Can the Prince save you now?

My solution involves the use of double integrals that are pretty easy to solve(believe me).

Assume the plate to be part of the xy- plane with the boundaries defined by the equations : $x=a/2,x=-a/2,y=b/2 ,y=-b/2$

We want to find the electric field due to the whole plate at the point $(0,0,d)$

It is pretty obvious that the resultant electric field would be in +ve z direction.

Hence we would be integrating for z-component of electric field.

Take a infinitesimal area element at a point $(x,y)$ having area $dxdy$ . Electric field due to that element is given by :

$d^{2}E = \dfrac { k\sigma dxdy\cos { \theta } }{ { x }^{ 2 }+{ y }^{ 2 }+{ d }^{ 2 } }$

Where $\theta$ is angle made by the net electric field by that element with the z-axis.

Now $cos(\theta) = \dfrac { d }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 }+{ d }^{ 2 } } }$

Putting the expression the double integral comes out to be :

$\displaystyle E = \int _{ -b/2 }^{ b/2 }{ \int _{ -a/2 }^{ a/2 }{ \frac { kd\sigma dxdy }{ { ({ x }^{ 2 }+{ y }^{ 2 }+{ d }^{ 2 }) }^{ 3/2 } } } }$

Solving the double integral and putting the value of $\sigma$ ,the closed form comes out to be :

$\displaystyle E = \frac { 4kq }{ ab } \tan ^{ -1 }{ (\frac { ab }{ 2d\sqrt { { a }^{ 2 }+{ b }^{ 2 }+4d^{2} } } ) }$

Where $q$ = Total charge, $a,b$ = Length and breadth of the plate respectively, $d$ = distance of the point from the centre of mass.

The problem is very similar to A charge oscillating above a charged square plate

Let $p=\dfrac{a}{2}$ and $q=\dfrac{b}{2}$

Consider two line charges parallel to each other and at same distance from center. We already know the formula for electric field of line charge. We take only vertical component.The small electric field due to these line charges is: $\displaystyle dE = \dfrac{\sigma h p}{\pi \epsilon_{0} (h^2+x^2) \sqrt{h^2+q^2+x^2}}$

Hence, the net electric field is: $E=\displaystyle \dfrac{\sigma h p}{\pi \epsilon_{0}} \int_{0}^{q} \dfrac{{\mathrm dx}}{(h^2+x^2)\sqrt{h^2+q^2+x^2}}$

Substituting $\dfrac{(h^2+p^2)}{x^2}+1 = t^2$ ,

$E = \displaystyle \dfrac{\sigma p}{\pi \epsilon_{0} h} \int_{\sqrt{\frac{h^2+p^2}{q^2}+1}}^{\infty} \dfrac{{\mathrm dt}}{t^2 h^2+p^2}$

= $\displaystyle \dfrac{\sigma}{\pi \epsilon_{0}} \arctan \bigg(\dfrac{ht}{p}\bigg) \Bigg|_{\sqrt{\frac{h^2+p^2}{q^2}+1}}^{\infty}$

= $\displaystyle \dfrac{\sigma}{\pi \epsilon_{0}} \arctan \Bigg(\dfrac{pq}{h \sqrt{h^2+p^2+q^2}}\Bigg)$

where $\sigma=\frac{q}{ab}$ , $p=\frac{a}{2}$ , and $q=\frac{b}{2}$

Notice that if $p$ and $q$ are very huge, the field approaches $\dfrac{\sigma}{2 \epsilon_{0}}$ , which is the one of an infinite plane.