# Electric field a distance z above flat circular disk.

• -Dragoon-

## Homework Statement

Find the electric field a distance above the center of a flat circular disk of radius R, which carries a uniform surface charge σ.

## The Attempt at a Solution

Basically, I want to solve this usually trivial problem without using symmetry arguments and the superposition principle but rather the more laborious method introduced in Griffith's section 2.1. Hence:

$\vec{E(r)} = \frac{1}{4\pi\epsilon_{0}}\int\frac{\sigma(r)}{r'^2}\hat{r'}da$

Where $\hat{r'} = \frac{z\hat{z} - r\hat{r}}{\sqrt{z^2+r^2}}$ and $r' = \sqrt{r^2 + z^2}$, thus:

$\vec{E(r)} = \frac{\sigma}{4\pi\epsilon_{0}}\int\frac{z\hat{z} - r\hat{r}}{(z^2 + r^2)^{3/2}}da$

Converting to polar coordinates we have:
$\vec{E(r)} = \frac{\sigma}{4\pi\epsilon_{0}}\iint\frac{z\hat{z} - r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta$

=> $\frac{\sigma}{4\pi\epsilon_{0}}\int_0^{2\pi}\int_0^r\frac{z\hat{z} - r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta$
Which gives:
$\frac{\sigma2\pi}{4\pi\epsilon_{0}}[z\hat{z}\int_0^r\frac{r}{(z^2 + r^2)^{3/2}}dr - \int_0^r\frac{r^2\hat{r}}{(z^2 + r^2)^{3/2}}dr]$

Evaluating the first integral gives me the correct answer for the electric field:
$\frac{\sigma2\pi z}{4\pi\epsilon_{0}}[- \frac{1}{\sqrt{z^2+r^2}}\Big|_0^r]\hat{z}$

Which means, the second integral must be zero: $\int_0^r\frac{r^2\hat{r}}{(z^2 + r^2)^{3/2}}dr = 0$

Except I am not exactly sure on how to show that is the case. The radial unit vector seems to be the problematic term, and it cannot be simply taken outside the integral. Would like to have some help and insights on this, thanks.

Last edited:

Go back and think about the ##\theta## integration in the expression $$\iint\frac{ r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta$$ Note that ##\hat{r}## is not a constant when integrating over ##\theta##.

• 1 person
Go back and think about the ##\theta## integration in the expression $$\iint\frac{ r\hat{r}}{(z^2 + r^2)^{3/2}}rdrd\theta$$ Note that ##\hat{r}## is not a constant when integrating over ##\theta##.

Ah, I see. Thanks. Is then ##\hat{r}## a constant when integrating with respect to ##r## then?

Last edited:
Yes. If you integrate over ##r## first, then you are keeping ##\theta## constant. So, the unit vector ##\hat{r}## has a fixed direction.

• 1 person