Electric field of infinite plane of charge is given by:
$$E=\frac{\sigma}{2\varepsilon_0}$$
This can be derived using Gauss's Law or integrating the contribution by circular loops of charge enclosing one another. But if I use square loops instead, it becomes:
$$E=\frac{2\sigma}{\pi\varepsilon_0}$$
Electric field of z component is
$$\mathrm{d} \hat{E}_{z}=\frac{k \mathrm{d} q}{R^2}\cos\theta$$
Substituting
$$\mathrm{d} q=\sigma (8r \mathrm{d} r)$$
$$R^2=z^2+r^2$$
$$\cos\theta=\frac{z}{\sqrt{z^2+r^2}}$$
Integrating
$$E_{z}=8k\sigma z\int_{0}^{r}\frac{r}{({z^2+r^2})^{1.5}}\:\mathrm{d} r$$
When r=$\infty$,
$$E=\frac{2\sigma}{\pi\varepsilon_0}$$
So difference is it will be a factor of $\frac{2}{\pi}$ instead of $\frac{1}{2}$, greater than the electric field that would be obtained using the correct formula. Why is this? Since it is an infinite plane, I don't think it matters regarding what enclosing shapes we are using, but it seems like different model yields different answers.
Best Answer
Your mistake is that you are assuming $E_z$ to have the same angular dependence as if you were considering circular loops. Note that as you move along the side of the square $E_z$ must vary because you are changing your distance with respect to the charge. The reason one takes circular loops is precisely due to this fact that $E_z$ is constant along circular loops, which makes the integration easier.