulating disk? $\endgroup$ May 6, 2013 at 16:58
  • $\begingroup$ It is a conducting disk. $\endgroup$ May 6, 2013 at 17:01

3 Answers 3


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

  • $\begingroup$ +1 it's always good to see another derivation. Illuminating closed form solution -- I've added it to my answer -- I hope you don't mind. $\endgroup$ May 7, 2013 at 0:04
  • $\begingroup$ Ah I hadn't refreshed the page and seen this. Thankyou very much for the help. $\endgroup$ May 7, 2013 at 0:39

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.

  • $\begingroup$ Have a look at my answer, there is a closed formula for the potential, I believe that you can resum the series for $\cos\theta=0$. $\endgroup$ May 6, 2013 at 19:38
  • $\begingroup$ Yep, I have done this in the end of my answer. $\endgroup$ May 6, 2013 at 20:05
  • $\begingroup$ @PeterKravchuk thanks for the insight, I've added the closed form solution to this answer. Full credit goes to you for noticing that. $\endgroup$ May 7, 2013 at 0:02
  • $\begingroup$ @JoshCalahan this is a well-known problem, I'm sure there are other solutions. There's no need to reference mine. Peter's answer is another way to derive it -- they are both valid. $\endgroup$ May 7, 2013 at 0:03
  • $\begingroup$ Thankyou very much for your answer, it has been a great help! $\endgroup$ May 7, 2013 at 0:40

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.

  • 4
    $\begingroup$ The math here may be correct (though I have not checked), but a conducting disk doesn't have a uniform surface charge density. $\endgroup$ May 6, 2013 at 17:57

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