# Integration problem in electric field

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

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

Thanks.

• 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. – 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}
• 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! – Dunkelheit Mar 26 at 5:37