Electric field from disc versus point charge A uniformly charged disc of radius R and net charge Q with an x-axis through the center of the disc will have an electric field in a point $x_0$ on the x-axis
$E=kQ(1-\dfrac{x_0}{\sqrt{x_0^2+R^2}})$ 
if my calculations arent wrong. 
If a point charge of charge Q replaces the disc, the electric field in $x_0$ will be 
$E=\dfrac{kQ}{x_0^2}$. 
Mathematically the field is stronger in $x_0$ from the point charge than from the disc, but I can't figure out the physical explanation for this. 
Can anyone explain this? 
 A: The charge Q which was concentrated at one point has now spreaded to a disc of radius R. For simplicity assume charge on disc is spreaded uniformly and divide the disc in little squares, say N. So charge of each square becomes $\frac{Q}{N}$. Now observe, other than the center the other squares are located at a greater distance than $x_0$ so their electric field will obviously less than if they were located at center. Adding electric field due to each little square results in an total electric field lesser than if all the charge was concentrated at single point. 
A: the electric field you calculated is wrong
let's say disc$(\sigma,R)$ is in $XY - plane $
$\vec E$- field  due to disc at $h$ distance above on it's central axis is $\mathbf {\vec{E_{above}}}= \dfrac{\sigma}{2\epsilon_{0}}\left(1-\dfrac{1}{\sqrt{1+\frac{R^2}{h^2}}}\right)\mathbf {\hat z} $ (in $+ z$-direction)
$\vec E$-field at $h$ distance below on it's central axis is 
$\mathbf {\vec{E_{above}}}= -\dfrac{\sigma}{2\epsilon_{0}}\left(1+\dfrac{1}{\sqrt{1+\frac{R^2}{h^2}}}\right)\mathbf {\hat z} $
(in $-z$-direction)
if you plot the graph you will get cusp at $h=0$
conclusions:
(1)
from infinte distance above as you come closer to the disc (along z-axis) you see more charge in your field of view and when you reach very close to the disc it will look like a infinte plane to you whose field is $\dfrac{\sigma}{2\epsilon_{0}}$
(2) as you move away from the disc's circumferance to $\infty$ .Less charge will be seen in your field of view and disc will look like point charge of magnitude $\sigma .\pi R^2$
similar two conclusions can be drawn for moving either towards $-\infty$ or away from it too.
A: This can be thought of using vector addition ,if u see carefully some components will cancel out giving a net lesser effect in case of a disk but u will see if u continuously decrease the distance the field strength will approach what we call not defined.
