# Electric field of a point very far from uniformly charged rectangle sheet

I was wondering what is the Electric field at a point which is very far from a rectangular sheet and it is also above the center of the rectangle. So form a mathematical perspective you get Electric field due to a finite rectangular sheet of charge on the surface $$S = \left\{(x,y,z)\in \mathbb{R}^3 \mid -a/2< x < +a/2; -b/2< y < +b/2 ; z = 0 \right\} .$$ is $$E(0,0,r) = \frac{\sigma r}{4\pi\epsilon_o} \int_{x=-a/2}^{x=+a/2}\int_{y=-b/2}^{y=+b/2} \frac{dx dy}{(x^2+y^2+r^2)^{3/2}}$$ so $$E(0,0,r) = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$. It seems very counter intutive that for $$r>>a$$ and $$r>>b$$ electric field is not $$E(0,0,r) = \frac{\sigma}{\pi \epsilon_0}\arctan\left( \frac{ab}{4r^2} \right)$$ but $$E(0,0,r) =k_e\frac{q}{r^2}$$ where $$q=\sigma ab$$. My question is shouldn't it behave like a point charge if it is very far away from the point where I am calculating electric field? Why is that not so? What am I doing wrong?

$$\arctan(\theta)\approx \theta-\frac{\theta^3}{3}$$ near $$\theta=0$$ so \begin{align} \frac{\sigma}{\pi\epsilon}\arctan\left(\frac{ab}{4r^2}\right) \approx \frac{\sigma}{\pi\epsilon}\frac{ab}{4r^2}\tag{1} \end{align} and since $$a\times b$$ is the area, $$\sigma\times a\times b=Q$$, the charge on your plate. At this level of approximation you then get \begin{align} E_z(0,0,r)\approx \frac{Q}{4\pi\epsilon r^2} \end{align} which is the field of a point charge.
The additional term $$\theta^3/3$$, which I did not include in (1), gives the leading correction due to the finite size of the plate. It is negative because, in $$Q/4\pi\epsilon r^2$$, you are concentrating all the charge at a single point whereas the actual field will be a little less since the charge is diluted over the entire area, and thus some of the charge is a slightly greater distance from $$(0,0,r)$$ than the centre of the plate, resulting in a slightly smaller contribution than if it was at the origin.
For $$a \ll r$$ and $$b \ll r$$ the argument of the arctan function (call it $$x \equiv ab/4r^2$$) is much less than 1. And for $$x \ll 1$$, we have $$\arctan x \approx x$$.