I would like to evaluate the electrostatic field of a disk on its axis using the definition; I have no trouble in evaluating it considering the infinitesimals, but I would like to use the definition to have a better mathematical knowledge of the thing.

Suppose the superficial density $\sigma$ is constant, let $R$ be the radius of the disk and let $q$ the charge uniformly distribuited over the disk; I will use as reference system the axis of the disk as the $x$ axis.

So if $\sigma$ is the superficial density of charge (which is constant), and $\hat{u}$ is a versor oriented like $\vec{r}$ (which is the distance between a point on the axis and a point on the disk) we have that

$$\vec{E}(x,y)=\frac{1}{4\pi\varepsilon_0}\int_\Sigma \frac{\sigma }{r^2}\text{d}S \ \hat{u}$$

Now I have to parametrize the disk: let $\theta$ be the angle that $\vec{r}$ forms with the $x$ axis, I've chosen to parametrize it with polar coordinates $(\rho \sin \theta, \rho \cos \theta)$ with $\rho \in [0,R]$ and $\theta \in [0,2\pi)$; so I get

$$\vec{E}(x,y)=\frac{\sigma}{4\pi\varepsilon_0 r^2}\int_0^{2\pi} \left(\int_0^R \rho \text{d}\rho\right)\text{d}\theta \ \hat{u}=\frac{\sigma R^2}{4 \varepsilon_0 r^2}$$

Now by this (if it is correct, if not please tell me where are the mistakes) I would like to deduce that

$$E(x)=\pm \frac{q}{2\pi \varepsilon_0 R^2} \left(1-\frac{|x|}{\sqrt{x^2+R^2}}\right)\hat{u}_x$$


  • 1
    $\begingroup$ No, you've taken $r^{-2}$ out of the integral, which just leaves constants, so the integral becomes just the constants times the area of the disk. You can't do that, $r$ depends on the integration variable. It makes no sense to have a result containing $r$ as a free variable, since $r$ refers to a point on the disk. $\endgroup$ – joriki Mar 26 at 4:07

I think you're being sloppy in a lot of ways. For example in the second line you are assuming that the distance $r$ from the source point to the field point doesn't change as we move around the source. This assumption leads to the far field of the disk, treating it like a point charge. Also your final expression should have been a vector. You could fix that by multiplying by $\hat r=\frac{\vec r}r$. What you need to do to get the near field is think about the source point $\vec r_s=\langle0,\rho\cos\theta,\rho\sin\theta\rangle$ and the field point $\vec r_f=\langle x,0,0\rangle$ so you can get the vector $\vec r=\vec r_f-\vec r_s=\langle x,-\rho\cos\theta,-\rho\sin\theta\rangle$ from the source point to the field point. Now you can get $$\begin{align}\vec E(x)&=\frac1{4\pi\epsilon_0}\int_0^R\int_0^{2\pi}\frac{\sigma\hat r}{r^2}d\theta\,\rho\,d\rho=\frac1{4\pi\epsilon_0}\int_0^R\int_0^{2\pi}\frac{\sigma\vec r}{r^3}d\theta\,\rho\,d\rho\\ &=\frac1{4\pi\epsilon_0}\int_0^R\int_0^{2\pi}\frac{\sigma\langle x,-\rho\cos\theta,-\rho\sin\theta\rangle}{(\rho^2+x^2)^{3/2}}d\theta\,\rho\,d\rho\\ &=\frac1{4\pi\epsilon_0}\int_0^R\frac{\sigma\langle2\pi x,0,0\rangle}{(\rho^2+x^2)^{3/2}}\rho\,d\rho=-\left.\frac{\sigma\langle2\pi x,0,0\rangle}{4\pi\epsilon_0(\rho^2+x^2)^{1/2}}\right|_0^R\\ &=\frac{\sigma x\hat i}{2\epsilon_0}\left(\frac1{|x|}-\frac1{\sqrt{R^2+x^2}}\right)=\frac{qx}{2\pi\epsilon_0R^2}\left(\frac1{|x|}-\frac1{\sqrt{R^2+x^2}}\right)\hat i\end{align}$$

  • $\begingroup$ You're right, I've been negligent in the writing and even if I thought that I couldn't transport $\frac{1}{r^2}$ outside the integral I didn't write it. However I've done my calculations again with your answer as guide and it is all clear now, thank you so much! $\endgroup$ – Dunkelheit Mar 26 at 5:37

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