Electric field of a circular plate with non uniform charge density.

[loops496]

Hey!

I need to calculate the electric field on the axis of a circular plate of radius a with the following charge distribution:
$$\sigma_0 \frac{r^2}{a^2} \delta (z), \; r\leq a$$$$0, \; r>a$$
where $\sigma_0$ is a constant.
I've already calculated the potential and taken its gradient to get the field, which is only in z and its given by,
$$E_z = \frac{\sigma_0}{2 \epsilon_0 a^2} \left[\frac{4}{3}z \sqrt{a^2+z^2}- 2z \sqrt{z^2} - \frac{z(a^2 - 2z^2)}{3 \sqrt{a^2+z^2}}\right]$$

Now I have to take the approximation $z>>a$ and describe the field. Intuitively it should be the like the field of a point charge, but i can not get the $1/r^2$ dependence after the approximation.
Is there something wrong?

Thanks in advance.

 

[Einj]

You just need to approximate your right hand side. Taking just the square brackets, it is equal to:
\begin{align}
&\frac{4}{3}z^2\sqrt{1+\frac{a^2}{z^2}}-2z^2-\frac{2z^2\left(\frac{a^2}{2z^2}-1\right)}{3\sqrt{\frac{a^2}{z^2}+1}}\simeq \frac{4}{3}z^2\left(1+\frac{a^2}{2z^2}\right)-2z^2-\frac{2}{3}z^2\left(1-\frac{a^2}{2z^2}\right)\left(1-\frac{a^2}{2z^2}\right) \\
&=-\frac{2}{3}z^2+\frac{2}{3}a^2+\frac{2}{3}z^2-\frac{2}{3}a^2+\frac{a^4}{6z^2}=\frac{a^4}{6z^2},
\end{align}
and so your field is:
\begin{align}
E_z\simeq \frac{\sigma_0a^2}{12\epsilon_0z^2}.
\end{align}

 

[loops496]

Oh my algebra was off on that one! thanks.