# Electric field at a point $P$

Problem
Charge is distributed uniformly over a large square of side $l$, as shown in the figure. The charge per unit area ($C/m^2$) is $\sigma$. Determine the electric field at a point $P$ a distance $z$ above the center of the plane, in the limit that $z<<l$. Assuming that $dQ=\sigma dxdy$ I said that

\begin{align} E_z=\dfrac{1}{4\pi \epsilon_0}\int\limits_{-l/2}^{l/2}\int\limits_{-l/2}^{l/2}\dfrac{dQ}{r^2}\cos\theta&=\dfrac{\sigma}{4\pi \epsilon_0}\int\limits_{-l/2}^{l/2}\int\limits_{-l/2}^{l/2}\dfrac{1}{x^2+y^2+z^2}\dfrac{z}{\sqrt{x^2+y^2+z^2}}dxdy\\ &=\dfrac{\sigma z}{4\pi \epsilon_0}\int\limits_{-l/2}^{l/2}\int\limits_{-l/2}^{l/2}\dfrac{1}{(x^2+y^2+z^2)^{3/2}}dxdy. \tag{1}\end{align}

To begin with I think that $E_x$ and $E_y$ are 0, but I dont know if $(1)$ is correct or not. If it is correct, how can I calculate this integral. And is there an alternative method for calculating the electric field at point $P$?( maybe by integrating only in one dimension $dy$?)

Your setup looks good, but you are making the problem too hard for yourself. As it stands, solving that integral will give you the exact answer. Remember, $z \ll l$, and you can use this assumption before solving any integrals.
If you really want to solve the integral as is, well, that's not a physics issue, but rather a math one. You could chop the corners off (an approximation, admittedly) to make a circular plate, in which case the integral is quite easy in polar coordinates. (Keep this in mind once you decide how to handle the $z \ll l$ condition.) For exactness in solving the equation as written, there's a trig substitution that makes it doable.