Proving limits of electric fields on infinite planes

[Cactus]

Homework Statement

Hey
So I've got this question here to work on, and I've so far managed to solve the first part (a) (Proving that at the center the field is equivalent to an infinite plane however this next part of the question has stumped me as I cannot figure out how they solved to the factor they've provided.
I realise that the factor is a division of answer a and b (So E(field) at x = L/2 divided by E(field) at x = 0) and thats where the 1/pi comes from, as the n/2e in both cases (formula for efield of a plane of charge) will cancel to just 1/pi, however I'm not sure how they get the rest of the formula in part b
Attached below is the current working I've done but I can't see a connection from there

Relevant Equations

Algebra

rt a)

 

[timetraveller123]

did you try expanding things to first order one of your terms in your calculation you set it to zero
$x \approx \frac{L}{2}$
not exactly equal to it
and $d<<L$
so you need to expand to first order this
$\frac{L}{2d} \frac{L-2x}{\sqrt{L^2+{(L-2x)}^2}}$
and
$tan^{-1}(\frac{L}{2d} \frac{L+2x}{\sqrt{L^2+{(L+2x)}^2}})$

 

[Cactus]

did you try expanding things to first order one of your terms in your calculation you set it to zero
$x \approx \frac{L}{2}$
not exactly equal to it
and $d<<L$
so you need to expand to first order this
$\frac{L}{2d} \frac{L-2x}{\sqrt{L^2+{(L-2x)}^2}}$
and
$tan^{-1}(\frac{L}{2d} \frac{L+2x}{\sqrt{L^2+{(L+2x)}^2}})$

I'm not exactly sure what you mean in saying that?

 

[timetraveller123]

i am saying that the S is not just
$\frac{E_{plate}(\frac{L}{2})}{E_{plate}(0)}$ which is what you tried to do
because x is not$\frac{L}{2}$ but very close to it but not exactly it
so you expand the terms to first order