I have a uniformly charged disk, radius $R$, with surface charge density $\sigma$. I want to find the electric field along the axis through the centre of the disk (which I've called the $x$ axis).
I am aware that one method is to consider small rings and integrate from $0$ to $R$, but that is not how I approached the problem and so I would like to know how to do it my way.
I considered a small section on the disk (I don't know how to formulate its charge), but it had position vector $r \hat{r}$ in cylindrical coordinates. Then I got stuck on how to express the charge of that small section. I think it is $dq = \sigma dz dr$ (again in cylindrical coordinates), but I am not sure.
Any help is appreciated!
Best Answer
If a small piece of your disk is located at $$ x=r\cos\theta\, ,\qquad y=r\sin\theta,\qquad z=0 $$ then the density would be $dq=\sigma dA=\sigma r dr d\theta$ where $\theta$ is the polar angle on the disk since $dA=r dr d\theta$ is the area of a small piece of your disk, with radius $r$ and arclength $rd\theta$.
For a point located on the $\hat z$ axis at $Z\ne 0$, this small amount of charge will produce the infinitesimal field $$ d\vec E=\frac{dq}{4\pi\epsilon_0} \frac{Z\hat z- r\cos\theta\hat x - r\sin\theta\hat y}{(r^2+Z^2)^{3/2}}\, . $$ The full field is obtained by integrating the infinitesimal fields.