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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 $z<<l$.

Figure 1

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 $dy$?)

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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.

Ask yourself: More or less, how big does the plate appear if I'm really close compared to its actual size? (The diagram is not drawn to scale, by the way.)

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.

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