Calculating the electric field of a disk
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!
$\endgroup$ 22 Answers
$\begingroup$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
$$E = \frac{\sigma}{2 \epsilon_0} \left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)$$
$\endgroup$ 0 $\begingroup$Actually the exact expression for the electric field is
\begin{equation} E = \frac{\sigma}{2\epsilon_0}\left( \frac{z}{|z|}-\frac{z}{\sqrt{z^2+R^2}}\right). \end{equation} Careful should be taken in simplifying $\sqrt{z^2}$, since this is equal to $|z|$, not $z$. This is important because the field should reverse its direction as we pass through $z=0$. This ensures that a positive electric charge is always repelled by the disc (or by the plane in the limit $R \rightarrow \infty $), not importing if the charge is above or below the disc.
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