# Electric field a distance z above flat circular disk.

1. Sep 27, 2013

### -Dragoon-

1. The problem statement, all variables and given/known data
Find the electric field a distance above the center of a flat circular disk of radius R, which carries a uniform surface charge σ.

2. Relevant equations

3. 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: Sep 27, 2013
2. Sep 27, 2013

### TSny

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 $\theta$.

3. Sep 27, 2013

### -Dragoon-

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

Last edited: Sep 27, 2013
4. Sep 27, 2013

### TSny

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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted