I'm having trouble regarding how to calculate the electric field of a disk. Here's the scheme:
The exercise states that the disk is uniformely charged. This is what I did:
Density charge : $\sigma = \frac{Q}{\pi*R^2}$
Thickeness ring centered on the disk : $dA = 2\pi ada$
So I have $dq = \sigma dA = 2\pi \sigma ada$
Because the charges along the x axis "cancelled" themselves, I'm only interested on the charge along the z axis. It follows that:
$dE_{z} = dEcos\theta$ or $cos\theta = \frac{z}{r}$ and $r = \sqrt{a^2 + z^2}$, hence $dE_{z} =\frac{1}{4 \pi \epsilon_{0}} * z * \frac{2\pi a \sigma da}{(z^2 + a^2)^\frac{3}{2}}$
Now I just integrate from $0$ to R.
$E_{z} = \int_{0}^{R}dE_{z} = … =\frac{z \sigma \pi}{4 \pi \epsilon_{0}} * [-\frac{1}{2(z^2 + a^2)^\frac{1}{2}}]_{0}^{R} =\frac{z \sigma \pi}{4 \pi \epsilon_{0}} * [-\frac{1}{2(z^2 + R^2)^\frac{1}{2}} + \frac{1}{2z}] = \frac{z \sigma \pi}{4 \pi \epsilon_{0}} * [\frac{1}{2z}*(1-\frac{z}{(z^2 + R^2)^\frac{1}{2}})] = \frac{ \sigma}{8 \epsilon_{0}} * [(1-\frac{z}{(z^2 + R^2)^\frac{1}{2}})] $
My problem comes now. I'm sure that my reasoning is correct at this point. But when $R >> z$, I should get the electric field of a plan (like in a condensator, i.e $\frac{\sigma}{2\epsilon_{0}}$) but with my result, I have $E=\frac{\sigma}{8\epsilon_{0}}$.
I'm stuck on this for two hours and I can't figure what I did wrong!
Best Answer
Starting with your expression (which after cancelling factors can be written)
$$dE = \frac{\sigma z}{2 \epsilon_0} \frac{ada}{(z^2+a^2)^{3/2}}$$
Now integrate to get
$$E = \frac{\sigma z}{2 \epsilon_0} \int_0^R\frac{ada}{(z^2+ a^2)^{3/2}}$$
To solve the integral let $w = z^2 + a^2$ then
$$E = \frac{\sigma z}{4 \epsilon_0} \int_{z^2}^{z^2+R^2}\frac{dw}{w^{3/2}} = \frac{\sigma z}{2 \epsilon_0} \left[-\frac{2}{\sqrt{\omega}}\right]_{z^2}^{z^2+R^2}$$
which gives the correct result
$$E = \frac{\sigma}{2 \epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)$$