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

Question

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?

Answer 1

$\arctan(\theta)\approx \theta-\frac{\theta^3}{3}$ 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.

Answer 2

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 $\arctan x \approx x$.

Take it from there.