# electric field

• March 30th 2008, 01:26 AM
heathrowjohnny
electric field
Find the electric field a distance $z$ above the center of a flat circular disk of radius $R$, which carries a uniform surface charge $\sigma$. What does your formula give in the limit $R \to \infty$. Also check the case $z >> R$.

So $dq = \sigma da'$. So I broke the surface up into circular rings (now we can consider line charges). Then $dq = \lambda dl'$. So $\sigma \cdot 2 \pi r \cdot dr = \lambda \cdot 2 \pi r$ and $\lambda = \sigma dr$. Could you also say that $\sigma \cdot \pi r^2 = \lambda \cdot 2 \pi r$?

Then $E_{\text{ring}} = \frac{1}{4 \pi \epsilon_{0}} \frac{(\sigma dr) 2 \pi r z}{(r^{2}+z^{2})^{3/2}}$ and so $E_{\text{disk}} = \frac{1}{4 \pi \epsilon_{0}} 2 \pi \sigma z \int_{0}^{R} \frac{r}{(r^2+z^2)^{3/2}} \ dr$.

Is this correct?
• March 30th 2008, 06:34 AM
topsquark
Quote:

Originally Posted by heathrowjohnny
Find the electric field a distance $z$ above the center of a flat circular disk of radius $R$, which carries a uniform surface charge $\sigma$. What does your formula give in the limit $R \to \infty$. Also check the case $z >> R$.

So $dq = \sigma da'$. So I broke the surface up into circular rings (now we can consider line charges). Then $dq = \lambda dl'$. So $\sigma \cdot 2 \pi r \cdot dr = \lambda \cdot 2 \pi r$ and $\lambda = \sigma dr$. Could you also say that $\sigma \cdot \pi r^2 = \lambda \cdot 2 \pi r$?

Then $E_{\text{ring}} = \frac{1}{4 \pi \epsilon_{0}} \frac{(\sigma dr) 2 \pi r z}{(r^{2}+z^{2})^{3/2}}$ and so $E_{\text{disk}} = \frac{1}{4 \pi \epsilon_{0}} 2 \pi \sigma z \int_{0}^{R} \frac{r}{(r^2+z^2)^{3/2}} \ dr$.

Is this correct?

Why not do it directly?

I have a disk of radius R centered on the origin and lying in the xy plane. I have an observation point (0, 0, z) and a constant surface charge density $\sigma$. Thus
$\bold{E} = \int \frac{\bold{x}~dq}{4 \pi \epsilon _0x^3}$
where x is the displacement from the charge element to the observation point.

Now,
$\sigma = \frac{Q}{A}$

$Q = \sigma A \implies dq = \sigma ~dA$

Note that the electric field will be pointed in the z direction, so we may automatically integrate over that component and not worry about a radial component.

The area element in the plane will be $r~dr~d \theta$, thus
$\bold{E} = \int_0^R \int_0^{2 pi} \frac{\sigma rz~d \theta ~dr}{4 \pi \epsilon _0 (r^2 + z^2)^{3/2}}~\bold{\hat{z}}$

$\bold{E} = \frac{\sigma}{2 \epsilon _0} \cdot \frac{\sqrt{R^2 + z^2} - z}{\sqrt{R^2 + z^2}} ~\bold{\hat{z}}$

-Dan