There is a particular paragraph in Electricity and Magnetism by Purcell that I'm not able to understand. It's the last para in section 1.13, pg-30 which goes like this
The field of an infinitely long line charge, we found, varies inversely as the distance from the line, while the field of an infinite sheet has the same strength at all distances. These are simple consequences of the fact that the field of a point charge varies as the inverse square of the distance.
If that doesn’t yet seem compellingly obvious, look at it this way: roughly speaking, the part of the line charge that is mainly responsible for the field at P in Fig. 1.24 is the near part – the charge within a distance of the order of magnitude r. If we lump all this together and forget the rest, we have a concentrated charge of magnitude $q \approx \lambda r$, which ought to produce a field proportional to $\frac{q}{r^2}$,or $ \frac{ \lambda}{r}$. In the case of the sheet, the amount of charge that is “effective,” in this sense, increases proportionally to $r^2$ as we go out from the sheet, which just offsets the $\frac{1}{r^2}$ decrease in the field from any given element of the charge
I don't get anything. First that near part approximation and then that lumping stuff.
Edit: The electric field due to the element $\lambda dx$ given by
$$dE_y\propto \frac{\lambda dx}{(r^2+x^2)}\cos\theta=\frac{\lambda dx\cdot r}{(r^2+x^2)^{3/2}}$$ $$\frac{dE_y}{dx}\propto \frac{ r}{(r^2+x^2)^{3/2}}$$ If you plot the function on the right, you get a plot that has a peak around $x=0$, So That's clear that the contribution is coming around this part. $$\frac{dE_y}{dx}\propto \frac{ r}{(r^2+x^2)^{3/2}}\approx \frac{1}{r^2}\left(1-\frac{3x^2}{2r^2}\right)$$
But still, I don't get the fact why we should take the magnitude of order $r$. If I take it for a grant then lumping can be understood. $$q = \int dq\approx \int_0^r\lambda dx=\lambda r$$