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The integrals are difficult but not impossible, unless I've made a mistake with WolframAlpha. The result is: $$E = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$ When $a,b \to \infty$ the whole arctangent goes to $\pi/2$ and we recover $E=\frac{\sigma}{2\epsilon_0}$, which is definitely encouraging. And I ...

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